*270*
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*English*
*Pages 265*
*Year 2014*

- Author / Uploaded
- Loukas Grafakos

Loukas Grafakos

Modern Fourier Analysis, Third Edition Solutions of all the exercises March 20, 2014

Springer

Γ ια την Iωα´ ννα, την Kωνσ ταντι´ να, και την Θ ε oδ ω´ ρα

vi

I would like to express my deep gratitude to the following people who helped in the preparation of the solutions of the books Classical Fourier Analysis, 3rd edition, GTM 249 and Modern Fourier Analysis, 3rd edition, GTM 250. I remain solely responsible for any error contained in the enclosed solutions. Mukta Bhandari, Jameson Cahill, Santosh Ghimire, Zheng Hao, Danqing He, Nguyen Hoang, Sapto Indratno, Richard Lynch, Diego Maldonado, Hanh Van Nguyen, Peter Nguyen, Jesse Peterson, Sharad Silwal, Brian Tuomanen, Xiaojing Zhang,

Contents

vii

1 Smoothness and Function Spaces

1.1 Smooth functions and tempered distributions Exercise 1.1.1 Given multiindices α, β show that there are constants C,C0 such that ρα,β (ϕ) ≤ C

∑ ∑

0 ργ,δ (ϕ) ,

∑ ∑

ργ,δ (ϕ) .

|γ|≤|α| |δ |≤|β | 0 ρα,β (ϕ) ≤ C0

|γ|≤|α| |δ |≤|β |

for all Schwartz functions ϕ. Hint: The first inequality follows by Leibniz’s rule. Conversely, to express ξ α ∂ β ϕ in terms of linear combinations of ∂ β (ξ γ ϕ(ξ )), proceed by induction on |α|, using that ξ j ∂ β ϕ = ∂ β (ξ j ϕ) − ∂ β ϕ − (β j − 1)∂ β −e j ϕ if β j ≥ 1 and ξ j ∂ β ϕ = ∂ β (ξ j ϕ) if β j = 0. Here β = (β1 , . . . , βn ) and e j = (0, . . . , 1, . . . , 0) with 1 in the jth entry. Solution. First of all, we notice that ∂ β (ξ α ϕ) is equal to a sum of terms of the form cγ ξ γ ∂ β −γ ϕ, so any ρα,β (ϕ) is bounded by a finite 0 (ϕ). sum of seminorms ργ,δ Conversely, we first verify the identity ξ j ∂ β ϕ = ∂ β (ξ j ϕ) − ∂ β ϕ − (β j − 1)∂ β −e j ϕ if β j ≥ 1 by induction on β j . Indeed, say that it holds for β j . Then ∂ β +e j (ξ j ϕ) =∂ j ∂ β (ξ j ϕ) = ∂ j (ξ j ∂ β ϕ) + ∂ j (∂ β ϕ − (β j − 1)∂ β −e j ϕ) =∂ β ϕ + ξ j ∂ β +e j ϕ + ∂ β +e j ϕ + (β j − 1)∂ β ϕ =ξ j ∂ β +e j ϕ + ∂ β +e j ϕ + β j ∂ β ϕ from which it follows that the claimed identity holds for β j + 1. Having verified the preceding identity, we conclude that ξ j ∂ β ϕ can be expressed as a linear combination of terms of the form ∂ γ (ξ j ϕ) and ∂ γ (ϕ). Then ξk ξ j ∂ β ϕ can be expressed as a linear combination of terms of the form ξk ∂ γ (ξ j ϕ) and ξk ∂ γ (ϕ). Applying the preceding conclusion, we deduce that ξk ξ j ∂ β ϕ can be expressed as a linear combination of terms of the form ∂ γ (ξk ξ j ϕ), ∂ γ (ξk ϕ), ∂ γ (ξ j ϕ), and ∂ γ ϕ. Continuing in this way, we conclude that any ξ α ∂ β ϕ is a linear combination of terms 0 (ϕ) is bounded by a finite sum of seminorms ρ (ϕ). of the form ∂ δ (ξ γ ϕ(ξ )), thus any ρα,β γ,δ

1

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Exercise 1.1.2 Suppose that a function ϕ lies in C ∞ (Rn \ {0}) and that for all multiindices α there exist constants Lα such that ϕ satisfies lim ∂ α ϕ(t) = Lα .

t→0

Then ϕ lies in C ∞ (Rn ) and ∂ α ϕ(0) = Lα for all multiindices α. Solution. By assumption we have that limt→0 ϕ(t) = L0 , and thus setting ϕ(0) = L0 , we have a continuous extension of ϕ on Rn . Suppose that for all |α| ≤ N we have a C N extension of ϕ on Rn for some N ∈ Z+ . We have ∂ α ϕ(te j ) − Lα ∂ α ϕ(te j ) − ∂ α ϕ(0) = lim = lim ∂ j ∂ α ϕ(ct e j ) t→0 t→0 t→0 t t

∂ j ∂ α ϕ(0) = lim

by the mean value theorem, where ct ∈ (−t,t) \ {0}. The application of the mean value theorem is allowed since the function s 7→ ∂ α ϕ(se j ) is continuous on the closed interval [−t,t] and differentiable on the open interval (−t,t). By assumption the preceding limit exists and is equal to Lα+e j . This is valid for all j = 1, . . . , n and thus ϕ has a C N+1 extension on Rn .

Exercise 1.1.3 0 n 0 0 Let that uN ∈ S (R ). Suppose that uN → u in S /P and uN → vin S . Then prove that u − v is a polynomial. Hint: Use Proposition 1.1.3 or directly Proposition 2.4.1 in [161].

Solution. Given a ϕ in S0 (Rn ) we have huN , ϕi → hu, ϕi and huN , ϕi → hv, ϕi by assumption. Then hu − v, ϕi = 0 for all ϕ in S0 (Rn ). Passing to the Fourier transform, it follows that hud − v, ψi = 0 for all functions ψ with ∂ α ψ(0) = 0 for all α. In particular, this holds for all Schwartz functions that are supported in Rn \ {0}. Then the distribution ud − v is supported at the origin and it follows that it has to be a linear combination of derivatives of Dirac masses (Proposition 2.4.1 in [161]). Then u − v must be a polynomial.

Exercise 1.1.4 Suppose that Ψ is a Schwartz function whose Fourier transform is supported in an annulus that does not contain the origin b (2− j ξ ) = 1 for all ξ 6= 0. Show that for functions g ∈ L1 (Rn ) with gb ∈ L1 (Rn ) we have ∑ j∈Z ∆ Ψ (g) = g and satisfies ∑ j∈Z Ψ j pointwise everywhere. Solution. We write for every x ∈ Rn Z

g(x) =

Rn

gb(ξ )e2πix·ξ dξ =

Z Rn

gb(ξ )e2πix·ξ

∑ Ψb (2− j ξ )dξ = ∑

j∈Z

Z

n j∈Z R

b (2− j ξ )dξ gb(ξ )e2πix·ξ Ψ

where the last step is justified by the Lebesgue dominated convergence theorem since gb is integrable. But the last expression is equal to ∑ j∈Z ∆ Ψ j (g), hence the conclusion follows.

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3

Exercise 1.1.5 Let Θ and Φ be Schwartz functions whose Fourier transforms are compactly supported and let Ψ , Ω be a Schwartz functions whose Fourier transforms are supported in annuli that do not contain the origin and satisfy ∞

b (ξ ) + ∑ Ψ b (2− j ξ )Ω b (2− j ξ ) = 1 b )Θ Φ(ξ j=1

for all ξ ∈ Rn . Then for every f ∈ S 0 (Rn ) we have ∞ Ω Φ ∗Θ ∗ f + ∑ ∆ Ψ j ∆j (f) = f j=1

where the series converges in S 0 (Rn ).

Solution. It suffices to show that the convergence holds in S . So we fix ϕ in S . We need to prove that N β b (ξ ) − ∑ Ψ b (2− j ξ )Ω b (2− j ξ ) ϕ(ξ b )Θ b )ξ α → 0 sup ∂ξ 1 − Φ(ξ

ξ ∈Rn

j=1

The function inside the square bracket is supported in a set of the form |ξ | ≥ c2N for some constant c. Moreover, the function b )ξ α ) is a Schwartz inside the curly brackets is bounded and all of its derivatives are bounded. But for any γ ≤ β , ∂ γ (ϕ(ξ function and thus it has rapid decay at infinity. It follows that the preceding supremum over the set |ξ | ≥ c2N tends to zero as N → ∞ by Leibniz’s rule.

Exercise 1.1.6 (a) Show that for any multiindex α on Rn there is a polynomial Qα of n variables of degree |α| such that for all ξ ∈ Rn we have 2

2

∂ α (e−|ξ | ) = Qα (ξ )e−|ξ | . (b) Show that for all multiindices |α| ≥ 1 and for each k in {0, 1, . . . , |α| − 1} there is a polynomial Pα,k of n variables of degree at most |α| such that |α|−1 1 ξ1 ξn −|ξ | P , . . . , e ∂ α (e−|ξ | ) = ∑ k α,k |ξ | |ξ | k=0 |ξ | for every ξ ∈ Rn \ {0}. Conclude that for |α| ≥ 1 we have α −|ξ | 1 ∂ (e ) ≤ Cα 1 + 1 + · · · + e−|ξ | |ξ | |ξ ||α|−1 for some constant Cα and all ξ 6= 0. Hint: For the two identities use induction on |α|. Part (b): use that the ∂ j derivative of a homogeneous polynomial of degree at most |α| is another homogeneous polynomial of degree at most |α| + 1 times |ξ |−1 . Solution. To prove 2

2

∂ α (e−|ξ | ) = Qα (ξ )e−|ξ |

start with the case α = 0 which is obviously valid. If it is also valid for a certain α, let e j = (0, . . . , 0, 1, 0, . . . , 0) and consider α + e j . We have

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2 2 ∂ j ∂ α (e−|ξ | ) = (∂ j Qα )(ξ ) − 2ξ j Qα (ξ ) e−|ξ | and note that the expression in the square bracket is a polynomial of degree |α| + 1. Next we prove that |α|−1 1 ξ1 ξn −|ξ | ∂ α (e−|ξ | ) = ∑ P , . . . , e , k α,k |ξ | |ξ | k=0 |ξ | where the polynomials Pα,k have degree at most |α|. This identity is certainly valid for any multiindex α with |α| = 1 by a simple calculation. Assuming that it holds for a given α, then differentiate in ∂ j . We first notice that for any monomial we have ∂j

ξ β −e j ξβ 1 ξ β +e j β = − |β | j |ξ | |ξ ||β | |ξ ||β |−1 |ξ ||β |+1

if β is a multiindex, thus ∂ j Pα,k

ξ1 ξn ,..., |ξ | |ξ |

1 ] ξn ξ1 = P ,..., |ξ | α,k |ξ | |ξ |

] where Pα,k is polynomial of degree at most |α| + 1. The induction hypothesis gives

∂ j ∂ α (e−|ξ | ) |α|−1 ξ1 ξn −|ξ | 1 P , . . . , e = ∂j ∑ k α,k |ξ | |ξ | k=0 |ξ | |α|−1 ξj 1 ξ1 ξ1 ξ1 k ξj ξn 1 1 ] ξn ξn Pα,k Pα,k ,..., + k P ,..., − ,..., e−|ξ | = ∑ − k+1 |ξ | |ξ | |ξ | |ξ | |ξ | |ξ |k |ξ | |ξ | |ξ | |ξ | |ξ | α,k |ξ | k=0 |α|−1 |α|−1 ξj ξ1 ξ1 ξ1 1 ξn ξn 1 ξj ξn −|ξ | ] −|ξ | − k Pα,k ,..., + Pα,k ,..., e − ∑ Pα,k ,..., e = ∑ k+1 k |ξ | |ξ | |ξ | |ξ | |ξ | |ξ | |ξ | k=0 |ξ | |ξ | k=0 |ξ | |α| |α|−1 ξj 1 ξ1 ξn ξn 1 ξj ξ1 ξn ξ1 ] −|ξ | =∑ (` − 1) P , . . . , + P , . . . , e − P , . . . , e−|ξ | − α,`−1 ∑ |ξ |k |ξ | α,k |ξ | α,`−1 |ξ | ` |ξ | |ξ | |ξ | |ξ | |ξ | |ξ | `=1 k=0 |α|+1−1 1 ξ1 ξn −|ξ | = ∑ Pα+e j ,m ,..., e . m |ξ | |ξ | m=0 |ξ | We examine the degrees of the polynomials inside the square brackets. They all have degree at most |α| + 1. Thus Pα+e j ,m has degree at most |α| + 1 = |α + e j |. The last assertion follows from the boundedness of Pα,k on the unit sphere.

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5

1.2 The Laplacian, Riesz Potentials, and Bessel Potentials Exercise 1.2.1 (a) Let 0 < s,t < ∞ be such that s + t < n. Show that Is It = Is+t as operators acting on S (Rn ). (b) Let Re s > 2 Re z. Prove the operator identities s

Is (−∆ )z = (−∆ )z Is = Is−2z = (−∆ )z− 2 on functions whose Fourier transforms vanish in a neighborhood of zero. (c) Prove that for all z ∈ C we have

(−∆ )z f | (−∆ )−z g = f | g whenever the Fourier transforms of f and g vanish to sufficiently high order at the origin. (d) Given s with Re s > 0, find an α ∈ C such that the identity

2

Is ( f ) | f = (−∆ )α f L2 is valid for all functions f as in part (c). Solution. (a) We have by definition, for f ∈ S (Rn ) \ −(s+t) b −( s+t 2 ) ( f )(ξ ) = (2π|ξ |) f (ξ ). I\ s+t ( f )(ξ ) = (−∆ ) Again, s Is\ (It ( f ))(ξ ) = (−∆\ )− 2 It ( f )(ξ )

= (2π|ξ |)−s I[ t ( f )(ξ ) = (2π|ξ |)−s (2π|ξ |)−t fb(ξ ) = (2π|ξ |)−(s+t) fb(ξ ) n \ and this function is locally integrable. Therefore we have I\ s+t ( f )(ξ ) = Is (It ( f ))(ξ ) for all f ∈ S (R ). This gives (a). (b) We note that Re (s − 2z) > 0. Then, if fb vanishes in a neighborhood of zero, it can be multiplied with |ξ |−N , where N > 0 and we have s−2z \ \ Is−2z ( f )(ξ ) = (−∆ )−( 2 ) ( f )(ξ ) = (2π|ξ |)−(s−2z) fb(ξ )

= (2π|ξ |)−s+2z fb(ξ ) = (2π|ξ |)2z−s fb(ξ ). Clearly, we have s (−∆\ )z− 2 ( f )(ξ ) = (2π|ξ |)2z−s fb(ξ ).

Again s \ Is ((−∆ )z ( f ))(ξ ) = (−∆ )− 2 (−∆ )z ( f ) b(ξ ) \ = (2π|ξ |)−s (−∆ )z ( f )(ξ ) = (2π|ξ |)−s (2π|ξ |)2z fb(ξ ) = (2π|ξ |)2z−s fb(ξ ).

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Similarly we can show, \ (−∆ )z Is ( f )(ξ ) = (2π|ξ |)2z−s fb(ξ ). s \ \ )z Is ( f )(ξ ). This gives us Is (−∆ )z = (−∆ )z Is = So we have, Is−2z ( f )(ξ ) = (−∆\ )z− 2 ( f )(ξ ) = Is ((−∆ )z ( f )) b(ξ ) = (−∆ s z− Is−2z = (−∆ ) 2 which proves (b). (c) Let f be as in part (b). If the Fourier transforms of f and g vanish to sufficiently high order at the origin we have h(−∆ )z ( f )|(−∆ )−z (g)i =

Z Z

=

(−∆ )z ( f )(x)(−∆ )−z (g)(x)dx ∨ \ (−∆ )z ( f )(ξ )(−∆ )−z (g) (ξ ) dξ

Z

\ \ )−z (g)(ξ ) dξ (−∆ )z ( f )(ξ )(−∆

Z

(2π|ξ |)2z fb(ξ )(2π|ξ |)−2z gb(ξ )dξ

= = Z

=

fb(ξ )b g(ξ )dξ Z

=

f (x)g(x)dx

= h f |gi. (d) If the identity holds we must have Z

Z

\ |(−∆ )α ( f )(ξ )|2 dξ

(2π|ξ |)−s fb(ξ ) fb(ξ )dξ =

Z

|(2π|ξ |)−2α fb(ξ )|2 dξ

Z

Z

(2π|ξ |)−4α | fb(ξ )|2 dξ

Z Z

Z

\ Is ( f )(x) f (x)dx = k(−∆ )α ( f )k2L2

b I[ s ( f )(ξ ) f (ξ )dξ =

(2π|ξ |)−s | fb(ξ )|2 dξ =

[(2π|ξ |)−s − (2π|ξ |)−4α ]| fb(ξ )|2 dξ = 0.

This gives us that (2π|ξ |)−s = (2π|ξ |)−4α . So we must have α = 4s .

Exercise 1.2.2 Prove that for −∞ < α < n/2 < β < ∞ and that for all f ∈ S (Rn ) we have

α/2 ββ−n/2

n/2−α −α β /2

β2 −αn ,

≤ C ∆ ( f ) ∆ ( f ) L∞ (Rn ) L2 (Rn ) L (R )

f

where C depends only on α, n, β . Hint: You may want to use Exercise 2.2.14 in [161]. Solution. We know by Exercise 2.2.14 in [161] that for −∞ < α < n2 < β < ∞, there exists a constant C = C(n, α, β ) independent of g ∈ S (Rn ) such that α

β − n2 β −α L2 (Rn )

kgkL1 (Rn ) ≤ Ck|x| g(x)k Replacing g by fb in the above inequality we get

β

n −α 2 β −α L2 (Rn )

k|x| g(x)k

.

Contents

7 β− n 2

n −α 2

k fbkL1 (Rn ) ≤ Ck|x|α fb(x)kLβ2−α k|x|β fb(x)kLβ2−α . (Rn ) (Rn ) Clearly k fbkL∞ ≤ k f kL1 . Consequently, β− n

k f kL∞

n −α

2 2 b β −α β b f (x)k = k fbkL∞ ≤ Ck|x|α fb(x)kLβ2−α k|x| . n (R ) L2 (Rn )

b since fb(x) = f˜(x) = f (−x). We also know that α (−∆ ) 2 f b(ξ ) = (2π|ξ |)α fb(ξ )

and thus we obtain the desired inequality.

Exercise 1.2.3 Show that when 0 < s < n we have sup

Is ( f )

n

L n−s (Rn )

k f kL1 (Rn ) =1

=

Js ( f )

sup

= ∞.

n

L n−s (Rn )

k f kL1 (Rn ) =1

n ). Thus both Is and Js are not of strong type (1, n−s

Solution. Put C := 2−s π −n/2Γ ((n − s)/2)/Γ (s/2) so that Is f (x) = C n L1 → L n−s , we notice that Z

Z

n

Is (χB(0,1) ) n−sn = L n−s

Rn

≥

Rn

|x|≥2

≥

|x − y|

n n−s

dx

n n−s

dy

dx

|y|≤1

|x|≥2

(|x| /2)s−n dy

n n−s

dx

B(0,1)

Z

= C0

f (y) |x − y|s−n dy. To see that Is is unbounded from

s−n

Z

Z

Rn

χB(0,1) (y) |x − y|s−n dy

Z

Z

R

|x|−n dx

|x|≥2

= ∞. n

Next, we show that Js is unbounded from L1 → L n−s . By Proposition 6.1.5, there is K > 0 such that Gs (z) ≥ K |z|s−n , whenever z ∈ B(0, 2). Let 0 < ε < 1/10. Then for |y| ≤ ε and |x| ≤ 2 − ε we have |x|/2 ≤ |x − y| ≤ 2 and

Js (χB(0,ε) )

n L n−s

Z

= ≥

Rn

Z Rn

n n−s χB(0,ε) Gs (x − y)dy dx

Z B(0,2−ε)\B(0,2ε)

≥

Z B(0,2−ε)\B(0,2ε)

≥

Z B(0,2−ε)\B(0,2ε)

Z B(0,ε)

Z

n n−s

Gs (x − y)dy K |x − y|s−n dy

dx

n n−s

dx

B(0,ε)

Z B(0,ε)

K(|x| /2)s−n dy

n n−s

dx

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Z

n K(|x| /2)s−n n−s dx

= B(0,2−ε)\B(0,2ε)

= C0 [ln(2 − ε) − ln(2ε)] . So, letting ε → 0 gives the desired result.

Exercise 1.2.4 s

1 − n (1+δ ) ) Let 0 < s < n. Consider the function h(x) = |x|−s (log |x| for |x| ≤ 1/e and zero otherwise. Prove that when 0 < δ

s > 0. n

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Exercise 1.2.5 For 1 < p < ∞ and 0 < s < ∞ define the Bessel potential space Lsp (Rn ) as the space of all functions f ∈ L p (Rn ) for which there exists another function f0 in L p (Rn ) such that Js ( f0 ) = f . Define a norm on these spaces by setting k f kLsp = k f0 kL p . Prove the following properties of these spaces: (a) k f kL p ≤ k f kLsp ; hence Lsp (Rn ) is a subspace of L p (Rn ). (b) For all 0 < t, s < ∞ we have Gs ∗ Gt = Gs+t and thus r Lsp (Rn ) ∗ Ltq (Rn ) ⊆ Ls+t (Rn ) ,

where 1 < p, q, r < ∞ and 1p + 1q = 1r + 1. (c) The sequence of norms k f kLsp increases, and therefore the spaces Lsp (Rn ) decrease as s increases. p (Rn ). onto isometry from Lsp (Rn ) to Ls+t (d) The map Jt is an one-to-onep and p n The Bessel potential space Ls (R ) coincides with the Sobolev space Ls (Rn ), introduced in Section 2.2. Solution. (a) By Young’s inequality, one has k f kL p = kJs ( f0 )kL p = k f0 ∗ Gs kL p ≤ k f0 kL p kGs kL1 . Since kGs kL1 = 1, it follows that k f kL p ≤ k f0 kL p = k f kLsp . (b) We have bs+t (ξ ). bs (ξ )G bt (ξ ) = (1 + 4π 2 |ξ |2 ) s+t 2 =G G This implies Gs ∗ Gt = Gs+t . Take f1 ∈ Lsp , g1 ∈ Ltq . By definition, there exist f ∈ L p , g ∈ Lq such that f 1 = f ∗ Gs ,

g1 = g ∗ Gt

Therefore, f1 ∗ g1 = f ∗ Gs ∗ g ∗ Gt = f ∗ g ∗ Gs ∗ Gt = f ∗ g ∗ Gs+t By Young’s inequality, one has f ∗ g ∈ Lr where 1 1 1 + = +1 p q r r or L p ∗ L q ⊂ L r . Therefore f1 ∗ g1 ∈ Ls+t s t s+t (c) Suppose that f = f0 ∗ Gs = f1 ∗ Gs+t ,

t ≥ 0.

It follows that f0 = f1 ∗ Gt . Thus, by Young’s inequality, one has k f kLsp = k f0 kL p = k f1 ∗ Gt kL p ≤ k f1 kL p kGt kL1 = k f1 kL p = k f kL p . s+t

This means that k f kLsp increases as s increases. (d) Take f ∈ Lsp . Then f = f0 ∗ Gs , f0 ∈ L p . One has kJt ( f )kL p = k f0 ∗ Gs ∗ Gt kL p = k f0 ∗ Gs+t kL p = k f0 kL p = k f kLsp . s+t

s+t

s+t

p Thus Jt is an isometry from Lsp to Ls+t . Moreover, Jt is clearly injective and onto.

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Exercise 1.2.6 For 0 ≤ s < n define the fractional maximal function M s ( f )(x) = sup t>0

1

Z

(vnt n )

n−s n

| f (x − y)| dy ,

|y|≤t

where vn is the volume of the unit ball in Rn . (a) Show that for some constant C we have M s ( f ) ≤ C Is ( f ) for all f ≥ 0 and conclude that M s maps L p to Lq whenever Is does. sp . Show that there is a constant C > 0 (depending (b) (Adams [1] ) Let 0 < s < n, 1 < p < ns , 1 ≤ q ≤ ∞ be such that 1r = 1p − ns + nq on the previous parameters) such that for all positive functions f we have

n

sp sp

Is ( f ) r ≤ C M p ( f ) nq f 1−p n . L L L Hint: For f 6= 0, write Is ( f ) = I1 + I2 , where Z

I1 =

Z

f (y) |x − y|s−n dy ,

I2 =

|x−y|≤δ n

f (y) |x − y|s−n dy .

|x−y|>δ

n

Show that I1 ≤ Cδ s M 0 ( f ) and that I2 ( f ) ≤ Cδ s− p M p ( f ). Optimize over δ > 0 to obtain sp 1− sp n n Is ( f ) ≤ C M p ( f ) n M 0 ( f ) , from which the required conclusion follows easily. Solution. (a) For f ≥ 0 , Is ( f )(x) = C(n,s)

Z Rn

f (x − y)|y|n−s dy

where n

C(n,s) = 2−s π − 2

Γ

n−s 2 Γ 2s

.

For |y| ≤ t and 0 ≤ s < n it follows that 1≤

t |y|

n−s

= t n−s |y|s−n .

Therefore, 1 (νnt n )

Z n−s n

|y|≤t

| f (x − y)|dy ≤

1

Z n−s n

| f (x − y)||y|−n+st n−s dy

|y|≤t (νnt n ) 1 = n−s Is ( f )(x). νn n

Now taking supremum over all t > 0 yields M s ( f )(x) ≤ CIs ( f )(x) for every x and for every f ≥ 0. Therefore, M s ( f ) ≤ CIs ( f ) for every f ≥ 0. Taking Lq norms we deduce that M s maps L p to Lq whenever Is does so. (b) For f 6= 0, set

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11

Is ( f )(x) =

Z

f (x − y)|y|−n+s dy

Rn

Z

=

f (y)|x − y|−n+s dy

Rn

=: I + II, where, f (y) dy, |x − y|n−s f (y) dy |x − y|n−s

Z

I= |x−y|≤δ

Z

II = |x−y|>δ

For every k ∈ Z+ define Ak (x) = {y : 2k δ ≤ |x − y| < 2k+1 δ }. Then, ∞ Z −n+s |I| = ∑ f (y)|x − y| dy k=1 A−k (x) ∞

≤

Z

∑

−k −k+1 δ k=1 2 δ ≤|x−y| 0. Define the operator Tλ by setting T λ ( f )(ξ ) = m(λ ξ ) f (ξ ). Show that there exists a constant C = C(n, s) such that for all f and u ≥ 0 and λ > 0 we have Z Rn

|Tλ ( f )(x)|2 u(x) dx ≤ C kmk2L2 s

Z Hint: Prove that |Tλ ( f )(x)|2 ≤ C kmk2L2 s

Rn

Z Rn

| f (y)|2 M(u)(y) dy .

λ −n (1 + 4π 2 |λ −1 (x − y)|2 )−s | f (y)|2 dy using the Cauchy-Schwarz inequality.

Solution. b = m, then since m ∈ Ls2 (Rn ), the function (1 + 4π 2 |x|2 ) 2s K(x) is in L2 (Rn ) with norm equal to kmkL2 . The kernel of Tλ If K s is λ −n K(λ −1 x). Then we have

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2 Z Z Z −n −1 2 λ K(λ (x − y)) f (y) dy u(x) dx |Tλ ( f )| udx = n n n R R R 2 s Z Z (1 + 4π 2 |λ −1 (x − y)|2 ) 2 λ −n K(λ −1 (x − y)) = f (y) dy u(x) dx s Rn Rn (1 + 4π 2 |λ −1 (x − y)|2 ) 2 Z Z Z λ −n 2 −1 2 s −n −1 2 2 ≤ |(1 + 4π |λ (x − y)| ) λ K(λ (x − y))| dy | f (y)| dy u(x) dx 2 −1 (x − y)|2 )s Rn Rn Rn (1 + 4π |λ by the Cauchy-Schwarz inequality. But the expression inside the first parenthesis is equal to kmk2L2 , so the preceding expression s is bounded by Z Z −n λ kmk2L2 | f (y)|2 dy u(x) dx s Rn Rn (1 + 4π 2 |λ −1 (x − y)|2 )s By Fubini’s theorem the above is bounded by C kmk2L2 s

Z

Z

Rn Rn

λ −n u(x) dx | f (y)|2 dy ≤ C0 kmk2L2 s (1 + |λ −1 (x − y)|)2s

Z Rn

M(u)(y)| f (y)|2 dy ,

since 2s > n and the function inside the integral is radially symetrically decreasing.

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33

Section 1.4. Lipschitz spaces Exercise 1.4.1 Fix k ∈ Z+ , α1 , . . . , αn ∈ Z+ ∪ {0}, and γ > 0. Set |α| = α1 + · · · + αn . (a) Let Q(x) = x1α1 · · · xnαn be a monomial on Rn of degree |α|. Define C(k, m) = ∑kj=0 kj (−1)k− j jm for m ∈ Z+ . Show that when |α| ≥ k for all x, h ∈ Rn we have α1 αn Dkh (Q)(x) = ∑ ··· C(k, |β |) hβ xα−β β β n 1 β ≤α |β |≥k |α|

and note that C(k, |β |) = 0 if |β | < k and C(k, |β |) = k! if |β | = k. Conclude that Dh (Q)(x) = |α|! Q(h). Also observe that n Dm h (Q)(x) = 0 if m > |α| for all h, x ∈ R . (b) Show that a continuous function f satisfies Dkh ( f )(x) = 0 for all x, h in Rn if and only if f is a polynomial of degree at most k − 1. . (c) Prove that a polynomial lies in Λγ if and only if it has degree at most [γ]. Hint: (a) Use the Fourier transform. (b) One direction follows by part (a) while for the converse direction, use a partition of unity as in the proof of Theorem 1.4.6 to show that fb is supported at the origin and use Proposition 2.4.1 in [161]. Solution. (a) Let Q(x) = x1α1 · · · xnαn be a monomial of degree |α| = α1 + · · · + αn ≥ k. For every ϕ ∈ S (Rn ) we have Z Rn

Dkh (Q)(x)ϕ(x) dx =

Z Rn

Q(x)Dk−h (ϕ)(x) dx

∂ α1 ∂ αn −2πiξ ·h 1 kb (e − 1) ϕ(ξ ) · · · ∂ ξnαn (−2πi)|α| ∂ ξ1α1 ξ =0 1 α1 αn β −2πiξ ·h k b = ··· ∂ (e − 1) ∂ α−β (ϕ)(0) ∑ βn (−2πi)|α| β ≤α β1 ξ =0 k Z α1 αn 1 k k− j β −2πi jξ ·h · · · = (−1) ∂ e (−2πi)|α−|β | xα−β ϕ(x) dx ∑ ∑ j n βn j=0 (−2πi)|α| β ≤α β1 R ξ =0

=

|β |≥k

Z

=

Rn

k k α1 αn k− j |β | β α−β ϕ(x) dx ∑ β1 · · · βn ∑ j (−1) j h x j=0 β ≤α |β |≥k

since when |β | < k, the terms ∂ β (e−2πiξ ·h − 1)k ξ =0 vanish. This proves the claim when |α| ≥ k. As a consequence, we conclude that the distribution Dkh (Q)(x) coincides with the constant function k!Q(h) if |α| = k. n Finally, notice that Dm h (Q)(x) = 0 for all h, x ∈ R when m > |α| by a similar calculation, which relies on the fact that when |β | ≤ |α| < m, we have that β −2πiξ ·h m ∂ (e − 1) = 0. ξ =0

(b) If f is equal to a polynomial P of degree at most k − 1 and Pk−1 is the sum of monomials of degree k − 1 in P, by part (a) k we have that Dk−1 h (P)(x) = (k − 1)! Pk−1 (h), thus Dh (P) = 0. k Conversely, if f is continuous and satisfies Dh ( f )(x) = 0 for all x, h in Rn , then we have for all h ∈ Rn \ k ( f ) = 0. (e2πi(·)·h − 1)k fb = D h We want to show fb is supported at origin. We introduce a partition of unity ψr , r ∈ F, where F is a finite set, such that • ∑r∈F ψ(ξ ) = 1

for all ξ 6= 0

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• each ψr is a homogeneous function of degree zero and smooth on the sphere Sn−1 . • each ψr is supported in a conical neighborhood of Rn . • for each r ∈ F there is a unit vector hr 6= 0 such that 14 |ξ | ≤ |ξ · hr | ≤ |ξ | for all ξ in the support of ψr . We fix a Schwartz function φ whose support does not contain the origin. Since ψr is not smooth at the origin and φ is 1 supported in Rn \ {0}, the function ψr φ is Schwartz. Then 12 ξ · hr is not an integer, and the function (1 − e2πi 2 hr ·ξ )−k ψr (ξ )φ (ξ ) is Schwartz. We now have h fb, φ i = ∑ h fb, ψr φ i r∈F

1

1

= ∑ h(e2πi(·)· 2 hk − 1)k fb, (e2πi(·)· 2 hk − 1)−k ψr φ i r∈F

1

= ∑ h0, (e2πi(·)· 2 hk − 1)−k ψr φ i = 0 . r∈F

We conclude that fb vanishes on Rn \ {0}. By Proposition 2.4.1 in [161], for some positive integer m, we have fb =

aα ∂ α δ0 .

∑

|α|≤m

Then !∨ f=

∑

α

=

aα ∂ δ0

|α|≤m

∑

aα x α .

|α|≤m

Therefore f is a polynomial, say of degree m. If m ≥ k, write f as f (x) =

∑

aα xα +

|α|≤k−1

aα xα = f1 (x) + f2 (x).

∑ k≤|α|≤m

f1 is a polynomial of degree at most k − 1, so Dkh ( f1 ) = 0 by part (a). It follows that Dkh ( f2 )(x) = 0 for all x, h ∈ Rn . Using the formula in part (a) we have that α1 αn k 0 = Dh ( f2 )(x) = ∑ aα ∑ ··· C(k, |β |) hβ xα−β β β n 1 k≤|α|≤m β ≤α |β |≥k

for all h, x ∈ Rn . Since h, x are arbitrary, and the preceding polynomial is identically equal to zero, it follows that all the coefficients aα = 0 for all α satisfying k ≤ |α| ≤ m. Therefore we have that f is a polynomial of degree at most k − 1. . [γ]+1 (c) If a polynomial P has degree at most [γ], then by part (a) we have Dh (P) = 0 for all h, thus P lies in Λγ . . Conversely, if a polynomial P of degree d lies in Λγ , suppose that d ≥ [γ] + 1 to reach a contradiction. Let R be the sum of monomials of degrees r with [γ] + 1 ≤ r ≤ d. Then if R(x) =

∑

cα x α

[γ]+1≤|α|≤d

then we have [γ]+1

Dh

[γ]+1

(P)(x) = Dh

(R)(x) =

cα

∑

It follows from this formula that

∑

β ≤α |β |≥[γ]+1

[γ]+1≤|α|≤d

[γ]+1

sup sup x∈Rn h∈Rn

|Dh

α1 αn ··· C([γ] + 1, |β |) hβ xα−β β1 βn

(P)(x)| = ∞. |h|γ

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35

.

This provides a contradiction, since P lies in Λγ . Thus we must have d ≤ [γ] which is what we needed to prove.

Exercise 1.4.2 (a) Show that for all continuous and bounded functions f we have

f ∞ ≤ f ≤ 3 f ∞ , L Λ L 0

hence the space (Λ0 (Rn ), k · kΛ0 ) can be identified with (L∞ (Rn ) ∩ C (Rn ), k · kL∞ ). (b) Given a continuous function f on Rn we define

f . ∞ = inf k f + ckL∞ : c ∈ C . L

.

Let L∞ (Rn ) be the space of equivalent classes of continuous functions whose difference is a constant, equipped with this norm. Show that for all continuous functions f on Rn we have

f . ∞ ≤ sup | f (x + h) − f (x)| ≤ 2 f . ∞ . L L x,h∈Rn

.

.

In other words, (Λ0 (Rn ), k · kΛ. ) can be identified with (L∞ (Rn ) ∩ C (Rn ), k · kL. ∞ ). 0

Solution. (a) Recall that a function f in Rn is said to be Lipschitz of order γ = 0 if it bounded, continuous, and | f (x + y) − f (x)| ≤ c for some constant c < ∞. In this case we have k f kΛ0 (Rn ) := k f kL∞ + sup sup | f (x + h) − f (x)| x∈Rn h∈Rn

and Λ0 (Rn ) := { f : Rn → C, continuous such that k f kΛ0 (Rn ) < ∞}. Obviously, we have k f kL∞ ≤ k f kΛ0 (Rn ) . Also, k f kΛ0 (Rn ) = k f kL∞ + sup sup (| f (x + h)| + | f (x)|) x∈Rn h∈Rn

≤ k f kL∞ + k f kL∞ + k f kL∞ = 3k f kL∞ . This yields,

.∞

k f kL∞ ≤ k f kΛ0 ≤ 3k f kL∞ .

(b) Here k f kL. ∞ = inf{k f + ckL∞ : c ∈ C} and L (Rn ) is the space of all functions f such that k f kL. ∞ < ∞. Now we identify all . . functions in L∞ whose difference is a constant. Let f1 , f2 ∈ L∞ such that f1 − f2 = c. Then we have, f1 = f2 + c. Thus k f2 kL.∞ ≤ k f2 + ckL∞ = k f1 kL∞ . Therefore, k f2 kL.∞ ≤ k f1 kL∞ . But k f1 kL. ∞ is the infimum of k f2 + ckL∞ , which gives k f1 kL. ∞ ≤ k f2 kL. ∞ . Similarly, we can show . that k f2 kL. ∞ ≤ k f1 kL. ∞ . Consequently, k f1 kL. ∞ = k f2 kL. ∞ . This gives that functions with constant difference have equal L∞ norms. Now we prove k f kL. ∞ ≤ sup | f (x + h) − f (x)| ≤ 2k f kL. ∞ . x,h∈Rn

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Consider | f (x + h) − f (x)| as a function of h and f (x) as a constant. Then we have, sup | f (x + h) − f (x)| = k f − f (x)kL∞ ≥ k f kL. ∞ . h

Consequently, it follows that

k f kL. ∞ ≤ sup | f (x + h) − f (x)| x,h

Now by approximation property of infimum we have, for a given ε > 0 there exists c ∈ C such that k f + ckL∞ ≤ k f kL. ∞ + ε. Then for x, h ∈ Rn | f (x + h) − f (x)| ≤ | f (x + h) − c| + | f (x) + c| ≤ k f − ckL∞ + k f − ckL∞ = 2k f − ckL∞ ≤ 2k f kL. ∞ + 2ε. Hence,

sup | f (x + h) − f (x) ≤ 2k f kL. ∞ + 2ε.

x,h∈Rn

Letting ε → 0 we have,

sup | f (x + h) − f (x) ≤ 2k f kL. ∞ .

x,h∈Rn

The preceding equations yield that

k f kL. ∞ ≤ sup | f (x + h) − f (x)| ≤ 2k f kL. ∞ . x,h∈Rn

.

.∞

This proves that Λ0 (Rn ) = L (Rn ) ∩ C (Rn ).

Exercise 1.4.3 Let f be a continuous function on Rn . (a) Prove the identity k+1

Dk+1 h ( f )(x) =

k+1−s

∑ (−1)

s=0

k+1 f (x + sh) s

for all x, h ∈ Rn and k ∈ Z+ ∪ {0}. + n (b) Prove that Dkh Dlh = Dk+l h for all k, l ∈ Z ∪ {0} and all h ∈ R . n (c) Prove that Dh1 Dh2 = Dh1 +h2 − Dh1 − Dh2 for all h1 , h2 ∈ R . (d) Suppose that |Dt2 ( f )(x)| ≤ C|t|M for all t, x ∈ Rn and some constants C, M > 0. Prove that |Dt Ds ( f )(x)| ≤ C0 (|t| + |s|)M for all t, s, x ∈ Rn and some other constant C0 that depends on C and M. Solution. (a) We use induction on k ∈ Z+ . By definition of the difference operator, it follows that it is true for k = 0, 1. Assume that the formula is true for k = m. That is, m+1 m+1−s m + 1 Dm+1 ( f )(x) = (−1) f (x + sh). ∑ h s s=0 We wish to establish the same formula for k = m + 1. (m+1)+1 Dh ( f )(x) = D1h Dm+1 ( f )(x) h

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37 m+1

m+1 f (x + sh) s s=0 m+1 m+1−s m + 1 = ∑ (−1) [ f (x + (s + 1)h) − f (x + sh)] s s=0 m+1 m+1−s m + 1 = ∑ (−1) f (x + (s + 1)h) s s=0 m+1 m+1−s m + 1 − ∑ (−1) f (x + sh) s s=0 m+2 m+1 m+2−s m + 1 m−s m + 1 = ∑ (−1) f (x + sh) + ∑ (−1) f (x + sh) s−1 s s=1 s=0 m+1 m−s m + 1 = f (x + (m + 2)h) + ∑ (−1) f (x + sh) s−1 s=1 m+1 m−s m + 1 f (x + sh) + (−1)m f (x) + ∑ (−1) s s=1 m+1 m+1 m+1 m−s = f (x + (m + 2)h) + ∑ (−1) + f (x + sh) + (−1)m f (x) s−1 s s=1 m+2 m−s m + 2 = ∑ (−1) f (x + sh) . s s=0

= D1h

∑ (−1)m+1−s

Thus the formula is true for k = m + 1 whenever it is true for k = m. Hence, by induction, the formula is true for all k ∈ Z+ ∪ {0}. This proves (a). (b) From the proof in (a) it follows that for every k ∈ Z+ ∪ {0} D1h Dkh = Dh1+k

for

every

k ∈ Z+ ∪ {0}.

Suppose by induction m+k k Dm h Dh = Dh

for

every

k ∈ Z+ ∪ {0}

is true for m = 0, 1, 2, . . . , l where l ≥ 1. Next, we wish to prove this for m = l + 1. (l+1)+k k 1 l k 1 l+k Dl+1 D = D D D . h h h h = Dh Dh = Dh h Thus it is true m = l + 1 whenever it is true for m = l. Therefore, by induction, it is true for every l ∈ Z+ ∪ {0}. Thus (b) follows. (c) We notice that Dh1 Dh2 ( f )(x) = f (x + h2 + h1 ) − f (x + h1 ) − ( f (x + h2 ) − f (x)) = Dh1 +h2 ( f )(x) − Dh1 ( f )(x) − Dh2 ( f )(x) , thus these expressions are equal. (d) Indeed, the hypothesis gives t + s M | f (x + t + s) − 2 f (x + t+s 2 ) + f (x)| ≤ C 2 and also

t − s M t−s ) − 2 f (x + s + ) + f (x + s)| ≤ C | f (x + s + 2 t−s . 2 2 2 Adding and using the triangle inequality |A − B| ≤ |A| + |B| we obtain the claimed inequality t + s M t − s M |Dt Ds ( f )(x)| = | f (x + t + s) − f (x + t) − f (x + s) + f (x)| ≤ C +C ≤ C0 (|t| + |s|)M , 2 2

as required.

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Exercise 1.4.4 For x ∈ R let

∞

f (x) =

k

∑ 2−k e2πi2 x . k=1

(a) Prove that f ∈ Λγ (R) for all 0 < γ < 1. (b) Prove that there is an A < ∞ such that sup | f (x + t) + f (x − t) − 2 f (x)| |t|−1 ≤ A ; x,t6=0

thus f ∈ Λ1 (R). (c) Show, however, that sup | f (t) − f (0)|t −1 = ∞ ; 0 0. Then we have | f (x + t) + f (x − t) − 2 f (x)| ∑∞ 2−k sin2 (π2k t) ≤ 4 k=1 , |t| t where t > 0. Now, for every t > 0, there exists an n ∈ N, such that 2n πt

0. Then k k k ∞ 2πi2k t − 1 | f (t) − f (0)| k0 −k e2πi2 t − 1 e2πi2 t − 1 k0 −k e2πi2 t − 1 ∞ −k e . = ∑2 + ∑ 2−k ≥ 2 − 2 ∑ ∑ t t t t t k=k +1 k=1 k=k +1 k=1 0

0

But the second sum is bounded by 2−k0 t −1 ≤ C0 while the imaginary part of the first sum is k0

∑ 2−k k=1

2 k0 2π2k t sin(2π2k t) ≥ ∑ 2−k π = 4k0 = 4([log(4t)−1 ] + 1) t t k=1

since 2π2k t ≤ π/2. Letting t → 0+ we deduce that | f (t) − f (0)|t −1 → ∞, hence f is not differentiable at zero.

Exercise 1.4.5 For 0 < a, b < ∞ and x ∈ R let ∞

gab (x) =

bk x

∑ 2−ak e2πi2 k=1

.

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Show that gab lies in Λ ab (R). bk Hint: Use the estimate |DLh (e2πi2 x )| ≤ C min 1, (2bk |h|)L with L = [a/b] + 1 and split the sum into two parts. Solution. First, we shall establish the following estimate: For L = [ ba ] + 1, bk x

|DLh (e2πi2

)| ≤ C min(1, (2bk |h|)L ).

bk

Let f (x) = e2πi2 x . Then | f 0 (x)| ≤ 2π min(1, 2bk ). Using the mean value theorem, we obtain, |e2πi2 min(2, | f 0 (ξ )h|) ≤ 2π min(1, (2bk |h|). Now, for any general L,

bk h

− 1| = | f (h) − f (0)| ≤

L L f (x + sh) |DLh ( f )(x)| = ∑ (−1)(L−s) s s=0 L bk L 2πi2bk sh e = e2πi2 x ∑ (−1)(L−s) s s=0 bk bk = e2πi2 x (e2πi2 h − 1)L ≤ (2π)L min(1, (2bk |h|)L ). Hence, the claim is true. ∞

Since a > 0, |gab (x)| ≤

∑ 2−ak < ∞. Hence, ||gab ||L∞ < ∞. Now for γ = ba , k=1 ∞

|DLh (gab )(x)| |h|γ

∑ 2−ak |DLh (e2πi2 ≤

bk x

)|

k=1

|h|γ ∞

∑ 2−akC min(1, (2bk |h|)L ) ≤

k=1

|h|γ ∞

≤ C ∑ 2−ak min(|h|−γ , 2bkL |h|L−γ ) k=1

= C min(|2bk h|−γ , |2bk h|L−γ ). Given |h| 6= 0, there exists a j ∈ Z such that 2−b j−1 < |h| ≤ 2−b j . Then for γ =

a b

< L,

|DLh (gab )(x)| ≤ C ∑ min(|2bk h|−γ , |2bk h|L−γ ) +C ∑ min(|2bk h|−γ , |2bk h|L−γ ) |h|γ k≥ j+1 k≤ j ≤C

∑ k≥ j+1

k≤ j

|DLh (gab )(x)| < ∞ and therefore, gab ∈ Λ ab (R). |h|γ x∈R h∈R\{0}

Thus, kgab kΛ a (R) = kgab kL∞ + sup sup b

2−(k− j−1)bγ +C ∑ 2(k− j)(L−γ) = C(a, b) < ∞.

Exercise 1.4.6 Let γ > 0 and let k = [γ]. (a) Use Exercise 1.4.3 (a) and (b) to prove that if |Dkh ( f )(x)| ≤ C|h|γ for all x, h ∈ Rn , then |Dk+l ( f )(x)| ≤ C2l |h|γ for all l ≥ 1. (b) Conversely, assuming that for some l ≥ 1 we have

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41

|Dk+l h ( f )(x)| < ∞, |h|γ x,h∈Rn

(0.0.5)

sup

.

show that f ∈ Λγ . Hint: Part (b): Use (1.4.10) but replace [γ] + 1 by k + l. Solution. (a) Using Exercise 1.4.3 (a) we have k

Dkh ( f )(x) =

k−s

∑ (−1)

s=0

k f (x + sh) s

for all x, h ∈ Rn and k ∈ Z ∪ {0}. Using the hypothesis (0.0.5) and the result of Exercise 1.4.3(b), one obtains l

l k |Dk+l h ( f )(x)| = |Dh (Dh ( f )(x))| ≤

l ∑ s |Dkh ( f )(x + sh)| s=0

l

l = C|h|γ (1 + 1)l = C2l |h|γ . ≤ C|h| ∑ s s=0 γ

(b) We want to show that k f kΛ. < ∞, where γ

|D[γ]+1 ( f )(x)| . |h|γ x∈R h∈Rn \{0}

k f kΛ. := sup γ

sup

To do this we choose a finite family of unit vectors {ur }r so that the annulus Ur = {ξ ∈ Rn :

1 2

1 1 ≤ |ξ | ≤ 2, ≤ |ξ · ur | ≤ 2}, 2 4

≤ |ξ | ≤ 2 is covered by the union of sets r = 1, 2, . . . , M(n).

Let us write M(n)

b= Ψ

∑ Ψb (r) ,

r=1

d (r) are smooth functions with supp Ψ b (r) ⊂ Ur . Letting hr := 1 2− j ur , one gets e2πiξ ·hr 6= 1 where Ψ 8 Therefore, h i∨ (r) d (r) (2− j ξ )(e2πξ ·hr − 1)−(k+l) (e2πξ ·hr − 1)k+l fb(ξ ) Ψ ∗f= Ψ

∀ j ∈ Z and 2− j ξ ∈ Ur .

2− j

is well defined for 2− j ξ ∈ Ur , where j ∈ Z. 1 d (r) (r) d (r) (2− j ξ )(e2πi2− j ξ · 8 ur − 1)−(k+l) . Since ζd is well defined and smooth with compact support, it follows that Let ζ2− j = Ψ 2− j (r)

(r)

Ψ2− j ∗ f = ζ2− j ∗ Dk+l − j 1 ( f ), 2

8 ur

2πiξ ·h − 1)k+l ]∨ . By our definition we have b where Dk+l h ( f ) = [ f (ξ )(e m

∆Ψ j ( f ) = Ψ2− j

∗f=

(r) ∑ Ψb2− j

! ∗ f.

r=1

Therefore, one has M(n)

k∆ Ψ j ( f )kL∞ ≤

.

Applying Theorem 1.4.6, we obtain that f ∈ Λγ .

(r)

( f )kL∞ ≤ C2− jγ . ∑ kζ2− j kL1 kDk+l 2− j 18 ur

r=1

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Exercise 1.4.7 Let Ψ and ∆ Ψ j be as in Theorem 1.4.6. Define a continuous operator Qt by setting Ψt (x) = t −nΨ (t −1 x) .

Qt ( f ) = f ∗Ψt , Show that all tempered distributions f satisfy

sup t −γ Qt ( f ) L∞ ≈ sup 2 jγ ∆ Ψ j ( f ) L∞ t>0

j∈Z

with the interpretation that if either term is finite, then it controls the other termby a constant multiple of itself. Ψ Ψ Ψ − j ≤ t ≤ 21− j . Hint: Observe that Qt = Qt (∆ Ψ j−2 + ∆ j−1 + ∆ j + ∆ j+1 ) when 2 Solution. Observe that when 2− j−1 ≤ t ≤ 2− j+1 , we have Ψ Ψ Ψ Ψ Qt = Qt (∆ Ψ j−2 + ∆ j−1 + ∆ j + ∆ j+1 + ∆ j+2 ).

Then, for 2− j−1 ≤ t ≤ 2− j+1 we have t −γ kQt ( f )kL∞ = t −γ kΨt ∗ f kL∞

= t −γ Ψt ∗ (Ψ − j−2 ∗ f +Ψ − j−1 ∗ f +Ψ − j ∗ f +Ψ − j+1 ∗ f +Ψ − j+2 ∗ f ) 2

2

2

2

2

2

≤ t −γ kΨt kL1 k

∑ Ψ2− j+i ∗ f kL∞

i=−2 2

≤ CkΨ kL1 t −γ

kΨ2− j+i ∗ f kL∞

∑

i=−2 2

≤ CkΨ kL1 2( j+1)γ

∑

kΨ2− j+i ∗ f kL∞

i=−2 2

= CkΨ kL1

∑

2(i+1)γ 2( j−i)γ k∆ Ψ j−i ( f )kL∞

i=−2

≤ CkΨ kL1 23γ

2

∑

∞ 2( j−i)γ k∆ Ψ ( f )k L j−i

i=−2

! 3γ

≤ 2 CkΨ kL1 = C1 sup 2

jγ

j∈Z

5 sup 2

jγ

j∈Z

k∆ Ψ j ( f )kL∞

k∆ Ψ j ( f )kL∞

where C1 = 5(23γ )CkΨ kL1 . Take the supremum over t > 0 to obtain sup t −γ kQt ( f )kL∞ ≤ C1 sup 2 jγ k∆ Ψ j ( f )kL∞ . t>0

j∈Z

Ψ − j−1 ≤ t ≤ 2− j+1 . Therefore, we have On the other hand, observe that ∆ Ψ j = ∆ j (Qt ), when 2 jγ Ψ jγ 2 jγ k∆ Ψ j ( f )kL∞ = 2 k∆ j (Qt f )kL∞ = 2 kΨ2− j ∗ (Qt f )kL∞

≤ 2 jγ kΨ2− j kL1 kQt ( f )kL∞ = 2 jγ kΨ kL1 kQt ( f )kL∞ ≤ 2γ t −γ kΨ kL1 kQt ( f )kL∞ = 2γ kΨ kL1 t −γ kQt ( f )kL∞ ≤ C2 sup t −γ kQt ( f )kL∞ , t>0

L∞

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43

where C2 = 2γ kϕkL1 . Take the supremum over j ∈ Z to obtain −γ sup 2 jγ k∆ Ψ j ( f )kL∞ ≤ C2 sup t kQt ( f )kL∞ . t>o

j∈Z

Therefore we conclude that sup t −γ kQt ( f )kL∞ ≈ sup 2 jγ k∆ Ψ j ( f )kL∞ t>0

j∈Z

for every f ∈ S 0 (Rn ).

Exercise 1.4.8 .

(a) Let 0 ≤ γ < 1 and suppose that ∂ j f ∈ Λγ for all 1 ≤ j ≤ n. Show that n

f .

Λγ+1

≤

∑ ∂ j f Λ.γ

j=1

.

and conclude that f ∈ Λγ+1 . . (b) Let γ ≥ 0. If we have ∂ α f ∈ Λγ for all multi-indices α with |α| = r, then there is an estimate

f . ≤ ∑ ∂ α f . , Λ Λ γ+r

γ

|α|=r

.

and thus f ∈ Λγ+r . (c) Use Corollary 1.4.8 to obtain that the estimates in both (a) and (b) can be reversed with the insertion of a multiplicative constant. Hint: Part (a): Write Z 1 n D2h ( f )(x) = ∑ ∂ j f (x + th + 2h) − ∂ j f (x + th + h) h j dt . 0 j=1

Part (b): Use induction. Solution. (a) Let x, h ∈ Rn with h 6= 0. By Proposition 1.4.5, we have [γ+1]+1 [γ]+1 Dh ( f )(x) = Dh Dh ( f ) (x) Z 1 n

=

[γ]+1

∑ h j ∂ j Dh

0 j=1 Z 1 n

=

∑ hj

0 j=1

( f )(x + th)dt

[γ]+1 Dh (∂ j f ) (x + th)dt.

Now, by the definition of ∂ j f Λ. we have γ

[γ]+1 γ Dy (∂ j f )(z) ≤ ∂ j f Λ.γ |y| , for all z, y ∈ Rn . In view of the preceding facts, with z := x + th and y := h, we have [γ+1]+1 ( f )(x) ≤ Dh

n

∑

Z 1 [γ]+1

j=1 0

Dh

(∂ j f )(x + th) h j dt

44

Contents n Z 1

≤

∑

∂ j f . |h|γ h j dt Λ

∑

∂ j f . |h|γ+1 dt Λ

γ

j=1 0 n Z 1

≤

γ

j=1 0

!

n

∑ ∂ j f Λ.γ

=

|h|γ+1 ,

j=1

and consequently, [γ+1]+1 ( f )(x) Dh

n

≤

γ+1

|h|

∑ ∂ j f Λ.γ .

j=1

∈ Rn

Since x, h with h 6= 0 are arbitrary, the required conclusion follows. (b) We proceed via induction . on r with the base case being handled by part (a). So, assume inductively that the result holds for r and suppose ∂ α f ∈ Λγ , for every multi-index α with |α| = r + 1. Then for each multi-index β with |β | = r and for each . . j = 1, ..., n, we have ∂ j ∂ β f ∈ Λγ . By part (a), ∂ β f ∈ Λγ+1 and n

β

∂ f .

≤

Λγ+1

β ∂ f

∂ ∑ j . , Λγ

j=1

for each multi-index β with |β | = r. Applying our induction hypothesis and using the preceding estimate, we obtain k f kΛ.

γ+(r+1)

= k f kΛ. (γ+1)+r

β ≤ ∑ ∂ f .

Λγ+1

|β |=r

≤

n

β ∂ f

∂ ∑ j .

∑

Λγ

|β |=r j=1

=

∑

|α|=r+1

k∂ α f kΛ. . γ

This completes the induction and proves the claimed conclusion. (c) For each α with |α| = r, let Cn,γ+r,α be the constant given in Corollary 1.4.8 such that k∂ α f kΛ. = k∂ α f kΛ. γ

= k∂ α f kΛ.

(γ+r)−r (γ+r)−|α|

≤ Cn,γ+r,α k f kΛ.

.

γ+r

By summing over all such α in the preceding equation, we obtain the conclusion with Cγ := ∑|α|=r Cn,γ+r,α .

Exercise 1.4.9 Introduce a difference operator D ( f )(x) = β

Z Rn

[β ]+1

|Dy

( f )(x)|2

|y|n+2β

1 2

dy

,

where β > 0. Show that for some constant c0 (n, β ) we have Z

β 2

D ( f ) 2 n = c0 (n, β ) L (R )

Rn

| fb(ξ )|2 |ξ |2β dξ

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45

.2

for all functions f in the Sobolev space Lβ (Rn ). Solution. By definition, kD

β

( f )k2L2 (Rn )

Z

=

Rn

Z

=

[β ]+1

|Dy

Z

Rn

( f )(x)|2

Rn

|y|n+2β

1

Z

dy |y|n+2β

[β ]+1

Rn

|Dy

! dy dx ( f )(x)|2 dx.

By Plancherel Theorem, Z

[β ]+1

Rn

|Dy

( f )(x)|2 dx =

Z Rn

2 [β ]+1 ( f ))∧ (ξ ) dξ , (Dy

and by definition, [β ]+1

(Dy

( f ))∧ (ξ ) = (e2πiy·ξ − 1)[β ]+1 fb(ξ ).

Hence, kD β ( f )k2L2 (Rn ) =

1

Z Rn

Z

=

Rn

2 2πiy·ξ − 1)[β ]+1 | fb(ξ )|2 dξ (e n R ! Z |e2πiy·ξ − 1|2([β ]+1) dy dξ . |y|n+2β Rn Z

dy |y|n+2β | fb(ξ )|2

Set Z

g(ξ ) :=

Rn

|e2πiy·ξ − 1|2([β ]+1) dy . |y|n+2β

We claim that g(ξ ) is integrable over Rn . It is true, for, when y is big, |e2πiy·ξ − 1|2([β ]+1) 22([β ]+1) ≤ , n+2β |y| |y|n+2β so g(ξ ) is integrable at ∞. When y is small, 2([β ]+1) 2πiy·ξ − 1 e |y|n+2β

≤

(2π|y · ξ |)2([β ]+1) |y|n+2β

=

(|y||ξ |)2([β ]+1) |y|n+2β

≤

|ξ |2([β ]+1) |y|2([β ]+1) |y|n+2β

=

|ξ |2([β ]+1) . |y|n−2+2(β −[β ])

Here |y|n−2+2(β −[β ]) < |y|n as β − [β ] < 1, thus g(ξ ) is integrable at 0. Now, take R an n × n orthogonal matrix (rotation), i.e., RRT = I, det R = 1. Then Z

g(Rξ ) =

Rn

|e2πiy·Rξ − 1|2([β ]+1) dy = |y|n+2β

Z Rn

|e2πiR

T y·ξ

− 1|2([β ]+1)

|y|n+2β

dy.

Let u = RT y, then du = dy, |u| = |y| and we have Z

g(Rξ ) =

Rn

|e2πiu·Rξ − 1|2([β ]+1) du = g(ξ ). |u|n+2β

(0.0.6)

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Also, for λ > 0, Z

g(λ ξ ) =

Rn

|e2πiy·λ ξ − 1|2([β ]+1) dy = |y|n+2β

Z Rn

|e2πiλ y·ξ − 1|2([β ]+1) dy. |y|n+2β

Let u = λ y, then du = λ n dy, |u| = λ |y| and we obtain Z

g(Rξ ) =

Rn

|e2πiu·ξ − 1|2([β ]+1) n+2β 1 λ du = λ 2β g(ξ ). |u| λn

(0.0.7)

So, for any ξ 6= 0, by (0.0.6) and (0.0.7) we have ξ ξ g(ξ ) = g |ξ | = |ξ |2β g(e1 ), = |ξ |2β g |ξ | |ξ | where e1 = (1, 0, ..., 0). Since g(e1 ) depends only on n and β , we obtain g(ξ ) = c0 (n, β )|ξ |2β . Therefore, kD β ( f )k2L2 (Rn ) = c0 (n, β )

Z Rn

| fb(ξ )|2 |ξ |2β dξ .

Moreover, by Plancherel theorem, Z Rn

∨

2

2 | fb(ξ )|2 |ξ |2β dξ = |ξ |β fb(ξ ) L2 = |ξ |β fb(ξ ) L2 = k f k2L2 . β

Consequently,

β 2

D ( f ) 2 n = c0 (n, β )k f k2 2 , L (R ) L β

which shows that the operator

Dβ

is an isometry between

L2

and

Lβ2 .

Exercise 1.4.10 Suppose that a continuous function f (x) on the real line satisfies: |D2h ( f )(x)| ≤ C |h|1+γ for some γ ∈ (0, 1) and all x, h ∈ R. Follow the steps below to show, without appealing to Theorem 1.4.6, that f is differentiable. (a) The hypothesis implies that |D2h ( f )(x) − 2Dh ( f )(x + ih)| ≤ C |h|1+γ for i = 0, 1. Iterate to obtain |D2 j h ( f )(x) − 2 j Dh ( f )(x + ih)| ≤ Cγ |2 j h|1+γ for all 0 ≤ i < 2 j . (b) Given a positive integer m find a j such that 2 j−1 ≤ m < 2 j and use the estimate in part (a) to conclude that |Dh ( f )(x + mh) − Dh ( f )(x)| ≤ 2γ+1Cγ |mh|γ |h| for all integers m. (c) Use the result of Exercise 1.4.3(d) and a telescoping argument to conclude that f (x + nh) − f (x) = n( f (x + h) − f (x)) + O(|nh|1+γ ). Conclude that for all h 6= 0 and n ∈ Z \ {0} we have f (x + h) − f (x) f (x + h/n) − f (x) ≤ C00 |h|γ . − h h/n (d) Deduce that for all nonzero rationals h, h0 and all x ∈ R one has f (x + h) − f (x) f (x + h0 ) − f (x) − = O(|h|γ + |h0 |γ ). h h0 Use the continuity of f to extend this identity to all reals and obtain that

f (x+h)− f (x) h

satisfies the Cauchy criterion.

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47

Solution. (a) Obviously we have | f (x + 2h) − f (x) − 2( f (x + h) − f (x))| ≤ C |h|1+γ and also | f (x + 2h) − f (x) − 2( f (x + h + h) − f (x + h))| ≤ C |h|1+γ Write a given i with 0 ≤ i < 2 j as i = i j−1 2 j−1 + · · · + i1 2 + i0 where i0 , . . . , i j−1 ∈ {0, 1}. Then D2 j h ( f )(x) − 2 j Dh ( f )(x + ih) = D2 j h ( f )(x) − 2D2 j−1 h ( f )(x + i j−1 2 j−1 h) = 2D2 j−1 h ( f )(x + i j−1 2 j−1 h) − 22 D2 j−2 h ( f )(x + i j−1 2 j−1 h + i j−2 2 j−2 h) .. = . = 2 j−1 D2h ( f )(x + i j−1 2 j−1 h + · · · + i1 2h) − 2 j Dh ( f )(x + ih) This expression is bounded by C[|2 j−1 h|1+γ + 2|2 j−2 h|1+γ + 22 |2 j−3 h|1+γ + · · · + 2 j−1 |h|1+γ ] ≤ C

|2 j h|1+γ . 21+γ − 2

(b) Now given a positive integer m find a j such that 2 j−1 ≤ m < 2 j and use that |D2 j h ( f )(x) − 2 j Dh ( f )(x + mh)| ≤ Cγ |2 j h|1+γ and also that |D2 j h ( f )(x) − 2 j Dh ( f )(x)| ≤ Cγ |2 j h|1+γ to conclude that |Dh ( f )(x + mh) − Dh ( f )(x)| ≤ 2Cγ |2 j h|γ |h| ≤ 2Cγ 2γ |mh|γ |h|. (c) We now add the following identities f (x + nh) − f (x + (n − 1)h) = f (x + h) − f (x) + O(|nh|γ |h|) f (x + (n − 1)h) − f (x + (n − 2)h) = f (x + h) − f (x) + O(|(n − 1)h|γ |h|) .. . f (x + 2h) − f (x + h) = f (x + h) − f (x) + O(|2h|γ |h|) f (x + h) − f (x) = f (x + h) − f (x)

to obtain f (x + nh) − f (x) = n( f (x + h) − f (x)) + O(|nh|1+γ ) . Replacing h by h/n we deduce f (x + h) − f (x) f (x + h/n) − f (x) = + O(|h|γ ) . h h/n (d) Given k/` and k0 /`0 rationals k, `, k0 , `0 ∈ Z and `0 6= 0 6= ` we can write: f (x + k/` f (x + k/`) − f (x) k`0 ) − f (x) = + O(|k/`|γ ) . k/` k/` 0 k`

Also

0

0

f (x + kk/` f (x + k0 /`0 ) − f (x) 0 ` ) − f (x) + O(|k0 /`0 |γ ) . = 0 /`0 0 0 k k /` 0 k`

Thus

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Contents

f (x + k/`) − f (x) f (x + k0 /`0 ) − f (x) = O(|k/`|γ ) + O(|k0 /`0 |γ ) . − k/` k0 /`0 Using the continuity of f we extend this identity to all h and h0 nonzero reals: f (x + h) − f (x) f (x + h0 ) − f (x) − = O(|h|γ ) + O(|h0 |γ ) . h h0 It follows that the sequence

f (x+h)− f (x) h

is Cauchy as h → 0 and thus f is differentiable.

Chapter 2. Hardy Spaces, Besov Spaces, and Triebel-Lizorkin Spaces

2.1. Hardy Spaces Exercise 2.1.1 Prove that if v is a bounded tempered distribution and h1 , h2 are in S (Rn ), then (h1 ∗ h2 ) ∗ v = h1 ∗ (h2 ∗ v). Solution. Since v is bounded,we have h2 ∗ v ∈ L∞ (Rn ). Now for f ∈ S (Rn ), we have by definition h(h1 ∗ h2 ) ∗ v, f i = hv, (h^ 1 ∗ h2 ) ∗ f i. e e We now claim that (h^ 1 ∗ h2 ) ∗ f = h2 ∗ (h1 ∗ f ). Indeed, (h^ 1 ∗ h2 ) ∗ f (x) =

Z Rn

(h^ 1 ∗ h2 )(x − y)dy

Z

(h1 ∗ h2 )(y − x)dy Z = h1 (y − x − z)h2 dz f (y)dy Rn Z Z = h2 (z) h1 (y − x − z) f (y)dy dz Rn Z Z = h2 (z) he1 (x + z − y) f (y)dy dz =

Rn

Z

Rn

Z

=

Rn

Z

=

Rn

Z

=

Rn

h2 (z)(he1 ∗ f )(x + z)dz h2 (u − x)(he1 ∗ f )(u)du

( using x + z = u)

he2 (x − u)(he1 ∗ f )(u)du

= he2 ∗ (he1 ∗ f )(x) e e Therefore, we have (h^ 1 ∗ h2 ) ∗ f = h2 ∗ (h1 ∗ f ). Now we have, h(h1 ∗ h2 ) ∗ v, f i = hv, (h^ 1 ∗ h2 ) ∗ f i = hv, he2 ∗ (he1 ∗ f )i

49

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Contents

= hh2 ∗ v, he1 ∗ f i = hh1 ∗ (h2 ∗ v), f i Thus for every f ∈ S (Rn ), we have h(h1 ∗ h2 ) ∗ v, f i = hh1 ∗ (h2 ∗ v), f i. This actually proves that (h1 ∗ h2 ) ∗ v = h1 ∗ (h2 ∗ v).

Exercise 2.1.2 (a) Show that the H 1 norm remains invariant under the L1 dilation ft (x) = t −n f (t −1 x). (b) Show that the H p norm remains invariant under the L p dilation t n−n/p ft (x) interpreted in the sense of distributions. Solution. (a) Here ft (x) =

1 tn

f ( xt ). Now by using

k ft kL1 =

Z

y t

= u and dy = t n du we have,

sup |(Ps ∗ ft )(x)|dx

Rn s>0

Z

Z sup

Ps (x − y) ft (y)dy dx n n R s>0 R Z Z y 1 dy dx = sup Ps (x − y) n f n n t t R s>0 R Z Z 1 = sup Ps (x − tu) n f (u)t n du dx n n t R s>0 R Z Z = sup Ps (x − tu) f (u)du dx Rn s>0 Rn Z Z x − tu 1 = sup P f (u)du dx n s s Rn s>0 Rn x Z Z 1 t −u = sup P f (u)du dx s n n n s R s>0 R t Z Z 1 1 x t −u = sup P f (u)du dx s n s n n n R s>0 R t (t ) t Z Z 1 x s = sup P − u f (u)du dx n t t Rn s>0 Rn t Z Z 1 x s = sup P − u f (u)du dx n t Rn t s>0 Rn t Z Z 1 x s ∗ f )( ) dx (P = sup n t Rn t s >0 Rn t t Z Z x sup (Ps ∗ f )(w) dw (using w = = t Rn s >0 Rn t t Z Z s = sup (Pr ∗ f )(w) dw (using r = ) n n t R r>0 R =

= k f kH 1 . n

(b) We show, k f kH p = kt n− p ft kH p . We have Ps ∈ L1 , using this we first claim that, x

t n δ t τ −x Ps = τ − t Pst .

and dx = t n dw)

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51

Then, t n δ t τ −x Ps (w) = t n δ t Ps (w + x) = t n Ps (tw + x) tw + x 1 = tn n P s s 1 tw + x = s nP (t ) s x = Pst (w + ) t − xt s = τ Pt (w). x

This shows that t n δ t τ −x Ps = τ − t Pst . Using this result we have, n

kt n− p ft kH p =

p 1p n− n sup |(Ps ∗ t p ft )(x)| dx

Z

Rn s>0

Z =

Rn

Z =

Rn

n p 1p n− p sup |(P ∗ f )(x)| t dx s t s>0

n p 1p n− p −x sup |h ft , τ Ps i| dx t s>0

n p 1p 1 . −x n− p = sup |h n f ( ), τ Ps i| dx t n t t R s>0 Z p 1p . −x − np = t sup |h f ( ), τ Ps i| dx n t R s>0 Z p 1p 1 − np −x = t sup |hδ t f , τ Ps i| dx Z

Rn

Z =

Rn

Z =

Rn

s>0

n p 1p −p n t −x t sup |h f ,t δ τ Ps i| dx s>0

n p 1p x −p t sup |h f , τ − t Pst i| dx s >0 t

1 n p x p −p = t sup |(Pxt ∗ f )( )| dx s >0 t Rn t Z 1 p p − np n = (using change of variables) t sup |(Pxt ∗ f )(u)| t du Z

Rn

Z =

Rn

s >0 t

p 1p = k f kH p . sup |(Pxt ∗ f )(u)| du s >0 t

Exercise 2.1.3 √

Show that the continuous function ψ(s) =

1

e 1 − 22 (s−1) 4 π se

sin

√ 1 2 4 2 (s − 1)

defined on [1, ∞) satisfies

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( Z ∞ 1 k s ψ(s) ds = 0 1

if k = 0, if k = 1, 2, 3, . . . . √ 2

√ 2

3 4 k−1 e− 2 z ei 2 z integrated over the boundary of the domain formed Solution. Consider the analytic function F(z) = 4e π z (z + 1) by the disc of radius R intersected with the quadrant Re z ≥ 0, Im z ≥ 0. Set

IRk =

4e π

Z R

t 3 (t 4 + 1)k−1 e−

√ 2 2 t

ei

√ 2 2 t

dt

0

and note that the integral over the part of the contour that lies on the imaginary axis is equal to −IRk . The integral over the circular part of this contour can be evaluated by the parametrization z = Reiθ , which produces an expression bounded by 4e π

Z π/4

R3 (R4 + 1)k−1 e−

√ 2 2 R cos θ

e−

√ 2 2 R sin θ

Rdθ → 0

0

as R → ∞. Note that when k ≥ 1, then F has no singularities and the integral of √ F over the contour described above is zero. √ √ When k = 0, we compute the residue of F(z) at the simple pole i = 22 + i 22 . We have √ √ √ √ 4e z − i 3 − 2 z i 2 z 2 e 2 = 2i . 2πi Res(F, i) = 2πi lim z e √ 4 z→ i π z + 1 Letting R → ∞ and using Cauchy’s residue formula we deduce that 2i Im (I∞k ) = I∞k − I∞k + 0 But

Z ∞

sk ψ(s) ds =

1

e π

Z ∞

(s + 1)k−1 e−

√ 1 2 4 2 s

sin

0

√ 1 2 4 ds 2 s

( 2i when k = 0 = 0 when k ≥ 1.

=

4e π

Z ∞

t 3 (t 4 + 1)k−1 e−

0

√ 2 2 t

sin

√ 2 2 t dt

= Im (I∞k ),

so the conclusion follows.

Exercise 2.1.4 Let Pt be the Poisson kernel. Show that for any bounded tempered distribution f we have Pt ∗ f → f

in S 0 (Rn ) as t → 0.

Hint: Fix a smooth function φ whose Fourier transform is equal to 1 in a neighborhood of zero. Show that Pt ∗ (φ ∗ f ) → φ ∗ f bt (1 − φb) fb → (1 − φb) fb in S 0 (Rn ) as t → 0. in S 0 (Rn ) and that P Solution. To show that Pt ∗ (φ ∗ f ) → φ ∗ f in S 0 (Rn ) as t → 0, we notice that Pt ∗ (φ ∗ f ) is a (uniformly in t) bounded function which converges pointwise (in fact, uniformly on compact subsets of Rn ) to φ ∗ f , since Pt is an approximate identity and φ ∗ f is a smooth function. The Lebesgue dominated convergence theorem yields that hPt ∗ (φ ∗ f ), ψi → hφ ∗ f , ψi for any ψ in S (Rn ). bt φb fb → φb fb in S 0 (Rn ) as t → 0. This implies that P bt (1 − φb) fb → (1 − φb) fb in S 0 (Rn ), it will suffice to prove that for all functions ψ in S (Rn ) we have To show that P bt (1 − φb) fb, ψi b → h(1 − φb) fb, ψi b hP which is equivalent to bt (1 − φb)ψi b → h fb, (1 − φb)ψi b . h fb, P

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bt (1 − φb)ψ b → (1 − φb)ψ b in S (Rn ). So, given multiindices α and β , we need The last assertion is a consequence of the fact that P to prove that b )ξ β → 0 sup ∂ α (e−2πt|ξ | − 1)(1 − φb(ξ ))ψ(ξ ξ ∈Rn

ξ

as t → 0. This function is supported in |ξ | ≥ c for some c > 0 and if some derivative falls on e−2πt|ξ | − 1 then a factor of t appears and this tends to zero. If no derivative falls on e−2πt|ξ | − 1, then by the mean value theorem, |e−2πt|ξ | − 1| ≤ 2πt|ξ | and the factor of t that appears together with the rapid decay of ψ at infinity yield the assertion. bt fb → φb fb+ (1 − φb) fb = fb in S 0 (Rn ) as t → 0, and thus the required conclusion Combining these two facts we deduce that P follows.

Exercise 2.1.5 Fix a smooth radial nonnegative compactly supported function θ on Rn such that θ = 1 on the unit ball and vanishing outside n+1 the ball of radius 2. Set Φ (k) (x) = θ (x) − θ (2x) (2−2k + |x|2 )− 2 for k ≥ 1. Prove that for all bounded tempered distributions f and for all t > 0 we have ∞ Γ ( n+1 2 ) Pt ∗ f = (θ P)t ∗ f + 2−k (Φ (k) )2k t ∗ f , ∑ n+1 π 2 k=1 n+1

n+1

2 (1 + |x|2 ) 2 is the Poisson kernel. where the series converges in S 0 (Rn ). Here P(x) = Γ ( n+1 2 )/π n Hint: Fix a function φ ∈ S (R ) whose Fourier transform is equal to 1 in a neighborhood of zero and prove the required conclusion for φ ∗ f and for (δ0 − φ ) ∗ f . In the first case use the Lebesgue dominated convergence theorem and in the second case the Fourier transform.

Solution. We begin with the identity ∞ P(x) = P(x)θ (x) + ∑ θ (2−k x)P(x) − θ (2−(k−1) x)P(x) k=1

= P(x)θ (x) +

Γ ( n+1 2 ) π

n+1 2

∞

∑ 2−k k=1

θ ( · ) − θ (2( · )) (2−2k + | · |2 )

n+1 2

(x) 2k

which is valid for all x ∈ Rn . Fix a function φ ∈ S (Rn ) whose Fourier transform is equal to 1 in a neighborhood of zero. Then f = φ ∗ f + (δ0 − φ ) ∗ f and we also have Pt ∗ f = Pt ∗ φ ∗ f + Pt ∗ (δ0 − φ ) ∗ f . Given a function ψ in S (Rn ) we need to show that N

Γ ( n+1 2 ) (θ P)t ∗ φ ∗ f , ψ + 2−k (Φ (k) )2k t ∗ φ ∗ f , ψ → hPt ∗ φ ∗ f , ψi ∑ n+1 π 2 k=1

and

N

Γ ( n+1

2 ) (θ P)t ∗ (δ0 − φ ) ∗ f , ψ + 2−k (Φ (k) )2k t ∗ (δ0 − φ ) ∗ f , ψ → hPt ∗ (δ0 − φ ) ∗ f , ψi ∑ n+1 π 2 k=1

as N → ∞. The first of these claims is equivalent to N

Γ ( n+1

2 ) φ ∗ f , ψ ∗ (θ P)t + 2−k φ ∗ f , ψ ∗ (Φ (k) )2k t → hφ ∗ f , ψ ∗ Pt i ∑ n+1 π 2 k=1

as N → ∞. Here φ ∗ f ∈ L∞ and the actions h·, ·i are convergent integrals in all three cases. This claim will be a consequence of the LDCT since: N Γ ( n+1 2 ) ψ ∗ (θ P)t + 2−k ψ ∗ (Φ (k) )2k t → ψ ∗ Pt ∑ n+1 2 π k=1

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pointwise (which is also a consequence of the LDCT) as N → ∞ and hence the same is true after multiplying by the bounded function φ ∗ f and also N Γ ( n+1 −k (k) 2 ) |φ ∗ f | ψ ∗ (θ P)t + 2 ψ ∗ (Φ )2k t ≤ |φ ∗ f | (|ψ| ∗ Pt ) ∈ L1 (Rn ). ∑ n+1 π 2 k=1 We now turn to the corresponding assertion where φ is replaced by δ − φ . Using the Fourier transform, this assertion is equivalent to N Γ ( n+1

2 ) \ (k) ) k (1 − φ [ b)ψ bt (1 − φb)ψi b + b → h fb, P b fb, (θ P)t (1 − φb)ψ 2−k fb, (Φ ∑ 2 t n+1 π 2 k=1 Since fb ∈ S 0 (Rn ), this assertion will be a consequence of the fact that N Γ ( n+1 2 ) \ (k) ) k (1 − φ [ b)ψ bt (1 − φb)ψ b+ b→P b (θ P)t (1 − φb)ψ 2−k (Φ ∑ 2 t n+1 π 2 k=1

in S (Rn ). It will therefore be sufficient to show that for all multiindices α and β we have N α Γ ( n+1 −k \ β 2 ) (k) [ b b b (θ P)t (ξ ) + 2 (Φ )2k t (ξ ) − Pt (ξ ) (1 − φ (ξ ))ψ(ξ )ξ → 0 sup ∂ξ ∑ n+1 ξ ∈Rn π 2 k=1 The term inside the curly brackets is equal to the Fourier transform of (θ (2−N x) − 1)Pt (x), which is Z Rn

(e−2πt|ξ −ζ | − e−2πt|ξ | )2Nn θb(2N ζ ) dζ

since θb has integral 1. Note that |ξ | ≥ c > 0 since 1 − φb(ξ ) = 0 in a neighborhood of zero. Let εN > 0. First consider the integral Z |ξ −ζ |>εN

(e−2πt|ξ −ζ | − e−2πt|ξ | )2Nn θb(2N ζ ) dζ .

Both exponentials are differentiable in this range and differentiation gives γ

∂ξ Z

= |ξ −ζ |>εN

Z |ξ −ζ |>εN

(e−2πt|ξ −ζ | − e−2πt|ξ | )2Nn θb(2N ζ ) dζ

ξ ξ −ζ −2πt|ξ −ζ | −2πt|ξ | Nn b N 1 1 − Qγ |ξ | , |ξ | e 2 θ (2 ζ ) dζ , Qγ |ξ −ζ | , |ξ −ζ | e

where Qγ is a polynomial of the following n + 1 variables Qγ

ξ ξ 1 ξn 1 1 , = Qγ ,..., , |ξ | |ξ | |ξ | |ξ | |ξ |

that depends on γ. Note that Qγ ξ −ζ , 1 e−2πt|ξ −ζ | − Qγ ξ , 1 e−2πt|ξ | ≤ C |ζ | . |ξ −ζ | |ξ −ζ | |ξ | |ξ | Thus the integral is bounded by Z |ξ −ζ |>εN

C 2Nn |ζ |θb(2N ζ ) dζ = C0 2−N ,

which tends to zero as N → ∞. Now consider the integral γ

∂ξ

Z |ξ −ζ |≤εN

(e−2πt|ξ −ζ | − e−2πt|ξ | )2Nn θb(2N ζ ) dζ = ∂ξ

γ

Z |ζ |≤εN

(e−2πt|ζ | − e−2πt|ξ | )2Nn θb(2N (ξ − ζ )) dζ

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in which |ξ | ≥ c > 0. We obtain another expression which is bounded by C εNn 2N|γ| ≤ C εNn 2N|α| , which tends to zero if we pick εN = 2−N|α|/n /N. Finally o α−γ n β 0 we have ∞ Γ ( n+1 2 ) Pt ∗ f = (θ P)t ∗ f + 2−k (Φ (k) )2k t ∗ f , ∑ n+1 π 2 k=1 n+1

n+1

2 (1 + |x|2 ) 2 is the Poisson kernel. where the series converges in S 0 (Rn ). Here P(x) = Γ ( n+1 2 )/π n Hint: Fix a function φ ∈ S (R ) whose Fourier transform is equal to 1 in a neighborhood of zero and prove the required conclusion for φ ∗ f and for (δ0 − φ ) ∗ f . In the first case use the Lebesgue dominated convergence theorem and in the second case the Fourier transform.

Exercise 2.1.6 Let 0 < p < ∞ be fixed. Show that a bounded tempered distribution f lies in H p if and only if the nontangential Poisson maximal function M1∗ ( f ; P)(x) = sup sup |(Pt ∗ f )(y)| t>0 y∈Rn |y−x|≤t

p ∗ lies in L , and in this case we have k f kH p ≈ kM1 ( f ;∗P)kL p . Hint: Observe that M( f ; P) can be replaced with M1 ( f ; P) in the proof of part (e) of Theorem 2.4.4. Solution. One inequality trivially holds:

k f kH p = kM( f ; P)kL p ≤ C1 kM1∗ ( f ; P)kL p . To obtain the converse inequality, we fix an f ∈ H p (Rn ). Then Pt ∗ f = (θ P)t ∗ f +

Γ ( n+1 2 ) π

n+1 2

∞

∑ 2−k (Φ (k) )2kt ∗ f , k=1

where the series converges in S 0 (Rn ) and Φ k (x) = (θ (x) − θ (2x))

1 (2−2k + |x|2 )

n+1 2

and θ is as in Theorem 2.4.4(e). We note that for some fixed constant c0 = c0 (n, N), the functions c0 θ P and c0 Φ (k) lie in FN uniformly in k = 1, 2, 3, .... Combining these facts, we conclude that sup sup |(Pt ∗ f )(x)| ≤ sup sup |((θ P)t ∗ f )(x)| + t>0 y∈Rn |y−x|≤t

t>0 y∈Rn |y−x|≤t

≤ C2 (n, N)MN ( f )(x). This yields the pointwise estimate

∞ 1 Γ ( n+1 2 ) 2−k sup sup |(c0 Φ (k) )2k t ∗ f (x)| ∑ n+1 c0 π 2 k=1 t>0 y∈Rn |y−x|≤t

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M1∗ ( f ; P)(x) = sup sup |(Pt ∗ f )(y)| ≤ C2 (n, N)MN ( f )(x) y∈Rn |y−x|≤at

t>0

which implies the norm estimate kM1∗ ( f ; P)kL p ≤ C2 (n, N)kMN ( f )kL p ≤ C3 (n, N)k f kH p .

Exercise 2.1.7 (a) Let 1 < q ≤ ∞ and let g in Lq (Rn ) be a compactly supported function with integral zero. Show that g lies in the Hardy space H 1 (Rn ). (b) when Lq is replaced by L log+ L. Prove the same conclusion ∞ Hint: Part (a): Pick a C0 function Φ supported in the unit ball with nonvanishing integral and suppose that the support of g is contained in the ball B(0, R). For |x| ≤ 2R weR have that M( f ; Φ)(x)≤ CΦ M(g)(x), and since M(g) lies in Lq , it also lies in L1 (B(0, 2R)). For |x| > 2R, write (Φt ∗ g)(x) = Rn Φt (x − y) − Φt (x) g(y) dy and use the mean value theorem to estimate this expression by t −n−1 k∇ΦkL∞ kgkL1 ≤ |x|−n−1CΦ kgkLq , since t ≥ |x − y| ≥ |x| − |y| ≥ |x|/2 whenever |x| ≥ 2R and |y| ≤ R.Thus M( f ; Φ) lies in L1 (Rn ). Part (b): You may use Exercise 2.1.4(a) in [161] to deduce that M(g) is integrable over B(0, 2R). Solution. (a) According to Theorem 2.4.4, to show g lies in H 1 (Rn ), it is equivalent to show that M(g; Φ) lies in L1 (Rn ). Pick a C0∞ function Φ supported in the unit ball with nonvanishing integral and, without loss of generality, suppose that the support of g is contained in the ball B(0, R). For |x| ≤ 2R, we have M(g; Φ)(x) ≤ kΦkL1 M(g)(x). Notice that kM(g)kLq ≤ Cq kgkLq , thus M(g) lies in Lq as g lies in Lq , and hence M(g) also lies in L1 (B(0, 2R)). Let CΦ := kΦk1 , then we have Z Z |M(g)(x)|dx < ∞. |M(g; Φ)(x)|dx ≤ CΦ B(0,2R)

B(0,2R)

For |x| > 2R, consider that (Φt ∗ g)(x) =

Z ZR

n

ZR

n

ZR

n

= =

Φt (x − y)g(y)dy Φt (x − y)g(y)dy + Φt (x)

Z Rn

g(y)dy

(Φt (x − y) − Φt (x)) g(y)dy

= B(0,R)

(Φt (x − y) − Φt (x)) g(y)dy.

Therefore, by the mean value theorem, whenever |y| ≤ R and t ≥ |x − y| ≥ |x| − |y| ≥ 12 |x|, we have |Φt ∗ g(x)| ≤

Z B(0,R)

t −n k∇ΦkL∞ |ty||g(y)|dy

≤ t −(n+1) k∇ΦkL∞ RkgkL1 ≤

Rk∇ΦkL∞ kgkL1 . 2n+1 |x|n+1

Hence, Z Rn \B(0,2R)

|Φt ∗ g(x)|dx ≤

Rk∇ΦkL∞ kgkL1 2n+1

Z Rn \B(0,2R)

dx |x|n+1

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= Cn k∇ΦkL∞ kgkL1 ≤ CnCΦ kgkLq . Thus

Z Rn \B(0,2R)

|M(g; Φ)(x)|dx < ∞

as kgkLq < ∞. So we conclude that Z Rn

|M(g; Φ)(x)|dx < ∞,

that is, M(g; Φ) lies in L1 (Rn ) and therefore g ∈ H 1 (Rn ). (b) According to Exercise 2.1.4(c), if g is integrable and supported in a ball B, then M(g) is integrable over B if and only if Z

|g(x)| log(1 + |g(x)|)dx < ∞.

B

Realize that g ∈ L log+ L means B |g(x)| log(1 + |g(x)|)dx < ∞, we conclude that M(g) ∈ L1 (B(0, R)). Then similarly as part (a), we conclude that M(g; Φ) ∈ L1 (Rn ). R

Exercise 2.1.8 Show that for every integrable function g with mean value zero and support inside a ball B, we have M(g; Φ) ∈ L p ((3B)c ) for p > n/(n + 1). Here Φ is in S . Solution. Suppose that B ⊂ B(0, R) for some R > 0. We have (Φt ∗ g)(x) =

Z Rn

Z

=

Rn

t −n Φ(t −1 (x − y))g(y)dy t −n [Φ(t −1 (x − y)) − Φ(t −1 x)]g(y)dy =: It (x).

For all x ∈ (3B)c , we have Z

It (x) =

Rn

≤R

t −n

Z 1

h∇Φ(t −1 x − t −1 ys), −t −1 yidsg(y)dy

0

Z

t −n−1 max |∇Φ(t −1 x − t −1 ys)|g(y)dy.

B

0≤s≤1

Here we have used the fact that |y| ≤ R, ∀y ∈ B. Since Φ ∈ S , ∇Φ ∈ S . Thus, there exists C > 0 such that |x|n+1 |∇Φ(x)| ≤

C , R

∀x ∈ Rn \(3B).

Thus, |It (x)| ≤ C

Z B

t −n−1 max |t −1 (x − ys)|−(n+1) |g(y)|dy 0≤s≤1

2 ≤ C max |x − ys|−(n+1) kgkL1 ≤ C|x|−(n+1) kgkL1 , 0≤s≤1 3

∀x ∈ (3B)c .

Hence, M(Φ, g) ≤ C1 |x|−(n+1) kgkL1 , This implies

∀x ∈ Rn \(3B),

M(Φ, g) p ≤ C1p |x|−p(n+1) kgkLp1 ,

2 C1 = C. 3

∀x ∈ Rn \(3B).

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Therefore, M(Φ, g) ∈ L p , for all p(n + 1) > n.

Exercise 2.1.9 Show that the space of all Schwartz functions whose Fourier transform is supported away from a neighborhood of the origin is dense in H p . Hint: Use the square function characterization of H p . Solution. We know that for f ∈ H p k f kH p

!1 2

2 ∑ |∆ j ( f )|

j∈Z

≈

0 there exists j0 ∈ Z such that !p 2

Z Rn

∑

2

|∆ j f |

≥

Rn

∑

2

|∆ j f |

!p 2

Z Rn

∑ |∆ j f |2

Rn

dx +

!p 2

∑ |∆ j f |2

dx →

j≥ j0

2

∑ |∆ j f |2

Rn

dx

j> j0

!p 2

Z

∑ |∆ j f |2

Rn

This implies that

dx .

j∈Z

!p 2

Z Rn

!p

Z

j≤ jo

From this we observe that Z

dx

j≤ j0

∑

2

|∆ j f |

→0

j≤ j0

This forces our assertion to be true. Fix this j0 .Finally,

!1

2

2

p k f − fε kH = ∑ |∆ j ( f − fε )|

+

p

j≥ j0 L

!1 2

2 ∑ |∆ j ( f − fε )|

.

p j≤ j0 L

By the choice of j0 we have the control over the second norm on the right. For j ≥ 0 we observe that f and fε coincide for j ≥ j0 by construction of fε . That is, fb ≈ fbε . So the first norm on the right vanishes. This completes the proof.

Exercise 2.1.10 (a) Suppose that f ∈ H p (Rn ) for some 0 < p ≤ 1 and Φ in S (Rn ). Then show that for all t > 0 the function Φt ∗ f belongs to Lr (Rn ) for all p ≤ r ≤ ∞. Find an estimate for the Lr norm of Φt ∗ f in terms of k f kH p and t > 0.

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59

(b) Let 0 < p ≤ 1. Show that for all f in H p (Rn ), fb is a continuous function and prove that there exists a constant Cn,p such that for all ξ 6= 0

n | fb(ξ )| ≤ Cn,p |ξ | p −n f H p . Hint: Part (a): use Proposition 2.4.7. Part (b): use part (a) with r = 1. Solution. (a) Put N := [n/p] + 1. Given x ∈ Rn and y ∈ B(x,t), the definition of M1∗ ( f ; Φ) and equation 6.4.34 in the text gives a constant C1 (depending on n and p) such that |(Φt ∗ f )(x)| ≤ M1∗ ( f ; Φ)(x) ≤ C1 MN ( f )(y). It follows that 1 |B(x,t)|

Z

C1p ≤ |B(x,t)|

Z

|(Φt ∗ f )(x)| p =

=

C1p vn t n

B(x,t)

B(x,t)

Z B(x,t)

|(Φt ∗ f )(x)| p dy [MN ( f )(y)] p dy

[MN ( f )(y)] p dy

C1p kMN f kLp p vn t n Cp ≤ 1n C2 k f kHp p , vn t ≤

for some constant C2 (depending on n, p and Φ), by Theorem 6.4.4. Since x ∈ Rn is arbitrary, taking pth roots yields n

kΦt ∗ f kL∞ ≤ C3t − p k f kH p ,

(0.0.8)

where C3 := C1 (C2 /vn )1/p . Since Φt ∗ f is obviously controlled by M( f ; Φ), it follows that Φt ∗ f ∈ L p (Rn ) and kΦt ∗ f kL p ≤ kM( f ; Φ)kL p ≤ C4 k f kH p ,

(0.0.9)

for some (universal) constant C4 , again by Theorem 6.4.4. Interpolating between (0.0.8) and (0.0.9) we deduce n r−p r

kΦt ∗ f kLr ≤ Cn,pt − p

k f kH p

for all r satisfying p ≤ r ≤ ∞. This completes the proof of (a). 2 (b) To show that fb is continuous, choose Φ(x) = e−π|x| . Taking r = 1, t = 1 in the preceding inequality we obtain b ) fb(ξ )| ≤ k(Φ ∗ f )bkL∞ ≤ kΦ ∗ f k 1 ≤ Cn,p k f k p . |Φ(ξ L H b ) fb(ξ ) = (Φ ∗ f )b is continuous and Φ b 6= 0, it follows that fb is continuous. Now choose Φ ∈ S (Rn ) such that But since Φ(ξ b = 1 on the unit sphere Sn−1 . Then we have as before |Φ| n

b ) fb(ξ )| ≤ k(Φt ∗ f )bkL∞ ≤ kΦt ∗ f k 1 ≤ Cn,p t − p +n k f k p |Φ(tξ L H b = 1 on Sn−1 , we deduce the claimed conclusion. for all t > 0 for all ξ ∈ Rn . Picking t = |ξ |−1 and using that |Φ|

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Exercise 2.1.11 1 n 2 1 n 2 Show that H p (Rn , `2 ) = L p (Rn , `2 ) whenever 1 < p < ∞ and that H (R , ` ) is contained in L (R , ` ). 2 + Hint: Prove these assertions for `L first for some L ∈ Z .

Solution. We prove the corresponding assertions for `2L in place of `2 and then we let L → ∞ and we use the Lebesgue monotone convergence theorem to obtain the assertion for `2 (Z). Let ~f ∈ H p (Rn , `2L ). The set {Pt ∗ ~f : t > 0} lies in a multiple of the unit ball of L p (Rn , `2L ) which is the dual of the separable 0 Banach space L p (Rn , `2L ). By the Banach-Alaoglu theorem there exists a sequence t j → 0 such that Pt j ∗ ~f converges to some L p (Rn , `2L ) function ~f0 in the weak∗ topology of L p (Rn , `2L ). On the other hand, we see that Pt ∗ ~ϕ → ~ϕ in S (Rn , `2L ) as t → 0 for all ~ϕ in S (Rn , `2L ). Thus, Pt ∗ ~f → ~f in S 0 (Rn , `2L ), (0.0.10) and it follows that the distribution ~f coincides with the L p (Rn , `2L ) function ~f0 . Since the family {Pt }t>0 is an approximate identity, we have kk(Pt ∗ ~f − ~f )(x)k`2 kL p → 0 as t → 0, L

from which it follows that k~f kL p (Rn ,`2 ) = kk~f (x)k`2 kL p ≤ k sup k(Pt ∗ f )(x)k`2 kL p = k f kH p (Rn ,`2 ) . L

L

L

t>0

L

(0.0.11)

Since k(Pt ∗ ~f )(x)k`2 ≤ |(Pt ∗ k~f k`2 )(x)|, L

L

the converse inequality is a consequence of the fact that sup |k(Pt ∗ ~f )(x)k`2 | ≤ sup |(Pt ∗ k~f k`2 )(x)| ≤ M(k~f k`2 )(x), L

t>0

L

t>0

L

where M is the Hardy-Littlewood maximal operator. (See Corrollary 2.1.12.) For the case p = 1, it is sufficient to show that k f kL1 (Rn ,`2 ) ≤ k f kH 1 (Rn ,`2 ) , L

L

∀ f ∈ H 1 (Rn , `2L ).

(0.0.12)

Embedding L1 (Rn , `2L ) into the space of finite Borel measures M whose unit ball is weak∗ compact, we can extract a sequence t j → 0 such that Pt j ∗ ϕ converges to some measure µ in the topology of measures. In view of (0.0.10), it follows that the distribution f can be identified with the measure µ. It remains to show that µ is absolutely continuous with respect to Lebesgue measure, which would imply that it coincides with some L1 function (by the Radon-Nykodim Theorem). Let |µ| be the total variation of µ. We show that µ is absolutely continuous by showing that for all subsets E of Rn we have that |E| = 0 implies |µ(E)| = 0. Given an ε > 0, there exists a δ > 0 such that for any measurable subset F of Rn we have |F| < δ ⇒

Z

sup k(Pt ∗ f )(x)k`2 dx < ε.

F t>0

L

Given E with |E| = 0, we can find an open set U such that E ⊂ U and |U| < δ . Then for any g continuous function supported in U we have Z Z = lim gdµ g(x)k(P ∗ f )(x)k dx 2 tj ` Rn

j→∞

≤ kgkL∞ But we have

L

Rn

Z

sup k(Pt ∗ f )(x)k`2 dx < εkgkL∞ .

U t>0

L

Z |µ(U)| = sup gdµ : g continuous supported in U with kgkL∞ ≤ 1 , Rn

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61

which implies that µ(U) < ε. Since ε is arbitrary, it follows that |µ|(E) = 0; hence µ is absolutely continuous with respect to Lebesgue measure. Finally, (0.0.12) is a consequence of (0.0.11), which is also valid for p = 1.

Exercise 2.1.12 For a sequence of tempered distributions ~f = { f j } j , define the following variant of the grand maximal function: fN (~f )(x) = M

sup

sup sup

fN ε>0 y∈Rn {ϕ j } j ∈F |y−x| n/p.

Solution. Fix Φ ∈ S (Rn ) with integral 1. Then for all t > 0, Φt (x) = t −n Φ(t −1 x) ∈ S (Rn ) and has integral 1. Applying lemma (s) (6.4.5), for (ϕ j )t ∈ S (Rn ) with m = b = [ np ] + 1 there exist Schwartz functions (Θ j )t , 0 < s < 1 such that Z 1

(ϕ j )t (y) = and

Z Rn

0

(s)

((Θ j )t ∗ Φts )(y) ds

(s)

(1 + |w|)b |(Θ j )t (w)| dw ≤ C(Φt , n, p)sb Nb ((ϕ j )t ).

fN , we have, Fix x ∈ Rn . Then for y ∈ B(x,t) and ~ϕ = {ϕ j } j ∈ F Z 1Z

(s)

|(Θ j )t (z)||(Φts ∗ f j )(y − z)| dz ds b Z 1Z |x − (y − z)| (s) ∗∗ ~ ≤ |(Θ j )t (z)|Mb ( f , Φ)(x) + 1 dz ds ts 0 Rn b Z 1 Z |x − y| |z| (s) −b ∗∗ ~ ≤ s |(Θ j )t (z)|Mb ( f , Φ)(x) + + 1 dz ds t t 0 Rn

|((ϕ j )t ∗ f j )(y)| ≤

0

b

≤2

Rn

Mb∗∗ (~f , Φ)(x)

≤ 2b Mb∗∗ (~f , Φ)(x)

Z 1

−b

s 0

Z 1 0

Z Rn

(s)

|(Θ j )(w)|(|w| + 1)b dw ds

s−bCsb Nb (ϕ j ) ds

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≤ 2bCMb∗∗ (~f , Φ)(x)Nb (ϕ j ). fN , over all t > 0, and over all y ∈ B(x,t) ⊆ Rn , we obtain Applying `2 norms and taking supremum over all ~ϕ = {ϕ j } j ∈ F f(~f )(x) ≤ C1 M ∗∗ (~f , Φ)(x). M b Then using (6.4.47), (6.4.46), (6.4.45) and (6.4.49) of Theorem 6.4.13 successively we obtain the following: fN (~f )|| p n 2 ≤ C1 (n, p, Φ)||M ∗∗ (~f , Φ)|| p n 2 ||M b L (R ,` ) L (R ,` ) ≤ C2 (n, p, Φ)||M1∗ (~f , Φ)||L p (Rn ,`2 ) ≤ C3 (n, p, Φ)||M(~f , Φ)||L p (Rn ,`2 ) ≤ C4 (n, p, Φ)||~f ||H p (Rn ,`2 ) ≤ C5 (n, p, Φ)||MN (~f )||L p (Rn ,`2 ) .

Exercise 2.1.13 Suppose that the family {K j }Lj=1 satisfies L

∑ ∂ α K j (x) ≤ A |x|−n−|α| < ∞

(0.0.13)

j=1

and also

L

∑ Kcj (ξ ) ≤ B < ∞ . n

sup

(0.0.14)

ξ ∈R j=1

and let Φ be a smooth function supported in the unit ball B(0, 1). If Φε (x) = ε −n Φ(x/ε), then the family {Φε ∗ K j }Lj=1 also satisfies (0.0.13) and (0.0.14) with constants A0 and B0 proportional to A + B and B, respectively. Hint: In the proof of (0.0.13) consider the cases |x| ≥ 2ε and |x| ≤ 2ε. In the second case write Z

p.v.

Rn

Φε (x − y)K j (y) dy =

Z Rn

Φε (x − y) − Φε (x) K j (y)Φ0 (y/ε) dy

Z + p.v.

Rn

K j (y)Φ0 (y/ε) dy Φε (x) ,

where Φ0 (y) is a smooth function which is equal to 1 on the ball |y| ≤ 3 and vanishes outside the ball |y| ≤ 4.

Solution. b L∞ B. To prove (0.0.13) we fix ε > 0 and we Obviously (0.0.14) is satisfied for the family {Φε ∗ K j } j with constant B0 = kΦk consider first the case |x| ≥ 2ε. Then since |y| ≤ ε we have |x − y| ≥ |x|/2 and thus L

Z α ∑ ∂x Φε (y)K j (x − y) dy =

j=1

≤

Z

L

Z Z L α ∑ Φε (y)∂x K j (x − y) dy ≤ |Φε (y)| ∑ |∂xα K j (x − y)| dy

j=1

A dy ≤ |Φε (y)| |x − y|n+|α|

j=1

Z

|Φε (y)|

A 2n+|α| C0 A dy ≤ |x|n+α |x|n+|α|

for all |α| ≤ N. Now fix x such that |x| ≤ 2ε. Pick a smooth function Φ0 which is equal to 1 on the ball |x| ≤ 3 and vanishes outside the ball |x| ≤ 4. Then

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63

Z L ∑ ∂xα

|y|≤3ε

j=1

Φε (x − y)K j (y) dy =

Z L α ∑ ∂x Φε (x − y)K j (y)Φ0 (y/ε) dy

j=1

L Z L Z α α ≤ ∑ ∂x Φε (x − y) − Φε (x) K j (y)Φ0 (y/ε) dy + ∑ K j (y)Φ0 (y/ε) dy ∂x Φε (x) . j=1

j=1

which is bounded by ≤

Z

∂xα

L Z L α K j (y)Φ0 (y/ε) dy ∂x Φε (x) . Φε (x − y) − Φε (x) ∑ |K j (y)| |Φ0 (y/ε)| dy + ∑ j=1

j=1

The preceding term is bounded by 1 ε n+|α|

Z |y|≤4ε

L Z ∂ α Φ L∞

|y| n

∇∂ α Φ ∞ A|y|−n Φ0 ∞ dy + ∑ K c cj (ξ )ε Φ0 (εξ )dξ ε n+|α| , L L ε j=1

and this is bounded by 00 000 000 n+|α|

α C0 A

1 ∂ Φ ∞ ≤ C (A + B) ≤ C (A + B)2

∇∂ α Φ ∞ Φ0 ∞ + C B c Φ , 0 L L L L ε n+|α| ε n+|α| ε n+|α| |x|n+|α|

where the last inequality is due to the fact that |x|/2 ≤ ε.

2.2. Function Spaces and the Square Function Characterization of Hardy Spaces Exercise 2.2.1 Let 0 < q0 ≤ q1 < ∞, 0 < p < ∞, ε > 0, and α ∈ R. Prove the embeddings 0 j Bα,q1 , Bα,q p p α,q0 j Fpα,q1 , Fp 0 j Bα,q1 , Bα+ε,q p p

Fpα+ε,q0 j Fpα,q1 , where p and q1 are allowed to be infinite in the case of Besov spaces. Solution. Recall that Bα,q p is the space of all tempered distributions f for which k f kBα,q = kS0 ( f )kL p + p

∞

∑ (2 jα k∆ Ψj ( f )kL p )q

1

q

j=1

α,q0

is finite. We have the following fact; if q0 ≤ q1 then `q0 j `q1 with k · k`q1 ≤ k · k`q0 . Given f ∈ B p

∞

∑ (2

jα

q0

k∆ j ( f )k ) Lp

1

q0

= k{2 jα |∆ j ( f )kL p } j k`q0 < ∞.

j=1

The preceding embedding gives k{2 jα k∆ j ( f )kL p } j k`q1 ≤ k{2 jα k∆ j ( f )kL p } j k`q0 < ∞

we have

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Contents

which implies that k f kBα,q1 ≤ k f kBα,q0 . p

p

We recall that Fpα,q is the space of all tempered distributions f for which the quantity, k f kFpα,q = kS0 ( f )kL p + k α,q0

is finite. Given f ∈ Fp

∞

∑ (2 jα k∆ Ψj ( f )|)q

1 q

kL p

j=1

we have

∞ 1 q

∑ (2 jα |∆ j ( f )|)q1 1

Lp

j=1

∞ 1 q0

jα q0

≤ ∑ (2 |∆ j ( f )|)

Lp

j=1

α,q

which implies that f ∈ Fpα,q1 . Thus we have Fp 0 j Fpα,q1 . α+ε,q0 α+ε,q0 1 1 Now we show that B p j Bα,q and we need f ∈ Bα,q Then, p . Assume f ∈ B p p

jα

{2 k∆ j ( f )kL p }∞j=1 q ≤ {2 jα k∆ j ( f )kL p }∞j=1 q ≤ {2 j(α+ε) k∆ j ( f )kL p }∞j=1

`q0

` 0

` 1

< ∞,

since ε > 0. α+ε,q0 Finally, we have that Fp j Fpα,q1 since

∞ 1 q

∑ (2 jα |∆ j ( f )|)q1 1

j=1

Lp

∞ 1 q0

jα q0

≤ ∑ (2 |∆ j ( f )|)

Lp

j=1

∞ 1 q0

j(α+ε) q0

. |∆ j ( f )|) ≤ ∑ (2

p L

j=1

Exercise 2.2.2 Let 0 < q < ∞, 0 < p < ∞, and α ∈ R. Prove that the embeddings α,min(p,q)

Bp

α,max(p,q)

⊆ Fpα,q ⊆ B p

hold with norm one, if the norms in the spaces are defined with respect to the same Schwartz function Ψ . Hint: When p ≥ q use Minkowski’s inequality for L p/q for one embedding and the embedding `q j ` p for the other. When p < q use the reverse Minkowski inequality for L p/q for one embedding and the fact (∑k |ak |) p/q ≤ ∑k |ak | p/q for the other. Solution. Note that ∞

k f kBα,q := kS0 ( f )kL p + p and k f kFpα,q

!1

q

∑ (2 jα k∆ j ( f )kL p )q

j=1

" #1

∞ q

jα q

. := kS0 ( f )kL p + ∑ (2 |∆ j ( f )|)

j=1

p L

We consider the case q ≤ p. We will show that k f kBα,p ≤ k f kFpα,q ≤ k f kBα,q . p p which implies α,q Bα,q ⊆ Bα,p p ⊆ Fp p .

We use the inequality k · k` p ≤ k · k`q (which is valid since p ≥ q) to obtain

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65

!1

p

∞

∑ (2 jα k∆ j ( f )kL p ) p

=

j=1

Z

"

#1qp

∞

p 2 jα |∆ j ( f )|

∑

Rn

p

q

dx

j=1

1p

≤

Z

!p q

∞

∑ (2 jα |∆ j ( f )|)q

Rn

j=1

dx

1 p

.

Adding kS0 ( f )kL p on both sides we obtain k f kBα,p ≤ k f kFpα,q . p Next we show that k f kFpα,q ≤ k f kBα,q . By Minkowski’s inequality we have p Z

"

∞

∑

Rn

#p q

q 2 jα |∆ j ( f )|

dx

j=1

1 p

=

Z

"

Rn

∞

∑

#p q

q 2 jα |∆ j ( f )|

dx

j=1

q 1 p q

( ≤

∞

∑

hZ Rn

j=1

h

q i qp i qp 2 jα |∆ j ( f )| dx

)1

q

Adding kS0 ( f )kL p on both sides we obtain we deduce that k f kFpα,q ≤ k f kBα,q . p Let us consider the case p < q. We are going to show that k f kBα,q ≤ k f kFpα,q ≤ k f kBα,p . p p which implies α,q Bα,p ⊆ Bα,q p ⊆ Fp p .

Since

p q

< 1, we apply the inequality (∑k |ak |) p/q ≤ ∑k |ak | p/q to obtain

!1 q

jα q ∑ (2 |∆ j ( f )|)

j=1

! p 1p "Z q jα q ≤ (2 |∆ ( f )|) j ∑

∞

∞

Z

=

Rn

#1

"

p

jα

∑ (2 |∆ j ( f )|)

p

=

Rn j=1

j=1

Lp

∞

#1

∞

jα

2 k∆ j ( f )kL p

∑

p

p

.

j=1

Adding kS0 ( f )kL p on both sides gives k f kFpα,q ≤ k f kBα,p . p The other inequality in this case follows from the reverse Minkowski inequality for nonnegative functions. We have !1

q

∞

∑ (2 jα k∆ j ( f )kL p )q

j=1

!

∞

=

∑ |2 jα ∆ j ( f )|q

p Lq

j=1

1 q

∞

1

q ≤ ∑ |2 jα ∆ j ( f )|q qp = L

j=1

"

∞

∑ 2 jα p

j=1

Z Rn

#1

p

|∆ j ( f )| p dx .

Adding kS0 ( f )kL p on both sides gives k f kBα,q ≤ k f kFpα,q . p

Exercise 2.2.3 Let −∞ < α < ∞ and 0 < p, β < ∞. Let 10 = ∞ and p0 = p/(p − 1) for p 6= 1. γ,q (a) Suppose that the Fourier transform of function g is C ∞ and is equal to |ξ |−α for |ξ | ≥ 10. Show that g lies in B p (Rn ) if and only if 0 < q < ∞ and γ < α − pn0 , or, q = ∞ and γ ≤ α − pn0 . α− pn0 ,q

(b) If the Fourier transform of function g is C ∞ and is equal to |ξ |−α (log |ξ |)−β for |ξ | ≥ 10, then show that g lies in B p if and only if q > 1/β .

(Rn )

b be supported in 6/7 ≤ |ξ | ≤ 2 and satisfying ∑ j∈Z Ψ b (2− j ξ ) = 1 for ξ 6= 0. Then for j ≥ 3 we Solution. Part (a). Let Ψ −α b − j − jα |2− j ξ |−α Ψ b (2− j ξ ), so taking inverse Fourier transforms and L p norms yields have [∆ Ψ j (g)]b(ξ ) = |ξ | Ψ (2 ξ ) = 2 0 Ψ − jα jn/p k∆ j (g)kL p = c2 2 . Obviously ∞

0

∑ (2 jγ c2− jα 2 jn/p )q < ∞

j=3

if and only if γ < α −

n p0 .

0

Also sup j≥3 2 jγ 2− jα 2 jn/p < ∞ if and only if γ ≤ α −

n p0 .

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−α (log |ξ |)−β Ψ b (2− j ξ ) = 2− jα j−β |2− j ξ |−α Ψ b (2− j ξ ), so taking inverse Part (a). Let Ψ be as before. Then [∆ Ψ j (g)]b(ξ ) = |ξ | − jα j −β 2 jn/p0 . Then Fourier transforms and L p norms yields k∆ Ψ j (g)kL p = c2

∞

∑ (2

j(α− pn0 )

0

c2− jα j−β 2 jn/p )q

1/q 1.

Exercise 2.2.4 α,q n n that the space of Schwartz functions is dense in all the spaces Bα,q p (R ) and Fp (R ). Let 0 < p, q < ∞ and α ∈ R. Show n Hint: Fix a function ϕ ∈ S (R ) whose Fourier transform has integral one and is supported in a ball of radius 1 centered at zero. Given f ∈ Fpα,q (Rn ) consider the family of Schwartz functions N

fN,δ (x) = S0Φ ( f )(x)ϕ(δ x) + ∑ ∆ Ψ j ( f )(x)ϕ(δ x) . j=1

for 0 < δ < 1/10.

Solution. Fix a function ϕ ∈ S (Rn ) whose Fourier transform is supported in a ball of radius 1 centered at zero which has integral one. Given f ∈ Fpα,q (Rn ) consider the family of Schwartz functions N δ fN,δ = S0Φ ( f )ϕ δ + ∑ ∆ Ψ j ( f )ϕ , j=1

where N ≥ 10 and 0 < δ < 1/100. We note that S0Φ ( f − fN,δ ) = S0Φ ( f − S0Φ ( f )ϕ δ − ∆1Ψ ( f )ϕ δ ) and this is equal to S0Φ f − S0Φ ( f ) − ∆1Ψ ( f ) + (S0Φ ( f ) + ∆1Ψ ( f ))(1 − ϕ δ ) = S0Φ (S0Φ ( f ) + ∆1Ψ ( f ))(1 − ϕ δ ) . This is bounded by Cr M(|(S0Φ ( f ) + ∆1Ψ ( f ))(1 − ϕ δ )|r )1/r and thus its L p norm is controlled by k(S0Φ ( f ) + ∆1Ψ ( f ))(1 − ϕ δ )kL p which can be made smaller than a given ε > 0 by taking δ < δ00 , in view of the Lebesgue dominated convergence theorem (|1 − ϕ δ | → 0 as δ → 0). Now for a fixed 1 ≤ k ≤ N + 1 we look at Ψ Ψ ( f ) + ∆kΨ ( f ) + ∆k+1 ( f ))ϕ δ ∆kΨ ( f − fN,δ ) = ∆kΨ f − (∆k−1 where we tacitly set ∆0Ψ = S0Φ . The preceding identity is equal to Ψ Ψ ∆kΨ (∆k−1 ( f ) + ∆kΨ ( f ) + ∆k+1 ( f ))(1 − ϕ δ ) Just as before, since the Fourier transform of the expression inside ∆kΨ is supported in a ball, we have that the preceding Ψ ( f ) + ∆ Ψ ( f ) + ∆ Ψ ( f ))(1 − ϕ δ )|r )1/r whose L p norm can be made smaller expression is pointwise bounded by Cr M(|(∆k−1 k k+1 than ε divided by any function of N. We combine these elements to obtain

∞

∞ 1/q 1/q N+1

kα Ψ q c Nα+α Ψ kα Ψ q p,q

. kS0 ( f − fN,δ )kL p + ∑ (2 |∆k ( f − fN,δ )| ∑ k∆k ( f − fN,δ )kL p + ∑ (2 |∆k ( f )|

≤ ε +N 2

k=1

Lp

k=1

k=N+1

Lp

So, given ε > 0 first find N0 such that the term on the right above is smaller than ε for all N ≥ N0 . Then pick δ0 < δ00 such for δ < δ0 we have that k∆kΨ ( f − fN,δ )kL p < ε/N c p,q +1 2Nα+α . This implies that for δ < δ0 and N ≥ N0 we have k f − fN,δ kFpα,q ≤ 3ε, hence the claimed density follows.

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67

Note that the argument for Bα,q p is essentially identical: the only difference in the proof is the step below: kS0 ( f − fN,δ )kL p +

N+1

kα Ψ

q 1/q c0p,q Nα+α Ψ

p 2 ∆ ( f − f ) ≤ ε +N 2 k∆ ( f − f )k + L p N,δ N,δ ∑ ∑ k k L ∞

k=1

k=1

∞

∑

kα Ψ

1/q

2 ∆ ( f − fN,δ ) q p . k L

k=N+1

Exercise 2.2.5 n ] + 1. Assume that m is a C N function on Rn \ {0} that satisfies Let α ∈ R, let 0 < p, q < ∞, and let N = [ 2n + min(p,q)

|∂ γ m(ξ )| ≤ Cγ |ξ |−|γ| for all |γ| ≤ N. Show that there exists a constant C such that for all f ∈ S 0 /P 0 we have

(m fb)∨ .α,q ≤ C f .α,q . B B p

p

b (2− j+1 ξ )+Ψ b (2− j ξ )+Ψ b (2− j−1 ξ )). Then ∆ Ψ ((m fb)∨ ) = Hint: Pick r < min(p, q) such that N > n2 + nr . Write m j (ξ ) = m(ξ )(Ψ j m∨j ∗ ∆ Ψ j ( f ). Obtain the estimate Ψ ∆ ( f )(x − y) Z ∨ Ψ n m j ∗ ∆ j ( f ) (x) ≤ C sup j |m∨j (y)|(1 + 2 j |y|) r dy n j n n r R (1 + 2 |y|) y∈R 1 Z 2 j ∨ 2 2N r 1r |m (2 ( · )) (y)| (1 + |y|) dy ≤ C0 M(|∆ Ψ ( f )| ) (x) . j j Rn

The hypothesis on m implies that the preceding integral is bounded by a constant. Solution. b (2− j+1 ξ ) + Ψ b (2− j ξ ) + Ψ b (2− j−1 ξ )). Then we write ∆ Ψ ((m fb)∨ ) = ∆ Ψ ((m j fb)∨ ) = m∨ ∗ ∆ Ψ ( f ). Write m j (ξ ) = m(ξ )(Ψ j j j j n n Set N = 2 + r + δ . We have (1 + 2 j |y|) nr |m∨j (y)| ∆ Ψ ( f )(x − y) n dy j Rn (1 + 2 j |y|) r

Z ∨ Ψ m j ∗ ∆ j ( f ) (x) ≤ C

Z ∆Ψ j ( f )(x − y)

≤ C sup

n y∈Rn (1 + 2 j |y|) r 1

r r ≤ C0 M(|∆ Ψ j ( f )| ) (x)

n

Rn

Z

1

Rn

Z

≤ C000

≤C

Rn

Z Rn

Z

dy

Rn

2n +n+2δ

|m∨j (y)|2 (1 + 2 j |y|) r

|m j (2 j ( · ))∨ (y)|2 (1 + |y|)2N dy

1 2

1

r r M(|∆ Ψ j ( f )| ) (x)

∑

∑

|∂yβ (m j (2 j ( · )))(ξ )|2 dξ

Rn |β |≤N

Z Rn |β |≤N

1

r r = Cr M(|∆ Ψ j ( f )| ) (x).

1 2

(1 + 2 j |y|)−n−2δ dy

dy

1 2 r 1r |yβ m j (2 j ( · ))∨ (y)|2 dy M(|∆ Ψ j ( f )| ) (x)

Z

0000

2n +n+2δ

|m∨j (y)|2 (1 + 2 j |y|) r

r r − jn ≤ C00 M(|∆ Ψ j ( f )| ) (x)2

= C00

|m∨j (y)|(1 + 2 j |y|) r dy

1

r r M(|∆ Ψ j ( f )| )

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Contents

b (2ξ ) + Ψ b (ξ ) + Ψ b (2−1 ξ )) is supported in the fixed annulus 3 ≤ |ξ | ≤ 4 and in view of the Each function m j (2 j ξ ) = m(2 j ξ )(Ψ 7 j hypothesis on m all derivatives of m j (2 ξ ) are bounded (and supported in the same annulus). This explains why the last integral above is a constant. Ψ jα and taking `q quasi-norms yields the b∨ Taking L p norms we obtain that k∆ Ψ j ((m f ) )kL p ≤ C k∆ j kL p . Multiplying by 2 required conclusion that

(m fb)∨ .α,q ≤ C f .α,q . B B p

p

Exercise 2.2.6 ([299]) Let m be as in Exercise 2.5.5. Show that there exists a constant C such that for all f ∈ S 0 (Rn )/P 0 (Rn ) we have

(m fb)∨ . α,q ≤ C f . α,q . F F p

p

Hint: Use the hint of Exercise 2.5.5 and Theorem 4.6.6 in [161]. Solution. Ψ r 1r Following the hint of the preceding exercise, we deduce that m∨j ∗ ∆ Ψ j ( f ) (x) ≤ Cr M(|∆ j ( f )| ) (x) for every j. We now q p apply the ` norm in j and then the L norm in x. We obtain that

(m fb )∨ . α,q ≤ C f . α,q . F F p

p

Exercise 2.2.7 α ,q

(a) Suppose that B p00 0 = Bαp11 ,q1 with equivalent norms. Prove that α0 = α1 and p0 = p1 . Prove the same result for the scale of F spaces. α0 ,q0 α1 ,q1 (b) Suppose that B p0 = B p1 with equivalent norms. Prove that q0 j= q1 . Argue similarly with the scale of F spaces. Hint: Part (a): Test the corresponding norms on the function Ψ (2 x), where Ψ is chosen so that its Fourier transform is b 1 − 2 j , ξ2 , . . . , ξn ), where ϕ is a Schwartz supported in 21 ≤ |ξ | ≤ 2. Part (b): Try a function f of the form fb(ξ ) = ∑Nj=1 a j ϕ(ξ function whose Fourier transform is supported in a small neighborhood of the origin. Solution. Before we prove the statements, let us set up some definitions. Let Ψ be a Schwartz function whose Fourier transform is nonnegative, is supported in the annulus 1 − 71 ≤ |ξ | ≤ 2, is equal to one on the smaller annulus 1 ≤ |ξ | ≤ 2 − 72 , and satisfies

∑ Ψb (2− j ξ ) = 1

ξ 6= 0.

(0.0.15)

j∈Z

We define the Littlewood-Paley operators ∆Ψ j ( f ) = Ψ2− j ∗ f . Let Φ be a Schwartz function so that

b )= Φ(ξ

It follows that

b (2− j ξ ), when ξ = 6 0; ∑ j≤0 Ψ 1, when ξ = 0.

(0.0.16)

∞

S0 + ∑ ∆ Ψ j = I, j=0

(0.0.17)

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69

where S0 ( f ) = Φ ∗ f . By our assumption, we have C1 k f kBα0 ,q0 ≤ k f kBα1 ,q1 ≤ C2 k f kBα0 ,q0 , p1

p0

∀f,

p0

(0.0.18)

where C1 and C2 are positive constants. We prove part (a) in the following way. (a) Let η be a Schwartz function whose the Fourier transform of η is supported in 1 ≤ |ξ | ≤

12 7 ,

such that

b (2− j ξ ) = δk, j 2−kn η bk (2k ξ )Ψ bk (2k ξ ), 2−kn η

(0.0.19)

where δk, j is the Kronecker δ . To show that p0 = p1 , fix k ≤ 0 and choose f (x) = η(2k x). Then k f kBαl ,ql = k f kF αl ,ql = kη(2k ·)kL pl = 2

− kn p l

pl

pl

kηkL pl , l = 0, 1.

By assumption, one has − pkn

C1 2

0

− pkn

− pkn

kηkL p0 ≤ 2

kηkL p1 ≤ C2 2

1

0

kηkL p0 ,

∀k ≤ 0.

From the above inequalities one gets C1 ≤ 2

kn(p1 −p0 ) p0 p1

kηkL p1 ≤ C2 kηkL p0

∀k ≤ 0.

Suppose p0 > p1 . Then there exists kM ≤ 0 such that C2 ≤ 2

kM n(p1 −p0 ) p0 p1

kηkL p1 . kηkL p0

This contradicts our assumption. In the case p0 < p1 we can choose km ≤ 0 such that 2

km n(p1 −p0 ) p0 p1

kηkL p1 ≤ C1 . kηkL p0

Thus, p0 = p1 . Now let us show q0 = q1 as follows. We will use the fact that p0 = p1 = p. Let us fix k ≥ 1 and choose f (x) = η(2k x). Then, kS0 ( f )kL p = kS0 ( f )kL p = 0,

(0.0.20)

!1

∞

∑ (2

jα0

q0

q0 k∆ Ψ j ( f )kL p )

= 2kα0 k∆kΨ ( f )kL = 2kα0 kη(2k ·)kL p

j=1

k α0 − np

=2 and

kηkL p ,

!1

∞

∑ (2

jα1

q1 k∆ Ψ j ( f )kL p )

q1

kα1

=2

(0.0.21)

k∆kΨ ( f )kL p

k α1 − pn 1

=2

kηkL p .

(0.0.22)

j=1

By (0.0.18), we obtain k α0 − np

C1 2

k α1 − np

kηkL p ≤ 2

k α0 − np

kηkL p ≤ C2 2

kηkL p .

From the above inequalities one gets C1 ≤ 2k(α1 −α0 ) ≤ C2

∀k.

Suppose α0 < α1 . Then α1 − α0 > 0. Since C2 is fixed, it follows from (0.0.23) that there exists kM > 1 such that C2 < 2kM (α1 −α0 ) .

(0.0.23)

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This contradicts our assumption. We also will get a contradiction in the case α0 > α1 . Indeed, we have α1 − α0 < 0 which implies that there exists km > 1 such that 2km (α1 −α0 ) < C1 . Thus, α0 = α1 . Now let us show that similar results hold in the scale of Triebel-Lizorkin spaces. To show p0 = p1 we choose the same function f and η as in the proof of inhomogeneous Besov spaces. Choosing k ≤ 0, we obtain "Z

∑ (2 jαl |∆ Ψj ( f )|)ql

Rn

# p1

! pl

∞

l

= 0,

dx

l = 0, 1

j=1

and kS0 ( f )kL pl =

Z R

|∆kΨ ( f )| pl dx n

1

pl

,

l = 0, 1.

Using the same arguments as in the inhomogeneous Besov spaces, one gets p0 = p1 . As before, to show q0 = q1 , we choose k ≥ 1 and get the similar results as in the inhomogeneous Besov spaces, i.e., Z

k f kF α0 ,q0 = p0

Rn

Z =

Rn

p0 p10 Z kα0 Ψ 2 |∆k ( f )(x)| dx =

Rn

p0 2−nk 2kα0 |η(x)| dx

and k f kF α1 ,q1 =

Z

p1

Rn

1 p0

p0 p10 kα0 k 2 |η(2 x)| dx (0.0.24)

− pnk +kα0

=2

0

kηkL p0

p1 p11 − nk +kα1 = 2 p1 2kα1 |∆kΨ ( f )| dx kηkL p1 .

(0.0.25)

Setting p0 = p1 = p in (0.0.24) and (0.0.25), and arguing in the same ways as in the inhomogeneous Besov spaces, one obtains α0 = α1 . Now let us prove part (b) as follows (b) Here we choose f such that N

fb(ξ ) =

∑ a j φ (ξ1 − 2 j , ξ2 , . . . , ξn ),

j=1

where a j ∈ C, φ is a C0∞ function whose Let ξ j := (ξ1 − 2 j , ξ2 , . . . , ξn ), U j := {ξ

is supported in a small neighborhood of the origin, namely supp(φ ) = {ξ : |ξ | ≤ ε}. : |(ξ1 − 2 j , ξ2 , . . . , ξn )| ≤ ε} and φ j (ξ ) := φ (ξ1 − 2 j , ξ2 , . . . , ξn ). Then φ j (ξ ) 6= 0 for

ξ ∈ U j . We have b (2− j ) fb(ξ ) = Ψ

N

∑ Ψb (2− j ξ )ak φk (ξ ) = k=1

a j φ j (ξ ), ξ ∈ U j , 1 ≤ j ≤ N; 0, otherwise.

(0.0.26)

Applying the inversion of Fourier transform of fb, one gets N

Z

f (x) =

∑ a j φ (ξ1 − 2 j , ξ2 , . . . , ξn )e2πix·ξ dξ Rn j=1

N

=

∑

N

Z

2πix·ξ

j

j=1 U j

a j φ (ξ1 − 2 , ξ2 , . . . , ξn )e

dξ =

(0.0.27) 2πi2 j x1 ∨

∑ a je

φ (x).

j=1

Then (0.0.26) and (0.0.27) give ∆Ψ j (f) =

a j e2πi2 0,

jx

1

φ ∨ (x), j = 1, 2, . . . , N; otherwise.

(0.0.28)

In showing q0 = q1 we will use the facts that p0 = p1 = p and α0 = α1 = α which have been proved in Part (a). By (0.0.26) and (0.0.28), one has kS0 ( f )kL p = 0

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71

and ∞

∑ 2 jqk α

Mk :=

Z

p |∆ Ψ j ( f )| dx

Rn

j=1

qk ! q1k =

2 jα |a j |

∑

Z Rn

j=1

|φ ∨ (x)| p dx

1 #qk ! q1k p

(0.0.29)

!1

N

qk

q ∑ 2 jα |a j | k

= kφ ∨ kL p

"

N

p

,

k = 0, 1.

j=1

The assumption k f kBα,q0 ≈ k f kBα,q1 gives Lp

Lp

" C1

#1

N

∑

q0

q 2 |a j | 0 jα

"

N

≤

j=1

jα

q1

∑ (2 |a j |) ≤ C2

#1

N

q 2 |a j | 0 jα

∑

q0

∀ a j ∈ C, j = 1, 2, . . . , N.

j=1

j=1

The above inequalities show that `q0 ≈ `q1 . Thus, q0 = q1 . Let us show that this also holds in the Triebel-Lizorkin spaces. We have Ml :=

! pl

∞

Z Rn

∑ (2 jαl |∆ j ( f )(x)|)ql

p1

l

dx

=

∑ (2 jαl |a j ||φ ∨ (x)|)ql

Rn

p1

! pl

N

Z

j=1

" = kφ ∨ kL pl

ql

l

ql

dx

j=1

#1

N

ql

∑ (2 jαl |a j |)ql

,

l = 0, 1.

j=1

Since p0 = p1 and α0 = α1 , and k f kF α,q0 ≈ k f kF α,q1 , it follows that p

p

" C1

#1

N jα

q0

∑ (2 |a j |)

j=1

"

q0

≤

#1

N jα

q1

∑ (2 |a j |)

j=1

"

q1

≤ C2

#1

N jα

q0

∑ (2 |a j |)

q0

∀a j ∈ C.

j=1

The above inequalities imply `q0 ≈ `q1 . Thus, q0 = q1 .

2.3. Atomic Decomposition of Homogeneous Triebel–Lizorkin Spaces Exercise 2.3.1 (a) Given N ∈ Z+ , prove that there exists a Schwartz function Θ supported in the unit ball |x| ≤ 1 such that Rn xγ Θ (x) dx = 0 b (ξ )| ≥ 1 for all ξ in the annulus 1 ≤ |ξ | ≤ 2. for all multi-indices γ with |γ| ≤ N and such that |Θ 2 2 (b) Prove there exists a Schwartz function Ψ whose Fourier transform is supported in the annulus 12 ≤ |ξ | ≤ 2 and is at least c > 0 in the smaller annulus 53 ≤ |ξ | ≤ 53 and which satisfies for all ξ ∈ Rn \ {0} R

∑ Ψb (2− j ξ )Θb (2− j ξ ) = 1 .

j∈Z

Hint: Part (a): Let θ be an even real-valued Schwartz function supported in the ball |x| ≤ 1 and such that θb(0) = 1. Then for b (ξ ) = some ε ∈ (0, 12 ) we have θb(ξ ) ≥ 12 for all ξ satisfying |ξ | < 2ε. Set Θ = (−∆ )N (θε ). Part (b): Define the function Ψ −1 1 − j − j b b (ξ ) ∑ j∈Z η b (2 ξ )Θ (2 ξ ) , where η b (ξ ) is a Schwartz function supported in the annulus 2 ≤ |ξ | ≤ 2 and equal to 1 on η 3 5 the smaller annulus 5 ≤ |ξ | ≤ 3 . Solution.

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(a) Let θ be an even real-valued Schwartz function supported in the ball |x| ≤ 1 and such that θb(0) = 1. Then θb is real and for some ε ∈ (0, 21 ) we have θb(ξ ) ≥ 21 for all ξ satisfying |ξ | ≤ 2ε. Set Θ = (−∆ )N (θε ). Since θε (x) = ε −n θ (ε −1 x) is supported in |x| ≤ ε and ε < 12 < 1, Θ (x) is supported in |x| ≤ 1. For 12 ≤ |ξ | ≤ 2 we have |εξ | ≤ 2ε, thus b (ξ )| = |(−∆\ |Θ )N θε (ξ )| = (4π 2 |ξ |2 )N |θbε (ξ )| = (4π 2 |ξ |2 )N |θb(εξ )| 1 2 1 2N 1 1 1 ≥ 4π = π 4N > . 2 4 2 2 2 Since ((2πix)γ Θ (x))∨ (ξ ) = ∂ γ (Θ ∨ )(ξ ) = ∂ γ 4π 2 |ξ |2N θ ∨ (εξ ) and notice that this vanishes at the origin for all |γ| ≤ N. Thus we have Z xγ Θ (x) dx = (xγ Θ (x))∨ (0) = 0. ≥ (4π 2 |ξ |2 )N

b (ξ ) is supported in the annulus 12 ≤ |ξ | ≤ 2 and is 1 on the smaller annulus (b) Let η(ξ ) be a Schwartz function such that η 3 5 n b −j b − j+2 ξ ) = 0, there exists 5 ≤ |ξ | ≤ 3 . Then, since for any fixed ξ ∈ R and for all j ∈ Z, we have either η (2 ξ ) = 0 or η (2 r ∈ Z such that ∑ ηb (2− j ξ )Θb (2− j ξ ) = ηb (2−r ξ )Θb (2−r ξ ) + ηb (2−r+1 ξ )Θb (2−r+1 ξ ) > 0, j∈Z

thus the last expression is always strictly positive. Define b (ξ ) = Ψ

b (ξ ) η . ∑ ηb (2 ξ )Θb (2− j ξ ) −j

j∈Z

b (ξ ) is supported in the annulus Then Ψ is a Schwartz function such that Ψ b (ξ ) = Ψ

1 2

≤ |ξ | ≤ 2. For

3 5

≤ |ξ | ≤

5 3

we have

1 −j

∑ ηb (2

b (2− j ξ ) ξ )Θ

j∈Z

=

1 1

∑

b (2− j ξ ) b (2− j ξ )Θ η

j=−1

≥

1 −j

b (2− j ξ ) b (2 ξ )Θ max ∑ η

3 ≤|ξ |≤ 5 3 5

=

1 > 0. c

j∈Z

Fix ξ ∈ Rn \{0}. Then,

∑ ηb (2− j ξ )Θb (2− j ξ )

∑ Ψb (2− j ξ )Θb (2− j ξ ) =

j∈Z

j∈Z

∑ ηb (2−k ξ )Θb (2−k ξ )

.

k∈Z

Hence, b (2−r ξ ) + η b (2−r+1 ξ ) b (2−r ξ )Θ b (2−r+1 ξ )Θ η

∑ Ψb (2− j ξ )Θb (2− j ξ ) = ηb (2−r ξ )Θb (2−r ξ ) + ηb (2−r+1 ξ )Θb (2−r+1 ξ ) = 1.

j∈Z

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73

Exercise 2.3.2 Let α ∈ R, 0 < p ≤ . 1, p ≤ q ≤ +∞. (a) For all f , g in Fpα,q show that

f + g p. α,q ≤ f p. α,q + g p. α,q . F F F p

p

p

(b) For all sequences {sQ }Q∈D and {tQ }Q∈D show that

{sQ }Q + {tQ }Q p.α,q ≤ {sQ }Q p.α,q + {tQ }Q p.α,q . f f f p

p

p

Notice that the statement in part (b) extends to countable sums. Hint: Use |a + b| p ≤ |a| p + |b| p and apply Minkowski’s inequality on Lq/p (or on `q/p ). Solution. (a) By definition, for all f , g in S 0 (Rn ) we have

1

f + g p. α,q = ( ∑ (2 jα |∆ j ( f + g)|)q ) q p p L F p

j∈Z

Z

=

Rn

Z

=

∑ (2 jα |∆ j ( f + g)|)q

p q

dx

j∈Z

Rn

q

∑ (2 jα |∆ j ( f + g)|) p p

p q

dx.

j∈Z

For every j, since p ≤ 1 we have p p p 2 jα |∆ j ( f + g)| ≤ 2 jα |∆ j ( f )| + 2 jα |∆ j (g)| . Therefore, Z

∑ (2 jα |∆ j ( f + g)|) p

q p

!p q

dx ≤

Z

p p q ∑ ( 2 jα |∆ j ( f )| + 2 jα |∆ j (g)| ) p

j∈Z

!p q

dx .

j∈Z

q

By Minkowski’s inequality on ` p , we have Z

p q

p p q ∑ ( 2 |∆ j ( f )| + 2 jα |∆ j (g)| ) p jα

dx

j∈Z

≤

Z

Rn

p q ∑ ( 2 |∆ j ( f )| ) p jα

∑

q 2 jα |∆ j ( f )|

.

Z

dx +

p q

Rn

Z

dx +

j∈Z

p

p

= f F α,q + g F α,q . p p

Therefore we obtain

q

j∈Z

Z

=

p

Rn

∑

p q ∑ ( 2 |∆ j (g)| ) p jα

p q

j∈Z

q 2 jα |∆ j (g)|

p q

dx

j∈Z

.

f + g p. α,q ≤ f p. α,q + g p. α,q . F F F p

p

p

(b) By definition, for every sequence {sQ }Q∈D , Z

{sQ }Q p. = gα,q (s) p p n = ( L (R ) f α,q

Q∈D

Then we have,

α

1

1

p

∑ (|Q|− n − 2 |sQ |χQ )q ) q

dx.

dx

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Contents

{sQ }Q + {tQ }Q p.α,q = f

Z

(

p

α

1

1

p

∑ (|Q|− n − 2 |sQ + tQ |χQ )q ) q

dx =

Z

(

Q∈D

q

1

α

p

∑ (|Q|− n − 2 |sQ + tQ |χQ ) p p ) q dx.

Q∈D

Since |a + b| p ≤ |a| p + |b| p , we have Z

(

∑

Z q p α 1 (|Q|− n − 2 |sQ + tQ |χQ ) p p ) q dx ≤

Q∈D

1

α

q

1

α

p

∑ [(|Q|− n − 2 |sQ |χQ ) p + (|Q|− n − 2 |tQ |χQ ) p ] p q dx. Q∈D

q

By Minkowski’s inequality on ` p , Z

α

1

q

1

α

p

∑ [(|Q|− n − 2 |sQ |χQ ) p + (|Q|− n − 2 |tQ |χQ ) p ] p q dx Q∈D

≤

Z

=

Z

α

q

1

p

Z

p

Z

∑ [(|Q|− n − 2 |sQ |χQ ) p ] p q dx +

Q∈D

(

α

1

1

∑ (|Q|− n − 2 |sQ |χQ )q ) q

dx +

q

1

α

p

∑ [(|Q|− n − 2 |tQ |χQ ) p ] p q dx

Rn Q∈D

(

Q∈D

α

1

1

p

∑ (|Q|− n − 2 |tQ |χQ )q ) q

dx

Q∈D

p

p = {sQ }Q f.α,q + {tQ }Q f.α,q . p

Therefore we have,

p

{sQ }Q + {tQ }Q p.α,q ≤ {sQ }Q p.α,q + {tQ }Q p.α,q . f f f p

p

p

Notice that this argument concerns nonnegative sums, so it is also valid for finite sums of sequences and by a limiting procedure (using the Lebesgue monotone convergence theorem) for infinite sums.

Exercise 2.3.3 Let Φ be a smooth function supported in the unit ball of Rn . Use the same idea as in Theorem 2.3.11 to show directly (without appealing to any other theorem) that the smooth maximal function M( · ; Φ) of an L2 -atom for H p lies in L p when p < 1. Recall that M( f ; Φ) = supt>0 |Φt ∗ f |. Solution. Let √ f (x) be an L2 -atom for H p . Assume without loss of generality that f is supported in a cube Q centered at the origin. Set ∗ Q = 2 nQ. Since Φ ∈ C0∞ , using Corollary 2.1.12 in [161], we obtain p Z Z p M( f ; Φ) (x)dx = sup |Φt ∗ f |(x) dx Q∗

Q∗

≤

t>0

Z

sup (|Φt |∗ ∗ | f |) (x) p dx

Q∗ t>0

≤ kΦkL1 Since 0 < p < 1, we have

2 p

Z Q∗

M( f )(x) p dx.

> 1. Then using H¨older’s inequality, Z Q∗

p

M( f ; Φ) (x)dx ≤ kΦkL1 ≤ kΦkL1 p

Z Q∗

Z R∗

2

p 2

M( f )(x) dx 2

| f (x)| dx

Using the property k f kL2 ≤ |Q|1− 2 for an L2 -atom for H p , we get

p 2

1 2 0

|Q∗ | ( p )

√ np p (2 n)n− 2 |Q|1− 2 .

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75

Z Q∗

1 1 p p M( f ; Φ)(x) p dx ≤ Cn,p |Q| 2 − p |Q|1− 2 = Cn,p .

Let k = [ np − n] + 1. In the following, we use the vanishing moments property of f (x) for all multiindices γ with |γ| ≤ k − 1. Next we have Φt ∗ f (x) =

Z Q

Φt (x − y) f (y)dy

= t −n

Z

Φ(t −1 (x − y)) f (y)dy " Z −n =t f (y) Φ(t −1 x − t −1 y) − Q

# (−t −1 y)β dy ∑ (∂ Φ)(t x) β ! Q |β |≤k−1 " # Z −1 y)β (−t = t −n f (y) dy ∑ (∂ β Φ)(t −1 x − t −1 θ y) β ! Q |β |≤k−1 −1

β

for some 0 ≤ θ ≤ 1, using Taylor’s Remainder formula. For y ∈ Q and x ∈ / Q∗ , |x − θ y| ≥ |x| − |y| ≥

|x| . 2

Using this inequality and the fact that ∂ β Φ are Schwartz functions, we obtain , for any N, |Φt ∗ f (x)| ≤ t −n

≤

Z

| f (y)|

Q

CN

∑

−1 |x| )N |β |=k (1 + t 2

CN,p,nt −(k+n) (1 + t −1 |x|)N

Z

f |y|2 dy

|t −1 y| dy β!

1 Z 2

Q

|y|2k dy

1 2

Q

t −(k+n)

≤

1 1 k 1 CN,p,n |Q| 2 − p |Q| n + 2 (1 + t −1 |x|)N

≤

CN,p,nt −(k+n) 1+ nk − 1p |Q| (1 + t −1 |x|)N k

1

= CN,p,n |Q|1+ n − p

(t −1 |x|)k+n |x|−k−n . (1 + t −1 |x|)N

In the two cases when t −1 |x| ≤ 1 and t −1 |x| > 1, N can be chosen accordingly so that we have sup t>0

Z (Q∗ )c

M( f ; Φ) p dx =

Z (Q∗ )c

p sup |Φ ∗ f |(x) dx t>0 pk

≤ CN,p,n |Q| p+ n −1 pk

≤ CN,p,n |Q| p+ n −1 pk

≤ CN,p,n |Q| p+ n −1

Z (Q∗ )c

|x|−pk−pn dx

Z

|x|−pk−pn dx

√ 3 n

B(0, 2 `(Q))

Z √ 3 n 2 `(Q)

|γ|−pk−pn γ n−1 dγ

pk

= CN,p,n |Q| p+ n −1 `(Q)−pk−pn+n pk

= CN,p,n |Q| p+ n −1 |Q|

−pk−pn+n n

(t −1 |x|)k+n ≤ C. Then, (1 + t −1 |x|)N

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Contents

= CN,p,n . Hence, kM( f ; Φ)kL p ≤ CN,p,n and M( f ; Φ) lies in L p (Rn ).

Exercise 2.3.4 Extend Theorem 2.3.11 to the case 1 < q ≤ ∞. Precisely, prove that there is a constant Cn,p,q such that every Lq -atom A for H p satisfies kAkH p ≤ Cn,p,q . Hint: If 1 < q < 2 use the boundedness of the square function on Lq while for 2 ≤ q ≤ ∞ use its boundedness on L2 . Solution. p Fix an Lq -atom √ for H and support in a cube Q. Without loss of generality, we assume that Q is centered at 0. ∗ Set Q := 2 nQ and k := [n/p − n] + 1. Define r via 1/p = 1/q + 1/r. We have 0 < p ≤ 1 < q ≤ ∞. If q < ∞, it follows from H¨older’s inequality that Z

h

Q∗

2 i 2p ∆ (A)(x) dx j ∑

1

Z

p

≤

h

Q∗

j∈Z

2 i q2 ∆ (A)(x) j ∑

1

1

q

r

Q∗

j∈Z

≤ C kAkLq |Q| 1

1 Z dx

1 r 1

≤ C |Q| q − p |Q| r =C,

where we used the boundedness of the square function on Lq since q < ∞. If q = ∞, then Z Q∗

h

2 i 2p dx ∆ (A)(x) j ∑

1

p

≤

Z

h

Q∗

j∈Z

2 i 22 ∆ (A)(x) j ∑

j∈Z

≤ C kAkL2 |Q|

1 Z

1−1

2

Q∗

p

2

dx

1−1 p 2 1

1

1

1

1

1

≤ C kAkL∞ |Q| 2 |Q| p − 2 1

≤ C |Q|− p |Q| 2 |Q| p − 2 =C, Now, for x 6∈ Q∗ , since A(y) has k − 1 vanishing moments, we have Z

∆ j (A)(x) =

Q

A(y)ψ2− j (x − y)dx

j y)β (−2 = 2 jn A(y) ψ(2 j (x − y)) − ∑ (∂ β ψ)(2 j x) dy β ! Q n |β |≤[ p −n] " # Z (−2 j y)β β jn j =2 A(y) ∑ (∂ ψ)(2 (x − θ y)) dy, β! Q |β |=k Z

for some θ ∈ (0, 1), by Taylor’s remainder formula. Furthermore, since ψ has decay greater than the reciprocal of any polynomial, being in S , we may choose N so large so that the preceding displayed expression is bounded by 2

jn

Z Q

|A(y)|

k CN 2 j |y| (1 + 2 j |x − θ y|)N

dy.

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77

Continuing, since x 6∈ Q∗ , y ∈ Q and θ ∈ (0, 1), it follows easily that |x − θ y| ≥ |x| /2, thus 2 jnCN

Z

|A(y)|

Q

2 jk |y|k

dy ≤ CN 2 j(n+k) (1 + 2 j |x|)N ≤ ≤ ≤

Z

CN 2 j(n+k) N

(1 + 2 j |x|) CN

|A(y)| |y|k dy

Q

2 j(n+k)

(1 + 2 j |x|)N CN,n 2 j(n+k) (1 + 2 j |x|)N

kAkLq 1

Z

0

|y|kq dy

10 q

Q 1

1

|Q| q − p [diam(Q)]k |Q| q 1

k

|Q|1− p + n .

In summary, combining the preceding estimates and and summing over j ∈ Z, we obtain the (2.3.26) in the text. Since this inequality is independent of q, the remainder of the solution follows as in the proof of Theorem 2.3.11.

Exercise 2.3.5 (a) Suppose that skQ ≥ 0 for all Q ∈ D and k = 1, 2, . . . . Prove that

∞ k p

{ ∑ s }Q .α,q ≤ Q f p

k=1

∞

p

∑ {skQ }Q f.pα,q .

k=1

.α,q

(b) Prove the completeness of the spaces f p when α ∈ R, 0 < p ≤ 1, p ≤ q < ∞. Hint: Part (b): You may want to use part (a) together with the fact that if a quasi-normed space (X, || · ||) has the property ||x + y|| p ≤ ||x|| p + ||y|| p for all x, y ∈ X, then (X, || · ||) is complete if and only if for every sequence xk ∈ X with the property p < ∞ there is an x such that || N x − x || → 0 as N → ∞. ||x || ∑∞ ∑ ∗ ∗ k k=1 k=1 k Solution. (a) It follows from Exercise 2.3.2 (b) by induction that the claimed inequality is valid for finite sums. To pass to infinite sums we use the Lebesgue monotone convergence theorem which applies since the underlying space is L p (`q ). (b) Recall the following fact: if a quasi-normed space (X, || · ||) is p-normable, i.e., ||x + y|| p ≤ ||x|| p + ||y|| p for all x, y ∈ X, ∞ p then (X, || · ||) that || ∑Nk=1 xk − x∗ || is complete if for every sequence xk ∈ X with the property ∑k=1 ||xk || < ∞ there is an x∗ such −n as N → ∞. Proof of this fact: extract a subsequence xkn of xk such that k1 = 1 and ||xkn+1 − xkn || ≤ 2 . Then the series ∑Nn=1 (xkn+1 − xkn ) = xkN+1 − xk1 converges to a point x∗ in X, hence the subsequence xkn converges in x and by the quasi-triangle inequality the sequence xn converges to x∗ as well. .α,q Now let {sk }∞ k=1 be a sequence in f p which satisfies ∞

∑ ksk k pf.pα,q < ∞ .

k=1

Consider the sequence t = {tQ }Q∈D

k = ∑∞ k=1 |sQ | Q∈D . We have that

ktk p.α,q ≤ fp

∞

p

∑ {skQ }Q f.pα,q < ∞

k=1

.α,q

by the triangle inequality of part (a); thus t ∈ f p . It follows that for a fixed dyadic cube Q we have α

1

1

|Q| n − 2 + p

∞

α 1

k |s | = ∑ Q |Q| n − 2

k=1

∞

k |s |χ Q ∑ Q

k=1

Lp

≤ ktk f.α,q < ∞ p

k thus for any Q ∈ D the series ∑∞ k=1 |sQ | converges absolutely and hence it converges. We set

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Contents ∞

sQ =

∑ skQ k=1

.α,q

.α,q

and we let s = {sQ }Q∈D . The clearly s ∈ f p , since t ∈ f p and |sQ | ≤ tQ for every Q ∈ D. We claim that ∑Nk=1 sk −s f.α,q → 0. p Indeed, ∞

N k

∞

∑ s − s p.α,q ≤ ∑ |sk | p.α,q ≤ ∑ {|sk |}Q p.α,q → 0 Q Q Q f f f k=1

p

p

k=N+1

p

k=N+1

as N → ∞. The required conclusion follows.

Exercise 2.3.6 Show that for all µ, j ∈ Z, all N, b > 0 satisfying N > n/b and b < 1, all scalars sQ (indexed by dyadic cubes Q with centers cQ ), and all x ∈ Rn we have |sQ | min( j,µ) |x − c |)N Q Q∈Dµ (1 + 2

∑

n ≤ c(n, N, b) 2max(µ− j,0) b M

∑

b1 |sQ |b χQ (x) ,

Q∈Dµ

where M is the Hardy–Littlewood maximal operator and c(n, N, b) is a constant. Hint: Fix x ∈ Rn and define F0 = Q ∈ Dµ : |cQ − x| 2min( j,µ) ≤ 1 and for k ≥ 1 Fk = Q ∈ Dµ : 2k−1 < |cQ − x| 2min( j,µ) ≤ 1/b 2k . Break up the sum on the left as a sum over the families Fk and use that ∑Q∈Fk |sQ | ≤ ∑Q∈Fk |sQ |b and the fact that S Q∈F Q ≤ cn 2− min( j,µ)n+kn . k Solution. Define F0 = Q ∈ Dµ : |cQ − x| 2min( j,µ) ≤ 1 Fk = Q ∈ Dµ : 2k−1 < |cQ − x| 2min( j,µ) ≤ 2k ,

k ≥ 1.

Summing over the families Fk we get

∑ Q∈Fk

|sQ | min( j,µ) |x − c | N 1+2 Q

≤ C2−kN

|sQ |

∑ Q∈Fk

!1/b −kN

≤ C2

b

|sQ |

∑ Q∈Fk

−kN

Z

= C2

∑ S

Q

Q∈Fk

χQ dy |sQ | |Q|

!1/b

b

Q∈Fk

−kN+µ nb

[

= C2

Q∈Fk

Z 1/b 1 Q S Q S Q∈Fk

!1/b

≤ C cn 2

M

|sQ | χQ dy

∑ Q

Q∈Fk

Q∈Fk

" − min( j,µ) nb +µ bn +k nb −kN

b

!

∑ Q∈Fk

b

#1/b

|sQ | χQ (x)

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79

" −kN+k nb max(µ− j,0) nb

≤ C2

2

#1/b

!

M

∑

b

|sQ | χQ (x)

.

Q∈Dµ

Now summing over all k from zero to infinity and recalling that N > n/b, yields the desired inequality.

Exercise 2.3.7 Let A be an L2 -atom for H p (Rn ) for some 0 < p < 1. Show that there is a constant C such that for all multi-indices α with |α| ≤ k = [ np − n] we have

2p k+1 1 b )(ξ ) ≤ C A −22−pn ( n + 2 )+1 . sup |ξ ||α|−k−1 (∂ α A L (R ) ξ ∈Rn

[Hint: Subtract the Taylor polynomial of degree k − |α| at 0 of the function x 7→ e−2πix·ξ . Solution. Let Q be a dyadic cube in Rn . By definition, A is an L2 -atom for H p (Rn ) means (i) A is supported in Q, 1 1 (ii) RkAkL2 ≤ |Q| 2 − p , (iii) xγ A(x)dx = 0 for all multiindices γ with |γ| ≤ [ np − n]. Let α be a multiindex constant in Rn . By condition (ii) it follows that A ∈ L1 . So, Z

b )= A(ξ

A(x)e−2πix·ξ dx.

Q

Consider the Taylor polynomial of degree k − |α| centered at cQ , where cQ is the center of dyadic cube Q, of the function x 7→ e−2πix·ξ . Then, by using condition (iii) and the Taylor’s reminder formula, we get Z

αb

∂ A(ξ ) =

A(x)(−2πix)α A(x)e−2πix·ξ dx

Q k−|α| i h 1 γ −2πix·ξ ∂x (e )(cQ )(x − cQ )γ dx A(x)(−2πix)α A(x) e−2πix·ξ − ∑ γ! Q |γ|=0

Z

=

Z

=

A(x) Q

1 β −2πix·ξ ∂x (e )(−θ (x − cQ )x)(x − cQ )β dx β ! |β |=k−|α|+1

∑

where

0≤θ ≤1

But, ∂x e−2πix·ξ = (−2πi)β ξ β e−2πix·ξ . Taking absolute value we get β

b )| ≤ Cn,p |∂ α A(ξ ≤ Cn,p

Z Q

Z Q

|x − cQ ||α| |A(x)|

1 (2π)β |ξ |k+1−|α| |x − cQ ||β | dx β ! |β |=k+1−|α|

∑

|A(x)||ξ |k−|α|+1 |x − cQ |k+1 dx .

1

Note that |x − cQ | ≤ c|Q| n . Using this and the Cauchy Schwarz inequality, we obtain 1

0 b )||ξ ||α|−k−1 = Cn,p |∂ α A(ξ |Q|

k+1 n

0 ≤ Cn,p |Q|

k+1 n

0 = Cn,p |Q|

k+1 +1− 1 n p

kAkL2 |Q| 2 1

1

1

|Q| 2 |Q| 2 − p .

by condition (ii)

80

Contents 1

− 2p

1

Condition (ii) implies that kAkL2 ≤ |Q| 2 − p . That is, |Q| ≤ kAkL22−p . Also notice that, by the definition of k, Consequently, |Q|

k+1 +1− 1 n p

= |Q|

k+1 n

+ 1 − 1p > 0.

k+1 + 1 +( 1 − 1 ) n 2 2 p

2p k+1 + 1 − 2−p − 2−p n 2 2p

≤ kAkL2

− 2p

= kAkL22−p

1 ( k+1 n + 2 )+1

Therefore, the desired inequality follows.

Exercise 2.3.8 Let A be an L2 -atom for H p (Rn ) for some 0 < p < 1. Show that for all multi-indices α and all 1 ≤ r ≤ ∞ there is a constant C such that

2p 2|α| 1

α 2

|∂ A| b r0 n ≤ C A −22−pn ( n + r )+2 . L (R )

L (R )

Hint: In the case r = 1 use theL1 → L∞ boundedness of the Fourier transform and in the case r = ∞ use Plancherel’s theorem. For general r use interpolation. Solution. Since A is an L2 -atom for H p (Rn ), there exists a cube Q in which A is supported. Without loss of generality, assume Q is centered at the origin. Moreover, we have 1 1 ||A||L2 ≤ |Q| 2 − p . And, equivalently, − 2p

|Q| ≤ ||A||L22−p . First, let us consider the case r = 1. For this case, we will use H¨oler’s inequality, (0.1) and (0.2) as follows. Z 2 b )|2 = ∂ α A(x)e−2πiξ ·x dx |∂ α A(ξ ξ Q

2 Z α −2πiξ ·x A(x)e dx = (−2πix) Q

2|α|

≤ C|x|

2

Z

|A(x)| dx

Q

≤ C|Q|

Z

2|α| n

|A(x)|2 dx |Q|

Q

= CkAk2L2 |Q|1+ 2

≤ C|Q|1− p +1+

2|α| n

2|α| n

2|α| − 2p (2− 2p )+ n

≤ CkAkL22−p

2|α| − 2p ( n +1)+2

≤ CkAkL22−p

.

Next, consider the case r = ∞. For this case, we will use Plancherel, (0.1) and (0.2) to get Z

α 2

|∂ A| b 1 = |∂ α A(ξ b )|2 dξ L Q

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81

Z

α A(x)(ξ )|2 dξ |x\

= Q

Z

|x|2|α| |A(x)|2 dx

= Q

≤ |Q| ≤ |Q|

2|α| n

kAk2L2

2|α| 2 n − p +1 2|α| − 2p ( n − 2p +1)

≤ kAkL22−p

2|α| − 2p ( n )+2

≤ CkAkL22−p For any general 1 < r < ∞, set r = q=

1−θ q0

1 + qθ

=

1

1 1− 1r

1 θ

.

and have 0 < θ < 1. Let s0 = 2, s1 = 2, q0 = 1 and q1 = ∞. Since s =

1 = 2 and + sθ 1 Ls1 = L2 into

1−θ s0

= r0 , using Riesz-Thorin interpolation theorem, the mapping of Ls0 = L2 into Lq0 = L1 and 0

Lq1 = L∞ can be extended to that of Ls = L2 into Lq = Lr as follows: θ (1−θ ) 2p 2|α| 2p 2|α|

α 2 − 2−p ( n )+2

|∂ A| b r0 ≤ C kAk−22−p ( n +1)+2 kAkL2 L L 2|α| − 2p ( n +θ )+2

= CkAkL22−p

2|α| − 2p ( n + 1r )+2

= CkAkL22−p

.

Exercise 2.3.9 Let f be in H p (Rn ) for some 0 < p ≤ 1. Then the Fourier transform of f , originally defined as a tempered distribution, is a continuous function that satisfies

n | fb(ξ )| ≤ Cn,p f H p (Rn ) |ξ | p −n for of f . some constant 2Cn,p independent Hint: If f is an L -atom for H p , combine the estimates of Exercises 2.3.7 and 2.3.8 with α = 0 (and r = 1). In general, apply Theorem 2.3.12. Solution. First, consider the case when f = A where A is an L2 -atom for H p (Rn ). From Exercise 2.3.7 with α = ~0 we obtain 1 − 2p ( k+1 n + 2 )+1

b )| = C|ξ |k+1 kAk 22−p |A(ξ L

=: B1 .

From Exercise 2.3.8 with α = ~0 and r = 1 we obtain − 2p (0+ 12 )+1

b )| = C|ξ |k+1 kAk 22−p |A(ξ L It follows that

α(θ )

b )| ≤ Bθ1 B1−θ = C|ξ |(k+1)θ kAk 2 , |A(ξ 2 L where

We have

=: B2 . for all θ ∈ [0, 1],

2p k + 1 1 2p 1 α(θ ) = − + +1 θ + − + 1 (1 − θ ). 2− p n 2 2− p 2

(0.0.30)

82

Contents

2p k + 1 p θ− (θ + 1 − θ ) + 1 2− p n 2− p 2p k + 1 p =− θ− +1 2− p n 2− p 2 − 2p 2p k + 1 θ+ . =− 2− p n 2− p

α(θ ) = −

Hence, α Setting θ =

1 n k+1 p

1 n 2p 1 − p 2 − 2p −n =− + = 0. k+1 p 2− p p 2− p

− n in (0.0.30), one obtains n

n

b )| ≤ C|ξ | p −n ≤ C1 kAkH p |ξ | p −n . |A(ξ

(0.0.31)

Here we have used the fact that for every L2 -atom A, there exists C2 = C2 (n, p) > 0 such that (cf. Theorem 6.6.10) C2 (n, p) ≤ kAkH p (Rn ) . Now suppose we are given a general distribution f ∈ H p with atomic representation: ∞

f=

A j : L2 -atom for H p ,

λ j ∈ C,

∑ λ jA j,

j=1

where ∑∞j=1 |λ j | p ≤ C0 k f kHp p . Since the series f = ∑∞j=1 λ j A j converges in H p , it converges in S 0 /P and since the Fourier transform is continuous on this space, we have ∞

fb(ξ ) =

∑ λ j Abj ,

j=1

where fb denotes the distributional Fourier transform of f . Each Abj is a continuous function and in view of (0.0.31) the series ∑∞j=1 λ j Abj converges uniformly over compact subsets of Rn to a given continuous function. Then the distribution fb should coincide with this continuous function. This fact, together with inequality (0.0.31) imply ∞

| fb(ξ )| ≤

∑ |λ j ||Abj (ξ )|

j=1

≤

∞

p

∑ |λ j | |Abj (ξ )| p

1

p

,

0< p j we have Z

2 jn 2( j−ν)D

1

∑ CN |Q|− p +1 2 jα (1 + 2 j |x − cQ |)N dx. Q

M2 :=

j 0 to be chosen. Then a j and b j are supported in I j and L j respectively, they have mean value zero, and they satisfy ka j kL2 ≤ 2 j/2 = |I j |−1/2 kb j kL2 ≤ 2 j/2 = |L j |−1/2 provided c > 0 is small enough. Thus a j and b j are L2 -atoms for H 1 (R). We also have

Contents

85 ∞

h=

1

∑ c j1+ε (a j + b j )

j=1

since AvgI j (hχI j ) + AvgL j (hχI j ) = 0, in view of the fact that h is an odd function. Since the sequence { j−1−ε }∞j=1 lies in `1 , it follows that h lies in H 1 .

2.4. Singular Integrals on Function Spaces Exercise 2.4.1 Let f be an integrable function on the line whose Fourier transform vanishes on (−∞, 0). Show that f lies in H 1 (R). Solution. Since fb is continuous and vanishes on (−∞, 0), it must also vanish on (−∞, 0]. [ We have that H( f )(ξ ) = fb(ξ )(−i sgn(ξ )) = −i fb(ξ ) since fb vanishes on (−∞, 0]. Since the Fourier transforms of H( f ) and −i f are equal, applying the distributional inverse Fourier transform we deduce that H( f ) = −i f . Hence H( f ) lies in L1 , since f does so. It follows from Corollary 2.4.7 that f lies in the Hardy space H 1 .

Exercise 2.4.2 (a) Let h be a function on R such that h(x) and xh(x) are in L2 (R). Show that h is integrable over R and satisfies

khk2L1 ≤ 8 khkL2 xh(x) L2 . (b) Suppose that g is an integrable function on R with vanishing integral and g(x) and xg(x) are in L2 (R). Show that g lies in H 1 (R) and that for some constant C we have

kgk2H 1 ≤ C kgkL2 xg(x) L2 . Hint: Part (a): split the integral of |h(x)| over the regions |x| ≤ R and |x| > R and pick a suitableR. Part (b): Show that both H(g) and H(yg(y)) lie in L2 . But since g has vanishing integral, we have xH(g)(x) = H(yg(y))(x). Solution. (a) Let B := {x : |x| < ε}. Write Z

|h(x)| dx =

Z

R

|h(x)| dx +

B

Z Bc

|h(x)| dx.

Applying H¨older inequality with the exponents p = q = 2 to the first term in the above equality, one gets Z

|h(x)|dx ≤

B

Z

|h(x)|2 dx

B

1

2

1 1 |B| 2 ≤ (2ε) 2 khkL2 = 2ε

khkL2 kyh(y)kL2

1 2

1

(khkL2 kyh(y)kL2 ) 2 .

The second term can be estimated as follows Z Bc

−1

|h(x)||x||x| dx ≤

Z

2

|xh(x)| dx

1 Z 2

−2

Bc

|x| dx

1

2

=

2 kyh(y)kL2 ε khkL2

1 2

1

(kyh(y)kL2 khkL2 ) 2 .

Therefore, Z

" |h(x)|dx ≤ 2ε

khkL2 k(y)h(y)kL2

1 2

2 kyh(y)kL2 + ε khkL2

1 # 2

1

(kyh(y)kL2 khkL2 ) 2 .

86

Contents kyh(y)kL2 khkL2 ,

Choosing ε = (b) We have

the estimate in part (a) follows. 1 H(g)(x) = lim π ε→0

Z |x−y|>ε

1 g(y)dy. x−y

c ) = −i sign(ξ )b Since Hg(ξ g(ξ ), it follows, by Plancherel, that kHgkL2 = kgkL2 . Using our assumption that g ∈ L2 , we conclude 2 Hg ∈ L . We have Z 1 1 yg(y)dy H(yg(y))(x) = lim π ε→0 |x−y|>ε x − y and xH(g)(x) = Since g has vanishing integral, i.e.,

R

R g(x)dx

1 lim π ε→0

Z |x−y|>ε

1 xg(y)dy. x−y

= 0, we claim that the following equality holds H(yg(y))(x) = xH(g)(x).

Indeed, H(yg(y))(x) − xH(g)(x) = lim

Z

ε→0 |x−y|>ε

1 (y − x)g(y)dy = − lim ε→0 x−y

Z

g(y)dy = 0. |x−y|>ε

By Plancherel’s identity, we have kH(yg(y))kL2 = kyg(y)kL2 . Thus yH(g(y)) ∈ L2 . Applying Theorem 2.4.6, we get kgkH 1 ≈ kgkL1 + kH(g)kL1 . Therefore, 1/2 ˜ kgkH 1 ≤ C1 (kgkL2 kyg(y)kL2 )1/2 +C2 (kH(g)kL2 kyHg(y)kL2 )1/2 = C(kH(g)k L2 kyHg(y)kL2 )

which proves the assertion. Here we have used the result of part (a) and Plancherel’s identity.

Exercise 2.4.3 (a) Let H be the Hilbert transform on the real line. Prove the identity H( f g − H( f )H(g)) = f H(g) + gH( f ) for all f , g real valued Schwartz functions. (b) Show that the bilinear operators ( f , g) 7→ f H(g) + H( f ) g , ( f , g) 7→ f g − H( f ) H(g) , 0

map L p (R) × L p (R) → H 1 (R) whenever 1 < p < ∞. R f (t) dt is holomorphic on the upper half space and has Hint: Part (a): Consider product U f (z)Ug (z), where U f (z) = πi R z−t boundary values f + iH( f ). Part (b): Use part (a) and Theorem 2.4.6. Solution. (a) We have that the holomorphic function on R × R+ U f (z) =

i π

f (t) dt R z−t

Z

has boundary values f (x) + iH( f )(x) if z = x + iy, and y → 0+. For Re z > 0 consider the product

Contents

87

U f (z)Ug (z) = −

1 π2

Z Z R R

f (t) g(s) 1 dtds = − 2 z−t z−s π

Z Z R R

f (t)g(s) 1 1 − dtds = iU f H(g) (z) + iUgH( f ) (z) . t −s z−t z−s

If z = x + iy, we let y → 0+ to obtain ( f + iH( f ))(g + iH(g)) = i f H(g) + iH( f H(g) + i gH( f ) + iH(gH( f )) Equating the real parts we deduce f g − H( f )H(g) = −H( f H(g)) − H(gH( f )) . \ Also, H f (ξ ) = −i sgn(ξ ) fb(ξ ) yields

H 2 = −I

hence the preceding identity transforms to H( f g − H( f )H(g)) = f H(g) + gH( f ). (b) Let f , g be functions in the Schwartz class. Define T ( f , g) := f g − H( f )H(g).

S( f , g) := f H(g) + gH( f ) and

Notice that in view of the identities in part (a) and that fact that H 2 = −I we have H(S( f , g)) = H( f H(g) + gH( f )) = H 2 ( f g − H( f )H(g)) = − f g + H( f )H(g) = −T ( f , g) and H(T ( f , g)) = H( f g − H( f )H(g)) = f H(g) + gH( f ) = S( f , g) . Using Theorem 2.4.6, kS( f , g)kH 1 ≤ C (kS( f , g)kL1 + kH(S( f , g))kL1 ) = C (kS( f , g)kL1 + kT ( f , g)kL1 ) kT ( f , g)kH 1 ≤ C (kT ( f , g)kL1 + kH(T ( f , g))kL1 ) = C (kT ( f , g)kL1 + kS( f , g)kL1 ) 0

0

Using the L p to L p and also L p to L p boundedness of H, and H¨older’s inequality we obtain that kT ( f , g)kL1 ≤ k f kL p kgkL p0 + kH( f )kL p kH(g)kL p0 ≤ C p k f kL p kgkL p0 and also kS( f , g)kL1 ≤ k f kL p kH(g)kL p0 + kH( f )kL p kgkL p0 ≤ C p k f kL p kgkL p0 .

Exercise 2.4.4 Follow the steps given to prove the following interpolation result. Let 1 < p1 ≤ ∞ and let T be a subadditive operator that maps H 1 (Rn ) + L p1 (Rn ) into measurable functions on Rn . Suppose that there is A0 < ∞ such that for all f ∈ H 1 (Rn ) we have

sup λ x ∈ Rn : |T ( f )(x)| > λ ≤ A0 f 1 H

λ >0

and that it also maps L p1 (Rn ) to L p1 ,∞ (Rn ) with norm at most A1 . Show that for any 1 < p < p1 , T maps L p (Rn ) to itself with norm at most 1− 1 p p1 1− p1 1

C A0

1− 1p 1− p1 1

A1

,

88

Contents

where C = C(n, p, p1 ). R (a) Fix 1 < q < p < p1 < ∞ and f and let Q j be the family of all maximal dyadic cubes such that λ q < |Q j |−1 Q j | f |q dx . Write 1 S Eλ = Q j and note that Eλ ⊆ M(| f |q ) q > λ and that | f | ≤ λ a.e. on (Eλ )c . Write f as the sum of the good function gλ = f χ(Eλ )c + ∑(Avg f ) χQ j Qj

j

and the bad function

bλ = ∑ bλj ,

bλj = f − Avg f χQ j .

where

Qj

j

n

(b) Show that gλ lies in L p1 (Rn ) ∩ L∞ (Rn ), kgλ kL∞ ≤ 2 q λ , and that kgλ kLp1p1 ≤

Z

| f (x)| p1 dx + 2

np1 q

| f |≤λ

λ p1 |Eλ | < ∞ .

n

(c) Show that for c = 2 q +1 , each c−1 λ −1 |Q j |−1 bλj is an Lq -atom for H 1 . Conclude that bλ lies in H 1 (Rn ) and satisfies

bλ 1 ≤ c λ ∑ |Q j | ≤ c λ |Eλ | < ∞ . H j

(d) Start with Z

T ( f ) p p ≤ p γ p L

0

∞

λ p−1 T (gλ )| > 12 γλ dλ

+ pγp

Z ∞ 0

λ p−1 T (bλ )| > 12 γλ dλ

and use the results in parts (b) and (c) to obtain that the preceding expression is at most C(n, p, q, p1 ) max(A1 γ p−p1 , γ p−1 A0 ). p1 p −1

− p 1−1

Select γ = A1 1 A0 1 to obtain the required conclusion. n n (e) In the case p1 = ∞ we have |T (gλ )| ≤ A1 2 q λ and pick γ > 2A1 2 q to make the integral involving gλ vanishing. Solution. (a) Fix 1 < q < p < p1 < ∞ and f ∈ L p (Rn ). Fix λ > 0. By the Calder´on-Zygmund decomposition lemma for | f |q at height q λ , there exist maximal dyadic cubes Q j such that (i) | f (x)|q ≤ Rλ q a.e. x ∈ (Eλ )c where Eλ = ∪ j Q j . (ii) λ q < |Q1j | Q j | f |q dx ≤ 2n λ q . (iii) |Eλ | = | ∪ j Q j | = | ∑ Q j |. j

By (i), we have | f | ≤ λ a.e. on (Eλ )c . Let x ∈ Eλ . Then x ∈ Q j for some j and from (ii), we get λ q

λ }. Then 1

|Eλ | ≤ |{M(| f |q )| q > λ }| ≤ |{| f | > λ }| ≤

1 λ

(0.0.33)

Z

| f (x)| dx ≤

| f |>λ

Write f as the sum of the good function gλ = f χ(Eλ )c + ∑ j

1 |Q j |

Z Qj

f dx χQ j

k f kL1 . λ

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89

and the bad function bλ = ∑ bλj ,

bλj =

where

j

f−

1 |Q j |

(b) If y ∈ (Eλ )c ,

Z Qj

f dx χQ j .

n

|gλ (y)| ≤ | f (y)χ(Eλ )c | ≤ λ ≤ 2 q λ . If y ∈ Eλ , then y ∈ Q j for some j and using Jensen’s inequality, we get q Z Z 1 1 |gλ (y)|q ≤ | f |(y) dy ≤ | f |q (y) dy ≤ 2n λ q |Q j | Q j |Q j | Q j Thus,

n

||gλ ||L∞ ≤ 2 q λ . Now Z Rn

p1

|gλ (y)| dy ≤ ≤

Z

p1

| f (y)| dy +

(Eλ )c

Z Eλ

Z

| f (y)| p1 dy + 2

|gλ (y)| p1 dy

np1 q

| f |≤λ

≤ λ p1 −p || f ||Lp p + 2 Hence, gλ ∈ L p1 (Rn ) ∩ L∞ (Rn ) and

Z

kgλ kLp1p1 ≤

np1 q

| f (y)| p1 dy + 2

k f kL 1 < ∞. λ

λ p1

np1 q

| f |≤λ

λ p1 |Eλ |

λ p1 |Eλ |.

n

(c) Let c = 2 q +1 . We claim 1 bj cλ |Q j | λ Denote fQ j :=

1 R |Q j | Q j

f (y) dy. Then since

1 R |Q j | Q j ( f

is an Lq -atom for H 1 .

− fQ j ) = 0, we see that

1 bj cλ |Q j | λ and

Z Qj

is supported in Q j

1 b j (x) dx = 0. cλ |Q j | λ

Also, Z Rn

1 |b j |q (x) dx cλ |Q j | λ

1 q

1 = cλ |Q j |

Z

2

n q +1

q

Qj

| f − fQ j | (x) dx

1

=

1

q

1− 1q

λ |Q j |

1 |Q j |

Z Qj

q

| f − fQ j | (x) dx

1 Z q 1 1 q ≤ n | f | (x) dx 1 2 +1 1− |Q j | Q j 2 q λ |Q j | q 1

≤ |Q j | q −1 . It follows from Exercise 2.3.4, there exists Cn,q such that

1

bλj 1 ≤ Cn,q cλ |Q j | H

1 q

90

Contents

Hence,

kbλ kH 1 = ∑ cλ |Q j | j

1 k f kL1

bλj 1 ≤ ∑ cλ |Q j | kbλj kH 1 ≤ Cn,q λ ∑ |Q j | ≤ Cn,q λ |Eλ | ≤ Cn,q < ∞. cλ |Q j | λ H j j

(d) Now, for any γ, write kT ( f )kL p =

γp p 2p

≤ γpp

Z ∞

Z

0 ∞

0

λ p−1 |{|T ( f )| > 12 γλ }| dλ

λ p−1 |{|T (gλ )| > 21 γλ }| dλ + γ p p

Z ∞ 0

λ p−1 |{|T (bλ )| > 12 γλ }| dλ

≤ Ig + Ib , say. Now, using the fact that T maps L p1 (Rn ) to L p1 ,∞ (Rn ) with norm at most A1 , we obtain Ig ≤ γ p p

Z ∞ 0

−p1 λ p−1 A1p1 kgλ kLp1p1 ( γλ dλ 2 )

≤ C p1 γ

p−p1

A1p1 p

≤ C p1 γ

p−p1

A1p1 p

= I1 + I2 ,

Z ∞ Z0 ∞

λ p−p1 −1 kgλ kLp1p1 dλ λ p−p1 −1

Z | f |≤λ

0

| f (x)| p1 dxdλ +C p1 γ p−p1 A1p1 p

Z ∞

λ p−p1 −1 2

np1 q

0

say.

Then using Fubini, I1 = C p1 γ p−p1 A1p1 p ≤ C p1 γ p−p1 A1p1 p

Z ∞ 0

Z Rn

Z

λ p−p1 −1 | f (x)| p1 dxdλ | f |≤λ Z ∞ | f (x)| p1 λ p−p1 −1 dλ dx | f (x)|

1 | f (x)| p−p1 dx = C p1 γ p−p1 A1p1 p | f (x)| p1 n p1 − p R = C p1 γ p−p1 A1p1 k f kLp p . Z

And, I2 = C p1 γ p−p1 A1p1 p = C p1 γ

p−p1

A1p1 p

Z ∞ Z0 ∞

λ p−p1 −1 2

np1 q

1 λ p1 {|M(| f |q )| q > λ } dλ

λ p |{| f | > λ }|dλ

0

= C p1 γ p−p1 A1p1 k f kLp p . Hence,

Ig = C p,n,q γ p−p1 A1p1 k f kLp p .

Next using the estimate of sup λ |{x ∈ Rn : |T (bλ )(x)| > λ }| for bλ ∈ H 1 (Rn ) from the hypothesis, λ >0

Z ∞

λ p−1 |{|T (bλ )| > 21 γλ }| dλ Z ∞ 2 1 = γpp λ p−1 γλ |{|T (bλ )| > 12 γλ }| dλ γλ 2 0 Z ∞ 2 ≤ γpp A0 kbλ kH 1 dλ λ p−1 γλ 0 Z ∞ p−1 λ ≤ Cγ p−1 A0 p λ |Eλ |dλ λ 0

Ib = γ p p

0

λ p1 |Eλ |dλ

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91

≤ Cγ p−1 A0 p ≤ Cγ

p−1

A0 p

Z ∞ Z0 ∞

1 λ p−1 {M(| f |q ) q > λ } dλ λ p−1 |{| f | > λ }|dλ

0

≤ CA0 γ p−1 k f kLpP . Since

a+b 2

≤ max(a, b) ≤ a + b, kT ( f )kL p ≤ Ig + Ib ≤ C(n, p, q, p1 ) max(A1p1 γ p−p1 , γ p−1 A0 )k f kLp p . p1 p1 −1

Now using γ = A1

− p 1−1

A0

1

, we have p1 (p−p1 ) p1 −1 +p1

− p −11 A0 1

p1 (p−p1 ) p1 −1

(p−p )

kT k p,p ≤ C(n, p, q, p1 ) max A1

= C(n, p, q, p1 ) max A1 p1 (p−1) p(p1 −1)

≤ CA1

1− 1 p p1 1− p1 1

= CA0

(p−p )

− p −11 A0 1

p1 (p−1) p1 −1

A0

1

p1 −p p1 −1

!

, A1

p1 (p−1) p1 −1

, A1

− pp−1 −1 +1

!

A0

p1 −p ) p(p1 −1

A0

1− 1p 1− p1 1

A1

.

(e) In the case p1 = ∞, using the boundedness of T from L∞ to L∞,∞ with norm at most A1 , we have n

|T (gλ )| ≤ A1 kgλ kL∞ ≤ CA1 2 q λ . n

Picking γ > 2A1 2 q λ , we obtain that n

|{|T (gλ )| > 12 γλ }| ≤ |{|T (gλ )| > A1 2 q λ }| = 0. Thus Ig vanishes while the estimate for Ib remains exactly the same. Hence, the result is true for 1 < p < p1 ≤ ∞.

Exercise 2.4.5 Let Pt be the Poisson kernel and K j be the kernel of the Riesz transform R j . Let ϕb ∈ S is equal to 1 in a neighborhood of the origin. Then δ0 = ϕ + (δ0 − ϕ) and for a bounded distribution f (cf. Section 2.1.1) and t > 0 write (Pt ∗ K j ) ∗ f = (Pt ∗ K j ) ∗ (ϕ ∗ f ) + (Pt ∗ K j ) ∗ (δ0 − ϕ) ∗ f . Since Pt lies in L1 and ϕ ∗ f in L∞ , (K j ∗ Pt ) ∗ (ϕ ∗ f ) = R j (Pt ∗ ϕ ∗ f ) is a BMO function. The Fourier transform of ξ b )), which is a Schwartz function. Thus (Pt ∗ K j ) ∗ (δ0 − ϕ) is also a Schwartz (Pt ∗ K j ) ∗ (δ0 − ϕ) is −i |ξj| e−2πt|ξ | (1 − ϕ(ξ function and (Pt ∗ K j ) ∗ (δ0 − ϕ) ∗ f is a smooth function. Hence (Pt ∗ K j ) ∗ f = Pt ∗ R j ( f ) is a well-defined function for all t > 0 and j = 1, . . . , n. Let n−1 n < p < 1. (a) Show that there are constants Cn ,C1 such that for any f ∈ H p (Rn ) we have h n

i

sup Pδ ∗ f L p + ∑ Pδ ∗ Rk ( f ) L p ≤ Cn f H p

δ >0

when n ≥ 2 and

k=1

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h

i

sup Pδ ∗ f L p + Pδ ∗ H( f ) L p ≤ C1 f H p

δ >0

when n = 1. (b) Prove that there are constants C1 ,Cn such that for any bounded tempered distribution f on Rn we have h n

i cn f H p ≤ sup Pδ ∗ f L p + ∑ Pδ ∗ Rk ( f ) L p δ >0

k=1

when n ≥ 2 and h

i

c1 f H p ≤ sup Pδ ∗ f L p + Pδ ∗ H( f ) L p δ >0

when n = 1. Hint: Part (a): this is a consequence of Theorem 2.4.5. Part (b): define Fδ = (Pδ ∗ u1 , . . . , Pδ ∗ un , Pδ ∗ un+1 ), where u j (x,t) = n+1 and continuous up (Pt ∗ R j ( f ))(x), j = 1, . . . , n, and un+1 (x,t) = (Pt ∗ f )(x). Each Pδ ∗ u j is a harmonic function on R+ p p to the boundary. The subharmonicity of |Fδ (x,t)| has as a consequence that |Fδ (x,t + ε)| ≤ |(Fδ (·, ε)| p ∗ Pt )(x) in view of (2.4.53). Letting ε → 0 implies that |Fδ (x,t)| p ≤ |(Fδ (·, 0)| p ∗ Pt )(x), by the continuity of Fδ up to the boundary. Since Fδ (x, 0) R= (Pδ ∗ R1 ( f ), . . . , Pδ ∗ Rn ( f ), Pδ ∗ f ), the hypothesis that Pδ ∗ f , Pδ ∗ R j ( f ) are in L p uniformly in δ > 0 yields supt,δ >0 Rn |Fδ (x,t)| p dx < ∞. Fatou’s lemma implies (2.4.42) for F(x,t) = (u1 , . . . , un+1 ) and then Lemma 2.4.10 yields the claim. Solution. For notational convenience, in the argument below we take n ≥ 2. The case n = 1 is contained in the argument if the notation is changed. (a) For any δ > 0 we have n

kPδ ∗ f kL p + ∑ kPδ ∗ R j ( f )kL p j=1

n

≤ sup kPδ ∗ f kL p + ∑ sup kPδ ∗ R j ( f )kL p δ >0

j=1 δ >0

n

(by the definition of H p )

≤ k f kH p + ∑ kR j ( f )kH p j=1

n

!

≤ C k f kH p + ∑ Cn k f kH p

(by Theorem 2.4.5)

j=1

˜ f kH p ≤ Ck (b) Without loss of generality, we may assume that f is real-valued distribution. This means that h f , ϕi is real if ϕ is a real-valued Schwartz function. This property will be useful in applying Lemma 2.4.9. Assume that the right hand side of the inequality we would like to prove is equal to A < ∞. Let u j (x,t) = (Pt ∗ R j ( f )) (x),

j = 1, . . . , n.

un+1 (x,t) = (Pt ∗ f ) (x) and define for δ > 0 Fδ (x,t) =(Pδ ∗ u1 (x), . . . , Pδ ∗ un+1 (x)) = Pt+δ ∗ R1 ( f )(x), . . . , Pt+δ ∗ Rn ( f )(x), Pt+δ ∗ f (x) = u1 (x,t + δ ), . . . , u2 (x,t + δ ), un+1 (x,t + δ ) . n+1 Note that Pt is harmonic and so is u j and Pδ ∗ u j for every j = 1, . . . , n; moreover, Pδ ∗ u j are continuous on R+ (up to the n+1 boundary) for each j = 1, . . . , n + 1; hence Fδ is harmonic in R+ and continuous up to the boundary. Then, by Lemma 2.4.9, n−1 n−1 we have that |Fδ |q is subharmonic on Rn+1 + for every q ≥ n . Take q such that n < q < p < 1.

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93

Note that Z sup

Rn

t>0

1/p

p

h n

i

≤ sup sup Pt+δ ∗ f L p + ∑ Pt+δ ∗ Rk ( f ) L p ≤ A < ∞ .

|Fδ (x,t)| dx

δ >0 t>0

k=1

Thus we can apply inequality (2.4.53) with Fδ in place of F and obtain |Fδ (x,t + ε)|q ≤ (|Fδ (·, ε)|q ∗ Pt ) (x) for every x ∈ Rn ,t, ε > 0. Let ε → 0 and since Fδ is continuous up to and including the boundary we obtain |Fδ (x,t)|q ≤ (|Fδ (·, 0)|q ∗ Pt ) (x)

for every x ∈ Rn , t > 0.

Since Fδ (x, 0) = ((Pδ ∗ R1 ( f ))(x), . . . , (Pδ ∗ Rn ( f ))(x), (Pδ ∗ f )(x)) and Z Rn

|Fδ (x,t)| p dx

1/p ≤

Z

p

Rn

p q

> 1, we obtain

1/p

|Fδ (·, 0) ∗ Pt (x)| q dx

1/q

1/q

≤ kPt kL1 k|Fδ (·, 0)|q kL p/q = Cn,q kFδ (·, 0)kL p n

≤ Cn,p,q

Z

∑

n j=1 R

!1/p |Pδ ∗ R j ( f )| p dx

Z +

|Pδ ∗ f | p dx

!

n

∑ kPδ ∗ R j ( f )kL p + kPδ ∗ f kL p

= Cn,p,q

Rn

1/p

.

j=1

Thus we find that Z

|Fδ (x,t)|

Rn

p

1/p

!

n

≤ Cn,p,q sup kPδ ∗ f kL p + ∑ kPδ ∗ Rk ( f )kL p δ >0

= Cn,p,q A < ∞

k=1

for every δ > 0, t > 0. Using this and Fatou’s lemma with δ → 0, we obtain Z Rn

|F(x,t)| p

1/p

n

≤ Cn,p,q sup kPδ ∗ f kL p + ∑ kPδ ∗ Rk ( f )kL p δ >0

! = Cn,p,q A .

k=1

We have now proved that Z sup t>0

Rn

|F(x,t)| p dx

1/p ≤ Cn,p,q A.

q Furthermore, F is harmonic on Rn+1 + , and consequently |F| is subharmonic. Thus the hypotheses of Lemma 2.4.10 are now applicable. The conclusion of the lemma yields that the function

|F|∗ (x) := sup sup |F(y,t)| t>0 |x−y|0 |x−y| 0. To obtain the first inequality, split the function |ξ + η|s as ∑ j∈Z ∑k< j |ξ + η|s ψ(2 s − j −k b b ξ )ψ(2 η), where ψ is as in the proof of Theorem 7.6.1 and show that the operator corresponding ∑ j∈Z ∑k≥ j |ξ + η| ψ(2 ∞ norm bounded by a multiple of (kDr f k L∞ + kDt f k L∞ ) kgk L∞ by considering the terms j ≤ 0 and to the first symbol has L j > 0 in the sum. Solution. Using the notation in the proof of Theorem 8.6.1 we have Z

Ds ( f g)(x) =

R2n

|ξ + η|s fb(ξ )b g(η)e2πi(ξ +η)·x dξ dη Z

=

∑∑

2n j∈Z k< j R

Z

+∑

∑

2n j∈Z j≤k R

b − j ξ )ψ(2 b −k η) fb(ξ )b |ξ + η|s ψ(2 g(η)e2πi(ξ +η)·x dξ dη b − j ξ )ψ(2 b −k η) fb(ξ )b |ξ + η|s ψ(2 g(η)e2πi(ξ +η)·x dξ dη

= Π ( f , g)(x) + Π˜ ( f , g)(x). Now, we split Π ( f , g) (and then, similarly, Π˜ ) as follows Z

Π ( f , g)(x) =

∑

2n j∈Z R

b − j ξ )ϕ(2 b − j η) fb(ξ )b |ξ + η|s ψ(2 g(η)e2πi(ξ +η)·x dξ dη

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Contents

Z

=

∑

2n j≤0 R

Z

+∑

2n j>0 R

Z

=

∑

2n j≤0 R

Z

+∑

b − j ξ )ϕ(2 b − j η) fb(ξ )b |ξ + η|s ψ(2 g(η)e2πi(ξ +η)·x dξ dη b − j ξ )ϕ(2 b − j η) fb(ξ )b |ξ + η|s ψ(2 g(η)e2πi(ξ +η)·x dξ dη |ξ + η|s r f (ξ )b d b − j ξ )ϕ(2 b − j η)D ψ(2 g(η)e2πi(ξ +η)·x dξ dη |ξ |r |ξ + η|s t f (ξ )b d b − j ξ )ϕ(2 b − j η)D ψ(2 g(η)e2πi(ξ +η)·x dξ dη |ξ |t

2n j>0 R =: Π1 (Dr f , g) + Π2 (Dt

f , g).

We now look at the bilinear kernel of Π1 (the kernel for Π2 will be dealt with in a similar way). Z

Π1 ( f , g)(x) =

R2n

K1 (x − y, x − z) f (y)g(z)dydz,

(0.0.52)

where Z

K1 (y, z) =

|ξ + η|s b − j ξ )ϕ(2 b − j η)e2πi(ξ ·y+η·z) dξ dη ψ(2 r j≤0 |ξ |

∑ R2n

2 js

Z

=

−j b − j η)e2πi(ξ ·y+η·z) dξ dη, −r (2 ξ )ϕ(2 ∑ 2 jr ϕbs (2− j (ξ + η))ψd R2n j≤0

−r b b and ψ d where ϕbs (·) = | · |s ϕ(·) Using a Fourier series expansion we write −r (·) = | · | ψ(·). m

∑ n cm e2πi 8 ·t χ[−4,4]n ,

ϕbs (t) =

m∈Z

with

1 8n

cm =

Z

m

[−4,4]n

e−2πi 8 ·t ϕbs (t) dt

b − j ξ )ϕ(2 b − j η). Then we have so that 2− j (ξ + η) ∈ [−4, 4]n on the support of ψ(2 Z

K1 (y, z) =

∑ ∑ n cm 2 j(s−r) e2πi2 R2n

− j (ξ +η)· m 8

−j d b − j η)e2πi(ξ ·y+η·z) dξ dη ψ −r (2 ξ )ϕ(2

− j (ξ +η)· m 8

−j d b − j η)e2πi(ξ ·y+η·z) dξ dη ψ −r (2 ξ )ϕ(2

j≤0 m∈Z

=

∑ ∑ n cm 2

j(s−r)

Z R2n

j≤0 m∈Z

=

∑ ∑ n cm 2 j(s−r) 22 jn

j≤0 m∈Z

=

e2πi2

Z

m

R2n

2πi2 d b e2πi(ξ +η)· 8 ψ −r (ξ )ϕ(η)e

j (ξ ·y+η·z)

dξ dη

∑ ∑ n cm 2 j(s−r) 22 jn ψ−r ( m8 + 2 j y)ϕ( m8 + 2 j z).

j≤0 m∈Z

Hence, Z R2n

|K1 (y, z)|dydz ≤

∑ ∑n

|cm |2 j(s−r)

Z

j≤0 m∈Z

R2n

! =

∑2

j≤0

j(s−r)

22 jn |ψ−r ( m8 + 2 j y)||ϕ( m8 + 2 j z)|dydz !

∑

|cm | kψ−r k L1 (Rn ) kϕk L1 (Rn ).

m∈Zn

Since r < s we have ∑ j≤0 2 j(s−r) < ∞ and we have ∑m∈Zn |cm | < ∞ since by Lemma 7.6.2 we have |cm | ≤ (1 + |m|)−s−n . Consequently, it follows from (0.0.52), that

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257

Z

|Π1 ( f , g)(x)| ≤ k f k L∞ (Rn ) kgk L∞ (Rn ) |K1 (x − y, x − z)|dydz R2n ! ! ≤

∑ 2 j(s−r)

∑ n |cm |

j≤0

kψ−r k L1 (Rn ) kϕk L1 (Rn ) k f k L∞ (Rn ) kgk L∞ (Rn ).

m∈Z

Similarly, one gets Z

|Π2 ( f , g)(x)| ≤ k f k L∞ (Rn ) kgk L∞ (Rn ) |K2 (x − y, x − z)|dydz R2n ! ! ≤

∑ 2 j(s−t)

∑

|cm | kψ−t k L1 (Rn ) kϕk L1 (Rn ) k f k L∞ (Rn ) kgk L∞ (Rn ),

m∈Zn

j>0

with s − t < 0. Then

kΠ ( f , g)k L∞ ≤ C kDr f k L∞ + Dt f L∞ kgk L∞ . Interchanging the roles of f and g to deal with Π˜ then yields

Π˜ ( f , g) L∞ ≤ C kDr gk L∞ + Dt g L∞ k f k L∞ , and thus kDs ( f g)k L∞ ≤ C1

h

i

kDr f k L∞ + Dt f L∞ kgk L∞ + (kDr gk L∞ + Dt g L∞ ) k f k L∞ .

Next, for λ > 0 set fλ (x) = f (λ x) and similarly with gλ . We have kDs ( fλ )k L∞ = λ s k f k L∞ so plugging in fλ and gλ in the inequality obtained we get

kDs ( f g)k L∞ ≤ C2 λ r−s kDr f k L∞ + λ t−s Dt f L∞ kgk L∞

+C2 λ r−s kDr gk L∞ + λ t−s Dt g L∞ k f k L∞ .

By optimizing in λ we obtain the claimed identity.

Exercise 7.6.5 (a) Let Θ , Ω be Schwartz functions whose Fourier transform is supported in an annulus that does not contain the origin. Let s > 0. Show that the function σ (ξ , η) =

∑

2−s j

j≥−2

∑ Θb (2−k (ξ + η))Ωb (2−( j+k) ξ )Ωb (2−( j+k) η) k∈Z

satisfies |∂ξα ∂η σ (ξ , η)| ≤ Cα,β (|ξ | + |η|)−|α|−|β | for all multiindices α, β satisfying |α| + |β | < s. (b) Let Ψ be a Schwartz function whose Fourier transform is supported in an annulus that does not contain the origin and let s > 0. Show that for 1 < r < ∞ and 0 < γ < s we have

b r n 2 < ∞, sup σ l Ψ L ((R ) ) β

l∈Z

γ

where σ l (ξ , η) = σ (2l ξ , 2l η ). Hint: Part (b). Use the 3-lines lemma to show that

b (2l−k (ξ + η))Ω b (2l−( j+k) ξ )Ω b (2l−( j+k) η)Ψ b (ξ , η) sup ∑ Θ

l∈Z

k∈Z

where the norm is taken in the variables (ξ , η).

Lγr ((Rn )2 )

≤ C 2 jγ ,

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Contents

Solution. b we obtain a factor 2− j−k ≈ |ξ |−1 ≈ |η|−1 ≈ (|ξ | + |η|)−1 . (a) Notice that the sum in k is finite. If a partial derivative falls on Ω b we obtain a factor of 2−k ≈ 2 j (|ξ | + |η|)−1 . So, if the differential operator ∂ α ∂ηβ falls If a derivative ∂ξi or ∂ηi falls on Θ ξ b we obtain a factor of 2−k(|α|+|β |) ≈ 2 j(|α|+|β |) (|ξ | + |η|)−|α|−|β | . So for the series in j to converge the total number of on Θ differentiations should be |α| + |β | < s. (b) In view of the support properties of these functions we have that the sum over k is over the set k = l − j + c, c lies in some finite set. Thus we are looking at a term of the form

∑

b (2 j−l−c (ξ + η))Ω b (2−l−c ξ )Ω b (2−l−c η) . 2−s jΘ

j≥−2

It will suffice to show that

∑

j≥−2

b (2 j−l−c (ξ + η))Ω b (2−l−c ξ )Ω b (2−l−c η)Ψ b (ξ , η) r n 2 < ∞, 2−s j sup Θ L ((R ) ) γ

l∈Z

b and Ω b we have that the only l ∈ Z that matter in this calculation are For a fixed j ∈ Z, in view of the support properties of Ψ 0 0 l = −c + c where c lies in a finite set. To compute the Sobolev Lγr norm of the function b (2 j−c0 (ξ + η))Ω b (2−c0 ξ )Ω b (2−c0 η)Ψ b (ξ , η) G j (ξ , η) = Θ we argue as follows: First, we notice that if N is an even nonnegative integer, then |(I − ∆ξ ,η )N/2 G j (ξ , η)| ≤ CN 2 jN χA where A is a fixed annulus. Picking two even nonnegative integers N1 and N2 such that N1 < s < N2 and applying the 3-lines lemma, using that the function (I − ∆ξ ,η )z/2 G j (ξ , η) is entire in z, we deduce that |(I − ∆ξ ,η )γ/2 G j (ξ , η)| ≤ C 2 jγ χA and then the Lr norm of the preceding expression is also bounded by C0 2 jγ . Thus the series converges when γ < s.

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