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.
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OF
DIFFERENTIAL OPERATORS MATHEMATICAL PHYSICS An
Introduction
Dr. rer. nat. GUNTER RheinischWestfalische
HELLWIG
Professor of Mathematics Technische Hochschule Aachen
Translated from the German by BIRGITTA
HELLWIG,
Fil. kand.
A vv ADDISONWESLEY Reading,
Massachusetts

Palo Alto

PUBLISHING London

Don
COMPANY Mills, Ontario
This book is in the ADDISONWESLEY
Copyright
©
SERIES
IN
MATHEMATICS
1964 by SpringerVerlag,
Berlin/G6ttingen/
Heidelberg
German edition al in ig or e th of n io at sl an tr English This is the only authorized hed by Springeris bl pu , ik ys Ph en ch is at em th der ma entitled Differentialoperatoren elberg. Verlag, Berlin/Gottingen Heid
L RIGHTS AL C. IN Y, AN MP CO NG HI IS SONWESLEY PUBL DI AD BY 67 19 © ANY FORM T GH IN RI PY ED CO UC OD PR RE BE T NO Y RTS THEREOF, MA PA OR , OK BO IS TH STATES OF . ED ED IT RV UN SE E RE TH IN D TE IN PR R. HE OF THE PUBLIS CATALOG SS RE NG WITHOUT WRITTEN PERMISSION CO OF Y AR BR LI . DA NA EOUSLY IN CA AMERICA. PUBLISHED SIMULTAN CARD NO. 6714209.
Preface
This
book
is intended
to give
an
introduction
to the field
of differential
operators. Students who have had a thorough course in advanced calculus and ordinary differential equations will be able to read it without serious difficulty. Hence, an introduction to the Hilbert space and its operators is included. The differential operators of physics are mostly partial differential operators. Among these, the interest today is focused on those partial differential operators whose independent variables x,,..., x, vary in the entire ,,, since the Schrédinger operators of quantum mechanics have this property. Therefore, we have used in this text such operators in preference to the classical operators. An introduction to the Hilbert space § is given in Part 1. Part 2 deals with the operators in §. Such partial differential operators and, above all, Schrédinger operators have been used as examples for symmetry and boundedness from below. Part 3 presents the spectral theory of completely continuous operators,
which is sufficient for the classical differential operators. In Part 4, the spec
tral theory of Schrédinger operators is developed and for this, the spectral theory of selfadjoint operators in § is indispensable. The central spectral theorem for such selfadjoint operators is stated with explanations and comments; however, a proof of the theorem is not given in this book, as most
contemporary textbooks contain such proofs. Part 5 discusses the spectral theory of the Weyl differential operator. Since this is an ordinary differential operator, our presentation will not be a complete one. Another reason for this lack of detail is that in recent years, some excellent textbooks on this topic have been published. On the other hand, we treat in more detail the question of which boundary conditions can be used, since those following directly from the theory have such a complicated structure that they do not lend themselves to applications. K. O. Friedrichs and later F. Rellich have, under additional assumptions, given boundary
conditions which can easily be applied; however, these have not so far been presented in the textbook literature. By giving several examples, we hope to iii
iv
PREFACE
increase the interest in this important problem which, as a matter of fact, arises even with respect to very simple differential operators, for example, the Bessel differential operator. This book does not discuss how to compute explicitly the spectrum of a differential operator. Such a computation can usually be carried out only if, by separating the variables, the eigenvalue equation of the partial differential operator can be converted to eigenvalue equations for ordinary differential equations. Moreover, for these, sufficient knowledge in the field of special functions must be available. An excellent treatment of this kind of problem can
be found
in the books
by
E. C. Titchmarsh,
which,
however,
require a
considerably larger number of auxiliary tools from the theory of ordinary differential operators (in particular the Titchmarsh formulas for the spectral function) than can be presented in this introduction. It is hardly necessary to mention that the reader need not have any knowledge of quantum mechanics. The occasional references to quantummechanical problems may be omitted by the reader without loss of continuity. The list of references has intentionally been made short since the space limitations do not allow a comprehensive survey of the literature in this field. An excellent survey is given in the books by Dunford and Schwartz. In the text, we will come back to these and other works which are not explicitly mentioned in the preface. I wish to thank my wife and coworker, fil. kand. Birgitta Hellwig, for her help, from the planning to the completion of this book. Her assistance has been an invaluable contribution to the simplifying and improving of many of the presentations. My thanks also go to Dr. H.W. Rohde, for his conscientious help in the proofreading, and to the publishing house, SpringerVerlag, for its excellent cooperation.
Berlin, April
1964
G.H.
Translator’s Note
For the American edition the author has made available a number of new problems; presented a generalization of the investigations about the symmetry of Schrédinger operators; and added theorems on the regularity of eigenpackets and the spectral family, and a new proof of the essential selfadjointness of the WeylStone operator.
Contents
Hilbert Space Linear Space, Metric Space, and Banach Space Linear space . Metric space . . Complete metric space Banach space
.
Hilbert Space 5 Definition of Hilbert space Complete Hilbert space. Separable Hilbert spaces Dense subspaces.
Orthonormal Systems in 26 28 32
Definition and Bessel’s inequality . Complete orthonormal systems , The E. Schmidt orthogonalization process
PART
2
Chapter 4
Linear Operators in § Eigenvalue and Inverse Operators
4.1 4.2 4.3 4.4
Definitions and formulation of problems. The SturmLiouville operator in %, Auxiliary tools from the theory of partial differential equations The SturmLiouville operator in &,
Chapter 5
Symmetric
Definitions Symmetry and
boundedness
from. below:
the Sturm Liouville
33
Symmetry
boundedness
from. below:
the Sturm Liouville
Chapter 6
6.1 6.2 6.3 6.4
operator
in ®,
operator
in ®,,
A
49
55
Operators and Operators Bounded from Below
5.1 5.2:
5.4
37 39
and
SturmLiouville
from below
operator
in
Ne, “which
is “not
bounded
59 61
65 WZ,
The Schrodinger Operators
Some principles of quantum mechanics Energy operators : Symmetry of the Schrédinger operators , Boundedness from below: the Schrodinger singular potential
‘
operators
with
76 80 82 98
vi
CONTENTS PART
3.
Spectral Theory of Completely
Chapter
7
Completely
oe ee NIANAW
7.1 7.2.
Chapter
8
8.1 8.2 8.3. 8.4.
Continuous Operators
Continuous Operators
Definitions . The expansion theorem ‘for completely
operators
continuous, symmetric
P
A stronger form of the expansion theorem The complete continuity of integral operators The complete continuity of integral operators, continued The general SturmLiouville eigenvalue problem in Xt, The SturmLiouville eigenvalue problem in %, .
The initialvalue and boundaryvalue problem for Au + u—=f The initialvalue and boundaryvalue problem for Au + ii =f Green’s function in initialvalue and boundaryvalue problems boundaryvalue and for initialvalue theorems Existence problems .
4
Spectral Theory of SelfAdjoint Operators
Chapter
9
Preliminaries
9.2
A new version of the expansion continuous and symmetric operators Projection operators
theorem .
for
completely
130 132 134
138
145 150
SelfAdjoint Operators
Chapter 10 10.1
10.2 10.3 10.4 10.5
Definitions
.
.
The spectral theorem for self adjoint operators The spectrum of a selfadjoint operator . Eigenpackets 2S Relation between eigenpackets and E,
:
:
154 157 159 162 166
Essentially SelfAdjoint Operators
Chapter 11
11.1
11.2 11.3
11.4 11.5 Chapter 12 12.1 12.2 12.3. 12.4 12.5
PART
109 iTS 116 123 125 127
InitialValue and BoundaryValue Problems
PART
9.1
105
5
Chapter 13. 13.1
13.2
Definitions
: Examples Criteria for essential ‘self adjointness
; ee
Acriterion for essential selfadjointness of differential operators Proof of the Weyl lemma SelfAdjointness of Differential Operators Schrédinger operators with singular potential Coulomb potentials with interaction... Differential operators bounded from below . Summary. . The lowest point of the spectrum of an operator ‘bounded from below . The WeylStone
Eigenvalue
Weyl’s Alternative Preliminaries
.
The first Weyl theorem
—
172 174 iil 185 193 200 204 206 211 214
Problem
221 223
aienieal
CONTENTS
13°3 13.4 13.3 Chapter
14
14.1 14.2 14.3 Chapter 15
15.1 153
The second Wey] theorem Weyl’s alternative a ee las A criterion for the limit point case at
SelfAdjointness of the WeylStone
ea
x = «
22] 233 234
oe
Operator
237 244 244
Fundamental theorem 3 e The SturmLiouville operator in ®, The expansion theorem . The Rellich Boundary Regular Singular Point
Conditions
Regular singular point . The Rellich initial numbers Application and examples . Appendix
1
Appendix 2 Appendix 3
Appendix 4 Bibliography Index
.
.
Vil
for
Limit
Circle
Case
and
FRANZ
RELLICH
in memoriam
PART
HILBERT
1
SPACE
CHAPTER
Tl
Linear Space, Metric Space, and Banach
1.1.
LINEAR
Definition 1.
Space
SPACE A set 2 of elements u, v, w,... is called a linear space if
1. for every pair of elements we %, ve X there is defined w=u+ ve &, which is called the sum of u and v;
an
element
2. for every complex number « and every element ue & there is defined an element z = au € &, which is called the product of « and u; 3. sum and product obey the follows laws: (i) (iil) (ili)
utv=v+u4, (utv)+w=ut+(v+w), given two elements ue &, ve &, there exists at least one element zéesuch thatu+z=v, (iy yeo la =u, (v) (Pu) = (aB)u for arbitrary complex numbers «, f, (vi) («+ B)u = au + Bu, (vil) a(u + v) = ou + av.
If we make the restriction that «, B,... same laws hold, then &
must be real numbers
and the
is called a real linear space.
From a course in elementary algebra the reader will see that the above definition implies the following: (viii) There is one and only one element z € & satisfying uw +z=v. We denote z by v—u. (ix) u—wis independent of u and is denoted by 0; © is called the null element. Instead of © — u we will usually write —w. (x) Ou = © for every ve 2, and «© = © for every complex number «. (xi) (—a)u =a(—u) = —(au) for every uwe2 and every complex number «. Definition 2.
An expression of the form OyUy
+ A2U2 + °°
3
+ OU),
4
LINEAR SPACE, METRIC SPACE, BANACH
SPACE
12
where the elements u,, u2,...,u,E€ and a,,...,%, are complex numbers, is called a Jinear combination of u,,u2,..., Uy. The elements wy, v2,..., u P are said to be /inearly independent if any relation of the form
Oyu, +O,U,+°°*+a,u,=0
implies that 7, =«%, =*::=a,=0. are said to be /inearly dependent.
If this is not the case, then u,, u2,..., u P
Definition 3. {& is said to be finite dimensional, and more precisely, ndimensional if there are n linearly independent elements in & and if any n+ 1
elements in £ are linearly dependent. & is said to be infinite dimensional if for any natural number m there exist m linearly independent elements. Theorem 1. Let & be ndimensional and let u,,u,,...,u, be linearly independent elements. Then every element we & can be written as
U= for suitable complex determined by uw.
yy + Yale F °° * + Yall
numbers
y,,...,7,
These
(1) numbers
are
uniquely
Proof. & is ndimensional, and hence the elements w,, v2,...,u,,u are linearly dependent. Thus there exist numbers o,,..., 4,4 ,, not all zero, such that
04U, +°°+0,U, + &,41u = O. Here a,4,+0, for o,4;=0 would imply that a, =a,==a, =0, because u;,..., u, are linearly independent. If we let y; = —a,/a,41, we get Eq. (1). Suppose that there exists another expression of the form
U= PU,
(2)
+ Pole +°°* + Mnttn
Subtraction of (2) from (1) gives O=
independence
The
linear
1.2.
METRIC
Yn = Pn
(fp = Jays
Pee
+ (1 — P2)la +t
of u,,...,u,
implies
that
(3)
y, =7,, 7. =72,>
SPACE
Definition 1. A set IN of elements uv, v, w, ... is called a metric space if to each pair of elements u, v € M there corresponds a number o(u, v), called the distance between u and v, with the following properties: 1. o(u, v) > 0, and e(u, v) = 0 if and only if vu = v,
2. Q(u, v) = o(v, u), 3. o(u, v) + a(v, w) = e(u, w) (triangle inequality).
2:
METRIC SPACE
5
EXAMPLE . The ndimensional Euclidean space consisting of all points
P: (X,,%2,..,%,) is denoted by ,. vectors,
and
x = (x,,.,,...,X,)
The points of ®,, are usually written as
denotes
the
vector
with
Z=(ZisZoy:e%5Z,)
and
2 =x; + ys
with
Y=(Vie Vasee+s¥,)
and
y, = ax
which
represents
the
point P. The components x,, x3,..., x, of the vector are real numbers. We define addition x + y, multiplication «x with a real number «, and length x in the usual way:
Z=x+y y= ax
[x=
(1)
) Sing j=l
If P, Q are points of ®,,, represented by the vectors x and y, respectively, then
Ix — yl /3 (x;— y,)? j=1
is, as usual, the distance between the points
P and Q.
o(x, y) = x — yl.
Hence
(2)
With this definition of distance, the set of all vectors corresponding to points in R,, forma metric space. It is easy to prove that g has the desired properties. Consider any three points O, P, Q in ®,, and choose O as the origin.
Suppose that O, P, O are represented by the vectors 0 = (0,0), x = (x;, x2), Y=()1, 2). Let O, P, O be vertices of a triangle. Then the lengths of the sides of this triangle are given by x — 0] = x, y — o = y, and x — y}.
These lengths satisfy the inequality
x + lyl=lx— yl, or equivalently,
Q(x, 0) + (0, y) = a(x, y).
(3)
This is the reason for calling the inequality in Definition  the triangle inequality. Finally, we consider the set of all vectors x = (x,, X2,...,X,) whose components are complex numbers and where the distance is defined as
o(x, y) = es
= ile
This set with the given distance definition also forms a metric space. With our definition we have a welldefined concept of distance. We can now define convergence of sequences of elements.
6
LINEAR SPACE, METRIC SPACE, BANACH
SPACE
153)
Definition 2. A sequence u,,¥,... of elements, where u; € M, is said to be convergent if, given any € > 0, there exists a positive number N(e) such that for a suitable uw € WM, O(u,, U)
N(e).
Then wu is called the /imit of the sequence 14, v>,..., and we write Jay
n> 00
Theorem 1. one limit.
10,
A convergent sequence of elements w,, u2,... has one and only
Proof. Letlim,.,., U, = uand lim,.,,, U, =v with u + v. Then o(u, v) > 0 and, say, @(u, v) =d. From the triangle inequality we obtain
d = Q(u, v) < Q(u, Uy) + Q(U,, 0).
(4)
By Definition 2 we have @(u,, u) < d/4 for all n > N(d/4) and @(u,, v) < d/4
for alln > N(d/4).
Hence o(u, v) < d/2, which is a contradiction.
Definition 3. A sequence u,, v5, .. . of elements, where u; € Mt, is said to be a fundamental sequence if, given any ¢ > 0, there exists a positive number N(e) such that Q(Uyj, Um)
N(e).
A convergent sequence of elements w,, v3, ... is a fundamental We
have
0(u,, uv) < e/2 for all n > N(e/2), and
hence
Q(Un» Um) S Q(Uq, U) + Q(U, Up) < € for all n,m > N(e) with N(e) = N(e/2).
Of great importance is the question whether the converse of Theorem 2
holds.
Unfortunately this is not the case.
EXAMPLE 2. The set u,v, w,... of all rational numbers with the usual distance definition o(u, v) = w —v forms a metric space Wt. The sequence Wy, Uz,... with u; = (1 + 1//)/ is a fundamental sequence which is not convergent, because there is no rational number for which lim,.,,, vu, =u holds. 1.3.
COMPLETE
METRIC
SPACE
Definition 1. t is said to be a complete metric space if every fundamental sequence in Wt converges. We shall prove that every metric space which is not complete can be extended to a complete metric space by adding suitable elements. The
1.3
COMPLETE
METRIC
SPACE
ij
procedure used is analogous to the Cantor procedure of extending the system of rational numbers to the system of real numbers.
Definition 2. ZT is said to be a dense subset of IN if (a) T is a subset (not necessarily a proper subset) of Mt and (b) for every we9 there exists a sequence uw, U,...€ 2 such that lim,.,, u, = u.
In particular,
T is always a dense subset of itself.
As sequence u, we
merely choose u, u, u,... Definition 3.
Let Yt be a metric space with elements
Om (u,v), and let Mt be a metric Om (u, v). If there is a onetoone the elements wu, v,.. . € Mand the Om(i, 5),..., then Mand Mi are In questions
dealing
uw, v,... and distance
space with elements @, #,... and distance correspondence, ui, vd, ..., between elements @, 3,... € M such that Qg,(u, v) = said to be isometric.
only with distance
of elements,
e.g. convergence,
completeness, etc., we can consider isometric spaces as equal.
Theorem 1. If Nt is a noncomplete metric space, then by adding appropriate elements, we can extend Ito a complete metric space MM in such a way that
IM is a dense subset of Mi.
Note. Wtis noncomplete if there is at least one fundamental in I which is not convergent. Proof.
sequence
We shall complete the proof in four steps.
Step 1. As an abbreviation we write {u,;} €%t for the fundamental sequence W,;, W@2,...€ Mt. The set of all fundamental sequences,
u={uj},
o=
{v;},...
with
u,, Ju U Cpa x € Wi,
(1)
is denoted by M. Two elements i, i in 9 are here considered to be equal if and only if
lim Qyy(Uy, U,) = 0,
(2)
n>
where Qgy(u, v) is the distance between uw and v in YN. the distance between two elements #7, & € Was
We will, as a trial, define
Om(U, V) = lim Ooy(U,,, V,)
(3)
n> co
This limit always exists because Qan(Uns Un) S Oo(Uny Um) FH Qon(Ulins Om) + Omn(Lms Un)
(4)
Oan(Uns On) = Qon(Uims Um) S Conny
(5)
and hence Um) F Qon(Cns On)
8
LINEAR SPACE,
METRIC
SPACE,
BANACH SPACE
153)
A change of indices gives Oon(Ums Um) — Oon(Uns Un) S Qon(Ums Un) + Con(Uns Um)
(6)
and from (5) and (6) it follows that lQan(Uns Pn) = Conny
Um) S Conlin» Um) + Osn(Uns Um) < E
for all n,m > N(e). By Cauchy’s convergence numbers 4, = Qgy(U,, U,) IS Convergent.
criterion
the
sequence
(7)
of
It remains to be proved that the limit in (3) is independent of the choice
of fundamental sequences, which are to be considered equal in M. Let {u;}, {u;} and {v,}, {vj} be fundamental sequences in Mt with lim, Qon(Mn» Un) = 0 and lim,,. , Qoy(U,, ¥,) = 0. Then we must prove that
Him Qyq(Uy, Un) = LIM Qg_(h, Up)
(8)
Qan(Un» On) S OmlUny Un) + Oon(Uns Un) + Oon(Up» Un)
(9)
Tim Qgn(Uy, Un) S 1M Qgy(Uy, Un) 5
(10)
Oslin, Un) S OonlUns Un) + Qan(Uns Un) + Qan(Ynr Un)
(11)
From
it follows that no
noo
and
gives
Lim Qyy (Uj, Up) S 1M Og_(Uy, Un)
n?o
(12)
n>o
Finally, (10) and (12) together give (8).
Step 2. It will now be shown that M with distance og(a, 6) is a metric From Qyy(u,,U,) 20 it follows, by using the formula (3), that space. ox, 6) > 0. The equality og(a, 6) =0 implies lim, .,, Qm(Up, l,) = 0, and according to our definition, this relation holds if and only if a= 0. Furthermore, from Qgy(U,, U,) = Osn(Uns Un) We also obtain ag(U, 6) = Om, H). The triangle inequality follows from
Osn(Uny Un) S Qan(Uns Wn) + Qon(Wns Un) by a limiting process: Om(U, 6) = lim Qo (u,,, V,) < iM Og_(Uy, Wa) + LIM Ogn(Wp, Vn)
= Om (U, W) + Q—(W, 0). Hence M is a metric space.
13
COMPLETE METRIC SPACE
9
We consider the set of all fundamental sequences of the form uemM,
with
{usu u;...}
U—
which are called stationary fundamental sequences. This set is a subset T of MN. Let 7, 6 be two such stationary fundamental sequences. Then from (3) it follows that Og (UZ, 0) = lim Og,(u, v) = Og (u,
Vv).
(13)
no
There is a onetoone correspondence, @ = {u,u,u,...} N(e), which implies that lim,..,, om(a, u,) = 0. Hence for every av € M there exists a sequence w,, v>,...€ MW for which lim,..,, uv, = a holds.
Hence 9 is dense in M.
Step 4. Now we will prove that IN is complete. Let a, #,... be an arbitrary fundamental sequence in I. The set Mt is dense in Mt, and hence for every #; there exists a u; € Wt such that OR(Uj,
Uj) 0 and u = 0 if and only if v= ©; 2. lo = a  for any complex number a; 3. lu + v < ul + v]] (triangle inequality).
B is said to be a real Banach space if ® is a real linear space and if property 2
is satisfied for real numbers only. Theorem 1. space.
When the distance is defined as Q(u, v) = u — v, B is a metric
Proof. The distance u—v has the properties that (a) w —v > 0, and  — v = Oif and only if u = v; (b) jw — v = v — ul]; and (c) u — vl] < uw — w + w — vl]. Properties (a) and (b) are obvious; (c) follows from
u — v = e — w+ w— vl = [[(u — w) + (w— v) I< llu — wl] + [lw — of. Thus the given distance definition satisfies the axioms of a metric space. As 8 becomes a metric space when o(u, v) = u — vl], everything that we have said about metric spaces is true also for this 8. In particular, the concepts of convergence, fundamental sequence, completeness, subset, dense subset are at our disposal.
Moreover, we
shall find it convenient to use the
norm instead of the distance. For example, a sequence u,, v2, . . . of elements u; € Bis said to be convergent if, given any ¢ > 0, there exists a positive number N(é) such that for a suitable ue B lu, —ull
N(e).
Then wu is again called limit of the sequence u,, u,,..., and we write Jbie i ses, Uy 8 We complete our discussion of Section 1.2 by the following theorem. Theorem 2.
If lim,..,, uv, =u, lim,.,. UV, =v, then
1. lim,...,(au, + Bv,) = au + Bv for arbitrary complex numbers «, f;
2. lim,
ll4nll = lle,
3. lim!
0,4 — ov for every
Proof.
ue Bit lim,) a, —«.
The first formula follows from
I[(ate, + Bv,) — (ae + Bv) < lol lu, — ul] + [BI [len — oll, and formula 3 follows from
oe,u — oxte  = Joe, —  lul
1.4
BANACH
SPACE
11
For 2, we use a stronger form of the triangle inequality,
 lull — llol  < Ju — of, which follows from
u] = lv + (u — v) < Jol] + Ju — of) and
lol] = e + (v
—u) < ull + lo — ull = ull + Ju — of.
Hence,  u, — lz  < , — ul], and formula 2 becomes obvious. To Definition 2 in 1.3 we shall add a few more definitions. Definition 2. Tis said to be a subspace of B if T is a subset of B and u, implies wu + fv € I for arbitrary complex numbers « and f.
ve T
In particular, every subspace of 8 is again a Banach space.
Definition 3. ZT is said to be a dense subspace of B if T is a subspace and T is a dense subset of 8. Definition 4. I is said to be a closed subspace of B if I is a subspace of B and if for every convergent sequence u,,u2,...¢ 2 with lim,_.,, vu, =u, we have we I. Theorem 3. Let 8 be a noncomplete Banach space. Then, by adding elements, we can extend 8 to a complete Banach space &, such that B is a dense subspace of 8. Proof. First, 8 is a metric space with distance g(u, v) u,v é 8. By the method described in 1.3, we can extend metric space B. It remains to be proved that addition u + au by an arbitrary complex number «, and the norm u of over to B, so that B
is again
a Banach
space.
That
= B v, 8
u — v for all to a complete multiplication can be carried
the distance
o(u, v) in
® is carried over to B was already shown in 1.3. Step 1.
Let u,v be two elements in B:
Wi
els
a. hs
UV = {js
where {u;}, {vj} are fundamental sequences in 8.
[oxtd, = OeUlyy = [at] [ty — I[(un
 he
(1)
Then we can see from
mall,
Un) = (ttm + Uma)ll = [(Un = Um) + (On = Um)
(2)
S lUa = Ul + [len — nll that {au,;} and {u,; + v;} are also fundamental sequences. The elements of the space B that are defined by these sequences we will denote by aw and
12
LINEAR
SPACE,
METRIC
SPACE,
BANACH
SPACE
1.4
u+v.* We have thus defined addition and multiplication by complex numbers in %, and it is easy to prove that with these definitions B is a linear space. Step 2.
In order to define a norm in &, we first note that for allu, v,
we B
o(au, av) = jau — av) = e] ju — ol] = alo(u, v), o(u + w,v + w) = \(u + w) — (v + w) = lu — v]] = o(u, v). The relations o(au, av) = ljalo(u,v) and e(u+w,v+w)=eo(u,v) easily proved to hold in 8 by a limiting process. Now we set
lull =o(u,©)
forall
2) can
ue.
Then the three axioms for the norm are satisfied. two; the triangle inequality follows from
. = o(u, ©) + o(v, ©) = lull + lol.
% is a complete Banach space, and ®
(4)
This is obvious for the first
lu + vl] = o(u + v, ©) < Q(u + v, v) + Q(v, O) Hence
be
(5)
is a dense subspace of B.
Since the concept of Banach space is not essential to the problems subsequently dealt with in this book, we shall mention only the standard work by Banach [2].f * It is easily seen that, as in Section 1.3, these definitions are independent of the special choice of the fundamental sequences which are to be considered to be equal in 8. + The number in the bracket is keyed to the Bibliography at the end of the book.
CHAPTER
®2
Hilbert Space
2.1.
DEFINITION
Definition 1.
A
OF
linear
HILBERT space
§
SPACE with
elements
u,v, w,...
is said
to
be
Hilbert space if to every pair of elements u, v € § there corresponds a complex
a
number (u, v), called the scalar product of u and v, which satisfies the following properties:
1. (u, v) = (v, u), and hence (uw, wu) is real;* 2. (u+v, w) = (u, Ww) + (v, W); 3. (au, v) = a(u, v) for every complex number ¢;
4. (u, u) > 0, where equality holds if and only if vu = ©.+
These properties imply: Theorem 1. 5. (u,v + w) = (u, v) + (u, Ww). 6. (u, av) = &(u, v) for every complex
number «.
7. (au, av) = a?(u, v).
Proof. 5.
(u,v + w) =(v'+ w, u) = (v, u) + (Ww, u) = (v, u) + (W, u) = (u,v) + (u, W).
6.
(u, av) = (av, u) = a(v, u) = a(u, v).
7. (au, av) = a(u, av) = a(av, u) =
a?(u, v).
Analogously we define a real Hilbert space as follows: Definition 2.
A real linear space § with elements wu, v, w,...
is said to be a
real Hilbert space if to every pair of elements u, v € § there corresponds a real * The expression (v,u) denotes the conjugate of (v, w). + Beginning in Section 2.4 we will consider only Hilbert complete and separable.
13
spaces
which
are
also
14
HILBERT SPACE §
eS
Zr
number (uw, v), called the scalar product of u and v, which satisfies the following ae il (u,
2
= (Us t)s
nee
w)= (u, w) + (v, w);
3. (au, v) = a(u, v) for every real number «; 4.
(u, u) > 0, where equality holds if and only if u = ©.
The reader can easily formulate the corresponding consequences (5), (6),
(7) for real Hilbert spaces. The properties listed in the remainder of this section always refer to a (complex) Hilbert space. The following theorem shows that we may set u = V (u, u), where ul of u.
can be interpreted as the norm
Hence, the Hilbert space § with norm
\u = V(w, u) is a Banach space. Theorem 2.
With
u = V (u, u) we have:
1. lul > 0, and u = 0 if and only if vu = O; 2.
\\au = a u for every complex number «;
3. lu + vl] < lul] + loll \ze — vl] < lu — wl] + lwo] 4.
(triangle inequalities) ;
(u, v) < lull lel]. (Schwarz’s inequality), only if wu and v are linearly dependent.
where
equality
holds
if and
Proof. Parts 1 and 2 follow immediately from properties 4 and 7, respectively, of the scalar product. Let us consider part 4. Let v = ©. Then Schwarz’s inequality holds with the equality sign because (u, v) = 0 and  v] = 0. Moreover, Ou + lv = 0, w=u+ av with an so that uw, v are linearly dependent. Now let v + ©. Set arbitrary complex «. Then we have 0 0. If =(u + v,u + v) =(u
If w+v= 0, w+ v+ O, then
then
tv, u) + (u +
v, v)
HILBERT SPACE
1S)
jw + vl] < u + lloll
< (u + v, u) + (u + v, v) < lu + off lull + [lu + off [oll by 4. Dividing both sides by w + vl], we get the first inequality in part 3. The second inequality follows from lu — v = ju If we set u = J(u that has been said about particular, the concepts of subspace, dense subspace, we can complete Theorem Theorem 3 lim n> Ho “n
—w+w—ovdl < lu — wll + lw—o]. u), $ becomes a Banach space. Hence everything Banach space is true also for Hilbert space. In convergence, fundamental sequence, completeness, closed subspace, etc., are at our disposal. Now 2 of Section 1.4 with the following
(Continuity of the scalar product). eathenlim)s \(@,, 0.) = (u, v).
If
lim,.,u, =u
and
Proof. We claim that v, < C for all n. To prove this, we set ¢ = 1. Then v, < v, — vl] + lol] < 1+ lvl] for all 2» > N(1). We define
C = max { (lv,, lleal,., llexll, 1 + loll}, where K is the largest natural number as claimed. Hence we have
< N(1).
Then
lv, < C for all n,
(Ups Up) — (Us 0) = [Cun — Uy V_) + (Us ¥, = »)
S lla — ull [lenll + ul fle, — ell < Clu, — ull + [lull lle, — vl] N(e), which is equivalent to lim (u,,0,) = (u,v).
n> oo
By
setting
u,=v,,
2.2.
COMPLETE
we
can
obtain
but this proof is valid only in §. HILBERT
another
proof
for
lim,.,,,u,] = lull,
SPACE
Definition 1. is said to be a complete Hilbert space if every fundamental sequence in § converges. With methods analogous to those used for Banach prove the following theorem.
space, we can now
16
HILBERT SPACE $)
ae
22
Theorem 1. Let § be a noncomplete Hilbert space. Then, by adding new elements, we can extend § to a complete Hilbert space § such that § is a dense subspace of §. Proof. § with u = Vu, u) is a Banach space. Hence, according to 1.4, can be extended to a complete Banach space by the addition of new elements. We use the extension obtained in Section 1.4 and we must prove that the definition of scalar product (u,v) in § can be extended so that \w\ =
(uw, u) holds for all we § and the algebraic laws for scalar product
also hold in §. If {u;}
and
{v;}
are
two
fundamental
sequences
convergent in $ because of the completeness of §.
lim
n>o
v, “n =v.
in ,
then
they
are
Let lim,..,, u, =u and
We consider the sequence of numbers a, = (u,, v,) and see that
lan — Gl = [Cn Un) — (tins Um)  =
(teins v,
Dra) +
(u,
Uns
Vm) as
(u,
—
Uns
Vp —
Vm)
S [uml en = mall + [Un = Ural (Om) + Hey — Mall Op — Omlle
CL)
From this it is clear that a, — a,, < € holds for all n,m > N(e). According to Cauchy’s convergence criterion, the sequence of numbers {aj} is convergent; that is, lim,.,,, a, =a. We set (u, v) = lim (u,, v,),
(2)
noo
and we must prove that the limit is independent of the choice of representative fundamental sequences. Let {uj}, {uj} and {vj}, {v;} be fundamental sequences in $) with lim,_.,, v, — u/ = 0 and lim,.,,, lv,— v; = 0. Then we can easily prove that lim (u,, v,) = lim (u/, v/). no
Let {w;}
we.
be another fundamental
We consider the relations
(3)
no
sequence with lim,_,,, w, = w and w ed,
(Uns Un) = (Uns Un),
(Ug + Ups Wn) = (Uns Wn) + (Uns Wn)s
(cttlyy Up) = 2E(Uys Vy)s [gl] = /(uys Uy)
(4)
Taking the limits of both sides in these formulas, we find that the scalar product defined in (2) satisfies properties 1, 2, 3, and 4 in 2.1. From 1.4 and
1.3 it follows that § is a dense subspace of §.
EXAMPLE 1. Consider the set of all vectors w,v,w,... with u= (Uy, U2,...), U= (U4, ,...),..., Which have denumerably many com
2.2
COMPLETE HILBERT SPACE
ponents, where inequalities
the
components
are
complex
io 6]
ye;? < 00,
numbers
and
satisfy
5
the
¥ ojl2< 0, ...
ian
j=1
We define uv + v, au, and (uv, v) in the following way: w=u+ov
with
w=(W,,W2,...)
and
w,=u;+
Z— OU
with
Z—
and!
“Ze atts
(Zi 25
(u, v) ape
ee)
0;
u 0;
j=1
(dS)
Then it follows that )7_,z;? < oo and )72,w,? < 00 because w,]? = lu; + vj? < 2u,? + 2I0,7.
(6)
The convergence of the series in (5) follows from u;0; = u lv, Ss 4{u,?
a lol
(7)
By verifying the axioms one by one, we easily conclude that the set of all vectors with the abovementioned properties and the given operations form a Hilbert space §. The null vector (0,0, ...) becomes the null element O.
The space § is complete. Let uv, u’,... be a fundamental sequence
in §, with
0 = (u™, u®, u®, ...).
(8)
Then ju”
u™
=
AGE
as
uh,
Te)
uh”)
=
ay
—
ue
e
N(e).
In particular, it follows from (9) that
For fixed j, the Cauchy
numbers
n,m > N(e) and every j.
for all
ju —u™ N(e).
(12)
18
HILBERT SPACE
For
m> ©,
yp)
:
k
HB WO7 —ujls 2
for all
n> N(e)
(13)
n> N(e).
(14)
g=1
by (11).
k > 00, we obtain
Finally, letting
=P
Lie
Ss Ee
for all
Hence
%. e ™ u — u , 4) (1 by en th If we set u = (1, U2>)s w= (u—u) +u € , that s ow ll fo it 4) (1 om fr and
ju
e s e l a — ” \u yy ‘ = l — ul j=1
Thus § is complete.
\(u, v) < , ty li ua eq in s ’ z r a w h c S rm o f e th ve ha , v \\ + l ul ju + vl] < [l
ull vl],
ea
in
=
ahi
does
and
j=1
(15)
n> N(e).
for all the
triangle
ae
z
inequality,
(16)
Ie ¥ ) + l b D sf ? o + uy F
. D is R, in t se t in po d e t c e n pen con o an be D t Le 2. e L EXaMp a ball in or , &, e l o h w e th be D ten let of ll wi e W . ®, in n i by a , R, in be a dom ts in po e th be ri re,we desc o f e b s ,a If 0. > r us center a and radi 1,:>7 x,) by x =f
X= (% or ct ve e th of th ng le and denote the by d e b i r c s e d is K ll ba d ne then the abovementio
K:xal ,%1 K( = x) Furthermore, let k( roduct by p ar al sc a ne fi de to y D. We tr n Oi > x) k( th wi D in (19) u(x)v(x)k(x) ax: (u, v) = /D and s n O a i G x d = ®,, dx in t n e m e l e e m u l o v e Here dx stands for th
Did
COMPLETE HILBERT SPACE
19
Jp ° ++ dx is the volume integral over D, dx
Ue
=  GO.
An

ts
meg
1
Xace sid Xee
(20)
D is an open point set, and hence the integral in (19) may not exist. Therefore we consider only the set of all complexvalued continuous functions u(x), v(x), ... with the additional property  u(x)?k(x) dx < co, “D
{ le« luc) Z K(X) dx
We assert that this set with the given operations From
u(x) +
o(x)?k(x) < 2{lu(x)7k(x)
= .cO) nw
forms
a Hilbert space 9.
+ [o(x)7k(x)}
(21)
it follows that u(x) € § and v(x) € § imply u(x) + v(x) € § and, of course, also au(x) € §. Finally, the existence of the integral in (19) is proved in the following way: From
(u(x)
k(x) — fox)
(22)
k(x)? = 0
we find that
2u(x) Jo(x)k(x)
(23)
< Ju(x)7k(x) + v(x) ?k(x).
Hence
 Ju(x)v(x)k(x) dx 0 there exists a function v(x) € § such that
ju — vl] = J Sime
— 0(x)7k(x) dx < «.
Henceforth by a Hilbert space we will always mean a complete Hilbert space. The most important example is the set of all complexvalued measurable functions for which Jolu(x)7k(x) dx exists in the Lebesgue sense.* This Hilbert space we will (with a change in notation) denote by
oe {us
f ecorace) dx < co, D
(u, v) =  u(x)o(x)k(x) dx,
(28)
D
where {u(x)*} means the set of all u(x) with property *. Then the set of functions denoted by § in Example 2 is a dense subspace of (28). Since we will almost always be working with dense subspaces, a knowledge of the theory of Lebesgue integrals will not be an absolute necessity for an understanding of this book. Problem
1.
Consider
the
set
of
all
continuous,
realvalued
functions
u(x)
in
—1 1 (and no larger than the number of elements in the sequence) let v,, v2,..., v, be linearly independent. Then there exists a finite or infinite orthonormal system Uy, Uz, ... € S which is equivalent to v,, 5,....
Proof. We first prove that all the elements v,, v,... must be different from the null element. Suppose, to the contrary, that v; = ©. Set n =j and note that the relation Ov,
+ 0v,
+°+0v,_,
+10,
=0
(1)
holds, so that v,,..., v; are linearly dependent, contrary to the assumption. We set uy = 0,/X,], so that (u,,u,)=1. We construct uv, in two steps. Let fly = V2 — (V2, Uy )Uy
and observe that (#,, u,) = 0.
(2)
Then uw, is obtained from @,:
uy a2
Ue
(3)
Here i, + ©, for otherwise from (2) we would obtain =
(v2,
02
uy)
Ilexll
ti
(4)
which implies that v,, v2 are linearly dependent. We now have (u;, 4) = 5), Withjk = 15 2: Finally, suppose that u,, v2,..., u, with (Uy, U,) = Op ysJ, k = 1, 2.22490, have already been constructed. We then let n
ns.
=Ungt
—
y
j=l
(Vn 41> u,;)u;
(5)
3.3
THE E. SCHMIDT ORTHOGONALIZATION
and note that (41,4) and set
=Ofork
= 1,2,...,n.
PROCESS
We find that again @,,, + ©
Un,
aeltn+rll Thus (u;,u,) = 6;, for j,k =1,2,...,2+1, and an orthonormal system with the desired properties.
Theorem 2.
33
(6) by
induction
we
obtain
In every Hilbert space § there is a complete orthonormal system.
Proof. § is separable; hence there exists a sequence w,, w3,...€ 8 such that for every u € § and every ¢ > 0 we can find an element w, in the sequence such that  — w, 0 we can find elements v,,...,vy and complex numbers a,,..., 4, such that lu — pee a,v; a must hold for every eigenvalue 2 with a constant a. In other words, the point spectrum of A in must not extend to —o. This requirement is satisfied by all operators of
classical physics and the technical sciences. It is also satisfied by many operators of quantum mechanics, for example, by all operators which are energy operators (see Sections 6.1, 6.2). Another important problem in analysis is the question of whether the inverse of an operator exists.
Definition 5. Let A in 2 have the property that every fe YB, corresponds The rule which assigns to fe YW, the to one and only one element we. corresponding we Y, is a (linear) operator A~' with domain of definition
a' =,
and W,1=%.
inverse of A in QL.
Ao!
in W~!
is called the inverse operator or
4.2
THE STURMLIOUVILLE OPERATOR IN 9,
39
Obviously 47'Au = u for all we Wand AA~'f=f for allfe U*. easily seen that A! is linear.
It is
Theorem 1. 4 in YI possesses an inverse operator A”! if and only if A = 0 is not an eigenvalue of A in YI.
Proof. Suppose that A~' in 2~' = MW, exists. with Au = ©. Then 47 ‘Au = © and A~'Au =u. is not an eigenvalue. Now assume that 2 = corresponds to exactly one if f corresponds to two A(u, — uz) = ©. But 2 = 0 1Ss tj — U5. This theorem supplies
Let we YW be an element Hence u= © and 1=0
0 is not an eigenvalue. Then every fe WU element we YW such that Au =f holds. elements u,,u,¢%, then Au, =f, Au, is not an eigenvalue, and hence u, — u, =
' = W, Indeed, =f, or O, that
us with a criterion for the existence of A~*. But in applications we need an exact characterization of 2{~', and this is not given by Theorem 1. It is true that {~’ = WB,, but usually YW, cannot be simply characterized. 4.2.
THE
STURMLIOUVILLE
OPERATOR
IN
i,
Let us consider the differential equation (p(x)u')’ + (Ak(x) — q(x))u= 0
(1)
l 2 and log x for n = 2. Correspondingly, for x +, the functions
1 s(a, x)
=
(n — 2)@,
ja—x?"
for
n> 2, (8)
:
— — log a — x 2n
for
n=2
are solutions of A,w=0. They are said to be singularity functions because they have a singularity for x =a. In (8), w, denotes the surface of the ndimensional unit ball. Of course we have w, = 27, w3 = 4n. Definition 1.
Let ae D be a fixed point.
y(a,x) = s(a,x) + D(x)
defined for xe D, x +a,
The function
with ®(x)e C'(D),eC?(D), A,B =O0inD
(9)
is said to be a fundamental solution with respect
to D.
Theorem 1.
Let u(x) € C'(D),
A,w=f(x)
€ C?(D) be a solution of
with
f(x) ¢ C°(D).
Then for any arbitrary point x € D, u(x) can be written in the form
u(x) = [ bes y)uy(y) — u(y)p (x, y)] dS — Rice y)f(y) dy.
(10)
Proof. The function y(x, y) has a singularity for y = x. Hence we delete from D the ball K: y — x < @ with sufficiently small g. Application of the
43
EIGENVALUES AND INVERSE OPERATORS
52
second Green formula to the doubly connected domain D — K
gives
 Cylx, yA) — wy) Aw(x, »)] dy [~D=K =
“@D+0K
Lyx, yay) — u(y), vy dS.
(11)
Because of (9), the integral over 0K can be written as a sum of two integrals, namely,

0OK
[yu, — uy,] dS = 
“0K
‘
(su, — us,) dS + 
(Du, — u®,) dS.
OK
(12)
The last integral tends to zero as g > 0. If we set p = —v, then pis the outer y= x + pa. Since dS = 9" * da, normal of 0K and OK can be described by where dw is the surface element of the unit ball, we obtain (here we have assumed
that n > 2)
[ (su, — us,) dS = ———— (n — 2), JoK
xf 
ul=1
Lule + neyo?)y — Q? "u(x + Hey] deo. (13)
Since (0?~"), = (g7~")) = (2 —n)g'~", we have lim i (su, — us,) dS = e70 0K
(14)
—u(x).
For @ 0, the lefthand side of (11) has the limit {py(x, y) f(y) dy, and the proof is complete. The proof for n = 2 is analogous. For our purposes, the following application is important. Theorem 2.
for
Let g(t)e C? (0 [Fn dt + 2c2(r — rg),
88
THE SCHRODINGER
OPERATORS
ae
6.3
for r— 00, this is certainly not true. Thus F(r) < 2c? for all r, where 0 be sufficiently large.

The integral
Y pu (x) Dj udu b  Az,ll,
so that Aw, < [2,44] \l1,], which also holds for w, = ©. lw, ? in (23) gives
IIw,[l? = floll?— y I(v, pI? < lell?. j=1
(25)
A computation of
(26)
114
COMPLETELY
Moreover,
CONTINUOUS OPERATORS.
12
since lim;..,, A; = 0,
lim  Aw, ) < lim A, 44] [lwall < lim A, 4 ;/lol] =0. ncx
Now
n>
oO
noo
(27)
we have n
Aw, = Av — by (v, P;)AQ;, j=1
and thus the expansion theorem
Av= follows.
Equation
¥ (v, 9))Ag; = ¥ (v, 0)AjQ; j=1 j=
(28)
(28) can be rewritten as oo
i
2
jal
gat
ig
Av= ¥ (v, 4,9); = ¥ (v, Apo; = ¥ (Av, 9/9;
(29)
which is the desired expansion theorem. If ~,,..., @, is a finite sequence, then we get from (23) and (24) Aw, =© and n
n
(30)
Av = ¥' (x, 9))A9; = > (Av, 9/9; Jae
j=2
which again gives us the expansion theorem. Now it only remains to be proved that this method value of A in 2 which is different from zero. Let 2 + 0 A in YW with corresponding eigenelement ~. Let / be previously constructed eigenvalues 1,, 2,,... Then @ is Q;, 249 = Ag, and
yields every eigenbe an eigenvalue of different from the orthogonal to every
Ap =F (AQ, 9/9; = > AY, 9/9; = ©, J)
vw,
which is a contradiction to the assumption that 7g + ©. Thus the theorem has been proved. We have not claimed that the orthonormal system @~,, @2, . . . constructed in the above theorem is complete. Neither can this be expected to be the case, since we have not taken into account that 2 = 0 may also be an eigenvalue. Theorem 5. Let A in § be completely continuous and symmetric, and let A be other than the null operator. Let 9, g2,...€9 with (9; p;) = 6; be the eigenelements corresponding to 2,,/,,..., respectively, and let Wi, W2,...€H with (;, ~,;) =6,, be the eigenelements corresponding to 2=0.* Then ~;, ~2,.,W,, %2,... form a complete orthonormal system in §. * The eigenvalue A = 0 may have infinite multiplicity.
A STRONGER
73
Proof.
EXPANSION
OF THE
FORM
According to Theorem 4, for arbitrary
ve 5
SA enes
ee
ADO
Bue
Ls}
THEOREM
or when W = v — 9’; (v, @,)@
© = Av— 2 (Av, 9j)o; = A(v — » (v, pj)p;) = AW. Since any yw
satisfying the condition
W = >) a; with a, = (W, W,), o—
Ay =©
can
be written
in
the
PP; = 2 ei
form
(31)
ad
Now, W, and 9, are eigenelements corresponding to different eigenvalues. Hence (W,, g;) = 0, and we obtain «; = >, (a), W;) = (v, W;), so that by (31) we get v= >); (v, 9p; + DX (v, Wi). By Theorem 2 of Section 3.2, the proof is now complete. This theorem shows that the orthonormal system constructed in Theorem 4 is complete in § if and only if 2 = 0 is not an eigenvalue of A in 9.
The investigation of the eigenvalue problem for completely continuous
symmetric operators was first done by D. Hilbert [38]. here is due to F. Riesz [69]. 7.3.
A STRONGER
Theorem 1.
FORM
OF
THE
EXPANSION
The form
presented
THEOREM
Suppose that
1. A in YL with W, < YW is symmetric and completely other than the null operator;
continuous, and A is
2. for every weYW there exists a real nonnegative number \u < [wv] for all uw € YW with a fixed number « > 0;
[wv] such
that
3. for every u = Av with v € YU, there exists a w € YW for which
lim » — Py aj ] =)
n>
a; = (u, @;).*
(1)
Then for all uw = Av, v € YU, the expansion theorem holds even in the stronger form: lim E 
n> ox
. (4,0,)0,
j=1
= lim Ju
x (v, A@;) =
= lim e
¥ ae, =;
n> 0
n> 00
(2)
j=1
* Here we are interested only in the case of an infinite orthonormal
system.
COMPLETELY
Proof.
CONTINUOUS
OPERATORS.
7.4
We have
w=
Y ajo;
FEA
2
fod
w—
> 4,0;
j=1
a
116
Then lim,..,,” — 41 @;@,; =0. But according to Theorem 4 of Section 7.2, lim, “u —Xj=1 4;9; =0. Since the limit element wu is unique, we obtain w = u. The other versions of (2) are analogous to those in Theorem 4 of Section 7.2. EXAMPLE
.
Let
 Ju(x)? dx < col,
“wD
(u, v) =
D
u(x)v(x)
dx.
We consider a suitable operator A in W& = {uwe C°(D)}. If we set max, .5u(x), then, with V(D) = volume of D, we find that for all ue
ull =  [ luca? de < /V(D) max u(x) = au],
[u] =
(3)
xeD
where a = JV(D). 7.4.
THE
COMPLETE
CONTINUITY
OF
INTEGRAL
OPERATORS
Here we consider
o [uts) { luooPkon ax < ,
(u, 0) =f ulektx) de,
1)
and the integral operator A in Y& with
Au = [Ks y)u(y)k(y) dy
forall
=
we M;
(2)
UW = {u(x)u  C(D)}. We make the following assumptions concerning D, the kernel K(x):
(3) K(x, y), and
1. D is a normal domain in ,*; 2. K(x, y) = a(x, y)/x — y%, where a is complexvalued and continuous for
xeD,yeD;a(x,y)+0and0 2,0 0 there is a point y whose coordinates are rational and which satisfies the condition y — x Ois sufficiently small, then y € D, and this y also
belongs
to the sequence
2"), 2"?’,...
Hence
the inequality
holds for every x € D and for a suitably chosen /.
2'” — x 0 there is a natural number K(g) such that the balls x —z“” < owithn = 1, 2,..., K cover the domain D, since if this were not the case, then for every rail naniber m there would be a point x”? in
D such that x‘ — z  > @ forn=1,2,...,m. Dis bounded, and hence, according to the WeierstrassBolzano theorem, there exists a limit point x* of the sequence x"'?, x',..., and this limit point belongs to D. But then we have x* — z
> 0 forn=1,2,...
This is the desired contradiction to the
fact that 2", 2"), ... is dense in D (and also in D). Step3. Now
subsequence
12,3).
we
will show
that
g,(x), 92(x),g3(x),...
from /,(x), f5(x),...
such
that
lim,..,, nee
Pirst, we have that/,(x)= C for/—1, 2)...
the sequence of numbers f, (2), f(z“),
we
can
select a
exists
for
Ifiwenow form
..., then this sequence is bounded.
According to the WeierstrassBolzano theorem, the sequence has at least one
118
COMPLETELY CONTINUOUS OPERATORS
limit
point, and
thus
there
is a subsequence f,,(2"), f,2(z™),...,
pel) which lim,,.. /,,(2"”) exists.
of functions f;(x), /2(x),
lim,
/in(2"'’)
7.4
This fact can be stated as follows: The sequence
... has a subsequence /; ,(x), f,2(x),...,
exists.
for
For
the
same
reasons,
from
the
for which
subsequence
Fis(x), f12(x),... we can again select a subsequence f,,(x), fo2(x),... for which lim,..,, /o,(2°?’) exists. Continuing this process, we get the following scheme: fir(™),
fr2(%),
fis),
foi(*),
fr2(*),
foal),
fai(x),
fa2(%),
f33(*),
lim f,,(z) According
to
the
exists for
wellknown
7@=1,2,...,0=7 fon(Z~) exists for the sequence/, ;(%),/f55(x), fa3(), ... for j= 1,.2,..0: We now denote this diagonal sequence of the scheme Where g,(%) — Jan):
(5) by g;(x), g2(x), ...
Step 4. We will show that for all xe D, the sequence g,(x), g2(x),... converges uniformly to a continuous limit function f(x). First, g,(x), g(x), ... is a subsequence of f;(x), f,(x),... Let an arbitrary ¢ > 0 be given. According to our assumptions, there is a 6(€) >0, such that g,(x) —g,(yv) N(e)
andall
xeD.
(7)
According to our hypothesis, the following inequality holds: lon(x) — g,(y)  with lim,;.,, A; =0 if there are infinitely many eigenvalues. The continuity of g(x, y, 1) implies that the expansion theorem holds even in the sense of uniformly absolute convergence. Every u = A~'v with v e YB, can be written in the form
u(x) = ¥ aj,(x) j=
with
a; = (u, @;).
(8)
If u(x) QU is an arbitrary element, then it follows from Au = Au that u=A~'v with v = Au. Hence (8) holds for all u(x) €. As a matter of fact, there must be infinitely many eigenvalues, for otherwise we would have x) — ey a, (x) for all u(x) € Ql, and this is certainly not true. Therefore
lim;_.. A; = 0.
According to the discussion above, we have 9,(x) el. If we set A; = (1/A,;) + p, then the A, are all eigenvalues of A in YW and ¢,(x) are the corresponding eigenfunctions. But since 2; > a and lim,_,,, A; = 0, we have also proved the statements about the possibility of ordering (after possibly renumbering the eigenvalues) and about the limit point. That every eigenvalue has at most multiplicity 2 simply follows from the fact that ¢,(x) is a solution of a homogeneous linear ordinary differential equation of the second order, which has exactly two linearly independent solutions. The investigation of the eigenvalue problem of integral operators and its application on eigenvalue problems for ordinary and partial differential operators was first carried out by E. Schmidt [72] and D. Hilbert [38].
7.7.
THE
STURMLIOUVILLE
EIGENVALUE
PROBLEM
IN
,,
We consider
Aue ey1 [—A,u + q(x)u] in
W={u(x)ueC*(D),
€ C7(D), Aue 5: u = O0forxe oD}
(1)
128
COMPLETELY
CONTINUOUS OPERATORS

el
under the same assumptions as in Section 4.4 with q(x) > 0 and assume that D is such that the Green function g(x, y) in Section 4.4 exists. Since according to Section 4.3, g(x,y) = s(x, y) + ®(y) for fixed xe D, g(x, y) and also g(x,)) will, by Section 4.4, satisfy estimates similar to those of s(x, y); that is,*
+c,
for
n> 2,
lg(x, y) 0
n>
Hence lim,.,.,.(Aw,,/) = (2,f) = 0, whence z = O.
Theorem 3.
Let A in YI be essentially selfadjoint.
Then A in Wis selfadjoint.
Proof. According to Theorem 2, the closure A in Y& exists. The subspace (A + iE) is dense in §, and hence for every v € § there exists a sequence U,, U>,...€ YW such that lim v, =v
n>
with
v, = (A+
iB)u,.
(4)
Now
[]0p = Umll? = A(t, — yp) + E(Ulg — Um)? = Attn — UmdI? + Un — Mall?
(5)
In particular, v,,v,,... form a fundamental sequence; thus from (5) it follows that u,,u2,... and Au,, Auy,... are also fundamental sequences. Since § is complete, the sequences are also convergent. If we set lim,..., u, = 4, then by the definition of closure lim,_,,, Au, = Au. Hence from (4) we obtain
v = Au + iu, so that (A + i£E)2 = §. Analogously we show that
(A — iE)M = §. Thus A in
is selfadjoint.
174
ESSENTIALLY
SELFADJOINT
OPERATORS 
H2
In general, an essentially selfadjoint operator A in 2 need not possess a spectrum, since the general spectral theorem only states that the spectral decomposition Au = '% A dE,u holds for selfadjoint operators and that the eigenelements and eigenpackets, which determine the spectrum, are in the domain of definition of the selfadjoint operator. However, the operators in physics usually have the extremely valuable property that all their eigenelements and eigenpackets are already in those usually considered subspaces of § in which the operators are only essentially selfadjoint. 1.
Problem
Problem 2.
1 on p. 60 is not essentially selfadjoint.
Show that A in & of Problem
Let © = {u(x)J6u(x)? dx < c0}. We consider the operators
Au=—u"
in
A=
{u(x)ueC? (0 2,
(3) for
n—?2)
For y € K, we define a function*
vly)= J w(x)AQPLo(w)s(y,x] dx +f s(y, xdeCoEntex) — atx)w(x)] dx, (4) which is not necessarily everywhere finite valued and is integrable over K, according to Fubini’s theorem. For, we have
A\(e(x)s(y, x)) < const y — x*~" > aw means13 D7 1(82/@x7).
(5)
194
ESSENTIALLY
SELFADJOINT
OPERATORS ~
ee)
since A,s(y, x) = 0 for y +x. Next we shall prove that this function v(y) coincides with w(y) almost everywhere in K. By (4) and changing the order of integration, we find that for all ue €(K,)
 v(y)u(y) dy = [
1K
“K,
wio( 
K2
u(p)AS?Lo(x)s(y, x)] ay) dx — a(x)w co( fa y)s(y, xy)
sae ot x)[n(x)
K2
ds.
Recalling that e(x) = 1 in K, and noting that AYs(y, x) =0 for y+x,
(6)
we
obtain from (6)
 o(y)u(y) dy = [
“Ki
+ 
— Ko
K
w(o(  u(y)AYLe(x)s(y, x)] dy) dx NK

o(x)[n(x) — a(x)w(x)]
K2
u(y)s(y, x) dy)
dx.
(7)
For xe K, — K2 fu yA” Lo(x)s(y, x)] dy = air(ovn

“K>
since the integrand has no singularities. Theorem
4 of Section 4.3, for
K2
s(y, x)u(y) ay),
If (x) € C'(K,), then, according to
x)= — J soe HO) dy = =] sty, 00) dy the equation A,@ = '¥ is satisfied.
(8)
Rewriting the equation, we obtain
W(x) = A,([KS
s(y, x)P(y) ay),
In particular, we may write u(x) for ‘¥(x).
Hence
u(x) + aso( f s(y, x)u(y) dy) = (0). K2
By adding zero to the righthand side, we obtain from (7) )
v(y)u(y) dy = ) w(x) (uls) + AM  s(y, x)u(y) dy) dx JK, K> K2
+[
Ky = Ka
w(siayn(as) ik
s(y, x)u(y) dy) dx 2
4  0(x)(n(x) — a(x)w(x)) “Ky
 s(y, xu(v) dy) dx. K2
(9)
11,5
PROOF
OF THE
WEYL
LEMMA
195
Again we remember that (x) =  for x € K, and finally obtain

Ke
v(y)u(y) dye

(x) (x) dx +f
w( x6?
(o (x f
+/ esos “Ky
“Kz
Ke
stv
u(y) dy) dx
s(y, x)u(y) ay) dx.
(10)
With W(x) = a(x), 5( x)u(y) dy we obtain from (10)
 Lo(y) — w(y)Ju(y) dy = — J [. sae )Dy — n(x)h(x)] dx. JK;
(11)
By Theorem 4 of Section 4.3, (x) € C?(K,), and w(x) = 0 in a neighborhood of 0K,. Hence W(x) and its first and second derivatives can be uniformly approximated in K, by functions u(x) ¢€ ((G) and their first and second derivatives. Thus the righthand side of (11) is zero by formula (2). However,
[ (u(y) — w(y))u(y) dy =0 1K2
for all
u(x) € €(K)
implies that v(y) = w(y) almost everywhere in K). Hence, by (4), the integral equation w(y) = lp w(x) AML o(x)s(y, x)] dx + iL s(y,
x)o(x)[En(x)
— a(x)w(x)] dx
(12) holds almost everywhere in K,.
with
fx,—x, w(x)
AY
Since g(x) = 1 in K3, the first term coincides
[o(x)s(y, x)] dx and is arbitrarily
many
times
differ
entiable for ye K,. The integral as5(¥, x)o(x)n(x) dx is in C?(K,) according to Theorem 4 of Section 4.3. Thus, if K,; is a ball concentric to K,, K; ¢ K), then, by (12),
y)= I; s(y, x)o@(x)a(x)w(x) dx + 9,(y) with g,(y)€C?(K3)
almost
everywhere
proved for the case 5 =0. For the case a(x) = 0 we find from
in K,.
Thus
(13)
the lemma
has
been
(13) that
lw(y)] < if Salle. Jk, ly —x
ies
(14)
almost everywhere in K3, where c,,c) are two constants and 1 = 2 —4,/2. For
n>
3 we
even
get
2 = 2.
Our
choice
of 2, which
makes
the estimate
(14) weaker, enables us to treat the cases n = 2 and n > 3 simultaneously and prevents the occurrence of the “logarithmic case”’ in further computations. Now, from (14) we get the desired properties of w(x) by iteration. If K, is a ball which is concentric to K; with K, < K3, then, analogously to
196
ESSENTIALLY
SELFADJOINT
OPERATORS
15
(14), we have
ean
sae
almost everywhere in K, with new constants c3, c,. in (14), we obtain
Beene
geil
(15) Renaming the variables
rae
mle zl"
(16)
almost everywhere in K3; inserting this in (15), we get
lw(y) w(y)
0 with y = y(u) > 0,
(3)
M = M(u) > O}.
The eigenvalue problem for A can be simplified by separating the variables. If we introduce spatial polar coordinates r, w, ~, with 0 < g < 2n, 0(x)2k(
“Xo
(22)
and since u,(x) and u(x) form a fundamental system of D,u = 0, we have has the proved that in the limit circle case every solution u(x) of D,u =0 property that
Juco) cl opt
252
WEYL’S ALTERNATIVE
ee
1333
Letting x > m, we obtain from (15)
lim {u, u}, = lim {uy, uy}.41 — Col? — 5} x>m x>m
= [lu s(o?k(x) att — Col? — 12).
(23)
If in u(x) = Cu,(x) + uz(x) we choose for ¢ all values on the limit circle ¢ — Col = ro and if for each such choice of € we set w(x) = Cu,(x) + up(x), then we obtain
lim {w, w}, = 0.
(24)
x>m
By (4) this also implies that lim,.,,,[w, w], = 0. third statement of the theorem.
Thus we have proved the
CASE 2 (Limit point case). Here we have lim,.,,,7, = Oandlim,.,,, 6, = Co, and hence the limit circle degenerates into one point. If in u(x) = Cu,(x) + u(x) we use only this special ¢value and set U(x) = Cou,(x) + u2(x), then,
since €, is contained in all disks {u, u}, < 0 with x9
< x < m, it follows that
[weorEce) dx < 0. From relation (18) we find that lim,.,,, {u,, u,}, = 00, which, together with
(12), then gives
 u4(x)2k(x) dx = 00. For the solutions U(x) = 0 and u,(x) of D,u = 0 we have
 U(x)?k(x) dx < oo xo
and
 u4(x)2k(x) dx = 00, xo
and hence U(x) and u,(x) are linearly independent.
system.
They form a fundamental
Thus every solution v(x) which satisfies J" v(x)?k(x) dx < oo must
have the representation v(x) = cU(x) with a suitable complex If we insert U(x) = Cou,(x) + u(x) in (15), we get by (18)
1
(U, U, U}U,} = —{(uyuy,, uyua}}? = —{Mo uy, tsuy}, Bud Since {u,, u,}, 2 00 for
x
aya Pun
number
tas
c.
. (25)
+m,
limfUl Ute — 0)
x>m
(26)
"so that lim... {V, v}, = 0 also. By (4) we also obtain lim,.,,, [v, v], = 0, which was the second assertion of the theorem.
WEYL’S ALTERNATIVE
13.4 13.4.
233
ALTERNATIVE
WEYL’S
Definition 1. We say that at to A occurs if for this A
x =m
(x =/) the limit circle case with respect
[ucoPkt) dx 0 for 0 0 for0i0,
= bop + byx +°°°
(9)
3: For large'x (x = R),
a(x) = ax
with
a>0,
(10)
b(x) = Bx +y.
By applying the Laplace transform* with respect to t on Eq. (8), with initial values lim,_,)9 U(x, t) = Uo(x) for 0 < x < 00, we get the corresponding *The formal calculus a complex parameter,
of the Laplace transform is as follows: Let s = € + in be and let fo e“'U(x, t) dt= u(x, s) be convergent for one
5=So. Then we say that u was constructed by applying the Laplace operator L=Jce() dt to U and we write LU=u. Then u(x, s) is analytic in s in the right halfplane € > > with so = &> + ino. Thus the functions U(x, t) are mapped by the Laplace operator on the functions u(x, s), which have best “smoothness properties” in s. Integrating by parts, we find the formal rule LU, = su — Uo(x), where we have left out the term arising from the upper limit 00, since we hope that this term turns out to be zero for £9 > 0. By the Fourier integral theorem we obtain formally the inverse operator 1
Unf
=114= rel 2
sitio
Jey 10
e*u(x, s) ds
with
Ei = Eo
(&)
ApplyingLZ on (8) with initial values lim,
U(x,t) = Uo(x) and a = a(x),b = b(x),we
find
to be differentiation
formally,
(d/dx),
with
prime
LU, = su— Uo(x),
(’) understood
with
respect
L((bU),) = (L(bU)). = (OLU), = (bu) = (bu)’,
L((—aU),x) = (L(—aU))5x = (—aLU) xx = —(au)xx = —(au)’,
to x
—
260
no.3
APPLICATION AND EXAMPLES
261
homogeneous equation —(a(x)u)” + (b(x)u)’ + su = 0, where s is a complex number.
O=— x < &,
(11)
Equation (11) can be written in the form
Be Oats te SE pee ee ee yr a S a= 6 a
a
(12)
Multiplying by
p(x) = exp( f=) a(x)
ds) = a?(x) exp( (ee dx),
and accordingly —(au)” + (bu)’ + su = Uo(x) Now,
for
0
ec:
(t)
for (8) we must impose not only initial conditions but also ‘boundary condi
tions” at x =0 and x= 00. In classical physics, the physicist usually finds such boundary conditions through the physical theory itself, and later the mathematician verifies that they are mathematically admissible also. In modern physics, it is not possible to adopt this procedure. For example, the quantum mechanical axiom which states that the operators of this theory must have a spectral decomposition or a complete set of eigenfunctions and eigenpackets determines the admissible boundary conditions—or better, the admissible domain of definition of the operator —and this domain is thus a consequence of the axiom. In order to find admissible boundary conditions for (8),we can proceed as follows. In (+) we set Uo(x) = 0. Then we get an eigenvalue equation with eigenvalue parameter s. With Uo(x) = 0 we put (+) in the form Au =u and thus determine the corresponding Weyl subspace 2. Now we say that those boundary condition for (8) are admissible, which after a formal application of the Laplace operator transform into boundary conditions admissible in the subspace &. Thus in the limit point case there are no boundary conditions; in the limit circle case [U, v:]o = 0, for example, with U = U(x, t), and in the limit circle case with regularsingular point, for instance, the boundary condition is
ao(t) cos 6 + a(t) sind = 0. Obviously, this procedure can be used only for equations of the form
— (p(x) Ux)x +q(x)U + k(x)(KoU,, +
1 U, + k2U) =0
(tT)
for x € {l, m},0 < t < «©, where ko, k1, K2 are constants. The eigenvalue equation corresponding to (ff) is Au =u with 4= —(Kkos? + Kis +2) if we take into account
LU,, = s?u — sUo(x) — U,(x)
with
Howe (x, £) = Us(x). 10
Here lim,.., U,(x, t) = U:(x) is the second initial condition, which occurs if and only if (tt) contains second derivatives in t. However, we must also require that p(x) > 0, k(x) > O in the interval {/, m} and that ko = 0, as well as x; > O if ko = 0. We
must not impose any conditions on «2, since we can include k(x)«2U in q(x)U.
262
THE RELLICH BOUNDARY
CONDITIONS
~
153)
we obtain
(p(x)u’)’ + (Ak(x) — q(x))u = 0,
qe le
gg
ee a(x)
0