Fourier Analysis - NPTEL Course

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Fourier Analysis R. Radha Department of Mathematics Indian Institute of Technology Madras Chennai - 600 036 S. Thangavelu Department of Mathematics Indian Institute of Science Bangalore - 560 012

Contents 1 Fourier series 1.1 1.2

1

Appearance of Fourier series

. . . . . . . . . . . . . . . . . . . . . .

1

The formal Fourier series and some special kernels . . . . . . . . . . .

5

1.2.1

Dirichlet kernel . . . . . . . . . . . . . . . . . . . . . . . . . .

10

1.2.2

Dirichlet problem for the disc and Poisson kernel . . . . . . .

13

1.2.3

The heat kernel . . . . . . . . . . . . . . . . . . . . . . . . . .

14

1.3

Uniqueness of Fourier series and some consequences . . . . . . . . . .

16

1.4

Convolution theory . . . . . . . . . . . . . . . . . . . . . . . . . . . .

20

1.5

An approximate identity . . . . . . . . . . . . . . . . . . . . . . . . .

25

1.5.1

29

1.5.2

Fej´er kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . Poisson kernel

. . . . . . . . . . . . . . . . . . . . . . . . . .

32

1.6

Summability of Fourier series . . . . . . . . . . . . . . . . . . . . . .

34

1.7

Fourier series of f ∈ L2 (T) . . . . . . . . . . . . . . . . . . . . . . . .

35

1.8

Existence of a continuous function whose Fourier series diverges . . .

43

1.9

Brief history of Fourier series

. . . . . . . . . . . . . . . . . . . . . .

43

1.10 Solved problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

44

1.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

68

2 Applications of Fourier series

69

2.1

The values of Riemann zeta function at even positive integers . . . .

69

2.2

Isoperimetric inequality . . . . . . . . . . . . . . . . . . . . . . . . . .

72

i

2.3

Jacobi identity for the theta function . . . . . . . . . . . . . . . . . .

76

2.4

Weierstrass approximation theorem . . . . . . . . . . . . . . . . . . .

77

2.5

Wallis’ product formula . . . . . . . . . . . . . . . . . . . . . . . . . .

79

2.6

Weyl’s equidistribution theorem . . . . . . . . . . . . . . . . . . . . .

80

2.7

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

83

3 Fourier transform

85

3.1

Definition and basic properties . . . . . . . . . . . . . . . . . . . . . .

85

3.2

Schwartz space on R . . . . . . . . . . . . . . . . . . . . . . . . . . .

87

3.3

Fourier transform on the Schwartz space . . . . . . . . . . . . . . . .

92

3.4

Fourier transform for L2 -functions on R . . . . . . . . . . . . . . . . .

99

3.5

Fourier transform for L1 -functions on R . . . . . . . . . . . . . . . . . 101

3.6

Solved problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

3.7

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

4 Tempered distributions

109

4.1

Topology on S(R) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

4.2

Convergence in S 0 (R) . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

4.3

Some properties of S 0 (R) . . . . . . . . . . . . . . . . . . . . . . . . . 111

4.4

Fourier transform on S 0 (R) . . . . . . . . . . . . . . . . . . . . . . . . 114

4.5

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

5 Some topics on Fourier transform

121

5.1

Poisson summation formula . . . . . . . . . . . . . . . . . . . . . . . 121

5.2

Uncertainty principle . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 ii

5.2.1

Heisenberg’s uncertainty principle . . . . . . . . . . . . . . . . 123

5.3

Paley-Wiener theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 128

5.4

Wiener’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

5.5

A multiplier theorem for L1 (R) . . . . . . . . . . . . . . . . . . . . . 138

5.6

Solved problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

5.7

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

6 Fourier analysis on Rn

147

6.1

Fourier transform on Rn . . . . . . . . . . . . . . . . . . . . . . . . . 147

6.2

Fourier expansion of Fourier transform . . . . . . . . . . . . . . . . . 163

6.3

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168

7 For further reading - Spherical harmonics and Fourier transform

171

7.1

Fourier transform of radial functions . . . . . . . . . . . . . . . . . . 171

7.2

Spherical harmonics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180

7.3

Zonal harmonics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

7.4

Action of the Fourier transform on hk . . . . . . . . . . . . . . . . . . 194

Appendix

197

A.1 Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 A.2 Banach spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 A.3 Banach algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 A.4 Topological vector spaces . . . . . . . . . . . . . . . . . . . . . . . . . 201 Bibliography

202

iii

Chapter 1 Fourier series 1.1

Appearance of Fourier series The birth of Fourier series can be traced back to the solutions of wave equation

in the work of Bernoulli and the heat equation in the work of Fourier. Consider an elastic string of finite length l fixed at the end points x = 0 and x = l. At time, say t = 0, it is distorted from the equilibrium position and allowed to vibrate. The problem is to find the vibrations of the string at any point x and any time t > 0. The vibration of the string is governed by the linear partial differential equation 2 ∂ 2u 2∂ u = c , ∂t2 ∂x2

u(x, 0) = f (x),

∂ u(x, 0) = g(x). ∂t

(1.1)

Here c2 denotes a physical constant, f (x) gives the initial position and g(x) gives the initial velocity. In 1747, D. Alembert obtained the solution of (1.1) in the form 1 1 u(x, t) = [f (x + ct) + f (x − ct)] + 2 2c

x+ct Z

g(y)dy. x−ct

However, in 1753, Bernoulli had a different idea of solving this equation which was based on the observation that the functions sin λct, sin λx, and cos λct, cos λx satisfy the equation for any λ ∈ R. If we choose λ = 1

nπ , l

n = 1, 2, 3, . . . they also satisfy the

boundary conditions. As the equation is linear, he argued that any superposition of such solutions will also be a solution. To explain this method further, let us use the method of separation of variables. Let u(x, t) = F (x)G(t). Substituting for u in (1.1), we get c2 F 00 (x)G(t) = F (x)G00 (t), which can be rewritten as 1 G00 (t) F 00 (x) = 2 . F (x) c G(t)

(1.2)

In (1.2), the left hand side is a function of x and the right hand side is a function of t. This is possible only if 00

F (x) 1 G00 (t) = k, 2 = k, F (x) c G(t) where k is a constant. Thus we end up with two differential equations, F 00 (x) − kF (x) = 0,

(1.3)

G00 (t) − kc2 G(t) = 0.

(1.4)

Since the string is fixed at end points x = 0 and x = l, we have the boundary conditions u(0, t) = 0; u(l, t) = 0, t > 0,

(1.5)

which are satisfied if we assume F (0) = F (l) = 0.

(1.6)

The vibrations of the string at time t depend upon the initial deflection and initial velocity. Let f denote the initial deflection and g denote the initial velocity. Thus the initial conditions are given by u(x, 0) = f (x), 2

(1.7)

∂u (x, 0) = g(x). ∂t

(1.8)

Notice that the constant k which appears in (1.3) and (1.4) is a real number. Hence it can be positive or negative or zero. Case 1. If k = 0, then (1.3) and (1.4) become F 00 (x) = 0; G00 (t) = 0. Solving for F , we get F (x) = Ax + B, where A and B are constants. The boundary conditions force A = B = 0 and hence F = 0. Case 2. Let k = µ2 be positive. Substituting in (1.3), we get F 00 (x) − µ2 F (x) = 0. In this case, the solution is given by F (x) = Aeµx +Be−µx , where A and B are constants. Once again the boundary conditions force F to be zero. Case 3. Let k = −λ2 . Then (1.3) becomes F 00 (x) + λ2 F (x) = 0. On solving, we get F (x) = A cos λx + B sin λx, where A and B are constants. Applying F (0) = 0, we get A = 0. Thus, F (x) = B sin λx. The equation F (l) = 0 leads to B sin λl = 0. Then, we get either B = 0 or sin λl = 0. In order to get a non-trivial solution, we assume that B 6= 0. Thus, sin λl = 0, which leads to λl = nπ, n ∈ Z. As sin 0 = 0 and sin (−n) = nπ , n = 1, 2, 3, . . . and − sin n, it is enough to take n ∈ N. Thus we have λn = l Fn (x) = Bn sin λn x. 00

Now (1.4) becomes Gn (t) + λ2n c2 Gn (t) = 0. On solving we get, Gn (t) = Cn cos (cλn t) + Dn sin (cλn t).

Let un (x, t) = Bn sin λn x(Cn cos (cλn t) + Dn sin (cλn t)). It is clear that for each n, un (x, t) satisfies the wave equation with the correct boundary conditions. 3

The function un also satisfies the initial conditions ∂un (x, 0) = cBn Dn λn sin λn x. ∂t

un (x, 0) = Bn Cn sin λn x,

Consequently, the superpositions u =

∞ P

un solves the wave equation with the bound-

n=1

ary conditions u(0, t) = u(l, t) = 0. However, the initial conditions will be satisfied only if f (x) =

∞ X

Bn Cn sin λn x,

n=1

and g(x) = c

∞ P

Bn Dn λn sin λn x. At this point, it is not clear whether any f or g

n=1

can be expanded in terms of sin λn x as above and hence the solution of the wave equation obtained by Bernoulli was incomplete. In 1807, Fourier was working on the initial value problem for the heat equation ∂u ∂ 2u (x, t) = (x, t), u(x, 0) = f (x). ∂t ∂x2 Let us assume the same boundary conditions as above, namely u(0, t) = u(l, t) = 0 which means the temperature at the end points is kept at zero throughout. Proceeding as in the case of wave equation, we get the elementary solutions 2

un (x, t) = An (sin λn x) e−λn t . The superposition u(x, t) =

∞ P

un (x, t), which solves the heat equation, satisfies the

n=1

initial condition provided f (x) =

∞ X

An sin λn x.

n=1

4

Fourier asserted that any f satisfying f (0) = f (l) = 0 can be expanded in terms of sin λn x as above.

1.2

The formal Fourier series and some special kernels In Section 1.1, we mentioned the idea of expanding a function f (x) satisfying

x, n = 1, 2, . . . Note that the functions sin nπ x are f (0) = f (l) = 0 in terms of sin nπ l l periodic with period 2l and hence f also has to be periodic in order to be expanded as above. Given f with f (0) = f (l) = 0, we can always consider it as a periodic function of period 2l defined on the whole of R and hence it is reasonable to expect an expansion of the above type. For the sake of simplicity, we take l = π and consider functions f that are 2πperiodic. Rather than restricting our attention to functions satisfying f (0) = f (π) = 0, we consider all 2π periodic functions. In this case, besides sin nx, we also need to include cos nx in the expansion of the type

f (x) =

∞ X

(an cos nx + bn sin nx).

n=0

Using the formulae, cos nx = 21 (einx + e−inx ), sin nx = the above expansion as

f (x) =

∞ X n=−∞

5

cn einx .

1 (einx 2i

− e−inx ), we can rewrite

This is a reasonable expansion as f is 2π-periodic and so is each einx . The family of functions einx have the interesting property, which we call orthogonality, Zπ einx e−imx dx = 0 or 2π, −π

depending on whether n 6= m or n = m. Therefore, if termwise integration is allowed, then the expansion f (x) =

∞ X

cn einx ,

n=−∞

leads us to 1 2π

Zπ

f (x) e−inx dx = cn .

−π

This formula, due to Fourier, allows us to calculate cn in the expansion. We can now formally introduce the Fourier series. Definition 1.2.1. Let f be a Lebesgue integrable function on [−π, π]. Then the nth Fourier coefficient of f is defined by Zπ

1 fb(n) = 2π

f (x) e−inx dx,

n ∈ Z.

−π

The Fourier series of f is formally defined by f (x) ∼

∞ P fb(n) einx . n=−∞

Example 1.2.2. Let f (x) = x, −π ≤ x ≤ π. Then, the Fourier coefficients of f are obtained as follows: for n 6= 0 1 fb(n) = 2π

Zπ

x e−inx dx

−π

6

1 = 2π

Zπ

 xd

e−inx −in



−π

−1 = 2πin

Zπ

x d(e−inx )

−π

  π Z   −1 π = x e−inx − e−inx dx  2πin  −π −π

−1 = {2π(−1)n − 0} , n 6= 0. 2πin (−1)n+1 1 Rπ Thus fb(n) = , n 6= 0. But fb(0) = x dx = 0. In this case, the Fourier in 2π −π series of f can be formally written as ∞ ∞ X X (−1)n+1 (−1)n+1 inx e =2 sin nx. f (x) ∼ in n n=−∞ n=1 n6=0

Example 1.2.3. Let f (x) = x2 , −π ≤ x ≤ π. Then the nth Fourier coefficient of f is given by 1 fb(n) = 2π 1 = 2π

Zπ −π Zπ

x2 e−inx dx

2

xd



e−inx −in

−π

Using integration by parts, we get, 2(−1)n fb(n) = , n 6= 0. n2 7

 .

But 1 fb(0) = 2π

Zπ

x2 dx =

π2 . 3

−π

In this case, the formal Fourier series of f can be written as

f (x) ∼

=

∞ X 2(−1)n inx π2 + e 2 3 n n=−∞ n6=0 ∞ X

π2 4(−1)n + cos nx. 3 n2 n=1

Figure 1.1: Fourier series of x2 .

8

Consider the Fourier series ∞ X

f (x) ∼

fb(n) einx

n=−∞

= fb(0) + = fb(0) +

∞ X n=1 ∞ X

fb(−n) e−inx +

∞ X

fb(n) einx

n=1



n=1 ∞ X

+i

 fb(−n) + fb(n) cos nx 

 fb(n) − fb(−n) sin nx,

n=1

using e±inx = cos nx ± i sin nx. If f is an even function, then fb will also be an even function. In fact, 1 fb(−n) = 2π =

=

1 2π 1 2π

Zπ −π Zπ

−π Zπ

f (x)einx dx

f (−x)einx dx

f (−x)e−(−inx) dx

−π

= fb(n), by applying a change of variable x = −y. In this case, the resulting Fourier series ∞ P fb(n) cos nx. On the other hand, if f is an odd function, becomes f (x) ∼ fb(0) + 2 n=1

then fb will also be an odd function and the Fourier series takes the form f (x) ∼ 2i

∞ X n=1

9

fb(n) sin nx.

1.2.1

Dirichlet kernel The formal Fourier series will represent the function f provided the partial

sums of the series converges to f . We take the symmetric partial sum

SN f (x) =

N X

fb(n)einx ,

n=−N

which can be written in the form 1 SN f (x) = 2π

Zπ f (y)DN (x − y)dy, −π

where the function DN , called the Dirichlet kernel, is given by

DN (x) =

N X

einx .

n=−N

Indeed, from the definition, inx

fb(n)e

1 = 2π

Zπ

f (y)e−in(y−x) dy,

−π

and hence N X n=−N

inx

fb(n)e

1 = 2π

Zπ f (y)

N X


e−in(y−x) dy.

n=−N

−π

The kernel DN can be explicitly calculated: sin (2N + 1) x2 DN (x) = . sin ( x2 )

10

(1.9)

Figure 1.2: Dirichlet kernel.

11

In order to establish (1.9), consider N X DN (x) = einx n=−N

=

−1 X

inx

e

+

N X n=1

n=−N

= −1 + = −1 +

einx + 1

N X n=0 N X

−inx

e

+ n

N X

einx

n=0 N X

(e−ix ) +

n=0

n

(eix ) .

n=0

Writing w = eix , we have DN (x) = −1 +

N X n=0

w−n +

N X

wn

n=0 −(N +1)

1−w 1 − w(N +1) + 1 − w−1 1−w −N 1 − w(N +1) w−w + = −1 + w−1 1−w w−N − wN +1 = 1−w −1/2 w (w−N − wN +1 ) = w−1/2 (1 − w) w−N −1/2 − wN +1/2 = w−1/2 − w1/2 sin (2N + 1) x2 = . sin ( x2 )

= −1 +

Observe that, from the very definition, the Fourier coefficients of DN are given by   1 b N (n) = D  0

if |n| ≤ N otherwise.

12

1.2.2

Dirichlet problem for the disc and Poisson kernel Let D denote the unit disc in the complex plane i.e. D = {z ∈ C : |z| < 1}.

Given a continuous function f on the boundary of D we are interested in finding a function u on D satisfying the Laplace equation ∆u = 0 with the boundary condition lim u(reiθ ) = f (θ). This is called the Dirichlet problem for the unit disc D. The

r→1

Laplace equation in polar coordinates takes the form 1 ∂ 2u ∂ 2 u 1 ∂u + + = 0. ∂r2 r ∂r r2 ∂θ2 Integrating the equation against e−inθ dθ we see that ∂ 1 ∂2 r u (r) + r u (r) + n n ∂r2 ∂r 2π 2

Zπ

∂ 2u (r, θ)e−inθ dθ = 0, ∂θ2

−π

where 1 un (r) = 2π

Zπ

u(r, θ)e−inθ dθ.

−π

Integrating by parts, the third term becomes −n2 un (r) and hence we get r2

∂ ∂2 un (r) + r un (r) − n2 un (r) = 0. 2 ∂r ∂r

Clearly, un (r) = an r|n| satisfies the above equation and the initial conditions lim u(r, θ) r→1

= f (θ) leads to lim un (r) = fb(n). Hence, an = fb(n) and the solution u(r, θ) has the r→1

formal expansion u(r, θ) =

∞ X n=−∞

13

fb(n)r|n| einθ .

As before, we can write this in a compact form. Let us define a kernel Pr (θ), called the Poisson kernel, by ∞ X

Pr (θ) =

r|n| einθ .

n=−∞

Then u(r, θ) takes the form 1 u(r, θ) = 2π

Zπ f (ϕ)Pr (θ − ϕ)dϕ. −π

The kernel Pr (θ) can also be calculated explicitly. Indeed, we have

Pr (θ) =

∞ X

r|n| einθ

n=−∞ ∞ X

=1+

n=1

n inθ

r e

+

∞ X

rn e−inθ

n=1

= 1 + reiθ (1 − reiθ )−1 + re−iθ (1 − re−iθ )−1 2r cos θ − 2r2 1 − 2r cos θ + r2 1 − r2 = . 1 − 2r cos θ + r2 It can be easily shown that the Fourier coefficients of Pr are given by =1+

Pbr (n) = r|n| .

1.2.3

The heat kernel Returning to the heat equation, let u(x, t) solve the problem ∂u(x, t) ∂ 2 u(x, t) = , ∂t ∂x2 14

−π ≤ x ≤ π.

u(x, 0) = f (x), For each n, let 1 un (t) = 2π

Zπ

u(x, t) e−inx dx.

−π

Then we have d 1 (un (t)) = dt 2π

Zπ

∂ 2 u(x, t) −inx e dx, ∂x2

−π

which after integration by parts, leads to d (un (t)) = −n2 un (t). dt 2

The solution of this equation is given by un (t) = an e−tn . The initial condition u(x, 0) = f (x) gives an = fb(n) and hence,

u(x, t) =

∞ X

2 fb(n)e−tn einx .

n=−∞

As before, we can rewrite this as 1 u(x, t) = 2π

Zπ f (y)ht (x − y)dy, −π

where ht (x), called the heat kernel, is given by

ht (x) =

∞ X

2

e−tn einx .

n=−∞

Unlike the Poisson or Dirichlet kernel, we do not have a closed form expression for ht (x). 15

1.3

Uniqueness of Fourier series and some consequences Let T denote the unit circle. A function on T can be treated as a function on

R which is 2π-periodic. Let C(T) denote collection of all continuous functions on T. The following theorem gives the uniqueness of Fourier series. Theorem 1.3.1. Let f ∈ C(T). Suppose that fb(n) = 0 ∀ n ∈ Z then f = 0. Proof. For 0 < δ < π, to be fixed soon, consider the function pδ (x) = (1+cos x−cosδ). Then we claim that pδ (x) ≥ 1 on [−δ, δ] and pδ (x) < 1 on [−δ, δ]c . Since cos x is an increasing function on [−π, 0] and a decreasing function on [0, π] we have for x ∈ [−δ, 0], cos x − cos δ ≥ 0. Then pδ (x) = 1 + cos x − cos δ ≥ 1 on [−δ, 0]. On the other hand, if x ∈ [0, δ] then cos x ≥ cos δ and hence pδ (x) ≥ 1 on [0, δ]. However, if −π < x < −δ then cos x < cos (−δ) = cos δ. Hence, pδ (x) < 1 on [−π, −δ]. Similarly, if δ < x < π, then cos x < cos δ which leads to pδ < 1 on [δ, π]. Let f = g + ih, where g and h are real valued functions on T. We prove that f (x) = 0 at every x. Without loss of generality, we take x = 0. Suppose f (0) 6= 0. Then, again without loss of generality, we can assume that g(0) > 0 and we will arrive at a contradiction using the continuity of g. 1 Choose δ > 0 such that g(x) > g(0) on [−δ, δ]. Define pδ with this δ and 2 take Pn (x) = pδ (x)n . By the hypothesis, Zπ f (x)Pn (x)dx = 0, −π

16

which gives Zπ g(x)Pn (x)dx = 0. −π

On the other hand,

R

g(x)Pn (x)dx goes to 0 by dominated convergence theorem

[−δ,δ]c

since pδ (x) < 1. This shows that Zδ lim

n→∞ −δ

g(x)Pn (x)dx = 0.

On the other hand, Pn (x) ≥ 1 on this interval and hence, Zδ g(x)Pn (x)dx ≥ δ g(0). −δ

This contradiction proves our claim that f (0) = 0. Corollary 1.3.2. Suppose f ∈ L1 (T) and fb(n) = 0 ∀ n ∈ Z. Then f (x) = 0 a. e. Proof. Let f ∈ L1 (T). Define g(x) :=

Rx

f (t)dt, −π ≤ x ≤ π, then g is continuous

−π

and g 0 = f a.e. Further, for all n 6= 0, gb(n) =

i fb n

so that gb(n) = 0.

Applying Theorem 1.3.1 to the continuous function g(x) − gb(0), we conclude that g(x) = gb(0) is a constant. Hence, f = 0 a.e. Theorem 1.3.3. Let f ∈ C(T) be such that fb ∈ l1 (Z). Then the Fourier series of f converges uniformly to f on T. ∞ N P P Proof. As fb ∈ l1 (Z), |fb(n)| < ∞. Define SN f (x) = fb(n)einx . We claim n=−∞

n=−N

that SN (f ) → f uniformly on T. 17

For N > M , we have, for all x ∈ T, N M X X fb(n)einx − fb(n)einx |SN f (x) − SM f (x)| = n=−N n=−M X inx b f (n)e = N ≥|n|>M X ≤ |fb(n)|. N ≥|n|>M

Thus by applying Cauchy’s criterion for uniform convergence, one can conclude that SN (f ) converges to a continuous function g (uniformly) on T. Further gb(n) = fb(n) ∀ n ∈ Z. In fact, for all n 1 gb(n) = 2π =

1 2π

Zπ −π Zπ

−π

g(x)e−inx dx ∞ X

! fb(m)eimx

e−inx dx

m=−∞ π

Z ∞ 1 X fb(m)ei(m−n)x dx = 2π m=−∞ −π

= fb(n).

Here, changing the order of integral and the sum was possible because the convergence is uniform. As (f − g)b(n) = 0 ∀ n ∈ Z, we conclude that f = g.

Corollary 1.3.4. If f ∈ C 2 (T), then the Fourier series of f converges uniformly to f on T. 18

Proof. By the theorem, it is enough to prove that fb ∈ l1 (Z). We shall show that   b f (n) = O |n|1 2 as |n| → ∞. Consider, for n 6= 0, Zπ

1 fb(n) = 2π

f (x)e−inx dx

−π

  Zπ   −inx π 1 e 1 0 −inx = f (x) + f (x)e dx  2π  −in −π in −π

=

1 2πin

Zπ

0

f (x)e−inx dx

−π

  Zπ   −inx π 1 e 1 0 00 = f (x) f (x)e−inx dx . +  2πin  −in −π in −π

Thus 1 fb(n) = − 2πn2

Zπ

00

f (x)e−inx dx.

−π

00 Let M = max |f (x)|. Thus |fb(n)| ≤

x∈[−π,π]

∞ X

M n2

for all n 6= 0. Consequently,

X b f (n) |fb(n)| ≤ fb(0) +

n=−∞

n6=0 ∞ X 1 b ≤ f (0) + 2M n2 n=1

< ∞,

which shows that fb ∈ l1 (Z).

19

1.4

Convolution theory Before defining convolution of two functions, we observe the following fact.

Proposition 1.4.1. If f is a periodic function of period 2 π, then Zπ

Zπ f (x + a)dx =

−π

Proof. Consider

Rπ

π+a Z

f (x)dx = −π

f (x)dx. −π+a

f (x+a)dx. Taking x 7→ x+a, we get

Rπ

f (x+a)dx =

−π+a

−π

−π

π+a R

If a > 0, then π+a Z

Zπ f (x)dx =

−π+a

π+a Z f (x)dx + f (x)dx

−π+a

π −π+a Z

Zπ

π+a Z f (x)dx + f (x)dx

f (x)dx −

= −π

−π

π

−π+a Z

Zπ

π+a Z f (x + 2π)dx + f (x)dx

f (x)dx −

= −π

−π

Zπ

π+a Z

f (x)dx −

= −π

π π+a Z

f (y)dy + π

Zπ

f (x)dx = π

f (x)dx. −π

Similarly, we can prove the result when a < 0. Definition 1.4.2. For f, g ∈ L1 (T), the convolution of f and g is defined as 1 (f ∗ g)(x) := 2π

Zπ f (x − y) g(y) dy. −π

20

f (x)dx.

Immediately we observe that f ∗ g = g ∗ f . In fact, 1 (f ∗ g)(x) = 2π

x+π Z

Zπ

1 f (t)g(x − t)dt = 2π

g(x − t)f (t)dt = (g ∗ f )(x). −π

x−π

(b) χ[−1,1] ∗ χ[−1,1] .

(a) χ[−1,1] .

Figure 1.3 Theorem 1.4.3. L1 (T) is a commutative Banach algebra. Proof. We know that L1 (T) is a Banach space. Let f, g ∈ L1 (T). Consider 1 kf ∗ gk1 = 2π

Zπ |f ∗ g(x)| dx −π

1 = (2π)2 1 ≤ (2π)2

Zπ Zπ f (x − y) g(y) dy dx −π Zπ

−π

−π

 Zπ



|f (x − y)| |g(y)| dy dx.

−π

21

Applying Fubini’s theorem we get,

kf ∗ gk1 ≤

Zπ

1 (2π)2 1 2π



Zπ

|f (x − y)|dx |g(y)|dy

 −π

=



−π

Zπ kf k1 |g(y)| dy −π

= kf k1 kgk1 < ∞. Thus f ∗g ∈ L1 (T) and kf ∗gk1 ≤ kf k1 kgk1 . We have already proved that f ∗g = g ∗f. In order to prove associativity, consider, for x ∈ R, 1 (f ∗ g) ∗ h(x) = 2π 1 = 2π

Zπ (f ∗ g) (y) h(x − y) dy −π Zπ

−π

1 = 2π =

1 2π

Zπ

  1 Zπ f (t) g(−t + y) dt h(x − y) dy 2π −π x+π 1 Z  f (t) g(−t − u + x) h(u) du dt 2π

−π Zπ

x−π

f (t) (g ∗ h) (x − t) dt −π

= [f ∗ (g ∗ h)] (x).

By the linearity of integral, it follows that (f + g) ∗ h = (f ∗ h) + (g ∗ h) and αf ∗ g = f ∗ αg. Thus L1 (T) is a commutative Banach algebra.

Theorem 1.4.4. Let f, g ∈ C(T). Then f ∗ g ∈ C(T). 22

Proof. Let f, g ∈ C(T). Let  > 0 be given. Let M = max |g(y)|. For s, t ∈ T, we have y∈T

1 |(f ∗ g)(s) − (f ∗ g)(t)| = 2π

Zπ

Zπ f (s − y) g(y) dy −

−π

≤

≤

1 2π M 2π

f (t − y) g(y) dy

−π

Zπ |f (s − y) − f (t − y)| |g(y)| dy −π Zπ

|f (s − y) − f (t − y)| dy. −π

Since f is uniformly continuous on T, there exists δ > 0 such that |s − t| < δ implies |f (s) − f (t)|
0, δ≤|x|≤π

If f ∈ C(T), then kn ∗ f converges to f uniformly on T .

Proof. As

1 Rπ kn (y) dy = 1, 2π −π 1 f ∗ kn (x) − f (x) = 2π

Zπ [f (x − y) − f (x)] kn (y) dy, −π

which gives 1 |f ∗ kn (x) − f (x)| ≤ 2π

Zπ |f (x − y) − f (x)| |kn (y)| dy. −π

26

(1.10)

Let  > 0 be given. Then by uniform continuity of f there exists δ > 0 such that |f (s) − f (t)| <  whenever |s − t| < δ. By (1.10), 1 |f ∗ kn (x) − f (x)| ≤ 2π

Zδ |f (x − y) − f (x)| |kn (y)| dy −δ

Z

1 + 2π

|f (x − y) − f (x)| |kn (y)| dy δ≤|y|≤π

=: I1 + I2 . We have, 1 I1 = 2π

Zδ |f (x − y) − f (x)| |kn (y)|dy −δ

 ≤ 2π

Zδ |kn (y)| dy ≤

M . 2π

−δ

On the other hand Z

1 I2 = 2π

|f (x − y) − f (y)||kn (y)| dy δ≤|y|≤π

kf k∞ ≤ π

Z |kn (y)| dy,

(1.11)

δ≤|y|≤π

which goes to zero as n → ∞. Here kf k∞ = sup |f (x)|. Thus f ∗ kn converges to f x∈T

uniformly on T.

Remark 1.5.2. Since C(T) is dense in L1 (T) and uniform convergence is stronger than L1 convergence, it follows from the above theorem that {kn } gives an approximate identity for L1 (T). 27

Remark 1.5.3. We shall show that the Dirichlet kernel does not satisfy the hypothesis of Theorem 1.5.1. Recall N X

DN (x) =

einx =

n=−N

sin (N + 12 )x . sin ( x2 )

1 Rπ 1 Rπ inx DN (x) dx = 1. Since e dx = 0 for all n 6= 0. However, 2π −π 2π −π Rπ property (ii) of Theorem 1.5.1 fails. We shall show that |DN (x)| dx ≥ c ln N, ∀ N. We have

−π

Consider Zπ −π

Zπ 1 sin (N + 2 )x |DN (x)|dx = dx sin ( x2 ) −π

Zπ ≥2

|sin (N + 12 )x| dx, |x|

−π

sin x2 since x ≤ 1. Setting y = (N + 21 ) x, we get, 2

(N + 12 )π

Zπ

Z

|DN (x)|dx ≥ 2 −π

sin y y dy

−(N + 12 )π (N + 12 )π

Z

=4

sin y y dy

0

   ZN π sin y =4 y dy +   0

28

  sin y  y dy  . 

(N + 12 )π

Z

Nπ

Since the second term on the right hand side is non-negative, we get, Zπ

ZN π sin y |DN (x)| dx ≥ 4 y dy

−π

0

=4

N −1 X k=0

For k π ≤ y ≤ (k + 1) π, we have

Zπ |DN (x)|dx ≥ 4

=

kπ

N −1 X

1 (k + 1)π

≥ c0

N −1 X k=0 N X l=1

(k+1)π Z

| sin y| dy kπ

N −1 1 4X π k=0 (k + 1)

≥ c0

1.5.1

sin y y dy.

1 1 ≥ . Hence |y| (k + 1) π

k=0

−π

(k+1)π Z

Zπ | sin(y)|dy, 0

1 (k + 1) 1 ≥ c00 log N, as N → ∞. l

Fej´ er kernel As an example of approximate identity, we introduce the kernel σn (x). This

is defined, using the Dirichlet kernel Dk , by N −1 1 X σN (x) = Dk (x). N k=0

29

Note that f ∗ σN (x) =

N −1 1 X f ∗ Dk (x) N k=0

N −1 1 X = Sn (f (x)), N k=0

(1.12)

and these are called the Fej´er means. Fej´er proved the following result. In fact, he has proved for f ∈ Lp (T), 1 ≤ p < ∞. However, we shall give the proof for p = 1 now and give the remaining separately in the solved problem.

Figure 1.4: Fej´er kernel.

30

Theorem 1.5.4. Let f ∈ L1 (T). Then f ∗ σN → f in L1 (T). Proof. The theorem will follow if we can show that {σN } is an approximate identity. In order to show this result, first we shall calculate N σN (x).

Consider N σN (x) = = = = = = =

N −1 X

N −1 X

 w−n − wn+1 Dn (x) = 1−w n=0 n=0 (N −1 ) N −1 X X 1 w−n − wn+1 1 − w n=0 n=0   1 1 − w−N 1 − wN −w 1−w 1−w 1 − w1   −N 1 (1 − wN ) (1 − w ) −w w 1−w w−1 1−w  −N w w − 1 − 1 + wN 2 (1 − w) w (wN + w−N − 2) (1 − w)2 w 2 (wN/2 − w−N/2 ) . 2 (1 − w) 

Thus 1 σN (x) = N

sin N x2 sin ( x2 )


!2 .

Note that σN (x) ≥ 0 for every x. Further 1 2π

Zπ −π

1 σN (x) dx = 2π

Zπ −π

N −1 1 X Dn (x) dx N n=0

N −1 Zπ 1 X = Dn (x) dx = 1. 2πN n=0 −π

31

Fix δ > 0. Let δ ≤ |x| ≤ π. Since sin2 sin2 (N x2 ) . N Cδ

x 2




≥ sin2

δ 2




= Cδ > 0 we get σN (x) ≤

Consider Z

1 |σN (x)|dx = N

δ≤|x|≤π

Z δ≤|x|≤π


sin2 N x2
dx sin2 x2

Z

1 1 ≤ N Cδ


sin2 N x2 dx.

δ≤|x|≤π

≤

1 2π → 0 as N → ∞. N Cδ

Thus σN satisfies the required properties of Theorem 1.5.1. Consequently, we conclude that whenever f ∈ C(T), then f ∗ σN (x) converges uniformly to f on T .

1.5.2

Poisson kernel Recall that for 0 ≤ r < 1, ∞ X 1 − r2 Pr (θ) = r|n| einθ . = 2 1 − 2r cos θ + r n=−∞

We shall show that {Pr } gives an approximate identity. First, notice that Pr (θ) ≥ 0. (i) 1 2π

Zπ −π

1 Pr (θ) dθ = 2π

Zπ  X ∞ −π

 r|n| einθ dθ

n=−∞

Zπ ∞ 1 X = r|n| einθ dθ 2π n=−∞ −π

32

Zπ ∞ 1 X |n| = r einθ dθ 2π n=−∞ −π

= 1.

Figure 1.5: Poisson kernel.

33

(ii) Fix δ > 0 and consider δ ≤ |θ| ≤ π and

1 2

≤ r < 1. Then

1 − 2r cos θ + r2 = (1 − r)2 + 2r(1 − cos θ)
= (1 − r)2 + 4r sin2 2θ
≥ 2 sin2 2δ = 2cδ . Hence, 1 − r2 , 1 − 2r cos θ + r2 1 − r2 ≤ , 2cδ

Pr (θ) =

and consequently Z Pr (θ)dθ ≤

1 − r2 · 2π → 0 as r → 1. 2cδ

δ≤|θ|≤π

Thus Pr satisfies the required properties for an approximate identity. Consequently, we conclude that whenever f ∈ C(T), f ∗ Pr (θ) converges uniformly to f on T.

1.6

Summability of Fourier series Suppose we have a series of complex numbers n P

the series is defined as Sn :=

∞ P

ck . The nth partial sum of

k=0

ck . If Sn → s as n → ∞, then we say that

k=0

ck

k=0

1 (S0 N

+ S1 + · · · + SN −1 ). The arithmetic mean σN is ∞ P partial Cesaro sum of the series ck . If σN converges to a limit σ as

converges to s. Let σN = called the N th

∞ P

N → ∞, then we say that

k=0

∞ P

ck is Cesaro summable to σ.

k=0

A series of complex numbers every 0 ≤ r < 1, the series

∞ P

∞ P

ck is said to be Abel summable to s if for

k=0

ck rk converges and lim

∞ P

r→1 k=0

k=0

34

ck rk = s.

Remark 1.6.1. Convergence ⇒ Cesaro summability ⇒ Abel summability. (See more details in Solved problems.) Consider Zπ

1 f ∗ Pr (θ) = 2π 1 = 2π =

∞ X

f (x) −π Zπ

! r|n| ein(θ−x) dx

n=−∞ ∞ X

−π n=−∞ ∞ Zπ

1 X 2π n=−∞

f (x)r|n| ein(θ−x) dx



 f (x)e−inx r|n| einθ dx

−π

f ∗ Pr (θ) =

1 2π

∞ X

fb(n)r|n| einθ .

(1.13)

n=−∞

Thus we see that f ∗ Pr are nothing but the Abel means of f . Similarly, f ∗ σN are nothing but the Cesaro means. Hence we have the following theorems. Theorem 1.6.2. If f ∈ C(T), then the Fourier series of f is uniformly Cesaro summable to f . Theorem 1.6.3. If f ∈ C(T), then the Fourier series of f is uniformly Abel summable to f .

1.7

Fourier series of f ∈ L2(T) Recall that L2 (T) is a Hilbert space with the inner product 1 hf, gi = 2π

Zπ

f (t)g(t)dt for f, g ∈ L2 (T).

−π

35

Theorem 1.7.1. (Bessel’s inequality) If f ∈ L2 (T), then ∞ X

1 |fb(n)|2 ≤ 2π n=−∞

Zπ

|f (t)|2 dt.

−π

Proof. Consider

hf − SN f, f − SN f i = hf, f i − hf, SN f i − hSN f, f i + hSN f, SN f i

Recall that SN f (x) =

(1.14)

N P fb(n) einx . If we denote en (x) = einx , x ∈ R, then we can n=−N

N P write SN f = fb(n)en and also fb(n) = hf, en i. Thus n=−N

* hf, SN f i =

N X

f,

+ fb(n)en

n=−N

=

N X

fb(n)hf, en i

n=−N

=

N X

fb(n)fb(n)

n=−N

=

N X

|fb(n)|2 .

n=−N

Further * hSN f, SN f i =

N X

fb(n)en ,

n=−N

=

N X

N X

n=−N m=−N

36

N X

+ fb(m)em

m=−N

fb(n)fb(m)hen , em i.

As hen , em i = δn,m , we can conclude that N X hSN f, SN f i = |fb(n)|2 . n=−N

Hence, (1.14) turns out to be hf − SN f, f − SN f i = hf, f i −

N X

|fb(n)|2 ≥ 0,

n=−N

from which it follows that N X

|fb(n)|2 ≤ hf, f i

n=−N

1 = 2π

Zπ

|f (t)|2 dt.

−π

Letting N → ∞, we get ∞ X

1 |fb(n)| ≤ 2π n=−∞ 2

Zπ

|f (t)|2 dt.

−π

This leads us to the following important result. Corollary 1.7.2. (Riemann-Lebesgue lemma) If f ∈ L1 (T), then fb(n) → 0 as Rπ Rπ n → ± ∞. Equivalently, if f ∈ L1 (T), then f (x) cos nx dx → 0 and f (x) −π

−π

sin nx dx → 0 as n → ± ∞. Proof. The result follows immediately from the theorem for f ∈ L2 (T). Let f ∈ L1 (T). Since L2 (T) = L1 (T) ∩ L2 (T) is dense in L1 (T), there exists a sequence of functions {fn } in L2 (T) such that fn → f in L1 -norm. Now lim fbn (m) = 0 for each m→∞

n. Consider 1 |fbn (m) − fb(m)| ≤ 2π

Z2π |fn (x) − f (x)| dx 0

37

= kfn − f k1 → 0, as n → ∞. This shows that fbn converges to fb uniformly on Z. Hence lim fb(m) = lim lim fbn (m)

m→∞

m→∞ n→∞

= lim lim fbn (m) = 0. n→∞ m→∞

Theorem 1.7.3. (Best approximation theorem) Let f ∈ L2 (T). Let SN (f ) N P denote the N th partial sum of the Fourier series, and tN (x) = cn einx be any n=−N

trigonometric polynomial. Then kf − SN f k2 ≤ kf − tN k2 and the equality holds iff fb(n) = cn . Proof. Consider kf − tN k22 = hf − tN , f − tN i = hf, f i − hf, tN i − htN , f i + htN , tN i.

But * hf, tN i =

N X

f,

+ cn e n

n=−N

=

N X

cn hf, en i

n=−N

=

N X

cn fb(n).

n=−N

Also * htN , tN i =

N X

n=−N

cn e n ,

m X

+ cm em

n=−m

38

(1.15)

N m X X

=

cn cm hen , em i

n=−N m=−m N m X X

=

cn cm δn,m

n=−N n=−m N X

=

|cn |2 .

n=−N

Hence (1.15) becomes N X

kf − tN k22 = hf, f i −

cn fb(n) −

N X

n=−N

n=−N

N X

N X

= hf, f i −

|fb(n)|2 +

n=−N

cn fb(n) +

N X

|cn |2

n=−N

|fb(n) − cn |2 .

n=−N

Thus kf − tN k22 = kf − SN f k22 +

N X

|fb(n) − cn |2 ,

n=−N

proving our assertion. Recall that a complex valued function f on T is said to be Lipschitz continuous on T if |f (s) − f (t)| < |s − t| for all s, t ∈ T. In the following theorem, we show that Lipschitz continuity leads to the uniform convergence of Fourier series. Theorem 1.7.4. If f is a Lipschitz continuous function on T, then SN (f ) converges to f uniformly on T. Proof. Consider sin (N +

  it −it 1 iN t 2 −iN t 2 t = e e −e e . As f is Lipschitz contin2i is bounded. Now, applying Riemann-Lebesgue lemma

f (x−t)−f (x) t sin 2 i f (x−t)−f (x) (x) e 2 t and f (x−t)−f t t sin sin 2 2

uous, the function to

1 )t 2

e

−i t 2 ,

we can conclude the result.

39

Theorem 1.7.5. Let f ∈ C(T). Suppose f is piecewise continuously differentiable function on T. Then the Fourier series of f converges uniformly to f on T. Proof. It is enough to prove that fb ∈ l1 (Z). Given that f is a piecewise continuously differentiable function on T. We know that T can be identified with [−π, π]. Then there exists a partition x1 , x2 , . . . , xm−1 ∈ [−π, π], such that −π = x0 ≤ x1 ≤ · · · ≤ xm−1 ≤ xm = π and f |[xj−1 ,xj ] is continuously differentiable for each j = 1, 2, · · · , n. Let gj denote the continuous derivative of f on [xj−1 , xj ]. Choose a periodic function g on [−π, π] such that g = gj on [xj−1 , xj ]. Then, clearly, g ∈ L2 (T). Using ∞ P Bessel’s inequality, we have |b g (n)|2 ≤ kgk22 . n=−∞

Consider, for n 6= 0 Zxj Zxj −inx xj 1 e 0 + e−inx f (x)dx. f (x)e−inx dx = f (x) −in xj−1 in xj−1

xj−1

Thus 1 fb(n) = 2π

Zπ

f (x)e−inx dx

−π x m Zj 1 X = f (x)e−inx dx. 2π j=1 xj−1

Now gb(n) |fb(n)| = in   1 1 2 ≤ + |b g (n)| . 2 n2 Thus, we get, ∞ X n=−∞

|fb(n)| = |fb(0)| +

X n6=0

40

|fb(n)|

 1 X 1 2 ≤ |fb(0)| + + |b g (n)| 2 n6=0 n2 < ∞,

from which it follows that fb ∈ l1 (Z). Theorem 1.7.6. For any f ∈ L2 (T), SN (f ) converges to f in L2 (T), i.e kSN f − f k2 → 0 as N → ∞. Proof. Let  > 0 be given. We know that C(T) is dense in L2 (T). Therefore, given f ∈ L2 (T), there exists h ∈ C(T) such that

kf − hk2 < .

(1.16)

Writing f − SN f = (f − h) + (h − SN h) + SN (h − f ) and noting that kSN (h − f )k2 ≤ kf − hk2 , it is enough to show that kh − SN hk <  for all N ≥ N0 for some N0 . As h ∈ C(T), there exists a trigonometric polynomial P of degree N0 such that kh − P k2 < . For N ≥ N0 , SN P = P and hence

kh − SN hk2 ≤ kh − P k2 + kSN (P − h)k2 ≤ 2.

This proves our assertion. Theorem 1.7.7. (Parseval’s Theorem) For any f, g ∈ L2 (T), 1 2π

Zπ f (x)g(x)dx =

∞ X n=−∞

−π

41

g (n). fb(n)b

In particular, 1 2π

Zπ

|f (x)|2 dx =

∞ X

|fb(n)|2 .

n=−∞

−π

Proof. We have shown in Theorem 1.7.6 that SN f → f in L2 . Hence, hSN f, SN gi → hf, gi which simply means ∞ X

1 g (n) = fb(n)b 2π n=−∞

Zπ f (x)g(x)dx. −π

Corollary 1.7.8. The functions {einx : n ∈ Z} forms an orthonormal basis for L2 (T). (See Appendix A.1.3.) Corollary 1.7.9. Let f ∈ L2 (T). Then the Fourier series of f can be integrated term by term in (a, b) ⊆ [−π, π]. Proof. Let g(t) = χ[a,b] (t). Then g ∈ L2 (T) and hence by Parseval’s theorem, Zπ

1 2π

f (t)g(t)dt =

∞ X

fb(n)b g (n).

n=−∞

−π

This means that 1 2π

Zb a

 b  Z ∞ X 1 fb(n)  eint dt , f (t)dt = 2π n=−∞ a

which is our assertion.

42

1.8

Existence of a continuous function whose Fourier series diverges We shall show in this section that, there exists a continuous function whose

Fourier series diverges at 0. In order to prove the theorem, we make use of the uniform boundedness principle from functional analysis. (See Appendix A.2.5.) Let X = C[−π, π], Y = R. Define AN f := SN f (0). Then |AN f | = |SN f (0)| = |DN ∗ f (0)| ≤ kf k∞ kDN k1 . In other words, the operator norm of AN is given by kAN k ≤ kDN k1 . We shall show that kAN k = kDN k1 . Take g(x) = sgnDN (x). Then kgk∞ = 1. Although g is not continuous, it can be approximated by a sequence of continuous functions gn . Choose gn such that kgn k∞ = 1. Then AN gn → kDN k1 1, by using Lebesgue dominated convergence theorem. Thus, kAN k = kDN k1 . But kDN k1 ≥ c log N (see Remark 1.5.3). Hence, kAN k → ∞ as N → ∞. Hence, it follows from uniform boundedness principle that there exists a continuous function f in [−π, π] such that |AN f | = |SN f (0)| → ∞ as N → ∞, which means the Fourier series of f diverges at 0.

1.9

Brief history of Fourier series The appearance of Fourier series can be traced back to the works of d ’Alem-

bet(1747), Euler (1748) and D.Bernoulli (1753) in the study of vibrating strings, but the theory of Fourier series truly began with the profound work of Fourier on heat conduction at the beginning of the nineteenth century. He studied the problem of heat flow in a thin wire fixed at two end points with length π and zero temperature 43

at the ends. He showed that the initial temperature can be expressed as the sum of an infinite series of sines and cosines, now popularly known as Fourier series. Even though Fourier did not give a convincing proof of convergence of such infinite series, he offered the conjecture that convergence holds for an arbitrary function. Subsequent work by Dirichlet, Riemann, Lebesgue, and others, throughout the next two hundred years was needed to show that any arbitrary periodic function can be expressed in terms of trigonometric series.

1.10

Solved problems 

k

1.10.1. Suppose that f ∈ C (T). Show that fb(n) = O

1 |nk |

 as n → ∞.

Solution. 1 fb(n) = 2π 1 = 2π

Zπ

f (x)e−inx dx

−π Zπ

 f (x)d

e−inx −in

 .

−π

Using Bernoulli’s formula for successive integration by parts, we get (−1)k k b (f ) (n). fb(n) = (−in)k Then 1 |(f k )b(n)| |n|k 1 kf (k) k1 . ≤ k |n|

|fb(n)| ≤

44

Zπ

1 1.10.2. If f ∈ L1 (T), then show that fb(n) = 4π

(f (x) − f (x + π/n)) e−inx dx. In

−π

 addition, if f satisfies the Holder’s condition of order α, then fb(n) = O Solution. We have, Zπ

1 fb(n) = 2π

f (x)e−inx dx.

−π iπ

Since e = −1, we can write −1 fb(n) = 2π =

−1 2π

Zπ −π Zπ

f (x)e−inx+πi dx π

f (x)e−in(x− n ) dx.

−π

−1 = 2π

π−π/n Z

f (y + πn )e−iny dy.

−π−π/n

Hence −1 fb(n) = 2π

Zπ

f (x + πn )e−inx dx.

−π

Thus 1 2fb(n) = 2π

Zπ

−inx

f (x)e

1 − 2π

−π

Zπ

f (x + πn )e−inx dx,

−π

from which it follows that, 1 fb(n) = 4π

Zπ


f (x) − f (x + πn ) e−inx dx.

−π

45

1 |n|α

 .

Then 1 |fb(n)| ≤ 4π ≤

1 4π

Zπ −π Zπ


f (x) − f (x + πn ) dx π α c · 2π n

−π

k = , |n|α where k is a constant.

ZR

1.10.3. Using Riemann-Lebesgue lemma, prove that lim

R→∞ 0

ZR Solution. Let I = lim

R→∞

sin x dx. Let R = (N + 21 )π, then x

0

1 (N + )π Z2

I = lim

N →∞

sin x dx. x

0

Applying change of variable x = (N + 12 )y, we get Zπ I = lim

N →∞

sin (N + 12 )y dy. y

0

From the property of Dirichlet kernel, we get Zπ 0

sin (N + 12 )y dy = π sin ( y2 )

46

sin x π dx = . x 2

Hence, Zπ  I − π/2 = lim

N →∞ 0

Zπ  = lim

N →∞ 0

Since,

1 1 − y 2 sin

y 2

sin (N + 12 )y sin (N + 12 )y dy − y 2 sin ( y2 ) 1 1 − y 2 sin ( y2 )



 dy

sin (N + 12 )y dy.


∈ L2 (T), from Riemann-Lebesgue lemma, it follows that I = π/2.

1.10.4. Prove the Binomial identity

n P k=0


n 2 k

=

2n n


.

Solution. Consider n   X n ek (x) = (1 + e1 (x))n . k k=0

By Parsaval’s inequality n  2 X n k=0

k

1 = 2π =

=

=

=

1 2π 1 2π 1 2π 1 2π

Zπ −π Zπ

−π Zπ

−π Zπ

−π Zπ

Zπ

1 |f (x)| dx = 2π 2

|(1 + e1 (x))n |dx

−π

(1 + e1 (x))n (1 + e1 (x))n dx

(1 + e1 (x))n (1 + e1 (x))n dx

(1 + e1 (x))(1 + e1 (x))n dx

(2 + 2 cos x)n dx

−π

47

1 = · 2n 2π

Zπ

(1 + cos x)n dx

−π

Zπ

22n · cos2n 2π −π   2n = . k

=

x 2




dx

1.10.5. By considering the Fourier series of f (x) = |x|, x ∈ [−π, π], find the sum of ∞ ∞ 1 ∞ 1 P P P 1 the series and . What is the sum of . 4 4 2 n=0 (2n + 1) n=0 n n=1 n Solution. We have 1 fb(n) = 2π

Zπ

|x|e−inx dx

−π

 π  Z Zπ   1 |x| cos nxdx − i |x| sin nxdx =  2π  −π −π  π  Z   1 = 2 x cos nxdx  2π  0   π Zπ  1 x sin nx sin nx  = − dx if n 6= 0  π n 0 n 0

=−

1 1 · π n

Zπ sin nxdx 0

π 1 = 2 cos nx 0 nπ 1 = 2 {(−1)n − 1}. nπ 48

For n 6= 0, we have   0 fb(n) =  −2 n2 π

1 fb(0) = 2π =

1 π

if n is even if n is odd

Zπ |x|dx

−π Zπ

xdx 0

1 x2 = π 2 π = . 2 

π 0

Thus the Fourier Series of f (x) = |x| on [−π, π] is given by

|x| = fb(0) +

∞ X

fb(n)einx

n=−∞ n6=0

=

X −2 π + einx 2 n=±1,±3.. n2 π

∞ X π 1 = + 2(−2) ei(2n−1)x , x ∈ [−π, π] 2 2 π(2n − 1) n=1 ∞ 4X 1 π |x| = − ei(2n−1)x , x ∈ [−π, π] 2 2 π n=1 (2n − 1)

On putting x = 0, we get ∞ π 4X 1 0= − 2 π n=1 (2n − 1)2

49

Hence, it follows that ∞ X 1 π2 . = 2 (2n − 1) 8 n=1 Now we shall calculate the sum of the series

∞ 1 ∞ 1 P P . Let s = . Then 2 2 n=1 n n=1 n

∞ ∞ X X 1 1 + s= 2 n n2 n=1,3 n=2,4 X 1 X 1 = + 2 (2m) (2m − 1)2 m∈N m∈N ∞ 1X 1 π2 = + 4 m=1 m2 8

Thus 1 π2 s= s+ . 4 8 This leads to ∞ X 1 π2 = n2 6 n=1

Using Parseval’s inequality, we have 1 2π

Zπ −π

∞ X π2 |fb(n)|2 . x dx = + 4 n=−∞ 2

n6=0

Hence, 1 π

Zπ 0

x2 dx =

X π2 4 + , 4 4 n π2 n=±1,±3,...

which means that 1 π3 π2 8 X 1 · = + 2 . π 3 4 π n=1,3,... n4 50

Thus it follows that π2 π2 8 X 1 − = 2 , 3 4 π n=0,1,... (2n + 1)4 from which we arrive at 8 X 1 π2 = 2 . 12 π n=0,1,... (2n + 1)4 This implies that ∞ X n=0

π4 1 = . (2n + 1)4 96

Next, we shall calculate the sum

∞ 1 P . 4 n=1 n

Let ∞ X 1 s = . n4 n=1 0

Then s0 = =

X 1 X 1 + n4 n odd n4 n even ∞ X n=1

∞

X 1 1 + 4 (2n) (2n + 1)4 n=0 ∞

∞

X 1 X 1 1 = + . 4 16 n=1 n (2n + 1)4 n=0 Hence s0 −

1 0 π4 s = . 16 96

51

This leads to ∞ X 1 π4 . = 4 n 90 n=1

1.10.6. Compute the Fourier coefficients of the function f given by   |x| |x| < 1 f (x) =  0 otherwise.

Solution. We have 1 fb(n) = 2π

Zπ

f (x) e−inx dx

−π

1 = 2π

Z1

|x| e−inx dx

−1

 0  Z Z1   1 −x e−inx dx + xe−inx dx =  2π  −1 0   Z0 −inx Z1 −inx   −inx 1 −inx 0 xe 1 xe − e + e =− dx − dx  2π  −in −1 −in −in 0 −in −1 0 (  −inx 0  −inx 1 ) in −in 1 e 1 e e −1 e − =− + + 2π (−in) in −in −1 (−in) in −in 0  
1 1 −in 1 ein e−in 1 in =− e −e + 2 − 2 − 2 + 2 2π in n n n n  

1 1 −in 2 1 in −in in =− e −e + 2 − 2 e +e 2π in n n   1 2 2 sin n 2 cos n =− − − 2 2π n n n2 52

1 =− nπ



1 cos n − sin n − n n

 .

1.10.7. Compute the Fourier coefficients of 1-periodic function f (t) = 1, 0 ≤ t ≤ 1. Solution. We have Z1 fb(n) =

e−2πint dt

0

1 e−2πint = 2πin 0  i  −2πin = e −1 2πn i = [(−1)n − 1] , 2πn

n 6= 0

But Z1 fb(0) =

dt = 1. 0

Hence i [(−1)n − 1] n 6= 0 b 2πn f (n) =  1 n = 0.  

1.10.8. Let f ∈ L1 (T) such that fb(0) = 0. Define F (t) =

Rt 0

fb(n) continuous, 2π-periodic and Fb(n) = , n 6= 0. in Solution. Let  > 0 be given. Then for t > s, t Z Zs |F (t) − F (s)| = f (τ ) dτ − f (τ ) dτ 0

0

53

f (τ ) dτ . Show that F is

t Z = f (τ ) dτ s Zt

|f (τ )| dτ.

≤ s

As f ∈ L1 (T), there exists a δ > 0 such that Zt |f (τ ) |dτ ≤  whenever |t − s| < δ. s

Thus, F is continuous. Consider t+2π Z

F (t + 2π) =

f (τ ) dτ 0

Z2π =

2π+t Z

f (τ ) dτ + 0

f (τ ) dτ 2π

Zt = 2π fb(0) +

f (τ ) dτ 0

= F (t). Thus F is 2π-periodic. By Radon-Nikodym theorem, F is differentiable with F 0 (t) = f (t) − f (0). Thus Z2π Fb(n) =

F (t)e−int dt

0 Z2π

=

 F (t) d

0

54

e−int −in



2π Z2π −int e e−int = F (t) − F 0 (t)dt −in 0 −in 0

1 = in

Z2π

f (t)e−int dt − f (0)

0

=

Z2π

e−int dt

0

1 b f (n), in

n 6= 0.

1.10.9. Calculate χ[−π,0] ∗ χ[0,π] in L1 (T). Solution. Note that 1 χ[−π,0] ∗ χ[0,π] (t) = 2π

Z0 χ[0,π] (t − s) ds. −π

Now, for this integral to be non-zero, s should belong to [−π, 0] and [−π + t, t]. If −π ≤ t ≤ 0 1 χ[−π,0] ∗ χ[0,π] (t) = 2π

Zt dt =

1 [t + π]. 2π

−π

If 0 < t ≤ π, then Z0

1 χ[−π,0] ∗ χ[0,π] (t) = 2π

dt =

1 [−t + π]. 2π

−π+t

Hence χ[−π,0] ∗ χ[0,π] (t) =

1 (π − |t|)χ[−π,π] (t). 2π

1.10.10. Suppose θ : R → T is a continuous group homomorphism then show that θ(t) = eitx for some real number x. Use this to derive the form of the continuous group homomorphism θ : T → T. 55

Solution. If θ is the trivial homomorphism, then we are done. If not, there exists a a > 0 such that Za θ(t)dt 6= 0.

c= 0

Thus Za cθ(x) = 0

Za+x θ(x + t)dt = θ(t)dt. 0

So, θ is differentiable and 1 1 θ0 (x) = [θ(a + x) − θ(x)] = [θ(a) − 1]θ(x) = c0 , c c 1 [θ(a) − 1]. Thus, θ satisfies the ordinary differential equation θ0 = c0 θ. c 0 Hence, θ(t) = ec t and since |θ(t)| = 1 ∀ t, c0 = 2πix for some x ∈ R. where c0 =

Since T ' R/Z, any continuous group homomorphism θ : T → T gives rise to a continuous group homomorphism θ0 : R → T given by θ0 = θ · π, where π is the canonical quotient map, π : R → R/Z. Also, ker θ0 contains Z. Thus by the form of the continuous homomorphism from R → T, θ(t) = eint for some integer n. 1.10.11. Let G be a closed proper subgroup of T. Show that G is finite. Solution. Since G is closed, T/G is an abelian group. Let θ : T/G → T be a continuous group homomorphism. Then the map θ0 : T → T defined as θ0 = θ · π, π : T → T/G the canonical quotient map, is a continuous group homomorphism of T to T whose kernel contains G. But by the previous problem, θ0 (t) = eint for some integer n. Therefore ker(θ0 ) is finite and hence G is finite. 1.10.12. If {kn } is a sequence of good kernels, show that for 1 ≤ p < ∞, kkn ∗ f − f kp → 0 as n → ∞. 56

Solution. Suppose that the conclusion is true for any g ∈ C(T) ⊆ Lp (T). Let Rπ 1 |kn (t)|dt = c < ∞. As C(T) is f ∈ Lp (T). Since {kn } forms a good kernels, 2π 0

dense in Lp (T), given  > 0, there exists g ∈ C(T) such that kf − gkp
0 such that kft − f kp < , whenever |t| < δ. Now, for this δ there exists an n0 ∈ N such that for all 2π−δ R 1 n ≥ n0 , 2π |kn (t)|dt < . Thus δ

 kkn ∗ f − f kp ≤

1   2π

Zδ

2π−δ Z

|kn (t)|dt + 2kf k∞

−δ



|kn (t)|dt δ

Z2π |kn (t)|dt + 2kf k∞ 

0    sgn(n) = 0 n=0     −1 n < 0 show that

Rπ

˜ N (x)|dx ≤ c ln N . |D

−π

Solution. We know that 1 x ˜ N (x) = i cos ( 2 ) − cos (N + 2 )x . D sin ( x2 )

Then Zπ Zπ cos ( x2 ) − cos (N + 12 )x ˜ dx DN (x) dx = sin ( x2 )

−π

−π Zπ

= −π Zπ

=

cos ( x2 ) − cos N x cos( x2 ) + sin N x sin ( x2 ) dx sin ( x ) 2

cos ( x2 )[1 − cos (N x)] dx + sin N x sin ( x )

−π Zπ

2

1 − cos (N x) dx + 2π ≤ sin ( x2 ) −π Zπ 2 N x
sin 2 =2 dx + 2π sin ( x2 ) −π

59

π

Z2 2 sin x =4 sin x dx + 2π −

π 2

π

Z2 2 sin (N x) dx + 2π =8 sin x 0 π Z2 2 sin N x ≤8 dx + 2π 2x 0 π π Z2 2 sin N x dx + 2π = 4π x 0 Nπ Z2

sin2 y

= 4π

|y| 0 Nπ Z2

= 4π

dy + 2π

|sin y| dy + 2π |y|

0 π  nπ 2 Z2 Z N X  sin x  1 dx + = 4π  dx + 2π   x  x n=2 0

=c+

N X n=2 N X

(n−1)

nπ 2 ln x|(n−1) π + 2π 2

n π2 π + 2π (n − 1) 2 n=2   N X n =c+ ln n−1 n=2   N n = c + ln Π n=2 n − 1

=c+

ln

60

π 2

= c + ln N ≤ c ln N.

1.10.15. Let f ∈ C(π). For each pair of integers m, n, where 0 ≤ m < n, define n P 1 Si (f, x), where Si (f, x) stands for i-th partial sum of f at x. σm,n (f, x) = n−m i=m+1

(a) Show that σkn,(k+1)n (f ; x) → f (x) uniformly as n → ∞, where k, n are integers. (b) If k, m, n are positive integers with kn ≤ m ≤ (k +1)n show that |σkn,(k+1)n (f ; x)− Sm (f ; x)| ≤

2A , k

where f is such that |fb(j)| ≤

A |j|

for j 6= 0.

(c) Using (a) and (b) show that Sn (f ; x) → f (x) uniformly on T whenever fb(n) = O( n1 ). Solution. For integers m and n, we have σm,n (f ; x) =

n P 1 Si (f ; x). Rewritn − m i=m+1

ing this in terms of the kernel, we obtain

σm,n (f ; x) =

1 [(n + 1)σn+1 (f ; x) − (m + 1)σm+1 (f ; x)] n−m

(1.17)

Using the Fourier expansion of the kernel, one obtains the following Fourier expansion for σm,n (f ) as follows:

σm,n (f ; x) = Sm (f ; x) +

X n + 1 − |j| fb(j)eijx . n−m

(1.18)

m 0 be given. |n|  A < . By (a) we can find n0 ≥ k such that for Choose an integer k > 0 such that k 4  all n ≥ n0 , |σkn,(k+1)n (f ; x) − f (x)| < . Let m ≥ kn0 such that for some n ≥ n0 , 2 2A  kn ≤ m < (k + 1)n. By (b), we have |σkn,(k+1)n (f ; x) − Sm (f ; x)| < < . Thus k 2 by the above two inequalities, (c) follows. ∞ P

1 (−1)n−1 is Cesaro summable to . 2 n=1   1 ∞ if n is odd P Solution. (−1)n−1 . Here an =  −1 if n is even. n=1   1 if n is odd Here sn is the n-th partial sum of an and sn =  0 if n is even.

1.10.16. Show that

Consider s1 + s2 + · · · + sn . n s1 + s2 + · · · + s2n = 2n n = 2n

σn = σ2n

62

1 ∀ n ∈ N. 2 1 = 2 s1 + s2 + · · · + s2n−1 = 2n − 1 n−1+1 n = = . 2n − 1 2n − 1

= lim σ2n

n→∞

σ2n−1

1 as n → ∞ 2

Therefore σ2n−1 →

1 lim σn = . n→∞ 2 ∞ P

Hence

(−1)n−1 is Cesaro summable to

n=1

1.10.17. Show that

∞ P

1 . 2

(−1)n (n + 1) is not Cesaro summable but Able summable to

n=0

1 . 4 ∞ P Solution. We know that (−1)n (n + 1)rn = 1 − 2r + 3r2 − · · · = (1 + r)−2 for n=0

0 < r < 1. Then ∞ X

1 (−1)n (n + 1)rn = lim(1 + r)−2 = . r→1 r→1 4 n=0 ∞ P 1 Therefore, (−1)n (n + 1) is Abel summable to . Now consider the series 4 n=0 ∞ ∞ X X (−1)n (n + 1) = (−1)n−1 n. lim

n=0

Here sn =

 

−n 2

if n is even



n+1 2

if n is odd.

n=1

Clearly, σ2n =

s1 + s2 + · · · + s2n = 0 for all n, 2n 63

s1 + s2 + · · · + s2n−1 2n − 1 2n − 1 + 1 = 2(2n − 1) n = . 2n − 1

σ2n−1 =

∞ P 1 as n → ∞. Hence (−1)n (n + 1) is not Cesaro summable. 2 n=0 ∞ ∞ P P 1.10.18. Prove that if ci converges to s, then ci is Cesaro summable to s.

Therefore, σ2n−1 →

i=1

Solution. Let sn =

n P

i=1

ci . The assumption is that sn → s as n → ∞. Given  > 0

i=1

there exists m ∈ N such that |sn − s| <  for all n > m. Now for all n > m, we can write s1 + s2 + · · · + sn |σn − s| = − s n (s1 − s) + · · · + (sm − s) + (sm+1 − s) + · · · + (sn − s) = n n m P P (si − s) (si − s) i=m+1 i=1 + ≤ n n A n−m  ≤ + n n A ≤ + , n m P

A <  for all n ≥ m1 . n i=1 ∞ P cn is Choose m0 = max{m, m1 }. Then |σn − s| <  for all n ≥ m0 . The series

where A =

(si − s). Now we can find m1 ∈ N such that

n=1

Cesaro summable to s.

64

1.10.19. Assume that that

∞ P

∞ P

an is Cesaro summable to l and lim nan = 0. Then show n→∞

n=1

an converges to l.

n=1

Solution. Assume that

∞ P

an is Cesaro summable to l and lim nan = 0. Consider, n→∞

n=1 s1 + s2

+ · · · + sn sn = ai and σn = . n i=1 By the assumption, lim σn = l n P

n→∞

s1 + s2 + · · · + sn n a1 + (a1 + a2 ) + · · · + (a1 + a2 + · · · + an ) = n  n  X i−1 = 1− ai . n i=1

σn =

Now σn − sn =

n  X i=1

i−1 1− n

n

=

 ai −

n X

ai

i=1

n

1X 1X ai − i ai n i=1 n i=1 n

sn 1X = − i ai n n i=1   n 1 1X 1+ sn = σ n + iai n n i=1

(1.20)

n 1P i ai = 0. Taking limit as n → ∞ in 1.20, we have Since lim nan = 0, lim n→∞ n→∞ n i=1
lim 1 + n1 sn = lim σn = l. Thus lim sn = l.

n→∞

n→∞

n→∞

Here we have used the following result. If lim an = m 6= 0, lim an bn = l then n→∞ n→∞ l lim bn exists and equals . n→∞ m 65

1.10.20. If

∞ P

Cn is Cesaro summable to l, then prove that

n=1

∞ P

Cn is Abel summable

n=1

to l. Solution. Assume that

∞ P

Cn is Cesaro summable to l. Define sn and σn as before

n=1

then lim σn = l. It can be easily seen that C1 = s1 ; Cn = sn − sn−1 for all n ≥ 2 and n→∞

s1 = σ1 , sn = nσn − (n − 1)σn−1 for all n ≥ 2. Consider 0 < r < 1. Now it can be ∞ P easily seen that nσn rn−1 converges absolutely for all 0 ≤ r < 1. Then n=1

(1 − r)

2

∞ X

nσn r

n−1

= (1 − r)

n=1

= (1 − r)

∞ X

[nσn rn−1 − nσn rn ]

n=1 " ∞ X

nσn rn−1 −

= (1 − r) σ1 + = (1 − r) s1 + " = (1 − r) s1 +

# nσn rn

n=1

" n=1

"

∞ X

∞ X (n − 1)σn−1 rn−1 nσn rn−1 −

∞ X n=2 ∞ X n=2 ∞ X

n=2

# (nσn − (n − 1)σn−1 ) rn−1 # sn rn−1

n=2

= (1 − r)

∞ X

sn rn−1

n=1

=

∞ X

sn r

n−1

n=1

= s1 +

−

∞ X

sn r n

n=1 ∞ X

sn rn−1 −

∞ X

n=2

= s1 + = c1 +

∞ X n=2 ∞ X

sn−1 rn−1

n=2

(sn − sn−1 ) rn−1 Cn rn−1 =

n=2

∞ X n=1

66

Cn rn−1 .

#

Since

∞ P

nσn rn−1 converges for 0 ≤ r < 1, it follows that

n=1

0 ≤ r < 1. Define f (r) =

∞ P

∞ P

Cn rn−1 converges for

n=1

Cn rn−1 for all 0 ≤ r < 1. Now we shall show that

n=1

lim f (r) = l. Choose  > 0. Then there exists m ∈ N such that |σn − l| <  for all

r→1

n ≥ m. Now choose δ = . Then for all 1 −  < r < 1, we have ∞ X Cn rn−1 − l |f (r) − l| = n=1 ∞ ∞ X X nrn−1 nσn rn−1 − l(1 − r)2 = (1 − r)2 n=1 n=1 ∞ X = (1 − r)2 n(σn − l)rn−1 n=1

≤ (1 − r)2

∞ X

n |σn − l| rn−1

n=1

= (1 − r)2

"m−1 X

nrn−1 |σn − l| + 

= (1 − r)2

# nrn−1

n=m

n=1

"m−1 X

∞ X

# n |σn − l| + (1 − r)−2

n=1 2

= k(1 − r) + , where k =

m−1 X

n|σn − l|

n=1

≤ k2 + . Now for any 0 > 0, there exists unique  > 0 (in fact other root is negative) such that k2 +  = 0 . Hence lim f (r) = l. r→1

1.10.21. Assume that f is periodic and monotone on [−π, π].Then show that fb(n) =   1 O |n| . 67

Solution. Assume that f is monotone. Then by Lebesgue-Young theorem f is differentiable a.e and f 0 ∈ L1 [−π, π]. Now 1 fb(n) = 2π

Zπ

f (t) e−int dt

−π



 Zπ −int 1  e + f 0 (t) e = f (t) dt . 2π −in −π in −int π

(1.21)

−π

The first term of the R.H.S of (1.21) is zero since f (π) = f (−π). Then fb(n) = 1 Rπ 0 f (t)e−int dt which implies 2πin −π b f (n) ≤

Zπ

1 2π|n|

|f 0 (t)| dt

−π

K , since f 0 ∈ L1 [−π, π] 2π|n|   1 b f (n) = O . |n| =

1.11

Exercises

1.11.1. Let f ∈ L1 (T). Show that lim

Rπ

δ→0 −π

|f (x + δ) − f (x)| dx = 0.

1.11.2. Show that Fourier series of the function   1 x is rational f (x) =  0 x is irrational is identically 0 and does not converge to f for any rational number x.

68

Chapter 2 Applications of Fourier series One of the applications of Fourier series is the evaluation of certain infinite sums. For ∞ 1 ∞ 1 P P example, , , are computed in Chapter 1 (see for example, Remark 2.4.1). 2 4 n=1 n n=1 n In this chapter, we study several other applications of Fourier series.

2.1

The values of Riemann zeta function at even positive integers The Riemann zeta function is defined by ζ(s) =

∞ 1 P , Re(s) > 1. It is s n=1 n

analytic in the region {s ∈ C : Re(s) > 1}.

Figure 2.1: The Riemann Zeta function. One of the applications of Fourier series is evaluation of the values of Riemann zeta function at even positive integers. In order to study this application, we make 69

use of Bernoulli numbers and polynomials. These Bernoulli numbers Bn are defined 1 n+1
as follows: We set B0 = 1 and for n ≥ 1, define Bn = − Bk . Bernoulli n+1 k polynomials are defined recursively as follows:

B0 (x) = 1, 0

Bn (x) = nBn−1 (x) for n ≥ 1, and Z1 Bn (x) dx = 0. 0 0

Let us compute B1 (x). By definition, B1 (x) = 1. On integration, we get B1 (x) = R1 x + c, where c is a constant. But B1 (x) dx = 0. Hence, c = − 12 . In other words, 0

B1 (x) = x− 12 . Similarly Bn (x) can be calculated for n ≥ 2. The Bernoulli polynomials ∀ n ≥ 2.

satisfy the equation Bn (0) = Bn (1) = Bn

Functions on [0,1] can be considered as 1-periodic functions, i.e functions satisfying f (x + 1) = f (x), on R and they can be expanded in terms of e2π ikx , k ∈ Z. (See, for example, Remark 2.4.1.) Thus Bernoulli polynomials have a well defined Fourier series. Z1 b1 (k) = B

(x − 12 )e−2πikx dx

0

=−

1 2πik

k 6= 0,

and Z1 b1 (0) = B

(x − 21 ) dx = 0.

0

70

Thus the Fourier series of B1 (x) can be written as X −1 B1 (x) = e2πikx . 2πik k6=0 For n ≥ 2, the Fourier coefficients of Bn (x) are calculated as follows: Z1 bn (k) = B

Bn (x)e−2πikx dx

0

1 Z1 1 Bn (x)e−2πikx + Bn0 (x)e−2πikx dx, k 6= 0 = −2πik 0 2πik 0

=

1 2πik

Z1

nBn−1 (x)e−2πikx dx

0

=

n b Bn−1 (k) k 6= 0, 2πik

bn (0) = 0. Thus by induction we have and B Bn (x) =

−n! X e2πikx . (2πi)n k6=0 k n

For even integers 2n the above takes the form: X e2πikx (2n)! (2π)2n (i)2n k6=0 k 2n ( −1 ) ∞ (−1)n+1 (2n)! X e2πikx X e2πikx = + 2n (2π)2n k k 2n k=−∞ k=1 ( ∞ ) ∞ (−1)n+1 (2n)! X e−2πikx X e2πikx + . = 2n 2n (2π)2n k k k=1 k=1

B2n (x) = −

71

Thus (−1)n+1 (2n)! B2n (x) = (2π)2n

(

) ∞ X 1 · 2cos(2πkx) . 2n k k=1

Evaluating at x = 0, we get ∞

(−1)n+1 2(2n)! X 1 B2n (0) = . (2π)2n k 2n k=1 Thus for even positive integers, m = 2n, n ∈ N, we get ζ(2n) = B2n ·

2.2

(−1)n+1 (2π)2n . 2(2n)!

Isoperimetric inequality Among all the regions in the plane enclosed by a piecewise C 1 curve with a

given perimeter L the disc has the maximum area. If A is the area of the region then we have 4πA ≤ L2 and this inequality is called the isoperimetric inequality. Equality holds if and only if the curve is a circle. In order to prove isometric inequality, first we state and prove Wirtinger’s inequality. Theorem 2.2.1. (Wirtinger’s inequality) Let f be a continuous and piecewise C1 function on T . Then Zπ

Zπ

2

|f (x) − fb(0)| dx ≤ −π

|f 0 (x)|2 dx.

−π

Equality holds if and only if f (x) = fb(0) + fb(1)eix + fb(−1)e−ix . 72

Proof. Given that f is a continuous and piecewise C1 function on T . We can write the Fourier series of f as

f (x) = fb(0) +

X

fb(n)einx .

n6=0

From Plancheral theorem (Parseval identity) applied to fb(x) − fb(0) we get 1 2π

Zπ 2 2 X b b f (n) f (x) − f (0) dx = . n6=0

−π

Notice that f 0 (x) ∼

P

infb(n) einx . Since f 0 is piecewise continuous, f 0 is Riemann

n6=0

integrable on T . Hence, (f 0 )2 is Riemann integrable on T . This shows that f 0 ∈ L2 (T ). Again applying Parseval identity, we get 1 2π

Zπ

|f 0 (x)|2 dx =

X

n2 |fb(n)|2 .

(2.1)

n6=0

−π

Now, 1 2π

Zπ

1 |f (x)| dx − 2π 0

2

−π

=

X

Zπ

|f (x) − fb(0)|2 dx

−π

n2 |fb(n)|2 −

n6=0

X

|fb(n)|2 ,

n6=0

X = (n2 − 1)|fb(n)|2 ≥ 0. n6=0

This proves Wirtinger’s inequality. Clearly, equality holds if and only if (n2 − 1) |fb(n)|2 = 0 for all n 6= 0. This means that fb(n) = 0 for all integers |n| ≥ 2. Thus f (x) = fb(0) + fb(1) eix + fb(−1) e−ix . 73

Now, we shall prove the isoperimetric inequality. Suppose that the boundary curve Γ having length L is parameterized by the arc length. In other words, the boundary curve Γ : z(s) = x(s) + i y(s) satisfies the equation 

dx ds

2

 +

dy ds

2 = 1.

Let  f (θ) = x

Lθ 2π



 , g(θ) = y

Lθ 2π

 .

Then f and g are 2π-periodic functions on R. Further, 2  2 L 0 L 0 x (θ) + y (θ) f (θ) + g (θ) = 2π 2π
L2 = 2 x0 (θ)2 + y 0 (θ)2 4π L2 = 2. 4π 0

0

2

2



Also note that Zπ

g 0 (θ) dθ = g(π) − g(−π) = 0.

−π

Now, let Ω be the region enclosed by Γ. Then Zπ A = Area (Ω) =

Zπ

0

f (θ)g (θ) dθ = −π

≤

(f (θ) − fb(0))g 0 (θ) dθ

−π

 π Z 1 2

(f (θ) − fb(0))2 dθ +

−π

Zπ −π

74

g 0 (θ)2 dθ

  

≤

 π Z 1 2

f 0 (θ)2 dθ +

−π

Zπ

g 0 (θ)2 dθ

 

,



−π

using “ab ≤ 21 (a2 + b2 )” followed by Wirtinger’s inequality. Thus 1 A≤ 2

Zπ


f 0 (θ)2 + g 0 (θ)2 dθ

−π

L2 , = 4π proving our assertion. Equality in isoperimetric inequality forces equality in Wirtinger’s Rπ inequality and equality in “ab ≤ 21 (a2 + b2 )” which yields (f − fb(0) − g 0 )2 dθ = 0. −π

This implies that f (θ) = fb(0) + fb(1) eiθ + fb(−1) e−iθ and f − fb(0) − g 0 = 0. Thus it follows that g(θ) =

fb(1) eiθ fb(−1) e−iθ + + c, i (−i)

where c is a constant of integration. Put fb(n) = α(n) + iβ(n). Then Zπ α(1) = Re

and

Zπ

f (θ) e−iθ dθ =

f (θ) cos θ dθ = α(−1),

−π

−π

Zπ

Zπ

β(1) = Im −π

f (θ) e−iθ dθ =

f (θ) sin θ dθ = −β(−1). −π

Thus f = fb(0) + 2α(1) cos θ − 2β(1) sin θ, and g = c + 2β(1) cos θ + 2α(1) sin θ. Thus  2 L2 f − fb(0) + (g − c)2 = 4[α(1)2 + β(1)2 ]. On the other hand, f 0 (θ)2 + g 0 (θ)2 = 4π 2,

75

which leads to 4[α(1)2 + β(1)2 ] =

L2 . 4π 2

Thus f − fb(0))2 + (g − c)2 =

L2 , 4π 2

showing that

the resulting curve is a circle.

2.3

Jacobi identity for the theta function ∞ P

The theta function is defined as θ(t) =

exp (−πn2 t), t > 0. The Jacobi

n=−∞ 1
identity states that θ(t) = t− 2 θ 1t . In order to prove this identity, consider f (x) =   ∞ P −(x−k)2 exp . The series converges uniformly on [0, 1] and defines a 1-periodic 2t n=−∞

function on R, ı.e f (x + 1) = f (x) for all x ∈ R. We compute the Fourier coefficients of f on the interval [0, 1]. We have

fb(n) =

Z1 X ∞

−(x − k)2 2t





−(x − k)2 2t



exp

k=−∞

0

=



∞ Z1 X

exp

k=−∞ 0 −k+1 Z ∞

=

X

exp

k=−∞ −k  Z∞

=

exp



−x2 2t



−x2 2t



e−2πinx dx

e−2πinx dx

e−2πinx dx

e−2πinx dx

−∞

=

√

2πt e−2π

2 n2 t

.

The last inequality will be proved in Proposition 3.3.2 later. Clearly, fb ∈ l1 (Z). Thus

f (x) =

√

2πt

∞ X


exp −2π 2 n2 t e2πinx .

n=−∞

76

Letting x = 0, and changing t to

t , 2π

we get

  ∞ ∞ X √ X −πn2 2 t exp (−πn t) = exp . t n=−∞ n=−∞ In other words, 1

θ(t) = t− 2 θ

2.4

1 t




.

Weierstrass approximation theorem Before stating this theorem, we observe the following:

Remark 2.4.1. Let {kn } be a sequence of functions defined on [−1, 1] such that R1 (i) kn (x) dx = 1 ∀ n ≥ 1. −1 R1

(ii)

|kn (x)| dx ≤ M

−1

(iii) For every δ > 0,

∀ n ≥ 1 for some M > 0. R |kn (x)| dx → 0 as n → ∞.

δ≤|x|≤1 R1

Let f ∈ C[−1, 1]. Then

f (x − t)kn (t) dt converges to f uniformly on [−1, 1]. The

−1

proof of this theorem is exactly the same as in Theorem 1.5.1. Theorem 2.4.2. (Weierstrass approximation) Let f ∈ C[a, b]. Then there exists a sequence of polynomials {pn } such that pn converges to f uniformly on [a, b]. R1 Proof. Let kn (x) = cn (1−x2 )n , x ∈ [−1, 1], where cn is chosen so that kn (x) dx = 1. −1

Thus Z1 1 = cn

(1 − x2 )n dx

−1

Z1 = 2cn

(1 − x2 )n dx

0

77

√1

Zn ≥ 2cn (1 − x2 )n dx 0 √1

Zn ≥ 2cn (1 − nx2 ) dx 0

4cn = √ , 3 n which shows that cn ≤

3√ n 4


0. Then kn (x) ≤

√

n(1 − δ 2 )n for

δ ≤ |x| ≤ 1, from which it follows that kn → 0 uniformly in δ ≤ |x| ≤ 1. In order to prove the theorem, it is enough to assume that [a, b] = [0, 1], as there is a bijection x 7→

x−a b−a

between [a, b] and [0, 1]. Further it is enough to prove the theorem for such

function g for which g(0) = g(1) = 0. Assume for a moment that this theorem has been proved for such functions. Consider g(x) = f (x) − f (0) − x[f (1) − f (0)], x ∈ [0, 1]. Clearly, g(0) = g(1) = 0. Thus g can be approximated by a sequence of polynomials which in turn will approximate f . Thus, we assume that [a, b] = [0, 1] and f (0) = f (1) = 0. We also define f (x) = 0 for x ∈ / [0, 1]. Then f is uniformly continuous on R. Define Z1 f (x + t)kn (t) dt,

Pn (x) = −1

Z1−x = f (x + t)kn (t) dt −x

Z1 f (t)kn (t − x) dt

= 0

78

x ∈ [0, 1]

Z1 f (t)kn (x − t) dt,

= −1

as kn (−x) = kn (x). This shows that Pn (x) is indeed a polynomial. By the observation made above, we conclude that Pn converges to f uniformly on [0, 1]. This proves the theorem.

2.5

Wallis’ product formula It is well known that Wallis’ product formula for

π 2

can be written as

2 2 4 4 6 6 8 8 π = ··· 2 1 3 3 5 5 7 7 9 This can be obtained by calculating the Fourier series of the functionf (x) = cos px on [−π, π], where p is a real number but not an integer. Thus 1 fb(n) = 2π

Zπ

cos px e−inx dx

−π

 π  Z Zπ 1  = ei(p−n)x dx + e−i(p+n)x dx 4π −π n

=

−π

(−1) p sin pπ. 2 π p − n2

As fb ∈ l1 (Z) it is clear that the Fourier series of f converges pointwise [−π, π]. Thus ∞ X p (−1)n inx cos px = sin (pπ) e π p2 − n2 n=−∞

79

" # ∞ X p 1 (−1)n inx = sin (pπ) 2 + 2 e 2 − n2 π p p n=1 Letting x = π, we get, " # ∞ X 1 1 p . cos (pπ) = sin (pπ) 2 + 2 π p p2 − n2 n=1 In other words, ∞ X n=1

p2

2p 1 = π cot (pπ) − . 2 −n p

Integrating with respect to p, we get, ∞ X

p2 ln 1 − 2 n n=1 



 = ln

sin (pπ) pπ



In other words, we get, sin (pπ) = pπ

∞  Y n=1

p2 1− 2 n

 .

Now put p = 12 . We get the Wallis’ product formula for π2 .

2.6

Weyl’s equidistribution theorem A sequence of real numbers ξ1 , ξ2 , . . . , ξn , . . . in [0,1) is said to be equidis-

tributed if for every interval [a, b) ⊂ [0, 1),

lim

N →∞

#{1 ≤ n ≤ N : ξn ∈ [a, b)} = b − a. n

Theorem 2.6.1. (Weyl equidistribution theorem) A sequence of real numbers ξ1 , ξ2 , . . . , ξn , . . . in [0,1) is equidistributed if and only if for all integers k 6= 0, one N P has lim N1 e2πikξn = 0. N →∞

n=1

80

In order to prove this theorem, first we shall prove the following lemma. Lemma 2.6.2. A sequence of numbers {ξn } in [0, 1) is equidistributed if and only if 1

Z N 1 X lim f (ξn ) = f (x) dx, N →∞ N n=1 0

for every f ∈ L1 ([0, 1]). N P 1 f (ξn ) N →∞ N n=1

Proof. First assume that lim

=

R1

f (x) dx. Let [a, b] be an interval in

0

[0,1]. Then Z1

Zb dx = b − a.

χ[a,b) (x) dx = 0

a

Further N X

χ[a,b) (ξn ) = #{1 ≤ n ≤ N : ξn ∈ [a, b)}.

n=1

Thus 1

Z N 1 X lim χ[a,b) (ξn ) = χ[a,b) (x) dx, N →∞ N n=1 0

this shows that the sequence {ξn } is equidistributed. Conversely, suppose the sequence {ξn } is equidistributed. Then as shown above 1

Z N 1 X lim χ[a,b) (ξn ) = χ[a,b) (x) dx. N →∞ N N =1 0

But any step function on [0,1] is a linear combination of functions of the form χ[a,b) . Hence, the result follows for every step function. If f ∈ L1 ([0, 1]), then given  > 0

81

there exist step functions s and t such that s ≤ f ≤ t and Z1 (t(x) − s(x)) dx < . 0

Since the lemma holds for s, 1

1

Z Z N 1 X lim s(ξn ) = s(x) dx ≤  + f (x) dx. N →∞ N n=1 0

0

This means that there exists an N0 ∈ N such that for all N ≥ N0 , 1

Z N 1 X t(ξn ) ≤ 2 + f (x) dx. N n=1 0

But 1

Z N N 1 X 1 X f (ξn ) ≤ t(ξn ) ≤ 2 + f (x) dx. N n=1 N n=1 0

Similarly, we can show that

N R1 1 P f (ξn ) ≥ f (x) dx − 2. Thus, for all N ≥ N0 , we N n=1 0

get, X Z1 N 1 f (ξn ) − f (x) dx < 2, N n=1 0

proving our assertion. Proof of Theorem. Since f : [0, 1] → C defined by f (x) = e2πinx belongs to L1 ([0, 1]). R1 Further e2πinx dx = 0 for n 6= 0. If {ξn } is equidistributed, then 0

N 1 X 2πikξn lim e = 0, N →∞ N n=1

82

(2.2)

for each non zero integer k, by using lemma. Conversely, suppose (2.2) holds for k 6= 0. Since any trigonometric polynomial f is a linear combination of functions of the form e2πikx , we can conclude by linearity that 1

Z N 1 X f (ξn ) = f (x) dx. lim N →∞ N n=1 0

But since every continuous periodic function can be approximated by a sequence of trigonometric polynomials, the result follows for such continuous periodic functions. But since such functions form a dense subset of L([0, 1]), the result follows. These details are left as an exercise to the reader.

2.7

Exercises In the following problems, ]x[ denotes the fractional part of x ∈ R. 1

2.7.1. Show that { ]n 2 [ : n = 1, 2, 3, . . .} is equidistributed in [0, 1).  √    n 1+ 3 2.7.2. Verify whether : n = 1, 2, 3, . . . is equidistributed in [0, 1] or 2 not. 2.7.3. Let f be a continuous function on [0, 1]. Suppose

R1

f (x) xn dx = 0, n =

0

1, 2, 3, . . . . Using Weierstrass theorem, show that f is identically zero on [0,1].

83

84

Chapter 3 Fourier transform 3.1

Definition and basic properties First, let us look at the definition of Fourier transform and some basic proper-

ties of it without getting into mathematical rigour. We shall make some observations in order to put the matter in a rigorous fashion. For f : R → C, the Fourier transform of f is defined by Z∞ fb(ξ) =

f (x) e−2πixξ dx.

(3.1)

−∞

The definition of Fourier transform of f can also be defined as Z∞ 1 b f (x) e−ixξ dx. f (ξ) = √ 2π

(3.2)

−∞

However, we stick onto the definition given in (3.1). Let τs f (x) = f (x − s), s ∈ R. Then Z∞ (τs f )b(ξ) = −∞

(τs f )(x)e−2πixξ dx =

Z∞

f (x − s)e−2πixξ dx =

−∞

Z∞

f (y)e−2πi(s+y)ξ dy,

−∞

so that (τs f )b(ξ) = e−2πisξ fb(ξ).

(3.3)

Let Ms f (x) = e2πixs f (x), s ∈ R. Then Z∞ Z∞ Z∞ b −2πixξ 2πixs −2πixξ (Ms f ) (ξ) = (Ms f )(x)e dx = e f (x)e dx = f (x)e−2πi(ξ−s)x dx, −∞

−∞

−∞

85

which shows that (Ms f )b(ξ) = τs fb(ξ).

(3.4)

For δ > 0, define fδ (x) = f (δx), x ∈ R. Then Z∞ fbδ (ξ) =

f (δx)e

−2πixξ

−∞

1 dx = δ

Z∞

y

f (y)e−2πi δ ξ dy

−∞

Thus
1 fbδ (ξ) = fb δξ . δ

(3.5)

Observations • The definition of (3.1) makes sense if f is assumed to be in L1 (R). • For the validity of (3.2−3.4), we require τδ f , Ms f , fδ ∈ L1 (R). In fact, the operators τs , Ms , s ∈ R and Dδ given by Dδ f (x) =

1 f xδ δ 1/p


, δ > 0 are

bounded operators on Lp (R) and kτs f kp = kMs f kp = kDδ f kp = kf kp , 1 ≤ p < ∞. The Fourier transform satisfies the following properties also: (f 0 )b(ξ) = 2πiξ fb(ξ).

(3.6)

b dn f (ξ) = (2πi)n ξ n fb(ξ). dxn

(3.7)

More generally,  If g(x) = −2πixf (x), then d b f (ξ). dξ More generally, if g(x) = (−2πix)n f (x), then  dn  gb(ξ) = n fb(ξ) . dξ gb(ξ) =

86

(3.8)

(3.9)

These four properties will be proved in section 3.2.

Observations ˆ The property (3.6) requires the differentiability of f on R and f 0 ∈ L1 (R). Sim-

ilarly, to obtain (3.7), we have to assume that f must be n times differentiable on R and the nth derivative of f belongs to L1 (R). ˆ The property (3.8) requires that both f and xf ∈ L1 (R). ˆ The property (3.9) requires the xk f ∈ L1 (R) for 0 ≤ k ≤ n.

Thus, if we assume the required conditions on f as observed, the definition and the properties (3.1−3.9) will be valid.

3.2

Schwartz space on R Based on the observations of section 3.1, we ask the following question. Do

we have a subspace of L1 (R) which contains both f (n) and xm f for m, n ∈ N and is invariant under Fourier transform ? The answer is the space of all rapidly decreasing functions on R known as the Schwartz space. Definition 3.2.1. Let C ∞ (R) denote the space of all infinitely differentiable functions on R. The Schwartz class of rapidly decreasing functions S(R), is defined to be the collection of all f ∈ C ∞ (R) for which m dn < ∞ for all m, n ∈ N0 . f (x) sup x dxn x∈R It is clear that if f ∈ S(R), then p · f ∈ S(R) for any polynomial p and f (n) ∈ S(R) for any n. 87

2

Examples 3.2.2. (1) φ1 (x) = e−ax , a > 0. (2)  1  e x2 −1 φ2 (x) =  0

if x ∈ [−1, 1] otherwise.

It can be easily shown that the functions in (1) and (2) belong to S(R). (3) Let Hk denote the Hermite polynomial Hk (x) = (−1)k

dk  −x2  x2 e e , k ∈ N0 . dxk

√ 2 Let hk (x) = Hk ( 2πx)e−πx denote the Hermite functions. Then the functions hk can easily be shown to be in S(R). Apart from verifying the examples, it is left as an exercise to the reader to show that the functions ex , e|x| ,

1 1+x2

do not belong to S(R).

Remark 3.2.3. Clearly S(R) is a subspace of Lp (R),

1 ≤ p < ∞. In fact, if

φ ∈ S(R), then (1 + x2 ) φ(x) ∈ L∞ (R). Thus Z

p

Z

|φ(x)| dx = R

(1 + x2 )p |φ(x)|p dx ≤ c (1 + x2 )p

R

Z

1 dx < ∞, (1 + x2 )p

R

showing that φ ∈ Lp (R). Definition 3.2.4. Let supp(f ) denote the support of a function f : R → R defined to be the closure of {x ∈ R : f (x) 6= 0}. The class of test functions, denoted by D(R), is defined to be the space of infinitely differentiable functions on R whose supports are compact.

88

(a) φ1 (x).

(b) φ2 (x).

(c) Hermite functions.

Figure 3.1: Schwartz space functions. 89

(a) ex .

(b) e|x| .

(c)

1 1+x2 .

Figure 3.2: Functions which are not in Schwartz space.

The function φ defined in 3.2.2 (2) above belongs to D(R). Clearly, D(R) ⊂ S(R). We shall show that D(R) is dense in Lp (R), 1 ≤ p < ∞. This in turn leads to the density of S(R) in Lp (R), 1 ≤ p < ∞. Proposition 3.2.5. The space D(R) is dense in Lp (R), 1 ≤ p < ∞. Proof. Let f ∈ Lp (R) (1 ≤ p < ∞). Choose φ ∈ D(R), φ ≥ 0 such that
1. For δ > 0 consider φδ (x) = 1δ φ xδ and define

R R

φ(x)dx =

Z f ∗ φδ (x) =

f (x − y)φδ (y)dy.

(3.10)

f (y)Dδ φ(x − y)dy,

(3.11)

R

Note that Z f ∗ φδ (x) = R

by making a change of variables. By applying Holder’s inequality, we get

|f ∗ φδ | ≤ kf kp kφδ kp0 , 90

where

1 1 + 0 = 1, p p

which shows that f ∗ φδ is well defined. An application of Minkowski’s integral inequality shows that  p1

 Z 

|f ∗ φδ |p dx ≤

R

 p1

 Z

Z 

R

|f (x − y)|p dx |φδ (y)|dy = kf kp kφδ k1 = kf kp .

R

(3.12) Again, using

R

φ(x)dx = 1, we can write Z f ∗ φδ (x) − f (x) = [f (x − y) − f (x)]φδ (y) dy, R

R

which in turn will imply that Z Z kf ∗ φδ − f kp ≤ kτy f − f kp Dδ φ(y) dy ≤ kτδy f − f kp φ(y) dy. R

R

Since the map g 7→ τs g is continuous for g ∈ Lp (R) (1 ≤ p < ∞), it follows that kτδy f − f kp → 0 as δ → 0. (See problem 3.7.1.) Now, applying Lebesgue’s dominated convergence theorem, we can show that kf ∗ φδ − f kp → 0 as δ → 0. Choose a cut off function ψ ∈ D(R), 0 ≤ ψ ≤ 1 and ψ = 1 on [−1, 1]. Define F δ (x) = ψ(δx)f ∗ φδ (x). Since for two functions h and k, supp hk ⊂ supp h ∩ supp k, F δ has compact support. Further, by using (3.11), one can show that f ∗ φδ is infinitely differentiable as φ ∈ D(R). This in turn will imply that F δ ∈ D(R). Using the properties of f ∗ φδ   p1 Z kf − F δ kp =  |f (x) − ψ(δx)f ∗ φδ (x)|p dx R

91

  p1 Z =  |f (x) − f (x)ψ(δx) + f (x)ψ(δx) − f ∗ φδ (x)ψ(δx)|p dx R

  p1   p1 Z Z ≤  |f (x)|p |1 − ψ(δx)|p dx +  |f (x) − f ∗ φδ (x)|p |ψ(δx)|p dx R

R

 p1

 Z

 |f (x)|p dx + kf − f ∗ φδ kp .

 ≤ |x|> 1δ

Since each term on the right hand side of above inequality goes to zero as δ goes to zero, we conclude that kf − F δ kp → 0 as δ → 0. Thus, D(R) is dense in Lp (R), (1 ≤ p < ∞).

3.3

Fourier transform on the Schwartz space First, we shall prove (3.5) of section 3.1. Let N ∈ N. Consider ZN

0

−2πixξ

f (x)e −N

ZN dx =

d (f (x)) e−2iπxξ dx = dx

−N

ZN −N

N = e−2iπxξ f (x) −N + 2πiξ

ZN

f (x)e−2iπxξ dx.

−N

But e

−2πixξ

f (x) → 0 as N → ± ∞. Thus Z ZN 0b 0 −2iπxξ (f ) (ξ) = f (x)e dx = lim f 0 (x)e−2πixξ dx R

e−2iπxξ d(f (x))

N →∞ −N

ZN = 2πiξ lim

N →∞ −N

= 2πiξ fb(ξ), 92

f (x)e−2πixξ dx

proving (3.6). An application of mathematical induction leads to (3.7). We shall now establish (3.8). Let f ∈ S(R). Define g(x) = −2πixf (x), so that g ∈ S(R). Consider for h > 0,   Z Z 1 fb(ξ + h) − fb(ξ) =  f (x)e−2πi(ξ+h)x dx − f (x)e−2πiξx dx h h R R Z h e−2πihx − 1 i = xf (x)e−2πiξx dx. xh R

e−2πihx − 1 = −2πi+O(h), using power series expansion of the exponential funcxh e−2πihx − 1 tion. We also have lim = −2πi. Then, by applying dominated convergence h→0 xh theorem, we get

But

Z∞ Z∞ fb(ξ + h) − fb(ξ) = −2πi xf (x)e−2πiξx dx = g(x)e−2πiξx dx = gb(ξ), lim h→0 h −∞

−∞

proving our assertion. As a consequence,  b (−2πix)2 f (x) (ξ) = [−2πix(−2πix)f (x)]b(ξ) = [−2πixg(x)]b(ξ) d h b0 i b00 d [−2πixf (x)]b(ξ) = f (ξ) = f (ξ). = dξ dξ Now, one can obtain (3.9) inductively. The equations in (3.6) to (3.9) lead to the following: Theorem 3.3.1. If f ∈ S(R) then fb ∈ S(R). In the next proposition, we shall show that the Fourier transform of a Gaussian function is itself. 2 Proposition 3.3.2. If f (x) = e−πx , then fb(ξ) = f (ξ).

93

Proof. We have Z∞ fb(ξ) =

−2πiξx

f (x)e

Z∞ dx =

−∞

R∞

In particular, fb(0) =

2

e−πx e−2πiξx dx.

−∞

R∞

2

e−πx dx. But

−∞


2 R∞ R∞ −π(x2 +y2 ) 2 e−πx dx = e dxdy. Chang-

−∞

−∞ −∞

ing into polar co-ordinates by setting x = r cos θ , y = r sin θ, we get 

2

Z∞



e

−πx2

Z2π Z∞

e

dx =

−∞

0

 A simple integration leads to

−πr2

R∞

Z∞ rdrdθ = 2π

0

e

0

−πx2

2 dx

= 1. Thus, fb(0) = 1. Also, by using

−∞

(3.8),  b d b f (ξ) = − 2iπxf (x) (ξ) dξ Z∞ = −2πi xf (x)e−2πixξ dx −∞ Z∞

= −2πi

2

xe−πx e−2πixξ dx

−∞

Z∞ =i

2

re−πr drdθ.

f 0 (x)e−2πixξ dx

−∞

= i(f 0 )b(ξ) = i(2πiξ)fb(ξ) (using (3.6)) = −2πξ fb(ξ).

94

d b f (ξ) = −2πξ fb(ξ) with fb(0) = 1. In dξ = −2πξ. On integration, we get ln fb(ξ) = −πξ 2 + ln c, where c

Thus, we end up with a differential equation other words,

(fb)0 (ξ) fb(ξ)

2 is the constant of integration. This leads to fb(ξ) = ce−πξ . Using fb(0) = 1, we get 2 c = 1. Thus fb(ξ) = e−πξ .

Definition 3.3.3. For f, g ∈ S(R), we define the convolution of f and g by R∞ f (x − y)g(y)dy. (f ∗ g)(x) = −∞

d d (f ∗ g) = f ∗ g and x(f ∗ g) = xf ∗ g, one can easily show that dx dx f ∗ g ∈ S(R). It is also clear that f ∗ g = g ∗ f by applying change of variables. Since

The following property is an important property of Fourier transform which states that the Fourier transform of convolution of two functions becomes the product of their Fourier transforms:

(f ∗ g)b = fb · gb. To see this, consider Z (f ∗ g)b(ξ) = R

Z = R

(f ∗ g)(x)e−2πixξ dx =

Z

  Z  f (x − u)g(u)du e−2πixξ dx

R R   Z g(u)  f (x − u)e−2πixξ dx du, R

by using Fubini’s theorem. Now letting x − u = t, we get Z (f ∗ g)b(ξ) = R

  Z g(u)  f (t)e−2iπ(u+t)ξ dt du R

95

(3.13)

Z

g(u)e−2iπuξ du = fb(ξ)b g (ξ),

= fb(ξ)

ξ ∈ R,

R

proving our assertion. Theorem 3.3.4. (Multiplication formula) If f, g ∈ S(R), then Z

Z f (x)b g (x)dx =

R

Proof. Consider

fb(y)g(y)dy. R

Z

Z f (x)b g (x)dx =

R

  Z f (x)  g(t)e−2πitx dt dx

R

R  Z Z = g(t)  f (x)e−2πitx dx dt R Z

R

g(t)fb(t)dt,

= R

by using Fubini’s theorem. 2

Proposition 3.3.5. Let φ(x) = e−πx . Then for any f ∈ S(R), f ∗ D√δ φ converges to f uniformly on R as δ → 0. Proof. Consider

D√

1 δ φ(x) = √ φ δ



x √ δ



1 2 = √ e−πx /δ . δ

Then Z∞ −∞

D√δ φ(x)dx =

Z∞

Z∞ φ(x)dx =

−∞

−∞

96

2

e−πx dx = 1.

(3.14)

As f ∈ S(R), f ∈ C0 (R). Therefore, given  > 0, there exists a compact set K = [−M, M ](say) in R such that |f (x)| < , for all x, |x| >

M 2

. Clearly, f is uniformly

continuous on K. Thus there exists η > 0 such that if |x−y| < η, then |f (x)−f (y)| <  ∀ x, y ∈ K. If |x − y| < η but either x or y lies outside K, then both |x| and |y| are bigger than

M 2

and hence |f (x) − f (y)| < . Thus f is uniformly continuous on R.

Consider

f ∗ D√δ φ(x) − f (x) =

Z

Z

f (x − y)D√δ φ(y)dy − f (x)

R

D√δ φ(y)dy.

R

Thus Z

  f (x − y) − f (x)] D√δ φ(y)dy

f ∗ Dδ φ(x) − f (x) = R

Z

[f (x − y) − f (x)]] D√δ φ(y)dy

= |y|η

Hence, Z |f ∗ Dδ φ(x) − f (x)| ≤  +



 f (x − y) − f (x)] D√δ φ(y))dy ∀ x ∈ R,

|y|>η

using the uniform continuity of f . Further

|f (x − y) − f (x)| ≤ 2 sup |f (x)| ≤ 2kf k∞ , x∈R

as f ∈ S(R). Thus for every x ∈ R, Z |f ∗ Dδ φ(x) − f (x)| ≤  + 2kf k∞ |y|>η

97

−πy 2 1 √ e δ dy. δ

As δ → 0 the above integral goes to 0. Thus f ∗ D√δ φ converges to f uniformly on R as δ → 0. Theorem 3.3.6. For f ∈ S(R), the following Fourier inversion formula is valid. Z∞ f (x) =

fb(ξ)e2πixξ dξ.

(3.15)

−∞

Proof. It is enough to show that Z∞ fb(ξ)dξ.

f (0) =

(3.16)

−∞

Once this is proved, we will obtain (3.15). In fact, if f ∈ S(R), then τ−x f ∈ S(R) for x ∈ R. Applying (3.16) to τ−x f , we get Z f (x) = τ−x f (0) =

Z (τ−x f ) (ξ)dξ = b

R

e2πixξ fb(ξ)dξ.

(3.17)

R


R 2 In order to prove (3.16), take φ(x) = e−πx . As φ is even f ∗ D√δ φ (0) = f (x) R

D√δ φ(x)dx. By Proposition 3.3.5, f ∗ D√δ φ converges to f uniformly on R as δ → 0. In particular, we have Z

f (x) D√δ φ(x)dx = f (0).

lim

δ→0

(3.18)

R 2

Further, we have [e−πδx ]b(ξ) =

−πξ √1 e δ δ

2


= D√δ φ (ξ). Thus, by multiplication

formula, (3.18) becomes Z lim

δ→0

2 fb(ξ) e−πδξ dξ = f (0).

R

98

(3.19)

2 As lim e−πδξ = 1 and fb ∈ S(R) ⊂ L∞ (R) it follows from Lebesgue’s dominated δ→0 R R 2 convergence theorem that lim fb(ξ)e−πδξ dξ = fb(ξ) dξ. Thus (3.16) follows from

δ→0 R

R

(3.19).

3.4

Fourier transform for L2-functions on R

Theorem 3.4.1. Let f ∈ S(R) then kfbk2 = kf k2 . Proof. Define f˜(x) := f (−x). Then, clearly f˜ ∈ S(R). Define h := f ∗ f˜, then h ∈ S(R). By Fourier inversion formula, we have Z b h(ξ)dξ.

h(0) =

(3.20)

R

Since h = f ∗ f˜, we get, b h = fb · fb˜. But fb˜(ξ) =

Z

Z

f˜(x) e−2πiξx dx =

f (−x) e

−2πiξx

Z dx =

R

R

f (y) e−2πiξy dy.

R

This shows that fb˜ = fb. Hence, b h = fb · fb˜ = fbfb, = |fb|2 . From (3.20), we have Z h(0) =

|fb(ξ)|2 dξ = kfbk22 .

(3.21)

R

On the other hand, as h = f ∗ f˜, h(0) = f ∗ f˜(0) =

Z

f˜(0 − y)f (y)dy

R Z

Z f (y)f (y)dy =

=

R

R

99

|f (y)|2 dy.

Thus h(0) = kf k22 .

(3.22)

From (3.21) and (3.22), we get kfbk2 = kf k2 . Consider F : S(R) → S(R) defined by Ff = fb. Then clearly F is linear. Further, using the Fourier inversion formula, it follows that F is one-one and onto. Thus the Fourier transform is a linear isomorphism on S(R). Theorem 3.4.2. (Plancherel) The map F initially defined on S(R) extends to the whole of L2 (R) and satisfies kf k2 = kFf k2 ∀ f ∈ L2 (R). Moreover, the Fourier transform is a unitary operator on L2 (R). Proof. Let f ∈ L2 (R). In order to define Fourier transform for a function f ∈ L2 (R) we use the density argument. Since S(R) is dense in L2 (R), there exists a sequence {fn } in S(R) such that kfn − f k2 → 0 as n → ∞. In particular, {fn } is a Cauchy sequence with respect to L2 -norm. As kfbn − fbm k2 = kfn − fm k2 , it follows that {fbn } is also a Cauchy sequence. Since L2 (R) is a Banach space, there exists g in L2 (R) such that fbn → g in L2 (R). Call g = fb. To see that it is well defined, let {gn } ∈ S(R) be any other approximating sequence for f in L2 (R). Then

kb gn − fbn k2 = k(gn − fn )bk2 = kgn − fn k2 = kgn − f + f − fn k2 ≤ kgn − f k2 + kf − fn k2 → 0, 100

as n → ∞. Thus, fb is well defined. We also have fbn → g = fb in L2 (R). Hence kfbk2 = lim kfbn k2 = kfn k2 = kf k2 ∀ f ∈ L2 (R). n→∞

In order to prove F is onto, let f ∈ L2 (R). Choose {fn } ∈ S(R) such that fn → f in L2 . Then there exists gn ∈ S(R) such that gbn = fn . Clearly, {gn } is Cauchy and hence gn → g for some g ∈ L2 . We take gb = f.

3.5

Fourier transform for L1-functions on R

Definition 3.5.1. Let f ∈ L1 (R). Then the Fourier transform of f is defined to be Z fb(ξ) =

f (x)e−2πiξx dx.

R

We can also extend the definition of fb from S(R). As S(R) is dense in L1 (R), we can choose {fn } ∈ S(R) such that fn → f in L1 for any given f ∈ L1 . Further, it can be easily shown that {fbn } is uniformly Cauchy. Then there exists g continuous such that fbn → g uniformly on R and we define fb = g. As in the L2 case fb is well defined. Moreover, Z∞ fbn (ξ) =

fn (x)e−2πixξ dx.

−∞

Then it follows from Lebesgue dominated convergence theorem that Z∞ fb(ξ) =

f (x)e−2πixξ dx.

−∞

101

(a) Fourier transform of χh

1 1 −2,2

(b) Fourier transform of e−2|x| .

i.

Theorem 3.5.2. (Riemann-Lebesgue Lemma) If f ∈ L1 (R), then fb ∈ C0 (R). This means lim fb(ξ) = 0. |ξ|→∞

Proof. We can prove the theorem immediately if we use the density of S(R) in L1 (R). However, we give an alternative elementary proof. First, we have to show that fb is continuous. (See problem 3.7.2.) If f = χ[a,b] then one can show that fb ∈ C0 . As a simple measurable function is a linear combination of characteristic functions, the result will follow for a simple measurable function. Since any non-negative measurable function can be approximated by a sequence of simple measurable functions, the result will be true for a non-negative measurable function. But any real valued function f can be written as f = f + − f − , the result will follow for a real valued function. As a complex valued function f = u + iv, the result will follow for the general function. 102

Definition 3.5.3. For f, g ∈ L1 (R), the convolution of f and g is defined as Z∞ (f ∗ g)(x) =

f (x − y) g(y)dy. −∞

One can show that L1 (R) is a Banach algebra. But L1 (R) has no multiplicative identity.

Figure 3.3: χh0, 1 i ∗ χh− 1 ,0i . 2

Theorem 3.5.4. Suppose φ ∈ L1 (R) be such that

2

R

φ(x)dx = 1. Let f ∈ Lp (R), 1 ≤

R

p < ∞. Then kf ∗ Dδ φ − f kp → 0 as δ → ∞. Proof. Consider Z (f ∗ Dδ φ)(x) − f (x) =

Z f (x − y)Dδ φ(y)dy − f (x)

R Z

Dδ φ(y)dy R

[f (x − y) − f (x)] Dδ φ(y) dy.

= R

103

Then, p  1 p Z [f (x − y) − f (x)]Dδ φ(y)dy  dx R R Z Z ≤ |Dδ φ(y)| dy |f (x − y) − f (x)|p dx

 Z  kf ∗ Dδ φ − f kp =

R

R

using Minkowski’s integral inequality. Thus Z kf ∗ Dδ φ − f kp ≤

kτδy f − f kp |φ(y)|dy. R

As δ → 0, kτδy f − f kp → 0 and kτδy − f kp ≤ 2kf kp . Hence kf ∗ Dδ (φ) − f kp → 0 as δ → 0 by dominated convergence theorem. Proposition 3.5.5. Let f, g ∈ L1 (R). Then Z

fb(ξ) e2πixξ g(δξ)dξ =

R

fb(ξ) e2πixξ g(δξ)dξ =

f (ξ) Dδ gb(ξ − x) dξ. R

Proof. Let h(ξ) = e2πixξ g(δξ). Then Z

Z

Z

Z

R

R

R

f (ξ) b h(ξ) dξ,

fb(ξ)h(ξ)dξ =

using multiplication formula. Since Z b h(ξ) =

−2πiξy

h(y) e R

Z dy =

e2πiξy g(δy) e−2πiξy dy = Dδ gb(ξ − x)

R

proving our assertion.

104

R Theorem 3.5.6. Let f, g ∈ L1 (R). Let φ = gb ∈ L1 (R) such that φ(x)dx = 1 and φ R R 2πixξ 1 b is even. Then f (ξ)g(δξ) e dξ converges to f in L -norm as δ → 0. R

Proof. It follows from the above proposition that Z

fb(ξ)g(δξ) e2πixξ dξ = (f ∗ Dδ φ)(x).

R

Then, by Theorem 3.5.4 we conclude that f ∗ Dδ φ → f in L1 -norm as δ → 0. Corollary 3.5.7. (Inversion formula) Suppose f and fb ∈ L1 (R). Then Z f (x) =

fb(ξ)e2πiξx dξ

a.e.

R

2

Proof. Let g(x) = e−πx . Then g ∈ L1 and gb ∈ L1 . By theorem 3.5.6,

R R

2 2 fb(ξ)e−πδ ξ

e2πiξx dξ → f in the L1 -norm as δ → 0. Hence, there exists a sequence δk → 0 R 2 2 such that fb(ξ)e−πδk ξ e2πiξx dξ → f (x) pointwise a.e. Applying Lebesgue dominated R

convergence theorem, Z

2 2 fb(ξ)e−πδk ξ e2πiξx dξ →

R

Z

fb(ξ)e2πiξx dξ.

R

Thus Z

fb(ξ)e2πiξx dξ = f (x) a.e.

R

Corollary 3.5.8. If f ∈ L1 (R) and fb = 0, then f = 0 a.e.

105

3.6

Solved problems

3.6.1. Calculate χh0, 1 i ∗ χh− 1 ,0i . 2

2

Solution. 1

Z∞ χh0, 1 i ∗ χh− 1 ,0i = 2

Z2 χh0, 1 i (y)χh− 1 ,0i (x − y)dy =

2

2

χh− 1 ,0i (x − y)dy.

2

−∞

2

0

Thus, for this integral to be non-zero, y should belong to [0, 21 ]∩[x, x+ 12 ]. If 0 ≤ x < 21 , then 1

  Z2 1 h i h i χ 0, 1 ∗ χ − 1 ,0 (x) = dy = − x. 2 2 2 x

If

−1 2

< x < 0, then x+ 12

Z

χh

0,

1 2

i

∗

χh

i −1 ,0 2

dy =

=

1 + x. 2

0

In the other regions

χh0, 1 i ∗ χh −1 ,0i = 0. 2

2

Thus

χh

0,

1 2

i

∗

χh

i 1 − ,0 2

=

  

106

1 2

− |x| 0

if |x| ≤

1 2

otherwise.

3.6.2. Show that Lp (R) ⊆ L1 (R) + L2 (R) for 1 ≤ p ≤ 2. Solution. Let f ∈ Lp (R). Define g and h as   f (x) |f (x)| ≤ 1 g(x) =  0 otherwise.   f (x) |f (x)| > 1 h(x) =  0 otherwise. Then f (x) = g(x)+h(x). Further |g(x)|2 ≤ f (x) and |h(x)| ≤ |f (x)|. Thus g ∈ L2 (R) and h ∈ L1 (R). 3.6.3. Find the Fourier transform of χ[− 1 , 1 ] 2 2  1 1  1 if x ∈ [− , ] 2 2 Solution. ϕ(x) =  0 otherwise Z∞ ϕ(s) b =

1

ϕ(x) e−2πixs dx =

−∞

Z2

e−2πixs dx =

sin πs , πs

− 12 1

Z2 ϕ(0) b =

ds = 1 − 12

 sin πs  if s 6= 0 πs ϕ(s) b =  1 if s = 0 107

s 6= 0

3.7

Exercises

3.7.1. Show that the map x 7→ τx f is continuous from R into L1 (R). 3.7.2. Let f ∈ L1 (R). Then show that fb is continuous on R. 3.7.3. Let

    1   exp − 1 + for 0 ≤ x ≤ a x2 (x − a)2 ϕ(x) =   0 otherwise.

Show that ϕ ∈ D(R).   R∞ sin x 2 3.7.4. Evaluate dx. (Hint: Use Plancherel theorem) x −∞ 3.7.5. Find the Fourier transform of f (x) = e−a|x| , x ∈ R. For a, b > 0, find R∞ dx . 2 2 2 2 0 (a + x )(b + x )

108

Chapter 4 Tempered distributions 4.1

Topology on S(R) We know that S(R) is a vector space. On S(R), we can introduce a family of

seminorms pn,m defined as follows: pn,m (f ) = sup xm f (n) (x) . x∈R

Then this family induces a locally convex topology on S(R). Equivalently, the topology can also be defined in terms of the increasing family of seminorms

pk (f ) = sup 1 + x


2 k

x∈R

k X (j) f (x) . j=0

In fact, each pk is a norm and we can use them to define a metric d on S(R) by

d(f, g) =

∞ X k=0

2−k

pk (f − g) . 1 + pk (f − g)

Notice that d is translation invariant in the sense that d(f, g) = d(f − g, 0). When equipped with the topology introduced by the above seminorms or equivalently, with the metric topology, the operations (f, g) 7−→ f + g and (α, f ) 7−→ αf become continuous making S(R) into a topological vector space. Let {ϕn } be a Cauchy sequence in S(R). Then pk (ϕn − ϕm ) → 0 as n, m → ∞ for all k ∈ N0 . This in s d ds turn will imply that xr s ϕn − xr s ϕm → 0 as n, m → ∞ for all r, s ∈ N0 . Then dx dx

109

  s r d it follows from Cauchy criterion that x ϕn converges uniformly on R for all dxs r, s ∈ N0 . Assume that ϕn converges to ϕ on R. An application of closed graph theorem will lead to the convergence of pr,s (ϕn ) to pr,s (ϕ) as n → ∞. This will lead to the convergence of {ϕn } with respect to the metric d. Thus S(R) is a complete metric space. Using this topology on S(R), we can define the class of tempered distributions S 0 (R) as the dual space of S(R). In other words, every element of S 0 (R) is a continuous linear functional on S(R). We say that Λ ∈ S 0 (R) if Λ : S(R) → C is linear and continuous. The continuity of Λ can be taken as follows: If {ϕn } is a sequence of functions converging to ϕ in S(R), then a function Λ : S(R) → C is said to be continuous if {Λϕn } converges to {Λϕ} as a sequence of complex numbers. Since Λ is linear, we can say that Λ is continuous if {ϕn } is a sequence of functions converging to 0 in S(R) then {Λϕn } converges to 0 as a sequence of complex numbers. Examples 4.1.1. (1) Fix x ∈ R. Define δx : S(R) → C by δx (ϕ) = ϕ(x). Then δx ∈ S 0 (R) and is called the Dirac delta function at x ∈ R. R (2) Let f ∈ Lp (R), 1 ≤ p ≤ ∞. Define Λf (ϕ) = f (x)ϕ(x) dx for all ϕ ∈ S(R). Then R

0

Λf ∈ S 0 (R). In fact, |Λf (ϕ)| ≤ kf kp kϕkp0 < ∞ as S(R) is contained in Lp (R). If {ϕk } converges to 0 in S(R), then 0 kϕk kpp0

Z =

1 + x2


lp0

0

|ϕ(x)|p 1 + x2

R

≤ c sup 1 + x2


lp0

x∈R

110

0

|ϕk (x)|p ,


−lp0

dx

0

tends to 0 as k → ∞ for every positive integer l. This means ϕk → 0 in Lp norm. Thus |Λf (ϕk )| ≤ kf kp kϕk kp0 → 0 showing Λf is continuous on S(R). Clearly, it is linear on S(R), proving that Λf ∈ S 0 (R).

4.2

Convergence in S 0(R)

Definition 4.2.1. A sequence {Λn } in S 0 (R) is said to converge to Λ in S 0 (R) if Λn (ϕ) → Λ(ϕ) as a sequence of complex numbers for each ϕ ∈ S(R). Theorem 4.2.2. Let {fn } be a sequence of nonnegative functions on R such that R R fn (x)dx = 1 for all n. Suppose it also satisfies fn (x) dx → 0 as n → ∞ for R |x|≥η R every η > 0. Define Λf n (ϕ) = fn (x)ϕ(x) dx. Then {Λf n } converge to the dirac R

delta function δ0 in S 0 (R) as n → ∞. ∼

∼

Proof. Define Λfn by Λfn ϕ(x) = f n ∗ ϕ(0), where f n (x) = fn (−x). Then, by theorem, ∼

f n ∗ ϕ(0) → ϕ(0). But ϕ(0) = hδ0 , ϕi. Thus the result follows.

4.3

Some properties of S 0(R)

Definition 4.3.1. Let Λ ∈ S 0 (R). The derivative Λ0 of Λ is defined as Λ0 (ϕ) = −Λ(ϕ0 ) for all ϕ ∈ S(R). More generally, for m ≥ 1, the m-th derivative of Λ is defined as Λ(m) (ϕ) = (−1)m Λ(ϕ(m) ) for all ϕ ∈ S(R). Clearly, Λ(m) is well defined and it is linear on S(R). We can show that Λ(m) ∈ S 0 (R). Suppose {ϕn } is a sequence in S(R) which converges to 0 in S(R). This means xj Dk ϕn (x) → 0 for all j, k ∈ Z (m) and x ∈ R. But sup xj Dk ϕn (x) = sup xj Dk Dm ϕn (x) = sup xj Dk+m ϕn (x) from x∈R

which it follows that

x∈R

(m) {ϕn }

x∈R

(m)

also converges to 0 in S(R). As Λ ∈ S 0 (R), Λ(ϕn )

converges to 0 as a sequence of numbers, which means {Λm (ϕn )} converges to 0 as a sequence of numbers. Thus, Λm ϕ ∈ S 0 (R).

111

Examples 4.3.2. (1) Fix x ∈ R. Consider δx , the Dirac delta function. By definition of derivative of a tempered distribution, δx0 (ϕ) = −δx (ϕ0 ) for all ϕ ∈ S(R). But δx (ϕ0 ) = ϕ0 (x). Thus δx0 (ϕ) = −ϕ0 (x). (2) Let   1 x≥0 H(x) =  0 x < 0. The function H is called the Heaviside function or unit step function. Let ϕ ∈ S(R). Then Z∞ H(ϕ) =

Z∞ H(x)ϕ(x) dx =

−∞

Z∞ ϕ(x) dx =

0

(1 + x2 )ϕ(x) dx 1 + x2

0

Then |H(ϕ)| ≤ M sup |(1 + x2 )ϕ(x)| and as H is linear on S(R), it follows that x∈R 0

H ∈ S (R). Consider, for ϕ ∈ S(R),

0

Z∞

0

H (ϕ) = −H(ϕ ) = −

Z∞

0

H(x)ϕ (x) dx = −

−∞

ϕ0 dx = ϕ(0) = δ0 (ϕ),

0

which shows that the derivative of H is the Dirac delta function δ0 . (3) Let   x x≥0 Λ(x ) =  0 otherwise. Then it can be easily shown that Λ ∈ S 0 (R). The derivative Λ0 of Λ is given by

0

0

Z∞

Λ (ϕ) = −Λ(ϕ ) = − 0

112

xϕ0 (x) dx

=

− [xϕ(x)]∞ 0

Z∞ +

ϕ(x) dx 0

Z∞ ϕ(x) dx = H(ϕ) ∀ ϕ ∈ S(R),

= 0

which shows that Λ0 is the Heaviside function H.

(a) Heaviside function.

(b) The function Λ.

Definition 4.3.3. Let V be an open set in R. We say that an element Λ ∈ S 0 (R) vanishes on V if Λ(ϕ) = 0 for every ϕ belonging to S(R) which has support in V . Definition 4.3.4. Let Λ ∈ S 0 (R). Let Ω denote the union of all open sets in R on which Λ vanishes. Then the complement Ωc of Ω in R is called the support of Λ and it is denoted by suppΛ. Examples 4.3.5. (1) Supp δx = {x}, x ∈ R. (2) Supp H = [0, ∞). It is clear that S 0 (R) is a linear space. In fact, (Λ1 + Λ2 ) (ϕ) = Λ1 (ϕ)+Λ2 (ϕ) ∀ ϕ ∈ S(R) for the tempered distribution Λ1 , Λ2 and (αΛ)(ϕ) = αΛ(ϕ) for all ϕ ∈ S(R) for α ∈ C and Λ ∈ S 0 (R). Let ϕ ∈ S(R) and Λ ∈ S 0 (R). Then 113

the product ϕΛ is defined by ϕΛ(ψ) = Λ(ϕψ) for all ψ ∈ S(R). For α, β ∈ C, ψ1 , ψ2 ∈ S(R)

ϕΛ(αψ1 + βψ2 ) = Λ [ϕ(αψ1 + βψ2 )] = Λ [α(ϕψ1 )] + Λ [β(ϕψ2 )] = αΛ(ϕψ1 ) + βΛ(ϕψ2 ) = α(ϕΛ)ψ1 + β(ϕΛ)ψ2 ,

showing that ϕΛ is linear on S(R). If {ψn } converges to 0 in S(R), then using Leibnitz’s formula one can show that {ϕψn } converges to 0 in S(R). This shows that {Λ(ϕψn )} converges to 0. Thus ϕΛ ∈ S 0 (R).

4.4

Fourier transform on S 0(R)

b Definition 4.4.1. Let Λ ∈ S 0 (R). The Fourier transform of Λ is defined by Λ(ϕ) = ∨

∨

∨

Λ(ϕ) b for all ϕ ∈ S(R). The inverse of Fourier transform Λ is defined by Λ(ϕ) = Λ( ϕ) ∨

for all ϕ ∈ S(R), where ϕ is the inverse Fourier transform of ϕ given by the inversion R ∨ formula ϕ(x) = ϕ(ξ)e2πixξ dξ. Since the Fourier transform is continuous from S(R) R

∨

onto itself, it also maps S 0 (R) onto S 0 (R). Similarly, Λ ∈ S 0 (R) for all Λ ∈ S 0 (R). ∨ b ∨ b One can easily verify that Λ = Λ = Λ for all Λ ∈ S 0 (R). Remark 4.4.2. The Fourier transform on S 0 (R) can be thought of as an extension of the Fourier transform on L2 (R). In fact, let f ∈ L2 (R). Then f defines a distribution Λf which belongs to S 0 (R). We have for every φ ∈ S(R), the Fourier transform of the tempered distribution Λf is given by Z b f (ϕ) = Λf (ϕ) = Λ

Z f (x)ϕ(x) b dx =

R

fb(x)ϕ(x) dx = (Λbf , ϕ), R

114

showing that Fourier transform treated as an L2 function definition coincides with Fourier transform treated as a tempered distribution. Examples 4.4.3. (1) Let ξ ∈ R. Then for all ϕ ∈ S(R), (δξ )b(ϕ) = δξ (ϕ) b = ϕ(ξ) b = R ϕ(x)e−2πixξ dx. In particular if ξ = 0, then δb0 = 1. R

(2) Let ξ ∈ R. Then for all ϕ ∈ S(R), (δξ0 )b(ϕ) = δξ0 (ϕ) b = − δξ ( ϕ b0 ) = − ϕ b0 (ξ) =
R R 2πi xϕ(x)e−2πixξ dx = 2πi xe−2πixξ ϕ(x) dx = gξ (ϕ), where gξ = 2πixe−2πixξ . R

R

The Fourier transform and inverse Fourier transform are continuous on S 0 (R). In fact, if {Λn } is a sequence in S 0 (R) such that Λn → Λ in S 0 (R), then Λn (ϕ) → Λ(ϕ) for all b n (ϕ) and Λ(ϕ) b ϕ ∈ S(R). This shows that Λn (ϕ) b → Λ(ϕ). b But Λn (ϕ) b =Λ b = Λ(ϕ). b n (ϕ) → Λ(ϕ) b cn → Λ b in S 0 (R). Thus Λ for all ϕ ∈ S(R) which in turn implies that Λ This shows that the Fourier transform is continuous on S 0 (R). Similarly, the inverse Fourier transform can also be shown to be continuous on S 0 (R). Proposition 4.4.4. Let Λ ∈ S 0 (R). Then for all ϕ ∈ S(R),  (m) b b Λ (ϕ) = (2πix)m Λ(ϕ), b [(2πiξ)m Λ]b(ϕ) = (−1)m (Dm Λ)(ϕ). Proof.  (m) b Λ (ϕ) = (Dm Λ)b(ϕ) = Dm Λ(ϕ) b = (−1)m Λ(D m ϕ) b   = (−1)m Λ [(−2πix)m ϕ(x)]b b [(−2πix)m ϕ(x)] = (−1)m Λ   m mb = (−1) (−2πix) Λ (ϕ)   mb = (2πix) Λ (ϕ).

115

[(2πi)m ξ m Λ]b(ϕ) = ((2πi)m ξ m Λ) (ϕ) b = Λ [(2πi)m ξ m ϕ] b b m ϕ) = Λ((Dm ϕ)b ) = Λ(D b [(−1)m Dm ϕ] = (−1)m Λ b = (−1)m (Dm Λ)(ϕ).

Definition 4.4.5. Let ϕ ∈ S(R) and Λ ∈ S 0 (R). Then the convolution of Λ and ϕ ∼

∼

is defined to be (Λ ∗ ϕ)(x) = Λ(τx ϕ) for all x ∈ R, where ϕ(x) = ϕ(−x), x ∈ R, and τx ϕ(ξ) = ϕ(ξ − x), ξ ∈ R. Proposition 4.4.6. Let ϕ ∈ S(R) and Λ ∈ S 0 (R). Then Λ∗ϕ ∈ C ∞ (R) and Dm (Λ∗ϕ) = (Dm Λ) ∗ ϕ = Λ ∗ (Dm ϕ), m ∈ N. Proof. Fix x0 ∈ R. Suppose x → x0 ∈ R. Then, it can be easily shown that ∼

∼

∼

∼

τx ϕ → τx0 ϕ as x → x0 . This in turn will imply that Λ(τx ϕ) → Λ(τx0 ϕ). This means (Λ ∗ ϕ)(x) → (Λ ∗ ϕ)(x0 ) which shows that Λ ∗ ϕ is continuous on R. We shall show that Λ ∗ ϕ is differentiable. Consider for x ∈ R and δ > 0, ∼

∼

Λ(τx+δ ϕ) − Λ(τx ϕ) (Λ ∗ ϕ)(x + δ) − (Λ ∗ ϕ)(x) = δ δ ∼ ∼ Λ(τx τδ ϕ) − Λ(τx ϕ) = δ !! ∼ ∼ τδ ϕ − ϕ = Λ τx . δ But ∼

∼

∼

∼

τx+δ ϕ(u) − τx ϕ(u) ϕ(u − x − δ) − ϕ(u − x) ∼0 lim = lim = −τx ϕ δ→0 δ→0 δ δ in S(R). Thus ∼ (Λϕ)(x + δ) − (Λ ∗ ϕ)(x) ∼0 = Λ(τx (−ϕ )) = Λ(τx (ϕ0 )) = (Λ ∗ ϕ0 )(x ). δ→0 δ

lim

116

Hence Λ ∗ ϕ is differentiable and (Λ ∗ ϕ)0 = Λ ∗ ϕ0 . Further for all x ∈ R, 0

0

∼



∼ 0

(Λ ∗ ϕ)(x) = Λ (τx ϕ) = −Λ (τx ϕ)



∼0

= Λ(τx (−ϕ )) = (Λ ∗ ϕ0 )(x).

This shows that (Λ ∗ ϕ)0 = Λ0 ∗ ϕ = Λ ∗ ϕ0 . By induction, this leads to Dm (Λ ∗ ϕ) = (Dm Λ) ∗ ϕ = Λ ∗ (Dm ϕ) for any m ∈ N. Proposition 4.4.7. Let Λ ∈ S 0 (R) and ϕ ∈ S(R), then Λ ∗ ϕ ∈ S 0 (R). Proof. We have, for any x, y ∈ R, 1 + (x + y)2 ≤ 2(1 + x2 )(1 + y 2 ). Therefore, k P ϕ(j) (x) , satisfies the following: pk (ϕ) = sup(1 + x2 )k x∈R

j=0

pk (τx ϕ) ≤ 2k (1 + x2 )k pk (ϕ), x ∈ R. Since, Λ is a continuous linear functional on S(R), the norms pk define the topology of S(R), there is a k and c < ∞ such that |Λ(ϕ)| ≤ cpk (ϕ). (See Appendix A.4.4.) Thus ∼ Λ is a tempered distribution. Thus |Λ ∗ ϕ(x)| = Λ(τx ϕ) ≤ c2k pk (ϕ)(1 + x2 )k . This shows that Λ∗ϕ has polynomial growth. If a continuous function on R has polynomial m

growth, then it defines a tempered distribution. In fact, if |g(x)| ≤ c (1 + x2 ) , then Z

Z

|Λg (ϕ)| ≤

|g(x)| |ϕ(x)| dx ≤ c R

(1 + x2 )m |ϕ(x)| dx

R

Z =c

(1 + x2 )m+1 |ϕ(x)|

1 dx ≤ c pm+1 (ϕ) < ∞. (1 + x2 )

R

Thus, it follows that Λ ∗ ϕ ∈ S 0 (R).

117

b Theorem 4.4.8. Let Λ ∈ S 0 (R), ϕ ∈ S(R). Then (Λ ∗ ϕ)b = ϕ b Λ. Proof. Since Λ ∗ ϕ ∈ S 0 (R), Λ ∗ ϕ has Fourier transform in S 0 (R). Let ψ ∈ D(R). Then ψ has compact support. Let K = supp(ψ). Then ∼ bb b = Λ ∗ ϕ(ψ) (Λ ∗ ϕ)b(ψ) = Λ ∗ ϕ(ψ) Z ∼ = (Λ ∗ ϕ)(x)ψ(x) dx R Z

(Λ ∗ ϕ)(x)ψ(−x) dx

= R Z

∼

Λ(τx ϕ)ψ(−x) dx

= R

Z =

∼

Λ(τx ϕ)ψ(−x) dx −K

 Z ∼ =Λ ψ(−x)τx ϕ dx −K

= Λ ((ϕ ∗ ψ)∼ )
b (ϕ ∗ ψ)b =Λ b b ϕ = Λ( bψ) b b ψ). =ϕ bΛ( Hence −K = {−x : x ∈ K}. Since D(R) is dense in S(R) and the class of Fourier transform of members of D(R) is also dense in S(R), the result follows for every b ϕ ∈ S(R). Thus (Λ ∗ ϕ)b = ϕ b Λ.

118

4.5

Exercises

4.5.1. Check whether ex , x ∈ R is a tempered distribution. 4.5.2. Show that ex cos ex , x ∈ R is a tempered distribution. 4.5.3. Show that every polynomial defines a tempered distribution. Z f (x) dx is a tempered distribution. 4.5.4. Show that lim →0 x |x|≥

4.5.5. Find the derivative and Fourier transform of sgn(x), where   1 for x > 0    sgn(x) = 0 for x = 0     −1 for x < 0. 4.5.6. Find the Fourier transform of δ(x − x0 ), x0 ∈ R.  4.5.7. For  > 0, let f = . Then show that f → πδx0 in S 0 (R) as (x − x0 )2 + 2  → 0.

119

120

Chapter 5 Some topics on Fourier transform 5.1

Poisson summation formula This formula due to S.D.P oisson gives a connection between the series of

functional values evaluated at certain period and the series of values of the function’s Fourier transform, evaluated at the same period. This is obtained as a corollary of the following theorem. Theorem 5.1.1. Let f ∈ S(R). Then

∞ P

f (x + 2πn) =

n=−∞

Proof. Define g(x) =

∞ P

∞ 1 P fb 2π n=−∞

f (x + 2πn). Then

n=−∞ ∞ X

g(x + 2π) = =

f (x + 2π(n + 1))

n=−∞ ∞ X

f (x + 2πm)

m=−∞

= g(x).

Thus g is a periodic function with period 2π. Consider 1 gb(n) = 2π 1 = 2π

Zπ −π Zπ

g(x)e−inx dx ∞ X

f (x + 2πk)e−inx dx

−π k=−∞

121

n 2π




einx .

π

∞ Z 1 X = f (x + 2πk)e−i (x+2πk)n dx 2π k=−∞ −π

(2k+1)π Z ∞ 1 X f (y)e−iyn dy = 2π k=−∞ (2k−1)π

=

1 2π

Z∞

n

f (y)e−2πi( 2π )y dy =

1 b f 2π

n 2π




.

−∞

Here, the interchange of sum and integral is permissible as f is assumed to be in S(R). Since f ∈ S(R), one can show that the Fourier series of g converges absolutely and hence ∞ X

g(x) =

f (x + 2πn) =

∞ X

inx

gb(n)e

n=−∞

n=−∞

∞ 1 X b n  inx = f e , 2π n=−∞ 2π

proving our assertion.

Putting x = 0 in Theorem 5.1.1, we obtain the following:

Corollary 5.1.2.

∞ P n=−∞

5.2

f (2πn) =

1 2π

∞ P fb n=−∞

n 2π




for f ∈ S(R).

Uncertainty principle In quantum mechanics, the famous Heisenberg uncertainty principle states

that the position and the momentum of a particle cannot be determined simultaneously with a reasonable good degree of accuracy or certainty. From the Fourier analysis point of view this leads to the uncertainty principle between a given function and its Fourier transform. In general, uncertainty principle can be broadly classi122

fied into two types: qualitative uncertainty principle and quantitative uncertainty principle. In quantitative uncertainty principle, the inequalities gives information about the relation of a function and its Fourier transform. In the case of qualitative uncertainty principle, the theorem can establish the simultaneous behavior of both a function and its Fourier transform. Towards quantitative uncertainty principle, we discuss Heisenberg’s uncertainty principle and towards qualitative uncertainty principle we discuss Hardy’s theorem named after Hardy in this section.

5.2.1

Heisenberg’s uncertainty principle

Theorem 5.2.1. Let ψ ∈ S(R) be such that kψk2 = 1. Then Z∞

2

Z∞

2

b 2 dξ ≥ ξ 2 |ψ(ξ)|

x |ψ(x)| dx · −∞

1 . 16π 2

−∞

2

Further, equality holds if and only if ψ(x) = Ae−Bx , where B > 0 and A2 = Proof. Consider Z∞

∞ |ψ(x)|2 −∞



2

xd(|ψ(x)| ) = x

Z∞ −

−∞

|ψ(x)|2 dx

−∞

= −kψk22 = −1. Hence Z∞ 1=− −∞ Z∞

=−

xd(|ψ(x)|2 ) i h 0 0 x ψ(x)ψ (x) + ψ (x)ψ(x) dx.

−∞

123

q

2B . π

Thus ∞ Z 0 0 1 = x[ψ(x)ψ (x) + ψ (x)ψ(x)]dx −∞ Z∞

≤2

|x||ψ(x)||ψ 0 (x)|dx.

−∞

Z∞

 ≤ 2

 21  x2 |ψ(x)|2 dx 

−∞

Z∞

 21 |ψ 0 (x)|2 dx ,

(5.1)

−∞

on applying Schwarz inequality. But using Plancherel formula, we get, Z

0

Z

2

|ψb0 (ξ)|2 dξ

|ψ (x)| = R

R

= 4π

Z

2

b 2 dξ. ξ 2 |ψ(ξ)|

R

From (5.1), it follows that  21 

 Z

 21

x2 |ψ(x)|2 dx 4π 2

1 ≤ 2 R

Z

b 2 dξ  ξ 2 |ψ(ξ)|

R

 21

 21 

 Z = 4π 

x2 |ψ(x)|2 dx 

Z

b 2 dξ  . ξ 2 |ψ(ξ)|

R

R

Squaring both sides, we get

1 ≤ 16π

2

Z

2

2

Z

x |ψ(x)| dx R

b 2 ξ 2 ψ(ξ) dξ,

R

proving our assertion. The equality holds if equality holds in Schwarz inequality. This leads to the following. We can write ψ 0 (x) = βxψ(x) for some constant β, 124

which means

ψ 0 (x) ψ(x)

= βx. On integration, we get ψ(x) = Ae

βx2 2 .

But ψ ∈ S(R),

which forces that β must be negative. Let r β = −2B, B > 0. As kψk2 = 1, we get ∞ R 2B 2 . e−2B dx = 1 which leads to A2 = A2 π −∞ Definition 5.2.2. A function f : C → C is said to be of exponential type T , 0 < T < ∞, if |f (z)| ≤ ceT |z|

∀ z ∈ C and a constant c.

In order to prove Hardy’s theorem, we make use of the following well known theorem in complex analysis. Theorem 5.2.3. (Phragmen-Lindelof theorem) If f is analytic and of exponential type in a sector D with opening less than π and if it is continuous in D, |f | ≤ M on ∂D, then |f | ≤ M throughout D. Theorem 5.2.4. (Hardy) Suppose f is a measurable function on R such that |f (x)| ≤ 2 2 ce−ax and |fb(ξ)| ≤ ce−bξ , where a, b and c are positive constants. Then

(i) f (x) = 0 if ab > π 2 . 2

(ii) f (x) = c e−ax if ab = π 2 . Proof. Assume that the result is proved for a = b = π. For r > 0, define fr (x) = f (rx). Take r2 = πa . Then 2 2

2

|fr (x)| = |f (rx)| ≤ c e−ar x = c e−πx . ab 2 −bξ2 1 |fbr (ξ)| = fb(ξ/r) ≤ c e r2 = c e− π2 ξ . r

(5.2)

2

If ab = π 2 , then fr (x) = c e−πx . If ab > π 2 , the inequality (5.2) is impossible unless c = 0. Therefore, it is enough to prove for a = b = π. Let a = b = π and w = ξ + iη.

125

Then fb has a holomorphic extension to C and it satisfies Z −2πixw dx |fb(w)| = f (x) e R Z −2πxiξ 2πxη e dx = f (x) e Z ZR 2 e−πx e2πxη dx ≤ |f (x)| e2πxη dx ≤ c R

R

Z

−π(x−η)2

≤c

e

e

πη 2

2

dx = ceπη .

R

Clearly, fb is continuous. For any simple closed curve C in C, an application of Fubini’s theorem leads to  Z fb(w)dw = C

 Z

Z  C

Z = R

R Z 

f (x) e−2πixw dx dw 
e−2πixw dw  f (x)dx = 0.

(5.3)

C

Thus, it follows from Morera’s theorem that, fb is an entire function. Now assume ∞ P that f is an even function. Then fb is also an even function. Let fb(w) = cn w2n . n=0

∞ P √ Define h(w) := fb( w) = cn wn . Then h(w) is an entire function. We claim that n=0 √ πw e h(w) is a constant on C. Suppose w = R > 0. Then |h(w)| = fb( w) ≤ c e−πR .

But, if we take w = Reiθ , we get √ 2 √ |h(w)| = |fb( w)| ≤ c eπ(Im w) .

126

Hence, |h(w)| ≤ c eπR sin

2

θ 2.

Therefore h is of exponential type. Now consider,

) ( δ iπR ei(θ− 2 ) iπw e−iδ/2
= exp exp sin 2δ sin 2δ ( ) iπR[cos(θ − δ/2) + i sin(θ − δ/2)]
= exp sin 2δ ) ( −πR sin(θ − δ/2)
. = exp sin 2δ If θ = 0, then exp

−δ

!

h(w) = eπR |h(w)| ≤ eπR · c e−πR = c.

−δ

!

2 h(w) = e−πR |h(w)| ≤ e−πR c eπR sin (δ/2) ≤ c if 0 < δ < π.

iπw e 2 sin(δ/2)

If θ = δ, then exp

iπw e 2 sin(δ/2)

An application of Phragmen-Lindelof theorem leads to  |h(w)| ≤ c exp

πR sin(θ − δ/2) sin(δ/2)

 for 0 ≤ θ ≤ δ.

Letting δ → π, we get, |h(w)| ≤ c exp(−πRcosθ) for, 0 ≤ θ ≤ π.

Similarly, h(w) will satisfy the same bound in the lower half plane. Thus

|h(w)| ≤ c exp(−πR cosθ) for all w ∈ C. 127

Hence, |eπw h(w)| ≤ c for all w ∈ C. By applying Liouville’s theorem, we conclude that eπw h(w) is a constant on C, which 2

implies that h(w) = c e−πw and consequently f (w) = c e−πw for w ∈ C. Now assume that f is odd. Then fb is also odd and fb(0) = 0. fb(ξ) 2 , then g is even. By the previous argument, g(x) = c e−πx and If g(ξ) = ξ 2 −πξ 2 b . But by the given hypothesis, |fb(ξ)| ≤ c e−πξ . This is impossible f (ξ) = c ξ e unless c = 0.

5.3

Paley-Wiener theorem Consider the analytic extension of a Fourier transform of a square integrable

function to the complex plane. This theorem named after P aley and W iener gives a necessary and sufficient condition for such extended function to become an entire function of exponential type. The proof of this theorem also uses complex analysis techniques. Theorem 5.3.1. Let f ∈ L2 (R). Then fb can be extended as an entire function of exponential type less than or equal to 2πT if and only if f (x) = 0 for |x| > T . Proof. Suppose f (x) = 0 for |x| > T . Then f is integrable on R. Consider for w = ξ + iη Z fb(w) = R ZT

=

f (x)e−2πiwx dx,

f (x)e−2πwx dx.

−T

128

Then ZT |f (x)||e

|fb(w)| ≤

−2πiwx

ZT | dx ≤

|f (x)|e2π|w|T dx

−T

−T

ZT

≤ e2π|w|T

|f (x)| dx = kf k1 e2π|w|T .

−T

Therefore, fb is an entire function of exponential type 2πT . Conversely, suppose fb can be extended to an entire function of exponential type 2πT. In order to show that f = 0 for |x| > T, consider the function |h(w)| = 1 2

R

fb(w + y) dy. Then h(w) is an entire function and of exponential type 2πT. In fact,

− 12

we have, 1 1 Z2 Z12 Z2 h(w) = fb(w + y) dy ≤ |fb(w + y)| dy ≤ ce2π|w+y|T dy, − 1 −1 −1 2

2

2π|w|T

Z

2

1 2

= ce

0

e2π|y|T dy, = c e2π|w|T .

− 12

Now h can be viewed as follows: 1

Z2 h(w) =



Z∞

 − 21

 f (x) e−2πi(w+y)x dx dy

−∞ 1

Z∞ Z2 =

f (x) e−2πi(w+y)x dy dx

−∞ − 1 2

129

Z∞ =

  f (x) e−2πiwx 

−∞

  e−2πiyx dy  dx

− 12

Z∞ =

1

Z2

−2πiwx



f (x) e

sin πx πx

 dx

−∞



b sin πx = f (x) (w) πx On the real line, w = ξ, we have 1

Z2 |fb(ξ + y)| dy.

|h(w)| = |h(ξ)| ≤ − 12

Applying Cauchy-Schwartz inequality, we get  1  21  1  21 Z2 Z2     2 b |h(ξ)| ≤  |f (ξ + y)| dy   dy  , − 21

− 12

showing that |h(w)| ≤ kfbk2 = kf k2 , and  1  Z∞ Z∞ Z2   2 khk22 = |h(ξ)|2 dξ, =  |fb(ξ + y)| dy  dξ. −∞

−∞

− 12

1

Z2 Z∞ =

|fb(ξ + y)|2 dξ dy

− 21 −∞ 1

Z2 =

kfbk22 dy = kfbk22 = kf k22 .

− 21

Let B > T, and consider the function h(w)e2πiBw . Then 0

0

|h(w)e2πiBw | ≤ |h(w)| e2πB|w| ≤ c e2π|w|T e2πB|w| ≤ c e2π(B+T )|w| . 130

Thus h(w) e2πiBw is of exponential type 2π(B + T ). On the positive real line, w = R, we have, |h(R)e2πiBR | = |h(R)| ≤ kf k2 . On the positive imaginary axis, w = ib, b > 0, we have, 0

0

|h(ib) e2πiB(ib) | = |h(ib)|e−2πBb ≤ e−2πBb (c e2πbT ) = c e−2π(B−T )b ,

which goes to 0 as b → ∞. Thus h(w) e2πiBw is bounded on the positive real axis and the positive imaginary axis. Therefore by Phragmen-Lindelof lemma, we have |e2πiBw h(w)| ≤ c in the first quadrant. Similarly, |e2πiBw h(w)| ≤ c in the second quadrant. In other words, for w = iθ

Reiθ , 0 ≤ θ ≤ π, we have |h(Reiθ ) e2πiBRe | ≤ c, which means |h(Reiθ )| ≤ c e2πBR sin θ, for 0 ≤ θ ≤ π. Now for a fixed A > 0 and x > B > T , consider the integral ZR I=

e2πiwx h(w) dw. 1 − Aiw

(5.4)

−R

Now using the Cauchy theorem for the semi circle to the function get, Zπ I = −i

exp(2πiReiθ )h(Reiθ ) iθ R e dθ. 1 − AiReiθ

0

Then Zπ |I| ≤

|h(Reiθ )|| exp(2πiRx(cosθ + i sin θ))| R dθ |1 − AiReiθ |

0

Zπ ≤

ce2πBR sin θ e−2πRx sin θ R dθ |1 − AiReiθ |

0

131

e2πiwx h(w), we 1 − Aiw

cR = AR − 1

Zπ

e−2πR(sin θ)(x−B) dθ.

0

Now letting R → ∞, and applying inverse Fourier transform, we can conclude from h(w) (5.4) that = 0. Now letting A → 0, we conclude that h(w) = 0. 1 − Aiw  b sin πx (w). Thus f (x) = 0 for a.e x, x > B > T . In a Recall h(w) = f (x) πx similar way, we can show that f (x) = 0 for x < −T a.e. Consequently f (x) = 0 for |x| > T a.e.

5.4

Wiener’s theorem We know that L1 (R) is translation invariant. Now start with a closed translation-

invariant subspace Y of L1 (R). How big this space can be? Wiener’s theorem says that if the zero set of Y (defined below) is empty, then Y coincides with L1 (R). Definition 5.4.1. Let Y be a subspace of L1 (R). The zero set of Y , denoted by Z(Y ) is defined to be the set of all ξ ∈ R such that fb(ξ) = 0 for all f ∈ Y . In other words, Z(Y ) = {ξ ∈ R : fb(ξ) = 0 ∀ f ∈ Y }. Theorem 5.4.2. (Wiener) If Y is a closed translation invariant subspace of L1 (R) and if Z(Y ) is empty, then Y = L1 (R). In order to prove the theorem we first prove the following lemmas. Lemma 5.4.3. Let f ∈ L1 (R) and t ∈ R. Then for a given  > 0, there exists h ∈ L1 (R), khk1 <  such that (f + h)b(ξ) = fb(t), for every ξ in some neighborhood of t.

132

Proof. Let ϕ ∈ L1 (R) be such that ϕ b = 1 in some neighborhood of the origin. Fix δ > 0. For x ∈ R, define, ϕtδ (x) = eitx Dδ ϕ(x). Using this, we define, hδ (x) = ϕtδ (x)fb(t) − (ϕtδ ∗ f )(x).

(5.5)

Then hδ (x) =

ϕtδ (x)

Z f (y) e

−ity

Z dy −

R

Z =

ϕtδ (x − y)f (y) dy

R

  f (y) ϕtδ (x) e−ity − ϕtδ (x − y) dy.

R

Thus Z |hδ (x)| ≤

|f (y)||ϕtδ (x) e−ity − ϕtδ (x − y)| dy.

R

Consider t ϕδ (x) e−ity − ϕtδ (x − y) = eitx Dδ ϕ(x) e−ity − e−it(x−y) Dδ ϕ(x − y)
= eit(x−y) δ −1 ϕ xδ − eit(x−y) δ −1 ϕ

= eit(x−y) δ −1 ϕ xδ − ϕ x−y δ

. = δ −1 ϕ xδ − ϕ x−y δ

x−y δ




Thus |hδ (x)| ≤ δ

−1

Z

|f (y)| ϕ

x δ




−ϕ

x−y δ


dy.

R

Hence, it follows that  Z Z Z khδ k1 = |hδ (x)| dx ≤ δ −1  |f (y)| ϕ R

R

133

R

 x δ




−ϕ

x−y δ


dy  dx

 Z =

|f (y)| δ −1

R

 Z

ϕ

x δ




−ϕ

x−y δ


dx dy,

R

by applying Fubini’s theorem. But Z Z x


x−y −1 ϕ δ − ϕ δ dx = ϕ(u) − ϕ u − yδ du ≤ 2kϕk1 , δ R

R

and Z


ϕ(u) − ϕ u − y du = kϕ − τ y ϕk1 → 0 as δ → ∞. δ δ

R

In other words khδ k1 → 0 as δ → ∞. Further, using (5.5)
b hbδ (ξ) = ϕtδ (ξ)fb(t) − ϕ bδ (ξ)fb(ξ) Since ϕ b = 1 in some neighborhood of the origin, (ϕtδ ) = 1 in some neighborhood Vδ b

of t. Thus, for every ξ ∈ Vδ ,

b hδ (ξ) = fb(t) − fb(ξ),

proving our assertion. Definition 5.4.4. Let V be an open subset of R. We say that f ∈ S 0 (R) vanishes in V if hϕ, f i = 0, ∀ ϕ ∈ S(R), with support in V . Let Ω be the union of all open sets V in R in which f vanishes. The complement of Ω is called the support of f . Lemma 5.4.5. Let ϕ ∈ L∞ (R). If Y is a subspace of L1 (R) such that f ∗ ϕ = 0 for every f ∈ Y . Then the support of the tempered distribution ϕ b is contained in Z(Y ). Proof. In order to prove that support of ϕ b is contained in Z(Y ), it is enough to show that Z(Y )c ⊂ (supp ϕ) b c. 134

In view of the definition of support of a tempered distribution, it is enough to show that any t ∈ Z(Y )c will have a neighborhood in which ϕ b will vanish. Fix t ∈ Z(Y )c . By definition of Z(Y ), there exists f ∈ Y such that fb(t) 6= 0. Without loss of generality, we can assume that fb(t) = 1. Then by lemma 5.4.3 for a given  > 0, there exists h ∈ L1 (R) such that khk1 <  and (f + h)b(ξ) = fb(t) = 1, for every ξ in some neighborhood V of t. Thus, it is enough to prove that ϕ b vanishes in V . In other words, we should show that hg, ϕi b = 0 for every g ∈ S(R) with the support of g in V . By inverse Fourier transform theorem on S(R), it is enough to b ϕi show that hψ, b = 0 for every ψ ∈ S(R) whose Fourier transform ψb has support in bb b ϕi V . But hψ, b = hψ, ϕi = (ϕ ∗ ψ)(0). Thus, it is sufficient to show that ϕ ∗ ψ = 0. Fix ψ ∈ S(R) such that support ψb ⊂ V . Construct a sequence of functions {un } as follows: Let u0 = ψ, u1 = h ∗ u0 , u2 = h ∗ u1 . Inductively, we have un = h ∗ un−1 , n ≥ 1. ∞ N P P Consider the series un . Let SN = un . n=0

n=0

Consider, for N ≥ M ,

M N

X X

un kSN − SM k1 = un −

n=0 n=0 1

N

X

= un

n=M +1

≤

N X

1

kun k1 ≤

n=M +1

= kψk1

N X

khkn1 kψk1

n=M +1 N X

khkn1 .

n=M +1

135

Since khk1 < 1,

N P

khkn1 → 0 as M, N → ∞. Thus, {SN } is a Cauchy sequence

n=M +1

in L1 (R). Since, L1 (R) is complete, {SN } converges to s in L1 (R). Thus, the series ∞ P un is summable to s in L1 (R). Since (f + h)b(ξ) = 1 ∀ ξ ∈ V , and support n=0

b Thus ψ(ξ) b fb(ξ) + ψ(ξ) b b ψb ⊆ V , we have fb(ξ) + b h(ξ) = 1 for all ξ ∈ supp ψ. h(ξ) b b b fb(ξ). Recall that for every ξ, = ψ(ξ) for all ξ ∈ V . Hence, [1 − b h(ξ)]ψ(ξ) = ψ(ξ)   ∞ P 1 b b b b b b b fb(ξ). |h(ξ)| ≤ khk∞ ≤ khk1 < 1. Thus, ψ(ξ) = ψ(ξ)f (ξ) = hm (ξ)ψ(ξ) b m=0 1 − h(ξ) ∞ P b b fb(ξ) = sbfb. Taking the inverse Fourier transform, we get, Thus, ψb = hm (ξ)ψ(ξ) m=0

ψ = s ∗ f . Consequently, ψ ∗ ϕ = s ∗ f ∗ ϕ = 0, proving our assertion. Proof of Wiener’s theorem. By the Hahn-Banach theorem, it is enough to show that Y ⊥ , annihilator of Y is {0}. Let φ ∈ L∞ (R) such that hf, ϕi = 0 ∀ f ∈ Y . This R means f ∗ ϕ(0) = f (x)ϕ(−x)dx = 0 ∀ f ∈ Y . Then it follows from Lemma 5.4.5 R

that supp ϕ b is contained in Z(Y ). But given that Z(Y ) is empty. Thus support ϕ b = 0. Hence ϕ b = 0. Since the Fourier transform from S 0 (R) into S 0 (R) is one-one, it follows that ϕ = 0 as a tempered distribution. Thus ϕ = 0 in L∞ (R), proving Y ⊥ = 0. Theorem 5.4.6. (Wiener-Tauberian Theorem) Suppose ϕ ∈ L∞ (R), b h(ξ) 6= 0 ∀ ξ ∈ R, and lim (h ∗ ϕ)(x) = ab h(0). Then lim (f ∗ ϕ)(x) = afb(0) for all x→±∞

x→±∞

1

f ∈ L (R). Proof. Let Y = {f ∈ L1 (R) : lim (f ∗ ϕ)(x) = afb(0)}. Let f, g ∈ Y and α, β ∈ R. x→±∞

1

Then clearly, αf + βg ∈ L (R). Now,

lim ((αf + βg) ∗ ϕ)(x) = α lim (f ∗ ϕ)(x) + β lim (g ∗ ϕ)(x)

x→±∞

x→±∞

x→±∞

= αafb(0) + βab g (0) = a(αf + βg)b(0). 136

Hence, αf + βg ∈ Y proving that Y is a subspace. Let s ∈ R, f ∈ Y . Then (τs f )b(0) = fb(0). Again lim ((τs f ) ∗ ϕ)(x) = lim τs (f ∗ ϕ)(x) = lim (f ∗ ϕ) x→±∞

x→±∞

x→±∞

(x − s) = afb(0) = a(τs f )b(0). This proves that Y is translation invariant. Now we claim that Y is closed. Let {fn } be a sequence in Y such that kfn − f k1 → 0 as n → ∞ for some f ∈ L1 (R). We have to show that f ∈ Y . We have (fn ∗ ϕ)(x) → afbn (0) as x → ± ∞. Consider, ∀ x ∈ R, Z Z |fn ∗ ϕ(x) − f ∗ ϕ(x)| = fn (x − y)ϕ(y)dy − f (x − y)ϕ(y)dy R Z ≤ |fn (x − y) − f (x − y)| |ϕ(y)| dy R

Z = kϕk∞

|fn (x − y) − f (x − y)| dy R

= kϕk∞ kfbn − f k1 → 0, as n → ∞. Further, |fbn (0) − fb(0)| ≤ kfbn − fbk∞ ≤ kfn − f k1 → 0 as n → ∞. Here kϕk∞ denotes the L∞ -norm of φ and kfbn − fbk∞ denotes the supremum norm of fbn − fb. Thus fn ∗ ϕ converges to f ∗ ϕ uniformly on R. Thus lim (f ∗ϕ)(x) = lim lim (fn ∗ϕ)(x) = lim lim (fn ∗ϕ)(x) = lim afbn (0) = afb(0),

x→±∞

x→±∞ n→∞

n→∞ x→±∞

n→∞

showing that f ∈ Y . Now, it follows from Wiener’s theorem that Y = L1 (R), thus proving our assertion.

137

5.5

A multiplier theorem for L1(R)

Definition 5.5.1. Let T : L1 (R) → L1 (R) be continuous linear transformation. If T commutes with translation, viz, T τs = τs T

∀ s ∈ R, then we say that T is a

multiplier for L1 (R). From the above definition, it is not really clear why it is called a multiplier. We shall show in (statement v of ) the following theorem that in the Fourier domains, T acts as an multiplication operator. Theorem 5.5.2. Let T: L1 (R) → L1 (R) be a continuous linear transformation. Then the following are equivalent. (i) T τs = τs T

∀ s ∈ R.

(ii) T (f ∗ g) = T f ∗ g = f ∗ T g

∀ f, g ∈ L1 (R).

(iii) There exists a unique µ ∈ M(R) such that, T f = µ ∗ f

∀ f ∈ L1 (R), where

M(R) denotes the Banach algebra of all finite Borel measures on R. (iv) There exists a unique µ ∈ M(R) such that (T f )b = µ bfb for every f ∈ L1 (R), where µ b denotes the Fourier-Stieltjes transform of µ. (v) There exists a unique φ ∈ L∞ (R) such that (T f )b = φfb for every f ∈ L1 (R). Proof. Assume (i). In order to show that T (f ∗ g) = T f ∗ g for every f, g ∈ L1 (R), it is enough to show that hT (f ∗ g), hi = hT (f ∗ g), hi ∀ h ∈ L∞ (R). Let T ∗ denote the adjoint of T . Consider Z hT f ∗ g, hi =

h(t)(T f ∗ g)(t) dt R

Z = R

  Z h(t)  g(s)T f (t − s)ds dt R

138

Z =

  Z g(s)  h(t)T f (t − s)dt ds

R

R  Z Z = g(s)  h(t)τs T f (t)dt ds R

 R Z Z = g(s)  h(t)T τs f (t)dt ds R  Z Z = g(s)  T ∗ h(t)τs f (t)dt ds R

R

Z =

R

  Z T ∗ h(t)  g(s)f (t − s)ds dt

R Z

R

h(t)T (f ∗ g)(t)dt

= R

= hT f ∗ g, hi , for every h ∈ L∞ (R), using the assumption of (i), Fubini’s theorem and change of variables. In a similar way, we can show that

T (f ∗ g) = f ∗ T g

∀ f, g ∈ L1 (R).

Thus (i) ⇒ (ii). Now we shall assume (ii). In order to prove (iii), let {gα } be an approximate identity in L1 (R) such that kgα k1 ≤ 1. Consider, kT gα ∗ f − T f k1 = kT (gα ∗ f ) − T f k1 ≤ kT kkgα ∗ f − f k which goes to zero as α → ∞. As kT gα k1 ≤ kT k, kgα k1 ≤ kT k, {T gα } is a bounded subset of L1 (R). Since L1 (R) ⊂ M(R) and kf kM(R) ≤ kf kL1 (R) for every f ∈ L1 (R), we get {T gα } is a norm bounded subset 139

of M(R). But M(R) can be viewed as a dual space of C0 (R). By applying Banach Alaoglu theorem, we get a unique µ ∈ M(R) and a subset {T gβ } of {T gα } such that T gβ → µ in the weak * topology and also hT f, gi = limhT gβ ∗ f, gi = hµ ∗ f, gi, for β

all f, g ∈ Cc (R). As Cc (R) is dense in C0 (R), we get T f = µ ∗ f as elements of M(R) and in particular as element of L1 (R). As Cc (R) is dense in L1 (R), we get T f = µ ∗ f for every f ∈ L1 (R), proving (iii). Now, (iv) and (v) are obvious. To prove (i) by assuming (v), let f ∈ L1 (R). For s ∈ R, consider b −2πisξ b b b (T τs f )b(ξ) = φ(ξ)(τ f (ξ) s f ) (ξ) = φ(ξ)e

b fb(ξ)e−2πisξ = (T f )b(ξ)e−2πisξ = φ(ξ) = (τs T f )b(ξ).

Then (i) follows from the uniqueness of the Fourier transform.

5.6

Solved problems

5.6.1. Let

  1 − |x| if |x| ≤ 1 g(x) =  0 otherwise.

(a) Find the Fourier transform of g. (b) Apply the Poisson summation formula to g to show that where α is real, but not an integer. ∞ P (c) Prove as a consequence that

1 π2 = , 2 (sin πα)2 n=−∞ (n + α) ∞ P

1 π = , where α is real, but not an tan πα n=−∞ (n + α)

integer.

140

Solution. (a) We have Z0 gb(ξ) =

(1 + x) e−2πixξ dx +

−1

Z1

(1 − x) e−2πixξ dξ

0

   1 (−1)  −2πiξ =− 1 − e2πiξ + −e +1 2 2 (2πiξ) (2πiξ)   1 =− 1 − e2πiξ − e−2πiξ + 1 2 (2πiξ) 2 e−2πiξ e2πiξ − + = (2πiξ)2 (2πiξ)2 (2πiξ)2  πiξ 2  πiξ   −πiξ   −πiξ 2 e e e e = −2 + 2πiξ 2πiξ 2πiξ 2πiξ  πiξ    2 2 e − e−πiξ sin πξ = = . 2πiξ πξ (b) Let gα (x) = e2πiαx g(x). Applying Poisson summation formula to gα , ∞ X

gbα (n) =

2 X ∞ sin πα

1 π2 1 = 1 ⇒ = . (n + α)2 (n + α)2 (sin πα)2 n=−∞ n=−∞

π2

(c) By (b)

gα (n)

n=−∞ ∞ X

n=−∞



∞ X

π2 1 = . Integrating both sides, we obtain 2 (sin πα)2 n=−∞ (n + α) ∞ P

∞ X

1 π = . n + α tan πα n=−∞ 5.6.2. (a) Suppose a < b, and f is the function such that f (x) = 0 if x ≤ a or x ≥ b −1

−1

and f (x) = e x−a e b−x if a < x < b. Show that f is infinitely differentiable on R.

141

(b) Prove that there exists an infinitely differentiable function F on R such that F (x) = 0 if x ≤ a, F (x) = 1 if x ≥ b and F is strictly increasing on [a, b]. (c) Let δ > 0 be so small that a + δ < b − δ. Show that there exists an infinitely differentiable function g such that g is zero if x ≤ a or x ≥ b, g is one on [a + δ, b − δ] and g is strictly monotonic on [a, a + δ] and [b − δ, b]. Solution. Given

 −1 −1  e x−a e b−x a < x < b f (x) =  0 otherwise.

In order to prove that f is infinitely differentiable it is enough to prove that the function g defined as   e− x1 x > 0 g(x) =  0 otherwise is infinitely differentiable. In order to prove this it is enough to verify only at x = 0. Note that g is continuous. We prove by induction that

g

(k)

(x) =

 

Pk (x) − 1 e x x2k

x>0

 0 otherwise, where Pk (x) is any polynomial in x. This is clearly true for k = 0. Assume that this is true for any k ≥ 0. By the product rule, we have g (k+1) (x) =

Pk0 (x) − 1 2kPk (x) − 1 Pk (x) − 1 x2 Pk0 (x) − 2kxPk (x) + Pk (x) − 1 x − x + x = e e e e x x2k x2k+1 x2k+2 x2k+1

which proves our claim. Further notice that lim g (k) (x) = 0 ∀ k ≥ 0. Hence g is x→0

differentiable.

142

(b) By (a) we have an infinitely differentiable function f given by  −1 −1  e x−a e b−x a < x < b f (x) =  0 otherwise. Rx

Note that supp(f ) = [a, b]. Define F (x) =

a Rb

f (x)dx

. This F satisfies all the requiref (x)dx

a

ments. (c) Define functions F and G as follows: Rx F (x) =

a a+δ R a Rb

G(x) =

−1

−1

e x−a e a+δ−x dx −1

−1

e x−a e a+δ−x dx −1

−1

e x−(b−δ) e b−x dx

x

Rb

. e

−1 x−(b−δ)

e

−1 b−x

dx

b−δ

Both F and G are infinitely differentiable. Define the function g as    F (x) a 0 there exists δ > 0 such that x, y ∈ R with |x − y| < δ implies kτx ϕ − τy ϕkp < . kT (x) − T (y)kp = kτx f − τy f kp ≤ kτx f − τx ϕkp + kτx ϕ − τy ϕkp + kτy ϕ − τy f kp = kτx (f − ϕ)kp + kτx ϕ − τy ϕkp + kτy (f − ϕ)kp = 2k(f − ϕ)kp + kτx ϕ − τy ϕkp < 2 +  = 3 kT (x) − T (y)k <  for |x − y| < δ. Thus T is continuous on R.

5.7

Exercises

5.7.1. Let f ∈ L1 (R). If fb is an entire function of the exponential type less than or equal to 2πT , then show that fb(w) ≤ Ce2πBR sin θ , for B > T on the closed upper half plane. x2 1 − 4t . If f ∈ S(R), then show that u(x, t) = (f ∗ Ht )(x) e (4πt)1/2 for t > 0, then show that u(x, t) = (f ∗ Ht )(x), for t > 0, solves the heat equation ∂u ∂ 2u = . Further show that ku(·, t) − f (·)k2 → 0 as t → 0. ∂t ∂x2 1 t 5.7.3. Let Pt (x) = , x ∈ R, t > 0. 2 π x + t2 (a) Show that the Fourier transform of Pt is e−2π|ξ|t .

5.7.2. Let Ht (x) =

(b) Let u = f ∗ Pt , for f ∈ S(R). Then show that ∆u = 0. (c) Show that ku(·, t) − f (·)k2 → 0 as t → 0. (d) Using Poisson summation formula show that

145

∞ ∞ P 1 P t = e−2πt|n| . π n=−∞ t2 + n2 n=−∞

146

Chapter 6 Fourier analysis on Rn 6.1

Fourier transform on Rn In this section, we define the Fourier transform on Rn and study some of its

properties. For f ∈ L1 (Rn ), its Fourier transform, denoted by fb, is defined as

fb(ξ) = (2π)

−n 2

Z

f (x) e−ix·ξ dx,

ξ ∈ Rn .

Rn

We warn the readers that the defintion taken here is different from the one defined in one dimension. As mentioned in Chapter 3, we take in the remaining chapter the definition of Fourier transform as the extension of the definition given in 3.2. Clearly, fb is a bounded function on Rn with kfbk∞

≤ (2π)−n/2 kf k1 . The

Riemann-Lebesgue lemma states that, for f ∈ L1 (Rn ), fb is a continuous function n Q 1 2 1 2 e− 2 xj , then vanishing at infinity. If we take φ(x) = e− 2 |x| , x ∈ Rn , i.e., φ(x) = j=1

b = e− 21 |ξ|2 = φ(ξ). φ(ξ) We now introduce some multi-index notation. For x ∈ Rn , α ∈ Nn , we  αj n  Q ∂ . Given a define xα = xα1 1 · · · xαnn ; |α| = α1 + α2 + · · · + αn and ∂ α = ∂xj j=1 P polynomial p(x) = aα xα , the corresponding differential operator is defined by |α|≤m P p(D) = aα ∂ α . An easy computation shows that |α|≤m

p(D)ϕ(ξ) b = (2π)

−n/2

Z (p(D)e

−ix·ξ

) ϕ(x) dx = (2π)

−n/2

Rn

Z Rn

p(D)ϕ(ξ) b = (p(−ix)ϕ(.))b(ξ) 147

e−ix·ξ p(−ix) ϕ(x) dx,

Similarly, using the formula (i)|α| ∂xα (e−ix·ξ ) = ξ α e−ix·ξ , and integrating by parts, we get (p(D)ϕ)b(ξ) = p(iξ)ϕ(ξ). b In general, when f ∈ L1 (Rn ), fb need not be integrable. For instance, f (x) = χ(−1,1) (x), is an integrable function on R, whose Fourier transform is not so. We are interested in obtaining an inversion formula for the Fourier transform. For this purpose, we need to study subspaces W of L1 (Rn ), for which fb ∈ L1 (Rn ) whenever f ∈ W . Such a subspace of L1 (Rn ) is provided by the subspace of smooth functions on Rn which is invariant under differentiation and multiplication by polynomials. We denote by S(Rn ), the space of all functions f ∈ C ∞ (Rn ) such that xα ∂ β f ∈ L∞ (Rn ) for every α, β ∈ Nn . Clearly, S(Rn ) ⊂ L1 (Rn ), Cc∞ (Rn ) ⊂ S(Rn ), and so S(Rn ) is dense in L1 (Rn ). By definition, S(Rn ) is invariant under the Fourier transform, i.e.,

f ∈ S(Rn )

implies fb ∈ S(Rn ). Indeed, for f ∈ S(Rn ), ξ α ∂ξβ fb(ξ) = ξ α ((−ix)β f (·)) b (ξ) = (i)|α| (∂xα (−ix)β f (.))b(ξ). Also note that if f ∈ Cc∞ (Rn ) then fb ∈ S(Rn ). For this space S(Rn ) we obtain the inversion formula:

Figure 6.1: Two dimension Gaussian.

148

Theorem 6.1.1. For every f ∈ S(Rn ), we have Z

−n 2

f (x) = (2π)

fb(ξ) eix·ξ dξ, x ∈ Rn .

Rn

Proof. It is enough to prove the formula for x = 0. For, we can apply it to the function τx f ( where (τx f )(y) := f (y − x)) to obtain the required formula. We now prove that f (0) = (2π)

−n/2

Z fb(ξ)dξ. Rn

1 2 For this, we make use of the fact that φb = φ when φ(x) = e− 2 |x| . For t > 0, we define 1 2 2 b = e− 2 t |ξ| . Also, by φt (x) = t−n φ( xt ). An easy calculation gives that φbt (ξ) = φ(tξ)

Fubini’s theorem, we have Z

Z f (x)b g (x) dx =

Rn

fb(ξ) g(ξ) dξ,

for f, g ∈ S(Rn ).

Rn

Taking gb = φt , we get Z Z x t2 2 −n t f (x)φ dx = fb(ξ) e− 2 |ξ| dξ t or

Rn

Rn

Z

Z f (tx)φ(x) dx =

Rn

2

t 2 fb(ξ)e− 2 |ξ| dξ.

Rn

Here, f ∈ S(Rn ), φ ∈ L∞ (Rn ) and f (tx) → f (0) as t → 0. By the dominated convergence theorem, taking limit as t → 0, we get Z f (0)

Z φ(x) dx =

Rn

fb(ξ) dξ. Rn

149

As (2π)−n/2

R

1

2

e− 2 |x| dx = 1, the above gives f (0) = (2π)−n/2

Rn

R

fb(ξ) dξ, proving the

Rn

result. From the inversion formula for S(Rn ), it is clear that f ∈ S(Rn ) if and only b if fb ∈ S(Rn ) ( since fb(ξ) = f (−ξ)). Denoting F f = fb, we observe that F 4 =Id on S(Rn ), where Id denotes the identity operator. Using the density of S(Rn ) in L1 (Rn ), we prove the following inversion formula: Theorem 6.1.2. (Inversion formula) Assume that both f, fb ∈ L1 (Rn ). Then −n/2

Z

fb(ξ) eix·ξ dξ,

f (x) = (2π)

for a.e. x ∈ Rn .

Rn

Proof. We start with Z

Z fb(ξ) g(ξ) dξ =

Rn

f (x)b g (x) dx, Rn

for any g ∈ S(Rn ). By the inversion formula for S(Rn ), we have

g(ξ) = (2π)

−n/2

Z

gb(x)eix·ξ dx, ξ ∈ Rn .

Rn

Using this in the earlier equation, we see that  Z Z Z  −n/2 ix·ξ b (2π) f (ξ)e dξ gb(x)dx = f (x) gb(x) dx. Rn

Rn −n/2

Taking h(x) = (2π)

R

Rn ix·ξ

fb(ξ) e dξ, gives Z (f (x) − h(x))b g (x) dx = 0, for all g ∈ S(Rn ).

Rn

Rn

150

As Cc∞ (Rn )b⊂ S(Rn ), by the inversion formula for S(Rn ), Z

(f (x) − h(x))ψ(x)dx = 0, for all ψ ∈ Cc∞ (Rn ).

Rn

This shows that

R

(f (x) − h(x)) ψ(x) dx = 0 for any ψ ∈ Cc∞ (Rn ) with support

K

contained in a compact set K. Note that f − h ∈ L1 (K) (since f ∈ L1 (Rn ) and h ∈ L∞ (Rn )). Since Cc∞ is dense in L1 (K), we get (f − h)(x) = 0 for a.e x ∈ K. As K is an arbitrary compact set, f (x) = h(x) for a.e. x ∈ Rn , which means R f (x) = (2π)−n/2 fb(ξ) eix·ξ dξ, for a.e. x ∈ Rn . Rn

For functions f, g ∈ L1 (Rn ), their convolution is defined by Z (f ∗ g)(x) =

f (x − y) g(y) dy, x ∈ Rn .

Rn

Fubini’s theorem gives that f ∗ g is also integrable on Rn with kf ∗ gk1 ≤ kf k1 kgk1 . It n is easy to check that (f ∗ g)b(ξ) = (2π) 2 fb(ξ) gb(ξ). If we define f ∗ (x) := f (−x) then

we can check that (f ∗ )b(ξ) = fb(ξ). For f, g ∈ S(Rn ), applying the inversion formula (at 0) to the function f ∗ g, we get

f ∗ g(0) = (2π)

−n/2

Z

Z (f ∗ g)b(ξ) dξ =

Rn

fb(ξ) gb(ξ) dξ. Rn

Taking g = f ∗ gives ∗

Z

f ∗ f (0) =

2

Z

|f (x)| dx i.e. Rn

Rn

151

|fb(ξ)|2 dξ =

Z Rn

|fb(ξ)|2 dx.

Theorem 6.1.3. (Plancherel) The Fourier transform, defined on S(Rn ) extends to L2 (Rn ) as a unitary operator. For every f, g ∈ L2 (Rn ), we have Z

Z

fb(ξ) gb(ξ) dξ.

f (x) g(x) dx = Rn

Rn

Proof. We have already proved that

R

|f (x)|2 dx =

Rn

R

|fb(ξ)|2 dξ, for f ∈ S(Rn ).

Rn

Given f ∈ L2 (Rn ) choose a sequence {fk } ∈ S(Rn ) such that fk → f in L2 (Rn ). As kfk − fj k22 = kfbk − fbj k22 , we see that {fbk } is Cauchy in L2 (R). So we define fb to be the L2 -limit of {fbk }. It is easy to see that fb is well defined. As kfk k22 = kfbk k22 , the continuity of norm implies that kf k22 = kfbk22 . By polarisation, we obtain Z

Z f (x) g(x) dx =

Rn

fb(ξ) gb(ξ) dξ. Rn

Writing F f = fb, f ∈ L2 (Rn ) the above argument gives that F : L2 (Rn ) → L2 (Rn ) is an isometry. To show that F is unitary, we are left with checking that F is onto. For this, given g ∈ L2 (Rn ), we choose a sequence {gk } ⊂ S(Rn ) such that gbk → g in L2 (Rn ). Then {gk } will converge to some f ∈ L2 (Rn ) for which fb = g. Remark 6.1.4. As F 4 = Id on S(Rn ), we see that F 4 = Id on L2 (Rn ) as well. Remark 6.1.5. The definition of Fourier transform can be extended to Lp (Rn ), 1 ≤ p ≤ 2. Given f ∈ Lp (Rn ) we can write f = g + h, where g = f χ|f |>1 and h = f χ|f |≤1 . Note that g ∈ L1 (Rn ), h ∈ L2 (Rn ), and so we can define fb := gb+b h ∈ L∞ (Rn )+L2 (Rn ). To see that fb is well-defined, suppose f = g1 +h1 where g1 ∈ L1 (Rn ) and h1 ∈ L2 (Rn ). Then g − g1 = h1 − h so that (g − g1 )b = (h1 − h)b or gb + b h = gb1 + b h1 . Thus fb is well-defined. We shall make use of the following interpolation theorem. 152

Theorem 6.1.6. (Riesz-Thorin Interpolation). Let 1 ≤ p0 , p1 , q0 , q1 ≤ ∞, and for 0 < t < 1 define p and q by 1−t t 1 = + , p p0 p1

1 1−t t = + . q q0 q1

If T is a linear operator from Lp0 + Lp1 to Lq0 + Lq1 such that kT f kq0 ≤ M0 kf kp0 , for f ∈ Lp0 , and kT f kq1 ≤ M1 kf kp1 , for f ∈ Lp1 , then kT f kq ≤ M01−t M1t kf kp , for f ∈ Lp . Theorem 6.1.7. (Hausdorff-Young Inequality). Let 1 ≤ p ≤ 2. If f ∈ Lp (Rn ), 0 then fb ∈ Lp (Rn ), and kfbkp0 ≤ ckf kp . (Here,

1 p

+

1 p0

= 1.)

n Proof. We already have kfbk∞ ≤ (2π)− 2 kf k1 and kfbk2 = kf k2 . Applying the previous

theorem gives the required inequality. We have already observed that S(Rn ) is invariant under the Fourier transform. This property utterly fails for Cc∞ (Rn ), as the following result states. Theorem 6.1.8. If f ∈ Cc∞ (Rn ), then fb ∈ / Cc∞ (Rn ) unless f = 0. This will follow from the properties of holomorphic functions once we have the following result. Theorem 6.1.9. (Paley-Wiener). If f ∈ Cc∞ (Rn ), then fb can be extended to Cn as an entire function which satisfies |fb(ζ)| ≤ CN (1 + |ζ|)−N eR|Im ζ| , for every N ∈ N and ζ ∈ Cn . Conversely, if F is an entire function satisfying the above estimate, then there is a function f ∈ Cc∞ (Rn ) such that F = fb. Proof. Let f ∈ Cc∞ (Rn ) be supported in |x| ≤ R. Let F (ζ) = (2π)

−n 2

Z

e−ix·ζ f (x)dx,

Rn

153

ζ = ξ + iη ∈ Cn .

Since f is compactly supported, F (ζ) is well-defined for any ζ ∈ Cn , and F (ξ) = fb(ξ). R −ix·ξ x·η n e e f (x) dx, and hence F is a continuous function. Also, F (ζ) = (2π)− 2 |x|≤R

By Morera’s theorem, it follows that F is actually holomorphic. Moreover, −n 2

Z

|F (ζ)| ≤ (2π)

|f (x)| ex·η dx

|x|≤R

≤ C exp R|η| (Since |x · η| ≤ |x||η| ≤ R|η| as |x| ≤ R.)

Hence |F (ζ)| ≤ CeR|Im ζ| . For any α ∈ Nn , we have α

|α|

−n 2

Z

f (x)∂xα (e−ix·ζ ) dx.

∂ F (ζ) = (i) (2π)

|x|≤R

Integration by parts gives α

Z

−n 2

|α|

∂ F (ζ) = (−i) (2π)

(∂xα f )(x) eix·ζ dx.

|x|≤R

(Note that the boundary terms vanish.) This gives |∂ α F (ζ)| ≤ Cα eR|Imζ| . As this is true for any α, we get |F (ζ)| ≤ CN (1 + |ζ|)−N eR|Im

ζ|

for any N .

Conversely, let F be an entire function satisfying |F (ζ)| ≤ CN (1 + |ζ|)−N eR|Imζ| for all N. In particular, |F (ξ)| ≤ CN (1 + |ξ|)−N for any N , and hence F ∈ L1 (Rn ). So we can define −n 2

Z

f (x) = (2π)

Rn

154

eix·ξ F (ξ) dξ.

If we can show that f is compactly supported in |x| ≤ R, then by Fourier Inversion, we would get fb = F , proving the theorem. Since the function f ∈ C ∞ (Rn ) and ξ α F (ξ) ∈ L1 (Rn ) for any α, we have ∂xα f (x)

−n 2

Z

= (2π)

eix·ξ (iξ)α F (ξ) dξ.

Rn

It remains to show that f is compactly supported. We prove this using Cauchy’s theorem. Look at the contour integral

R Γ1

···

R

eix·ξ F (ζ) dζ1 · · · dζn , where Γj is the

Γn

contour bounded by the real axis, the lines |Re z| = r and the line Im z = ηj . By Cauchy’s theorem, this integral is zero, and so we have Zr

Zr ···

−r

ixξ

e

Zr ···

F (ξ)dξ =

−r

Zr

−r Zη1

−

e

−r Zηn

···

e

0

0

Zη1

Zηn ···

+ 0

i

i

i

e

n P

xj ·(ξj +ηj )

j=1

n P

xj ·(r+iθj )

1

n P

F (r + iθ1 , · · · r + iθn ) (i)n dθ1 · · · dθn

xj ·(r+iθj )

j=1

F (ξ1 + iη1 , · · · , ξn + iηn ) dξ1 · · · dξn

F (−r + iθ1 , · · · − r + iθn ) dθ1 · · · dθn .

0

The second integral on the right is bounded by Zη1

Zηn ···

0

e|x||θ| (1 + nr2 + |θ|2 )−N eR|θ| dθ1 · · · dθn

0

≤ C(1 + r2 )−N

Zη1 ··· 0

155

Zηn 0

eR|θ| e|x||θ| dθ1 · · · dθn .

This goes to zero as r → ∞ and so does the third integral on the right. Thus we are left with Z

ix·ξ

e

Z

eix·(ξ+iη) F (ξ + iη) dξ.

F (ξ) dξ = Rn

Rn

We claim that above integral is zero whenever |x| > R. For this, let x = rω, ω ∈ S n−1 , and consider −n 2

Z

f (x) = (2π)

eirω·ξ e−rω·η F (ξ + iη) dξ.

Rn

Choose η = tω, where t > 0. Then Z |f (x) ≤ CN e−rt (1 + |ξ|)−N eRt dξ ≤ CN0 e−(r−R)t . Rn

If r > R, we can take t → ∞ to conclude that f (x) = 0 for |x| > R. Hence f ∈ Cc∞ (Rn ) with support in |x| ≤ R. Note that when f (6≡ 0) ∈ Cc∞ (Rn ), fb cannot vanish on any open set in Cn . We know that fb ∈ S(Rn ) whenever f ∈ Cc∞ (Rn ). Hence |fb(ξ)| ≤ CN (1 + |ξ|)−N for any N . The above is the best possible decay we can get for fb when f ∈ Cc∞ (Rn ). For instance, if f ∈ Cc∞ (Rn ) and |fb(ξ)| ≤ Ce−a|ξ| for some a > 0, then f = 0. To see this, R n R observe that f (x) = (2π)− 2 fb(ξ)eix·ξ dξ. When |fb(ξ)| ≤ Ce−a|ξ| , fb(ξ) ei(x+iy)·ξ dξ Rn

Rn

converges and defines a holomorphic function on Ωa = {x + iy : |y| < a}. Since f ∈ Cc∞ (Rn ), this is possible only if f ≡ 0. If fb ∈ S(Rn ) and |f (x)| ≤ Ce−a|x| , then fb has a holomorphic extension to the tube domain Ωa = {x + iy : |y| < a}. Let us look at a function f on Rn having 2

Gaussian decay, i.e., |f (x)| ≤ Ce−a|x| , for some a > 0. 156

Then fb extends to Cn as an entire function. Let us look at fb(ζ) = (2π)

−n 2

Z

f (x) e−ix.ζ dx,

ζ = ξ + iη ∈ Cn

Rn

= (2π)

−n 2

Z

f (x) ex.η e−ix.ξ dx

Rn

= fbη (ξ), where fη (x) = f (x)ex.η. Then fη ∈ L2 (Rn ), and so by Plancherel theorem, Z

|fbη (ξ)|2 dξ =

Rn

Z

Z Rn

|fb(ξ + iη)|2 dξ =

i.e.,

|fη (x)|2 dx

Rn

Z

|f (x)|2 e2x·η dx

Rn

Integrating the above against φ(η) dη for a sufficiently nice function φ gives Z Z

Z Z

2

|fb(ξ + iη)| φ(η) dξ dη = Rn Rn

|f (x)|2 e2x·η φ(η) dx dη.

Rn Rn 2

Taking φ(η) = e−δ|η| for some δ > 0, we get Z Z

|f (x)|2 e2x·η φ(η)dx dη < ∞.

Rn Rn

But we do not know yet if Z Z

|fb(ξ + iη)|2 φ(η) dξ dη < ∞.

Rn Rn

157

n Recall that fb(ξ + iη) = (2π)− 2

R

e−ix·ξ f (x) ex.η dx, which gives

Z

Z

2

|f (x)|2 e2x·η dx.

|fb(ξ + iη)| dξ = Rn

Rn

Since |f (x)| < C e

−a|x|2

, we get Z Z |η|2 2 2 e2x·η e−2a|x| dx ≤ C e a . |fb(ξ + iη)| dξ ≤ C Rn

Rn

Hence Z Z

1 |fb(ξ)|2 φ(η) dξ dη < ∞ provided δ > . a

Rn Rn

We have Z Z

2 |fb(ξ + iη)|2 e−δ|η| dξ dη =

Rn Rn

Z Z

2

|f (x)|2 e2x·η e−δ|η| dx dη.

Rn Rn

We know that (2π)

−n 2

n Z Y

2

e2xj ·ηj e−δ|η| dηj = e

|x|2 δ

.

j=1

Thus we have Z Z

2 |fb(ξ + iη)|2 e−δ|η| dξ dη =

Z

|f (x)|2 e

|x|2 δ

dx, δ > 1/a.

2 The above argument gives that if |fb(ξ)| ≤ Ce−a|ξ| , then f extends to Cn as an entire

function, and Z Z

2

|f (x + iy)| e

−δ|y|2

Z dx dy =

Rn Rn

|fb(ξ)|2 e

|ξ|2 δ

1 dξ , δ > . a

Rn

Given a function g ∈ L1 (Rn ) can we find f (related to g) satisfying the inequality 2

2

|fb(ξ)| ≤ Ce−a|ξ| ? The inequality |b g (ξ)| ≤ Ce−a|ξ| does not hold for all g ∈ L1 (Rn ). For example, if we take g(x) = χ(−1,1) (x), then gb does not have Gaussian decay. We 158

know that (f ∗ g)b(ξ) = (2π)n/2 fb(ξ) gb(ξ). If we start with a function f ∈ L1 (R) and n

consider f ∗ pa , where pa is given by pa (x) = (4πa)− 2 e−

|x|2 4a

, then |(f ∗ pa ) b (ξ)| ≤

2

C e−a|ξ| . Actually, we have n n n 2 2 (f ∗ pa )b(ξ) = (2π) 2 (4πa)− 2 fb(ξ) e−a|ξ| = (2π)− 2 fb(ξ) e−a|ξ| .

Suppose we start with f ∈ L2 (Rn ) (instead of f ∈ L1 (Rn )), then we can extend f ∗ pa as an entire function. Note that the function (f ∗ pa )b ∈ L2 ∩ L1 (L2 by H¨ older’s inequality). By the inversion formula, we have −n 2

Z

f ∗ pa (x) = (4πa)

2 eix.ξ fb(ξ) e−a|ξ| dξ.

Rn

Can we holomorphically extend f ∗ pa to Cn ? Even if we start with f ∈ L2 (Rn ) the same is true for the following reason. The integral Z f (u) pa (z − u) du = (4πa) Rn

(where (z − u)2 =

−n 2

Z

1

2

f (u)e− 4a (z−u) du

Rn n P

(zj − uj )2 , z ∈ Cn ), makes sense and defines a holomorphic

j=1

function, giving the required extension of f ∗ pa to Cn . We also have f ∗ pa (x) = (2π)

−n/2

Z

2 fb(ξ) e−a|ξ| eix.ξ dξ

Rn

so that the extension is also given by Z 2 −n/2 f ∗ pa (z) = (2π) fb(ξ) e−a|ξ| ei(x+iy)·ξ dξ, z = x + iy. Rn

159

By Plancherel theorem, Z

2 Z Z 2 −a|ξ| −y·ξ ix·ξ e e dξ dx |f ∗ pa (x + iy)| dx = fb(ξ) e 2

Rn Rn

Rn

Z =

2

|fb(ξ)|2 e−2a|ξ| e−2y·ξ dξ.

Rn

n

We can integrate both sides against p a2 (y) = (2πa)− 2 e− Z Z

|y|2 2a

, to get

|f ∗ pa (x + iy)|2 p a2 (y) dx dy

Rn Rn −n/2

Z Z

= (2πa)

2

|fb(ξ)|2 e−2a|ξ| e−2y.ξ pa/2 (ξ)dξdy.

Rn Rn

n

Since (2πa)− 2

R

e−2y.ξ e−

|y|2 2a

2

dy = e2a|ξ| , we get

Rn

Z Z

2

Z

|f ∗ pa (x + iy)| p (y) dx dy = a 2

Rn Rn

2 2 |fb(ξ)|2 e−2a|ξ| e2a|ξ| dξ =

Rn

Z

|f (x)|2 dx.

Rn

Thus we have proved. Theorem 6.1.10. Let f ∈ L2 (Rn ). Then f ∗ pa extends to Cn as a holomorphic function and satisfies Z Z

Z

2

|f ∗ pa (x + iy)| pa/2 (y) dx dy =

|f (x)|2 dx.

Rn

Rn Rn

The transform which takes f into the entire function f ∗ pa (z) is called the Segal-Bargmann transform. We define Ba (Cn ) to be the space of all entire functions

160

F satisfying kFa k2Ba

Z Z :=

|F (x + iy)|2 p a2 (y)dx dy < ∞

Rn Rn

This forms a Hilbert space with the inner product given by Z Z hF, Gi =

F (x + iy) G(x + iy) pa/2 (y) dx dy. Rn Rn

The theorem says that the Segal-Bargmann transform takes L2 (Rn ) into Ba (Cn ). Actually, the Segal-Bargmann transform is onto Ba (Cn ). In order to prove this, without loss of generality, we consider the case a = 21 . If we take F ∈ B1/2 (C n ) then 1 2

the function G(z) = F (z)e 4 z , where z 2 = z12 + · · · + zn2 , satisfies Z

2

− 21 |z|2

|G(z)| e

Z Z dz =

Cn

1

2

1

2

1

2

1

2

|F (z)|2 e 2 |x| e− 2 |y| e− 2 |x| e− 2 |y| dx dy

Rn Rn

Z Z =

2

|F (z)|2 e−|y| dx dy < ∞.

Rn Rn 1 2

Thus we see that F ∈ B1/2 (Cn ) if and only if F (z)e 4 z belongs to the space F(Cn ), R 1 2 consisting of all entire functions G which satisfy |G(z)|2 e− 2 |z| dz < ∞. Also, Cn

the Segal-Bargmann transform f 7→ f ∗ p 1 gives rise to the transform Bf (z) = 2

e

1 2 z 4

2

n

f ∗ p 1 (z), which takes L (R ) into F(Cn ). In order to show that the Segal2

Bargmann transform is onto B 1 (Cn ), it is enough to show that the transform B : 2

2

n

n

L (R ) → F(C ) is onto. Theorem 6.1.11. B is a unitary operator from L2 (Rn ) onto F(Cn ).

161

We already know that B is an isometry. So it is enough to prove that B is √ 1 onto. In F(Cn ), consider the functions ζα (z) = (2|α| α! π)− 2 z α , α ∈ Nn . Then Z

2

ζα (z) ζβ (z) e−1/2|z| dz

hζα , ζβ iF(Cn ) = Cn

√ √ 1 = (2 α! π)− 2 (2|β| β! π)−1/2 |α|

Z

1

2

z α z −β e− 2 |z| dz.

Cn

Integrating in polar coordinates, we can check that hζα , ζβ i = δαβ . In fact, {ζα : α ∈ Nn } is an orthonormal system in F(Cn ). If F ∈ F(Cn ) then we have its Taylor P expansion F (z) = cα z α , which converges uniformly over compact subsets of Cn . α∈N nP It can be shown that cα z α actually converges in F(Cn ). Hence {ζα : α ∈ Nn } is α∈Nn

an orthonormal basis for F(Cn ). The surjectivity of B will be proved if we can find n Q functions Φα such that BΦα = ζα . We claim that if we take Φα (x) = hαj (xj ), j=1

where hαj (xj ) are Hermite functions, then BΦα = ζα . To prove this we can assume that n = 1. All we need is √ 1 1 2 Proposition 6.1.12. If hk (x) = (2k k! π)− 2 Hk (x)e− 2 x , where Hk is the k th Hermite polynomial, then Bhk (z) = ζk (z). √ 1 1 2 Proof. We need to show that e 4 |z| (hk ∗ p 1 )(z) = (2k k! π)− 2 z k , or equivalently, 2

1 √ 2π

Z

1

2

1

2

2

Hk (z − u) e− 2 (z−u) e− 2 u du = z k e−1/4z .

The above can be simplified to 1 √ 2π

Z

2

1 2

Hk (z − u) e−u ezu du = z k e 4 z .

162

Since both sides are holomorphic, it is enough to prove 1 √ 2π

Z

1 2

2

Hk (x − u) e−u exu du = xk e 4 x , x ∈ R

It is enough to prove that Z

1 √ 2π

1 2

2

Hk (u) e−u e−ixu du = (−i)k xk e− 4 x , x ∈ R.

From the definition, we have (by integration by parts) 1 √ 2π

Z 

d du

k

−u2

e

−ixu

e

(−i)k k x du = √ 2π

Z

1 2

2

e−u eixu du = (−i)k xk e− 4 x

Thus we have proved that BΦα = ζα , where Φα (x) =

n Q

hαj (xj ). Since {Φα : α ∈

j=1

Nn } is an orthonormal basis for L2 (Rn ), we see that for F ∈ F(Cn ), B −1 F (x) =

X

hF, ζα iF Φα (x)

α∈Nn

defines a function f in L2 (Rn ) and Bf = F . This proves the surjectivity of B.

6.2

Fourier expansion of Fourier transform In this section, we consider the Fourier transform of functions on R2 , which

we identify with C. For an integrable function f on C, we can write the Fourier transform in the form 1 fb(w) = 2π

Z

¯ f (z) e−iRe(z·w) dz, w ∈ C.

C

163

In polar coordinates, we write w = λeiϕ , λ > 0, ϕ ∈ R, so that 1 fb(λeiϕ ) = 2π

Z

f (z) e−iλRe(ze

−iφ )

dz.

C

For any λ > 0, ϕ 7→ fb(λeiϕ ) is a 2π- periodic function of ϕ. Hence we can expand fb(λeiϕ ) into a Fourier series fb(λeiϕ ) =

∞ X

Fm (λ) eimϕ ,

m=−∞

where 1 Fm (λ) = 2π

Z2π

fb(λeiϕ ) e−imϕ dϕ.

0

Using the definition of fb(w), we have  2 Z2π Z  1 −iλRe(ze−iϕ ) Fm (λ) = f (z) e dz e−imϕ dϕ. 2π 0

C

Writing z = reiθ , we have  2 Z∞ Z2π Z2π 1 i(θ−ϕ) ) Fm (λ) = f (reiθ ) e−iλrRe(e e−imϕ dϕ dθ r dr. 2π 0

0

0

Look at 1 2π

Z2π

−iλrRe(ei(θ−φ) )

e

e−imφ

1 dφ = e−imθ 2π

0

Z2π 0

164

e−iλr cos φ e−imφ dφ = e−imθ Km (λr)

Thus we have

Z∞ Fm (λ) =

fm (r) Km (λr) rdr, 0

where

Z2π

1 fm (r) = 2π

f (reiθ )e−imθ dθ.

0

We have 1 2π

Z2π

fb(λeiϕ ) e−imϕ dϕ =

0

Z∞ 

Z2π

1 2π

−imθ

iθ

f (re ) e

0

 dθ

Km (λr) rdr.

0

By the above calculations, we have proved Theorem 6.2.1. If f satisfies f (zeiθ ) = eimθ f (z), then fb also satisfies the same ∞  R iϕ imϕ b equation, and we have f (λe ) = e f (r) Km (λr) rdr . In other words, the 0

set of all functions f satisfying f (zeiθ ) = eimθ f (z) is invariant under the Fourier transform. If we denote by ϕλ,m (z) the function 1 ϕλ,m (z) = 2π

Z2π

eiλRe(ze

iφ )

e−imφ dφ,

0

then we have 1 Fm (λ) = 2π

Z2π

fb(λeiφ ) e−imφ dφ =

Z f (z) ϕλ,m (z) dz. Cn

0

Note that ϕλ,m (zeiθ ) = e−imθ ϕλ,m (z), so that ϕλ,m (reiθ ) = e−imθ ϕλ,m (r) = e−imθ Km (λr). 165

Consider the function 1 ϕλ,m (z) = 2π

Z2π

e−iλRe(ze

−iφ )

e−imφ dφ.

0

Clearly, 1 ϕλ,m (r) = 2π

Z2π

e−iλr cos φ e−imφ dφ

0

has a zero of order m at the origin. So we can define Jm (λr) = (λr)−m ϕλ,m (r). When Z2π Z1 1 1 1 m = 0, J0 (r) = e−ir cos φ dφ = e−irt (1 − t2 )− 2 dt. We claim that 2π 2π −1

0

(r/2)m √ Jm (r) = m Γ( 2 + 1) π

Z1

1

e−irt (1 − t2 )m− 2 dt.

−1

In order to prove this, let us look at d dr



Jm (r) rm



  1 Z   1 d −irt 2 m− 21 e (1 − t ) dt = m√  2 πΓ(m + 12 ) dr  −1  1  Z   1 d 2 m− 12 = m√ (cos rt) (1 − t ) dt  2 πΓ(m + 12 ) dr  −1

=

−1 √ 2m πΓ(m + 12 )

Z1

1

(sin rt) t (1 − t2 )m− 2 dt.

−1

Integrating by parts, we get   Z1 1 1 1 d Jm (r) =− (cos rt)(1 − t2 )m+ 2 dt 1 √ m m+1 dr r r 2 Γ(m + 1 + 2 ) π −1

=−

Jm+1 (r) rm 166

We now consider     Z2π d ϕ1,m 1 d 1 −ir cos φ −imφ = e e dφ dr rm 2π dr rm 0

= −

m 1 2π rm+1

Z2π

e−ir cos φ e−imφ dφ

0

Z2π

1 + (−i) 2πrm

e−ir cos φ (cos φ) e−imφ dφ.

0

ϕ1,m+1 (r) d  ϕ1,m  =− , then we can conclude that ϕ1,m (r) = m dr r rm c Jm (r). The function Jm (r) is called the Bessel function of order m. We have If we can show that

1 2π

Z2π

fb(λeiφ ) e−imφ dφ =

∞

Z 0

0

 1 2π

Z2π

 f (reiθ ) e−imθ dθ Jm (λr) rdr.

0

We can write the above equation as

λ−m Fm (λ) =

Z∞

r−m fm (r)

Jm (λr) 2m+1 r dr (λr)m

or

F˜m (λ) = Hm f˜m (λ),

0

where

Z∞ Hm g(x) =

g(r)

Jm (λr) 2m+1 r dr. (λr)m

0

Hm is called the Hankel transform of order m. By writing down the Fourier series of f as iθ

f (re ) =

∞ X

rm f˜m (r) eimθ ,

m=−∞

167

we see that fb(λeiφ ) =

∞ X

λm Hm f˜m (λ) eimφ

m=−∞

Hence we can reduce the study of Fourier transform to the study of Hankel transforms. We can generalise the definition of Bessel functions Jm to define (r/2)δ √ Jδ (r) = Γ(δ + 12 ) π

Z1

1

e−irt (1 − t2 )δ− 2 dt

−1

for any δ > − 21 . Then we can also define Z∞ Hδ g(λ) =

g(r)

Jδ (λr) 2δ+1 r dr, (λr)δ

g ∈ L2 (R+ , r2δ+1 dr).

0

6.3

Exercises

6.3.1. We can define the space of tempered distribution S 0 (Rn ) to be the dual space of S 0 (Rn ) as in the case of R. Show that Lp (Rn ), 1 ≤ p ≤ ∞ is a subspace of S 0 (Rn ). 6.3.2. Let s ∈ R. Let H s (Rn ) denote the collection of all f ∈ S 0 (Rn ) such that s

(1 + |ξ|2 ) 2 fb(ξ) ∈ L2 (Rn ). Then H s (Rn ) is called the Sobolev space of order s. Show that H s (Rn ) is a Hilbert space with the inner product given by Z hf, gi =

1 + |ξ|2


s

fb(ξ) gb(ξ) dξ.

Rn

6.3.3. Show that for m ∈ N0 , H m (Rn ) can be written as the collection of all f ∈ L2 (Rn ) for which Dα f ∈ L2 (R), α ∈ Nn0 with |α| ≤ m. Here, if α = (α1 , α2 , . . . , αn ),

168



∂ then |α| = α1 + α2 + · · · + αn and D f = ∂x1 derivative in the sense of tempered distribution. α

6.3.4. Show that S(Rn ) is dense in H s (Rn ).

169

α1 

∂ ∂x2

α2

 ···

∂ ∂xn

 αn denotes

170

Chapter 7 For further reading - Spherical harmonics and Fourier transform Our main aim is to study the action of the Fourier transform on functions on Rn . For this, we wish to obtain a decomposition of the space L2 (Rn ) as a direct sum of subspaces, which are invariant under the Fourier transform and the action of the Fourier transform on these spaces is simpler. More precisely, we are interested in finding subspaces {Vm } of L2 (Rn ) with the following properties:

1. L2 (Rn ) =

∞ L

Vm .

m=0

2. Vm is invariant under Fn , the Fourier transform on functions on Rn . 3. On Vm , Fn reduces to a simpler transform (may be Hankel).

First we look the action of the Fourier transform on tha class of radial functions on Rn .

7.1

Fourier transform of radial functions

Definition 7.1.1. A function f defined on Rn is is said to be radial, if f (x) = f0 (|x|) for some function f0 defined on [0, ∞). Or equivalently, if f (kx) = f (x) for 171

all k ∈ SO(n), the special orthogonal group consisting of orthogonal matrices with determinant 1. Lemma 7.1.2. If f ∈ L1 (Rn ) is radial, so is its Fourier transform. Proof. For ξ ∈ Rn , Z

n fˆ(ξ) = (2π)− 2

f (x) e−ix·ξ dx

Rn

For any k ∈ SO(n), we have n fˆ(kξ) = (2π)− 2

Z

f (x) e−ix.kξ dx

Rn

Making a change of variables x 7→ kx and noting that kx·kξ = x·ξ and f (kx) = f (x), we get fˆ(kξ) = fˆ(ξ). Let V0 = {f ∈ L2 (Rn ) : f (kx) = f (x) for all k ∈ SO(n)}, i.e., V0 is the radial subspace of L2 (Rn ). The above lemma states that Fn V0 = V0 . Let W0 be the orthogonal complement of V0 in L2 (Rn ). Given f ∈ L2 (Rn ), define f0 (x) = R f (kx) dk, which belongs to V0 . Let us look at hg, f0 i where g = f − f0 . S0(n)

Z

Z hg, f0 i =

f (x)f0 (x) −

(f (x) − f0 (x)) f0 (x) dx = Rn

Z

Rn

|f0 (x)|2 dx

Rn

But Z

Z f (x)f0 (x) = Rn

SO(n)

  Z  f (kx)f 0 (x) dx dk. Rn

As dx is invariant under SO(n), the above becomes   Z Z Z Z   f0 (x) f (kx) dk dx = |f0 (x)|2 . Rn

SO(n)

Rn

Rn

172

Hence hg, f0 i = 0. So we have L2 (Rn ) = V0 ⊕ W0 . We wish to study the action of the Fourier transform on these invariant subspaces. First we consider the one dimensional case. Here, the radial functions are the even functions and L2 (R) decomposes as the direct sum of the subspace of even functions and that of the odd functions. If f ∈ L2 (R), then f = fe + fo , where

fe (x) =

f (x) + f (−x) , 2

and fo (x) =

f (x) − f (−x) 2

are respectively the even and odd parts of the function f. As can be easily checked these two subspaces are inviariant under the Fourier transform. Moreover, for even functions 1 fˆ(ξ) = (2π)− 2

Z f (x) cos(xξ) dx R

is the cosine transform. Similarly, for odd functions the Fourier transform reduces to the sine transform. Now we proceed to obtain a decomposition of L2 (R2 ). For this, we identify a point (x, y) ∈ R2 with the complex number z = x + iy = reiθ . Let V0 = {f ∈ L2 (C) : f (eiθ z) = f (z), for all z ∈ C, eiθ ∈ S 1 } be the space of all radial functions on R2 . Let Vm = {f ∈ L2 (C) : f (eiθ z) = eimθ f (z), for all eiθ ∈ S 1 }.

Then for each m, the space Vm is invariant under the Fourier transform, i.e., if f ∈ Vm , then fb ∈ Vm . For an integrable function f on C, we can write its Fourier transform

173

in the form 1 fˆ(w) = 2π

Z

¯ f (z) e−iRe(z·w) dz, w ∈ C.

C

In polar coordinates, we write w = λeiφ , λ > 0, φ ∈ R, so that 1 fˆ(λe ) = 2π iφ

Z

f (z) e−iλRe(ze

−iφ )

dz.

C

For any λ > 0, the map φ 7→ fˆ(λeiφ ) is a 2π- periodic function of φ. Hence we can expand fˆ(λeiφ ) into a Fourier series fb(λeiφ ) =

∞ X

Fm (λ) eimφ ,

m=−∞

where 1 Fm (λ) = 2π

Z2π

fˆ(λeiφ ) e−imφ dφ.

0

Using the definition of fˆ(w), we have 1 Fm (λ) = (2π)2

Z2π Z f (z) e 0

−iλRe(ze−iφ )

 dz

e−imφ dφ.

C

iθ

Writing z = re , we have Z∞ Z2π Z2π 1 i(θ−φ) ) Fm (λ) = f (reiθ ) e−iλrRe(e e−imφ dφ dθ r dr. 2 (2π) 0

0

0

Look at Z2π Z2π 1 1 i(θ−φ) ) −imφ e−iλrRe(e e dφ = e−imθ e−iλr cos φ eimφ dφ 2π 2π 0

0

= e−imθ Km (λr) 174

Z∞

Thus we have

Fm (λ) =

fm (r) Km (λr) rdr, 0

1 fm (r) = 2π

where

Z2π

f (reiθ )e−imθ dθ.

0

We have 1 2π

Z2π

fˆ(λeiφ ) e−imφ dφ =

0

Z∞ 

1 2π

0

Z2π

iθ

f (re ) e

−imθ

 dθ

Km (λr) rdr.

0

By the above calculations, we have proved Theorem 7.1.3. If f satisfies f (zeiθ ) = eimθ f (z), then fˆ also satisfies the same   Z∞ iϕ imϕ equation, and we have fˆ(λe ) = e f (r) Km (λr) rdr . In otherwords, the 0

set of all functions f satisfying f (zeiθ ) = eimθ f (z) is invariant under the Fourier transform. 1 By defining ϕλ,m (z) = 2π

Z2π

eiλRe(ze

iφ )

e−imφ dφ, we can write

0

1 Fm (λ) = 2π

Z2π

fb(λeiφ ) e−imφ dφ =

Z f (z) ϕλ,m (z) dz.

(7.1)

Cn

0

Note that ϕλ,m (zeiθ ) = e−imθ ϕλ,m (z), so that ϕλ,m (reiθ ) = e−imθ ϕλ,m (r) = e−imθ Km (λr). Consider the function 1 ϕλ,m (z) = 2π

Z2π −iφ e−iλRe(ze ) e−imφ dφ. 0

175

Clearly, 1 ϕλ,m (r) = 2π

Z2π

e−iλr cos φ e−imφ

1 dφ = (−i)m 2π

0

Z2π

eiλr sin φ e−imφ dφ

0

has a zero of order m at the origin. So we can define Jm (λr) = (λr)−m ϕλ,m (r).

The function Jm (r) is called the Bessel function of order m. Lemma 7.1.4. (Poisson representation of Bessel functions) If m is a nonnegative integer then

Jm (r) =

2m

rm Γ(m + 1/2) Γ(1/2)

Z1

1

eirs (1 − s2 )m− 2 ds.

−1

Proof. Let rm ∗ Jm (r) = m 2 Γ(m + 1/2) Γ(1/2)

Z1

1

eirs (1 − s2 )m− 2 ds.

−1

By definition, 1 J0 (r) = 2π

Z2π

e−ir cos s

1 ds = 2π

0

Z2π

1

e−irs (1 − s2 )− 2 ds = J0∗ (r).

0

Let us compute d dr



Jm (r) rm



= −r−m

m r

0 (r) Jm (r) − Jm

176



 = −r−m 

m 2πr

Z2π

eit sin θ e−imθ dθ −

i 2πr

0

r−m = − 2π = −

r−m 2π

Z2π

 eit sin θ sin θ e−imθ dθ

0

Z2π 

 d eir sin θ e−imθ ir sin θ −imθ ir sin θ −imθ i ( ) + cos θe e − i sin θe e dθ dθ r

0 Z2π

eit sin θ e−i(m+1)θ dθ

0

= −r

−m

Jm+1 (r).

On the other hand  ∗   Z1  (r) d Jm 1 d −irt 2 m− 12 = m√ e (1 − t ) dt dr rm 2 π Γ(m + 21 ) dr −1

1 d = m√ 1 2 π Γ(m + 2 ) dr =

2m

−1 √ π Γ(m + 21 )

Z1

 Z1

2 m− 12

(cos rt) (1 − t )

 dt

−1 1

(sin rt) t (1 − t2 )m− 2 dt.

−1

Integrating by parts, we get d dr



∗ Jm (r) m r



1 1 √ =− m+1 r 2 Γ(m + 1 + 21 ) π =

Z1

1

(cos rt)(1 − t2 )m+ 2 dt

−1

J ∗ (r) − m+1 rm

∗ Thus both Jm and Jm satisfy the same recursion relation. As J0 = J0∗ , this gives ∗ Jm = Jm for all non-negative integers m.

177

∗ makes sense even for non-integral values of m. Thus The integral defining Jm

for any real δ > −1/2, we define the Bessel function Jδ as Z1

rδ Jδ (r) := δ 2 Γ(r + 1/2) Γ(1/2)

1

eirs (1 − s2 )δ− 2 ds, r ≥ 0.

−1

From (7.1), we have 1 2π

Z2π

fˆ(λeiφ ) e−imφ dφ =

∞

Z 0

0

 1 2π

Z2π

 f (reiθ ) e−imθ dθ Jm (λr) rdr.

0

We can write the above equation as

λ−m Fm (λ) =

Z∞

r−m fm (r)

Jm (λr) 2m+1 r dr (λr)m

0

or F˜m (λ) = Hm f˜m (λ),

where

Z∞ Hm g(x) =

g(r)

Jm (λr) 2m+1 r dr. (λr)m

0

Hm is called the Hankel transform of order m. By writing down the Fourier series of f as f (reiθ ) =

∞ X

rm f˜m (r) eimθ ,

m=−∞

we see that fˆ(λeiφ ) =

∞ X

λm Hm f˜m (λ) eimφ

m=−∞

178

Hence we can reduce the study of Fourier transform on radial funcions on R2 to the study of Hankel transforms. For any δ > −1/2, we can define Hankel tranforms Hδ by the equation Z∞ Hδ g(λ) =

g(r)

Jδ (λr) 2δ+1 r dr, (λr)δ

g ∈ L2 (R+ , r2δ+1 dr).

0

Theorem 7.1.5. The Fourier transform of a radial function f in L1 (Rn ) is radial, and reduces to the Hankel transform Hδ f0 where δ = n/2 − 1 and f (x) = f0 (|x|). Proof. For radial f ∈ L1 (Rn ), by Lemma 7.1.2, its Fourier tranform is also radial. Let us write F (λ) = fˆ(λω), ω ∈ S n−1 , so that Z

−n 2

f (x) e−iλx·ω dx

F (λ) = (2π)

Rn

Z∞ i.e., F (λ) =

 f0 (r)

(2π)

0

Z

−n 2

−iλx·ω

e

 dσ

rn−1 dr,

S n−1 n−1

where dσ is the normalised surface measure on S . The function Z −n ϕλ,ω (x) = (2π) 2 e−iλx.ω dσ S n−1

is radial and independent of ω as dσ is invariant under the action of SO(n). Thus we have −n 2

Z

ϕ(λ,ω) (x) = ϕλ (r) = (2π)

0

e−iλrx ·ω dσ(x0 ).

S n−1

As ϕλ,ω is independent of ω, we can choose ω = (1, 0, · · · , 0) so that Z 0 −n ϕλ,ω (x) = (2π) 2 e−iλrx1 dσ(x0 ) S n−1

179

Z1 =C

1

0

e−iλrx1 (1 − x21 )n/2−1− 2 dx1

−1

=

Cn J n2 −1 (λr) n

(λr) 2 −1

Thus when f is radial, fˆ is also radial and we have Z∞ F (λ) =

f0 (r)

J n2 −1 (λr) n

(λr) 2 −1

rn−1 dr

0

= H n2 −1 f0 (λ) Thus the Fourier transform of a radial function reduces to a Hankel transform.

7.2

Spherical harmonics Now that we have studied the action of the Fourier transform on radial func-

tions in all dimensions, we would like to obtain the decomposition stated in the ∞ L beginning. In the case of L2 (C), we have the decomposition L2 (C) = Vem , where m=0 L Vem = Vm V−m . Note that the elements of Vem are functions of the form f (z) = eimθ g(r) + e−imθ h(r), where g and h are functions on R+ = [0, ∞). We can think of eimθ and e−imθ as restrictions of z m and z¯m to S 1 . Note that z m and z¯m are harmonic and homogeneous of degree m. Taking this as a clue, we want to look at the higher dimensional analogues of z m and z¯m . Before proceeding, we recall some multi-index notation. For a point 180

x = (x1 , x2 , · · · , xn ) ∈ Rn and α = (α1 , α2 , · · · , αn ) ∈ Nn , we let |α| = α1 + α2 + · · · + αn , α! = α1 !α2 ! · · · αn ! and xα = xα1 1 xα2 2 · · · xαnn . Definition 7.2.1. A polynomial p(x) on Rn is said to be homogeneous of degree m, P if p(x) = cα xα , or equivalently, if p(λx) = λm p(x), for all x ∈ Rn , and all |α|=m

λ > 0. Let Pm stand for the set of all polynomials which are homogeneous of degree P m. By definition, an element of Pm has the form P (x) = cα xα . The set {xα : |α|=m

|α| = m} forms a basis for Pm . Let dm denote the dimension of Pm . Then, dm is given by the number of ways of choosing an n-tuple (α1 , α2 , · · · , αn ) ∈ Nn satisfying n P αj = m. Let us label (n − 1) boxes as B1 , B2 , · · · , Bn−1 among a linearly ordered j=1

array of (n+m−1) boxes. Label each of the remaining unmarked boxes as W. Writing the labels of the boxes, we get a sequence of the letters B1 , B2 , · · · , Bn−1 and W. Let α1 be the number of W 0 s which precede B1 . Let α2 be the number of W 0 s which are between B2 and B3 , and so on. Let αn be the number of W 0 s following Bn−1 . Thus, we obtain an n-tuple α = (α1 , α2 , · · · , αn ) with |α| = m. The number of such n-tuples is the number of ways we can select (n − 1) boxes among the (n + m − 1) boxes, which is given by  dm = 

n+m−1 n−1



n+m−1

 m



 = (n + m − 1)! . (n − 1)! m!

cα xα . The differential operator P (D) |α|=m P associated to this polynomial is the operator defined by P (D) = cα ∂ α , where Let P be a polynomial of the form P (x) =

P

|α|=m

181

∂ |α| . For P, Q ∈ Pm , define hP, Qi = P (D)Q. Then h·, ·i satisfies the ∂xα11 ∂xα22 · · · ∂xαnn following properties. ∂α =

(i) As P and Q are homogeneous of the same degree, hP, Qi ∈ C, for all polynomials P, Q ∈ Pm . For n-tuples α = (α1 , α2 , · · · , αn ), (β1 , β2 , · · · , βn ) ∈ Nn , we have 

∂ αn ∂ α1 ∂ α2 · · · α ∂xα11 ∂ αx22 ∂xαnn

Therefore, if hP, P i = 0, then



n P

xβ1 1 xβ2 2 · · · xβnn = δαβ α1 !α2 ! · · · αn !.

|cα |2 α! = 0. This can happen only when cα = 0 for

j=1

all α, which implies that P = 0. (ii) h·, ·i is linear in the first variable and conjugate linear in the second variable. (iii) Moreover, h·, ·i is hermitian symmetric. From the above properties, we infer that h·, ·i defines an inner product on the space Pm . Let Hm ⊂ Pm be the subspace consisting of harmonic polynomials, i.e., p ∈ Hm n P ∂2 means p ∈ Pm and ∆p = 0, where ∆ = . The elements of Hm are called ∂x2 j=1

j

solid harmonics of degree m. Note that Pm and Hm are invariant under the action SO(n), i.e., if p ∈ Pm (resp. p ∈ Hm ), then for every σ ∈ S0(n), the polynomial em = {p|S n−1 : p ∈ Hm }. An pσ (x) := p(σ.x) belongs to Pm (resp. to Hm ). Let H em is called a (surface) spherical harmonic of degree m. Every polynomial element of H em , there exists p ∈ Hm is uniquely determined byits restriction to S n−1 . For Y ∈ H x x a polynomial P ∈ Hm such that Y = P |S n−1 . Then, Y ( |x| ) = P ( |x| ) = |x|−m P (x). If

e m onto H em . Y = 0, then P = 0, and so the map P 7→ P |Sn−1 is an isomorphism of H Thus, em = dim Hm = dim Pm − dim Pm−2 = dm − dm−2 . dim H

182

The dimension of Hm is denoted by am . em , we can define an inner product (by considering H em as a subspace of On H Z L2 (S n−1 )), given by (p, q) = p(ω) q(ω) dσ(ω). S n−1

em ⊥ H ek if m 6= k. Lemma 7.2.2. H em , then p(x) = rm p(ω), Proof. For, if p ∈ H

x = rω, r > 0, ω ∈ S n−1 . Since

∆p = 0, we get 

∂ n−1 ∂ 1 + + 2 ∆S 2 ∂r r ∂r r



(rm p(ω)) = 0,

where ∆S is the Laplacian of S n−1 . This gives m(m − 1) r(m−1) p(ω)+

1 n−1 m rm−1 p(ω) + 2 rm ∆S p(ω) = 0 r r

i.e., ∆S p(ω) = (−m(m − 1) − m(n − 1)) p(ω) = −m(m + n − 2) p(ω)

em is an eigenfunction of ∆S with eigenvalue [−m(m + Thus, we see that every p ∈ H em , q ∈ H ek , with m 6= k, n − 2)]. Note that ∆S is a self-adjoint operator. If p ∈ H then −m(m + n − 2)(p, q) = (∆S p, q) = −k(k + n − 2)(p, q). em ⊥ H ek . Since m 6= k, this is possible only if (p, q) = 0, i.e., H Theorem 7.2.3. Every P ∈ Pk has the form P (x) = P0 (x) + |x|2 P1 (x) + · · · + |x|2l Pl (x).

183

Proof. As any polynomial of degree less than 2 is harmonic, we assume that k ≥ 2. Let n P ∂2 ∆= be the Laplacian on Rn . Then the map ϕ, defined by ϕ(P ) = ∆P maps ∂2 j=1

xj

Pk onto Pk−2 . For if ϕ is not onto, there exists a polynomial Q ∈ Pk−2 such that Q is orthogonal to ∆P for all P ∈ Pk , i.e., h∆P, Qi = h∆Q, P i = 0. In particular, taking P (x) = |x|2 Q(x), we get 0 = hQ, ∆P i = Q(D)∆P = P (D)P = hP, P i, a contradiction, since P 6= 0. Let Bm = {P (x) ∈ Pm : P (x) = |x|2 Q(x), for some Q ∈ Pm−2 }. Claim: Pm is the orthogonal direct sum of Hm and Bm . For, if R(x) = |x|2 Q(x) for Q ∈ Pm−2 , then hR, P i = 0 for all Q ∈ Pm−2 if and only if Q(D)∆P = 0 for all Q ∈ Pm−2 . This happens if and only if hQ, ∆P i = 0 for all Q ∈ Pm−2 , which is true if and only if ∆P = 0. In particular, if m = k and P ∈ Pk , then P (x) can be written as P (x) = P0 (x) + |x|2 Q(x), where P0 is harmonic and Q ∈ Pk−2 . Applying the above claim for m = k − 2, there exist a harmonic polynomial P1 and Q ∈ Pk−4 such that Q(x) = P1 (x) + |x|2 Q1 (x) = P0 (x) + |x|2 P1 (x) + |x|4 Q1 (x). Thus the result follows by induction. The following are some of the interesting consequences of the above theorem: Corollary 7.2.4. The restriction to the surface of the unit sphere S n−1 of any polynomial of n−variables is a sum of restrictions to S n−1 of harmonic polynomials. e be the collection of all finite linear combinations of the Corollary 7.2.5. Let H ∞

em . Then elements of cup H k=0

e is dense in the space of all continuous functions on S n−1 with respect to the (i) H L∞ norm; 184

e is dense in L2 (S n−1 ). (ii) H Proof. (i) By the Weierstrass Approximation theorem, any continuous function g on S n−1 can be uniformly approximated by polynomials restricted to S n−1 . Corollary e thus 7.2.5 gives these restrictions as finite linear combinations of elements of H, proving (i). (ii) For f ∈ L2 (S n−1 ) and  > 0, let g be a continuous functions on S n−1 satisfying e such kf − gk2 < /2. By (i), there is a finite linear combination h of elements of H that kg − hk∞
0, i.e., F is homogeneous of degree 0. Also, F (ρx) = |x|−l Pl (ρx) = |x|−l Pl (x) = F (x), i.e., F is invariant under rotations. This together with its homogeneity forces F to be a constant function, F (x) = cl 0 , for all x ∈ Rn . This implies that Pl (x) = cl 0 |x|l . As Pl is a polynom P mial, cl 0 6= 0 only when l is even. Thus P (x) = ck |x|2k , where ck = c02k , for k=0

k = 0, 1, · · · , m, and m is the largest integer less than or equal to (j/2).

The following result gives an important characterisation of the zonal harmonics. 188

ek is constant on the parallels of Theorem 7.3.5. Let e ∈ S n−1 . A function Y ∈ H (k)

S n−1 orthogonal to e, if and only if, there is a constant c such that Y = cZe . Proof. Let e ∈ S n−1 and ρ be a rotation that fixes e, ie., ρe = e. By Lemma 7.3.1(c), (k)

(k)

for any x0 ∈ S n−1 , we have Ze (x0 ) = Ze (ρx0 ). Since the point e is fixed under the action of ρ, the parallel of S n−1 orthogonal to e is left invariant by ρ. Let x01 , x02 belong to this parallel of S n−1 . Choose a rotation ρ which fixes e and maps x01 to x02 , i.e., ρe = e and ρx01 = x02 . Then Ze(k) (x02 ) = Ze(k) (ρx01 ) = Ze(k) (x01 ), (k)

i.e., Ze

is constant on the parallel of S n−1 orthogonal to e.

ek is constant on the parallels of S n−1 orthogonal Conversely, suppose that Y ∈ H to e. Let τ be a rotation such that e = τ e1 , for the point e1 = (1, 0, · · · , 0) ∈ S n−1 . Define W (x0 ) := Y (τ x0 ). Then W is a spherical harmonic which is constant on the parallels of S n−1 orthogonal to e1 . x ), x 6= 0, and P (0) = 0. Let ρ be a rotation which Define P (x) = |x|k W ( |x|

fixes e1 . Then P (ρx) = P (x) for all x ∈ Rn . Also, all polynomials of the form xm 1 , for m = 0, 1, 2, · · · , are invariant under the action of ρ. We write P (x) = k P k−j x1 Pj (x2 , · · · , xn ). Then we get that the rotation ρ maps (x1 , x2 , · · · , xn ) 7→ j=0

(x1 , x02 , · · · , x0n ), where (x2 , x3 · · · , xn ) 7→ (x02 , · · · , x0n ) is a rotation of the (n−1)dimensional space x1 = 0. As ρ varies over all rotations fixing x1 , these mappings give rise to all rotations of the space x1 = 0. Since P is invariant under the rotation ρ, so are all the polynomials P0 , P1 , · · · , Pk . This gives by Lemma 7.3.4 that Pj = 0 for j odd, and

189

j

1

that Pj (x2 , · · · , xn ) = cj (x22 + · · · + x2n ) 2 , for j even. Let R = (x22 + · · · + x2n ) 2 . Then we have 2 k−2l 2l P (x) = c0 xk1 + c2 xk−2 R . 1 R + · · · + c2l x1

(7.1)

Since W is a spherical harmonic, by definition, P is a solid harmonic, and so ∆P = 0. This gives l−1 X

k−2(j+1)

[c2j aj + c2(j+1) bj ] x1

R2j = 0

j=0

where aj = (k − 2j)(k − 2j − 1), and bj = 2(j + 1)(n + 2j − 1). This gives c2(j+1) = a

− bjj c2j , j = 0, 1, · · · , l − 1, i.e., P depends only on the coefficient c0 , and so any two polynomials of the form (7.1) differ by a constant multiple. But we have proved that any homogeneous polynomial of degree k whose restriction to S n−1 is constant on the (k)

x parallels of S n−1 orthogonal to e1 has the form (7.1). We also know that Ze1 ( |x| )|x|k

satisfies this property, and hence W (x0 ) = P (x0 ) = cZe(k) (x0 ), for all x0 ∈ S n−1 . 1 Using Lemma 7.3.1(c), we get (k)

Y (y 0 ) = W (τ −1 y 0 ) = cZe(k) (τ −1 y 0 ) = cZτ −1 e (τ −1 y 0 ) = cZe(k) (y 0 ). 1

Corollary 7.3.6. Suppose Fy0 (x0 ) is defined for all x0 , y 0 ∈ S n−1 and satisfies the following properties: (a) Fy0 is a spherical harmonic of degree k for each y 0 ∈ S n−1 , (b) Fρy0 (ρx0 ) = Fy0 (x0 ), for all rotations ρ. 190

(k)

Then, there is a constant c such that Fy0 (x0 ) = cZy0 (x0 ), for all x0 , y 0 ∈ S n−1 . Proof. Let y 0 ∈ S n−1 and ρ be a rotation that fixes y 0 . Then, by hypothesis (b), for any x0 ∈ S n−1 , we have Fρy0 (ρx0 ) = Fy0 (x0 ) = Fy0 (ρx0 ), ie., Fy0 is a spherical harmonic which is constant on parallels to S n−1 at y 0 ∈ S n−1 . Applying Theorem 7.3.5, we get (k)

a constant c(y 0 ) such that Fy0 = c(y 0 )Zy0 . Let y10 , y20 , ∈ S n−1 , and σ be a rotation such that σy10 = y20 . Again by hypothesis (b), we get (k)

(k)

2

1

c(y20 )Zy0 (σx0 ) = Fy20 (σx0 ) = Fσy10 (σx0 ) = Fy10 (x0 ) = c(y10 )Zy0 (x0 ), (k)

(k)

2

1

i.e., c(y20 )Zy0 (σx0 ) = c(y10 )Zy0 (x0 ).

(7.2)

But by Lemma 7.3.1(c), we have (k)

(k)

(k)

Zy0 (x0 ) = Zσy0 (σx0 ) = Zy0 (x0 ). 1

1

(7.3)

2

From (7.2) and (7.3), we get that c(y10 ) = c(y20 ), proving the result. For k = 0, 1, 2, · · · , let hk be the subspace of L2 (Rn ) generated by the functions of the form f (r)P (x), where f is a radial function and P is a solid harmonic of degree k. The following result describes the decomposition we are looking for: ∞ L hk holds in the sense Theorem 7.3.7. The direct sum decomposition L2 (Rn ) = k=0

that

(a) each subspace hk is closed, (b) the hk ’s are mutually orthogonal, (c) every element of L2 (Rn ) is a limit of a finite linear combination of elements belonging to the spaces hk , and 191

(d) for each k = 0, 1, 2, · · · , the space hk is invariant under the Fourier transform. ek , Proof. (a) The space of solid harmonics of degree k is isomorphic to the space H and so is of dimension ak . If P1 , P2 , · · · , Pak is an orthonormal basis of this space, then ak R P fj (r) Pj (x), with |f (x)|2 dx = every element f of the space hk has the form Rn

j=1 ∞ ak R P

|fj (r)|2 r2k+n−1 dr, and so, hk is closed.

j=1 0

(b) Integrating in polar coordinates and using Lemma 7.2.2 gives (b). (c) Suppose there is a function f ∈ L2 (Rn ) which is orthogonal to hk , for all k. By Corollary 7.2.5, the function f vanishes almost everywhere on almost every sphere centered at the origin, and so f = 0 in L2 (Rn ). (d) Let f ∈ L1 ∩ L2 (Rn ) be of the form f (u) = f0 (ρ)P (u) = ρk f0 (ρ)Y (u0 ), with ek , where ρ = |u|, u = ρu0 . Suppose we prove that fb ∈ hk whenever the function Y ∈H f is of the above form. Since finite linear combinations of such functions are dense in hk , this will prove that hk is invariant under the Fourier transform. For r = |x|, x = rx0 , we have n fb(x) = (2π)− 2

n

= (2π)− 2

Z Rn Z∞

e−ix·u f (u) du 

 f0 (ρ) ρk+n−1 

0

Z

0

0

e−irρx ·u Y (u0 ) du0 

S n−1

For s ≥ 0, Z

−isx0 ·u0

e S n−1

0

0

Y (u ) du

Z

−isx0 ·u0

Z

e

= S n−1

S n−1

192

(k)

Y (v 0 ) Zu0 (v 0 ) dv 0 du0

(7.4)

Z =

Y (v ) S n−1

Let Fx0 (v 0 ) =

0

Z

0

0

0

(k)

e−isx ·u Zv0 (u0 ) du0 dv 0 .

(7.5)

S n−1

(k)

0

e−isx ·u Zv0 (u0 ) du0 . By Fubini’s theorem and the orthogonality

R S n−1

ek , we get that Fx0 is orthogonal to all the spaces H ej , for j 6= k. So of the spaces H ek . Let u0 , w0 ∈ S n−1 and σ be a rotation such that u0 = σw0 . Then, by Lemma F x0 ∈ H 7.3.1(c), Z

0

Fσx0 (σv ) =

0

0

(k)

e−isσx ·u Zσv0 (u0 ) du0

S n−1

Z =

0

0

(k)

e−isσx ·σw Zσv0 (σw0 ) dw0

S n−1

Z =

0

0

(k)

e−isx ·w Zv0 (w0 ) dw0 = Fx0 (v 0 ).

S n−1

(k)

By Corollary 7.3.6, there is a number c = ϕ(s) such that Fx0 (v 0 ) = cZx0 (v 0 ), for all x0 , v 0 ∈ S n−1 . Therefore (7.5) becomes Z e

−isx0 ·u0

0

0

Y (u ) du =

S n−1

Y (v 0 ) Fx0 (v 0 ) dv 0

S n−1

Z i.e.,

Z

e

−isx0 ·u0

0

0

Z

Y (u ) du =

S n−1

(k)

Y (v 0 ) ϕ(s) Zx0 (v 0 )dv 0 = ϕ(s) Y (x0 )

S n−1

Substituting the above in (7.4), we get  n fb(x) = (2π)− 2 

Z∞ 0

which proves that fb ∈ hk .

193

 f0 (ρ)ϕ(rρ) ρk+n−1 dρ Y (x0 ),

7.4

Action of the Fourier transform on hk As we have discussed earlier, the Fourier transform of a radial function is

radial. In the two dimensional case, the Fourier transform of a radial function was expressed in terms of the Bessel functions. We proved that this was the case in higher dimensions too. We now study the action of the Fourier transform on the invariant subspaces hk , defined in the previous section. 2

Theorem 7.4.1. Suppose f (u) = e−π|u| P (u) for u ∈ Rn , where P (u) is a solid harmonic of degree k, then fb(v) = i−k f (v), for all v ∈ Rn . Proof. Since P is harmonic, its satisfies the mean value property. This gives, for a fixed t ∈ Rn , Z

Z∞

2

e−π|u| P (u + t) du =

Rn

 2 rn−1 e−πr 

0



Z∞

=

 Z

P (t + ru0 ) du0  dr

n−1

S  2

rn−1 e−πr dr ωn−1 P (t)

0

Z∞ = P (t)

 r

n−1

−πr2

e

i.e., P (t) =

0

du



0

Z

 Z

Z dr = P (t)

S n−1

2

e−π|x| dx

Rn

2

e−π|u| P (u + t) du

Rn

The polynomial P (t) has an analytic continuation P (z) = P (z1 , z2 , · · · , zn ), for z = (z1 , z2 , · · · , zn ) ∈ Cn . From the above equation we get

Z P (z) =

2

e−π|u| P (u + z) du, z = x + iy ∈ Cn .

Rn

194

In particular, for z = −iv, we get P (−iv) =

R

2

e−π|u| P (u − iv) du. Since P is

Rn

homogeneous of degree k, P (−iv) = (−i)k P (v). Using this and applying Cauchy’s theorem n times, we get Z

−π(u+iv)·(u+iv)

e

Z P (u) du =

e−πu·u P (u − iv) du = (−i)k P (v).

Rn

Rn

2 Multiplying both sides by e−π|v| , we get fb(v) = (−i)k f (v) = i−k f (v).

We now obtain an extension of the above result. Theorem 7.4.2. Let f ∈ L2 (Rn ) be of the form f (x) = p(x)g(|x|) where p ∈ Hm . The fb(ξ) = p(ξ)G(|ξ|). Thus the subspace of functions of the form f (x) = p(x)g(|x|), p ∈ Hm is invariant under the Fourier transform. Proof. When f (x) = p(x)g(|x|), with p ∈ Hm , is square integrable, the function g satisfies

Z∞

|g(r)|2 rn+2m−1 dr < ∞.

0

Let Dn+2m denote the space of all such functions with the obvious norm. We claim 2

that the subspace W consisting of finite linear combinations of e−πt|x| as t runs through positive reals is dense in Dn+2m . For, if there is a function g ∈ Dn+2m satisfying

Z∞

2

e−πtr g(r) rn+2m−1 dr = 0, for all t > 0.

0

Differentiating the above equation k times and evaluating at t = 1 gives Z∞

2

e−πr r2k g(r) rn+2m−1 dr = 0.

0

195

1

2

As the above equation is true for all k ∈ N, the function g(r)rn+2m−1 e− 2 πr is or1

2

thogonal to all functions of the form P (r2 )e− 2 πr , as P runs through all polynomials. Since these functions form a dense class in L2 ((0, ∞), dr) we get that g = 0. So it suffices to prove that W is invariant under the Fourier transform. 2 2 For t > 0, let ft (x) = tn f (tx), so that fbt (ξ) = fb(t−1 ξ). If f (x) = p(x)e−πt |x| , 2 take g(x) = p(x)e−π|x| , and consider fb(ξ) = t−n−m gbt (ξ) = t−n−m gb(t−1 ξ). Since gb(ξ) =

(−i)m g(ξ), we get −2 2 fb(ξ) = t−n−2m (−i)m p(x)e−πt |x| .

This proves that W is invariant and hence the theorem follows. Theorem 7.4.3. (Hecke-Bochner formula for the Fourier transform) Let f ∈ L2 (Rn ) be of the form f (x) = p(x)g(|x|), with p ∈ Pm . Then Fn (f ) = (−i)m p Hn/2+m−1 g. Proof. From the proof of the previous theorem, we get an operator Tmn on the space Dn+2m defined as follows. If g ∈ Dn+2m , and p ∈ Pm , the function p(x)g(|x|) ∈ L2 (Rn ) whose Fourier tranform is of the form p(x)G(|x|). As the Fourier transform is unitary, it follows that G ∈ Dn+2m . We can also think of g(|x|) as a radial function on Rn+2m whose (n + 2m)-dimensional Fourier transform will be a radial function, say G0 (|x|). We define another operator T0n+2m on Dn+2m as T0n+2m g = G0 . It is also clear that kT0n+2m gk2 = kgk2 . If we denote the Fourier transform on Rn by Fn , then T0n+2m g = Fn+2m g = Hn/2+m−1 g. From the proof of the above theorem we get that Tmn g = (−i)m T0n+2m g, for all g ∈ W. As W is a dense subspace of L2 (Rn ), this proves the result.

196

Appendix A A.1

Hilbert spaces

Definition A.1.1. Let H be an inner product space over C. Define kxk2 = hx, xi for all x ∈ H. Using this, we define

d(x, y) = kx − yk, ∀ x, y ∈ H.

Then H will be a metric space with metric d. If every Cauchy sequence in H converges to an element in H then H is called a Hilbert space. In other words, H is complete under the metric d induced by the inner product h·, ·i. Definition A.1.2. Let {uα : α ∈ A} be a set of vectors in a Hilbert space H. This set is said to be orthonormal if huα , uβ i = 0 for all α 6= β, α, β ∈ A and kuα k = 1 for each α ∈ A. Theorem A.1.3. Let {uα : α ∈ A} be an orthonormal set in a Hilbert space. Then the following conditions are equivalent. (a) {uα } is a maximal orthonormal set in H. (b) The set of all finite linear combination of member of {uα } is dense in H. P (c) |hx, uα i|2 = kxk2 ∀x ∈ H. α∈A P (d) hx, uα ihy, uα i = hx, yi x, y ∈ H. α∈A

197

Definition A.1.4. A set of vectors {uα : α ∈ A} in a Hilbert space H is said to be an orthonormal basis if {uα : α ∈ A} is an orthonormal set and satisfies one of the equivalent condition of above theorem.

A.2

Banach spaces

Definition A.2.1. Let X be a vector space over C. Then X is called a normed linear space if to each x ∈ X, there is associated a nonnegative real number kxk called the norm of x, such that (i) kxk = 0 if and only if x = 0. (ii) kαxk = |α|kxk ∀ α ∈ C, x ∈ X. (iii) kx + yk ≤ kxk + kyk ∀ x, y ∈ X. A normed linear space X will be a metric space under the metric d given by

d(x, y) = kx − yk ∀ x, y ∈ X.

Definition A.2.2. If a normed linear space X is complete in the metric defined by the norm, then X is called Banach space. Theorem A.2.3. Every Hilbert space is a Banach space. Let T be a linear transformation from a normed linear space X into a normed linear space Y . Then the norm of T is defined by

kT k = sup{kT k : x ∈ X, kxk ≤ 1}.

198

Remark A.2.4. For a linear transformation T , boundedness and continuity are equivalent. Theorem A.2.5. Uniform boundedness principle. Let X be a Banach space and Y be a normed linear space. Let {Tα : α ∈ A} denote a collection of bounded linear transformation of X into Y. Then either there is a ball B in Y with radius 1 and center at 0 such that Tα maps the unit ball of X into B or there exists x ∈ X such that no ball in Y contains Tα x for all x. Theorem A.2.6. Open mapping theorem Let U and V be open unit balls of the Banachs space X and Y, respectively. Let T be a bounded linear transformation of X into Y. Then there exists a δ > 0 such that {δy : y ∈ U } ⊂ T (U ).} Theorem A.2.7. Let X and Y be Banach spaces. Let T be a bounded linear transformation of X onto Y, which also one-to-one. Then there exists a δ > 0 such that kT xk ≥ δkxk

∀x ∈ X. In other words, T −1 is a bounded linear transformation of Y

onto X. Theorem A.2.8. Closed graph theorem. Let X and Y be Banach spaces. Let T be a linear mapping of X into Y with the following properties. If {xn } converges to x in X and {T xn } converges y in Y, then y = Tx. If such property holds, then T is continuous. Theorem A.2.9. Hahn-Banach theorem. If M is a subspace of a normed linear space X and if f is a bounded linear functional on N, then f can be extended to a bounded linear functional F on X such that kF k = kf k.

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Let X be a normed linear space. Let X ∗ denote the collection of all bounded linear functional on X. The elements of X ∗ can be denoted as x∗ and x∗ (x) can be denoted by hx, x∗ i, called the duality relation. The set X ∗ is also a vector space. Theorem A.2.10. Suppose B is a closed unit ball of a normed linear space X. Define kx∗ k = sup{|hx, x∗ i| : x ∈ B} for every x∗ ∈ X ∗ . Then X ∗ turns out to be Banach space under this norm. Let B ∗ be the closed unit ball of X ∗ for every x ∈ X, kxk = sup{|hx, x∗ i| : x∗ ∈ B ∗ }. Consequently, x∗ 7−→ hx, x∗ i is a bounded linear functional on X ∗ , of norm kxk. Theorem A.2.11. Let X and Y be normed linear spaces, let T be a bounded linear transformation of X into Y. Then kT k = sup{|hT x, y ∗ i : kxk ≤ 1, ky ∗ k ≤ 1.

Theorem A.2.12. Suppose X and Y are normed linear spaces. To each bounded linear transformation T of X into Y, there exists unique T ∗ , which will be bounded linear transformation of Y ∗ into X ∗ that satisfies the duality relation hT x, y ∗ i = hx, T ∗ y ∗ i for all x ∈ X y ∗ ∈ Y ∗ . Further, T ∗ satisfies kT ∗ k = kT k.

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Definition A.2.13. Let M be a subspace of a Banach space X. The annihilator of M is defined to be the set of all x ∈ X such that hx, x∗ i = 0 for every x ∈ M .

A.3

Banach algebras

Definition A.3.1. Let A be a vector space over C. A is called an algebra if A is a ring and α(xy) = (αx)y = x(αy) for every x, y ∈ A and α ∈ C. If, in addition, A is a Banach space with respect to a norm that satisfies kxyk ≤ kxkkyk ∀x, y ∈ A then A is called a Banach algebra. If A has multiplicative identity e with kek = 1, A is called a Banach algebra with identity. If the multiplication operation is commutative, A is called a commutative Banach algebra. Example A.3.2. (i) C[0, 1], the space of all complex valued continuous functions on [0, 1] with supremum norm is a commutative Banach algebra with identity. (ii) Let H be a Hilbert space. Let B(H) denote the collection of all bounded linear operators on H. Then B(H) is a non commutative Banach algebra with identity under the operator norm.

A.4

Topological vector spaces

Definition A.4.1. Let X be a vector space. If τ is a topology on X such that every singleton set in X is closed and the vector space operations are continuous with respect to τ , then X is called a topological vector space. In other words, the map ϕ : X × X → X (x, y) → x + y and ψ : F × X → X (α, x) → αx are continuous. The topological space X is said to be locally convex if there is a local base B whose members are convex. 201

Definition A.4.2. Let p be a real valued functions on X then p is called a semi norm on X if it satisfies the following: (i) p(x + y) ≤ p(x) + p(y) (ii) p(αx) = |α|p(x) for every x, y ∈ X, α ∈ C. A family P of semi norms on X is said to be separating if to each x 6= 0 corresponds at least one p ∈ P with p(x) 6= 0. Theorem A.4.3. Suppose P is a separating family of semi norms on a vector space X. Associate to each p ∈ P and to each positive integer n the set  V (p, n) =

1 x : p(x) < n

 .

Let B be the collection of all finite intersection of the sets V (p, n). Then B is a local base for a topology τ on X, which turns X into a locally convex space. Theorem A.4.4. Suppose P is separating family of semi norms on X. Let calQ be the smallest family of semi norms on X that contains P and satisfies the following: If p1 , p2 ∈ Q, p = max(p1 , p2 ), then p ∈ Q. Let Λ be a linear functional on X. Then Λ is continuous if and only if there exists p ∈ Q such that |Λx| ≤ M p(x) for all x ∈ X and some constant M < ∞.

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Bibliography [1] H. Dym and H. P. McKean, Fourier series and integrals, Academic Press, 1985. [2] G. B. Folland, Fourier analysis and its applications, American Mathematical Society, 2009. [3] Y. Katznelson, An introduction to harmonic analysis, Cambridge University Press, 2004. [4] W. Rudin, Functional analysis, McGraw Hill Publications, 1991. [5] E. M. Stein, Singular integrals and differentiability properties of functions, Princeton University Press, 1971. [6] E. M. Stein and Rami Shakarchi, Fourier analysis, Princeton University Press, 2003. [7] E. M. Stein and G. Weiss, Introduction to Fourier analysis on Euclidean spaces, Princeton University Press, 1971. [8] A. Torchinsky, Real variable methods in harmonic analysis, Dover Publications, 2004.

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