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8-1-1973

Functional equations of the second kindNed WilliamsAtlanta University

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Recommended CitationWilliams, Ned, "Functional equations of the second kind" (1973). ETD Collection for AUC Robert W. Woodruff Library. Paper 525.

H.H iAkHh IIHh~~dh~Ht HI HIH fl HU~~

FUNCTIONAL EQUATIONS OF THE SECOND KIND

A THESIS

SUBMITTED TO THE FACULTY OF ATLANT~L UNIVERSITY

IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR

THE DEGREE OF MASTER OF SCIENCE

BY

NED WILLIAMS

DEPARTMENT OF MATHEMATICS

ATLANTA, ‘GEORGIA

AUGUST 1973

• c. ‘~

\H

1 bh~* II U~jI[I!I~. !b~:~it~,

TABLE OF CONTENTS

] ritrodiiction

List of Symbols 2

Chapter

I. Basic Concepts 3l.lElementaryConcepts 3

rI. Equations With Completely Continuous Kernels 82.1 Properties of Completely Continucus Operators. 8

.111. Spectra 2~43.lSpectrumn o~ a Bounded Linear Operator 2~3.2 Resolvents 33

IV. PredholmAlternative ~48

V. Applications to Integral Equations 56

Bibliography 61

~ll~i ~ ~h ~a~I ~

I NTRODTJCTION

It will be observed how completely continuous operators

behave in reference to the null space, range space and the

resolvent set of a bounded linear operator.

Chapter one deals with the basic concepts that are used

to prove theorems throughout the thesis. Chapter two covers

the development of the properties of completely continuous

operators.

Chapter three is a study of the spextrum and rèsolvent

sets. This chapter is developed in accord with •the following

concepts: eigenvalue, eigenelement, and eigen—subspace.

Fredhblm alternative and application to the integral equation

o±~ second kind are discussed in chapters four and five

respectively.

1

li ~ ~ ~

LIST OP SYMBOLS

°°, infinity

s, is an element of

~>, implies that

~, there exists

i.e., that is

such that

.., therefore

V, for all

I H I I~ norm

E, su~miat i.o n

1 , absolute value

-~, converges to

R(U), range space of the operator U

N(U), null space of the operator U

X*, the space of all bounded linear. .functionals on X

jX—* Y], the space of bounded linear operators from X into Y

jX], the space of bounded linear operator on X

= ~[x, T0 is the restriction of T to X

2

~M~h ~k~ ~

CHAPTER I

BASIC CONCEPTS

1.1 Elementary Concepts:

Definition 1.1.1: Let ~X be a Banach—s~ace and Xo a

closed proper subspace of X, then X/X0 is a linear subspace.

Le~ X = = = x + X0: x e X}.

Let the norm on X be defined by I lxii = inf I lxii. With

this definition ]C becomes a normed linear space.1 Let

~: ~C -~ X/X0 be defined by ~(~c) = x. From this we have

I I~Cx1 ii = I I~ci ≤ infj Ixi which implies that I I~i I ≤ 1 Vx.

By the property of infinium, for each ~ e X, there

e~cists ~c e X such •that ~(~c) ]C and lxii ~≥ 1/2 I lxii.

Definition I .1.2: U is a completely continuous operator

if and only if it maps bounded sets onto relatively compact

sets.

Eénce for each sequence {xn} such that I Ix~I i~≤ 1, there

lAngus Taylor, IhtroductIon to ~ühct’iohäl AnalysisCNewY~ork, 1967), p. 105..

3

dI~~ 1 ~ ~

exists •a subsequence {x~} such that U(x~.) converges.

Theorem 1.1.3: If U is completely continuous, then U

is continuous.

Proof: Let x~ converge to zero. If U is not continuous

then there exists a subsequence {xn.} such that IIUx~~

converges to infinity. But by ccmplete continuity U(xn)

converges. This is a contradiction.

Theorem 1.1.14: If R(U), the range of a continuous linear

o.perator U, is finite—dimensional, then U is a completely

continuous operator.

Proof: Let R(U) be finite—dimensional. Let X such.

that I ~ J. ~ 1. Letting ~n = U(x~) we have

I Iy~I I = I IU(~)I I ~ I lUll I Jx~ I ~ JU~ [.

This implies that {y~} is bounded, there. exists a subsequence

•such that ~ co.nverges because R(U) is. finite—1 1

dimensional. Hence U is completely continuous.2

Theorem 1.1.. 5:. Let V be. continuous and U be completely

continuous. .Then VU and UV are completely ccntinuous.

Proof: . Let W = VU. Let I ~xf ~ 1. So there exists

such that U(x• ) converges to z. Then we have W(± : ) =

nj

- 2Allen A. Goldstein, Constr~3tj~ye Real ~aI~sis(New York, 196.7), p. 12.

5

V(Ux ) —÷ V(~) since V is continuous.ni

Let y~ = VXn~ Then I lynI I = I I~~)I I ~ 1 hIll I IX~I I ~S

I lvii. Hence by ccmplete continuity of U, there e~cists {y~•}such that iJ(y~.) cc:nverges, i.e., iJ(Vxn.) converges. Hence

UV is 2ompletely continuous.

The’o~’ëm 1.1.6: If EU and V are completely continuous,

then W = a~J + ~V is completely continuous.

Pi~oof: Let I ~n’ ~ ~ ~• Since U is comole~tely con

tinuous, there e~cists’~i~.} such that Ux. converges. Since1

iI~ II. ~ 1, there exists ~ } such that V(~ ) converges.1 ii

We ob~erVe that U~c~1. converges. He.nce-ii

wc~. ) = aEUx + ~Vx converges.n1. n1~

TheoPe~I.L7.:’ If EU is completely continuous and 2~ + 0

then N(21 — IT) is finite—dimensional.

P~op~f: Let B = : [ = 1 and (XI - U)x = 0}.

We want to show that B is compact. Let cn e B so that

• ~ ‘sn’ I = 1 and 2~r U~n~ By complete contThuity of U there

exist;s Xn. such that U(.Xn) converges to y, and hence x1 1 n1

converges to y/2~ =z. Obviously lizi I = Il~im ~n~11

By taking the limit :0f ~ = ~ we have1 1

Az •= Uz which implies tha~t (X1 — U)z = 0.

4 4~1 I ~]I~ ~LI~ .U.-~

6

Hence z~ B and B is compact. This proves that N(2L1 — Ti) is

finite—dimensional.

Corollary’ 1.1.8: If ii is completely co~tinuous and 2~ 4 0

then N~XI — is finite—dimensional.

proof: We have

(.~I — = — n~~U +“+ (—l)~U~.

— IJY, where

V n11 + + (i)flU~1

H~nce V IX~ and IJ is completely continuous..

Hence. ~V. Is. completely continuous by. the previous theorem,

and th~ result follows.

T~o~’e~ I.I.~9 Ri:esz’s:’ “I’e~±~’).: Suppose ~ is, a normed

li:near space. Let X~ be a subs~ace of DC such that DC~ is~ clo:sed

and proper. Then for each e such that 0, ~ & ‘~ 1 there. e~ists.

a yect-or DC su2h that I 1 and ~ ~I I ~ 8

if~ eDC0.3

3Angus Taylor, Tht~od~c:tior~ tb: ctior~aI A~aI~sis~(New York, 196:7), p. 96..

Lth ~ ~

CHAPTER II

EQUATIONS WITH COMPLETELY CONTINUOUS KERNELS

2.~ Pro~p.eftie.s of Com.p;Iet.e.]~y C.ontinuoU.S Operators.:

Let U map X into itself be a completely continuous

operator where X is •a Banach-spaCe. Hence its. conjugate. U*

maDping X~ into itself is also completely. continuou5.~ We

consider the, following equations,

(2.1.1) x — U(x) = y, (x,y e

and

(2.1.2). g — U*(g.) = f, (f,g s X*).

We also use notation T = I — U whe.re I is the identity

operator on the space X.

Lemma 2.1.3.: Let ‘T = I — U •.where U is a completely

continuous operator. Then the set T(X) is Diosed.

Proof.: Denote ~ = N(T.). = T l{~}, {x.’: T(~) = O}..’ Let

X = X/10 and .def~ne T : X~ X by’ T (~) TC~), where ~ •E x.

We show T is well defined.

~A’ngus Taylor, Irftroducti.o’n to Pth-~ctic’häl Analysis(New York, 1967), p. 275.

7,

J~~Jl~ ~ ~

8

Let ~ ~2’ ~ This. implies that — E Xo, and

T(x1 — 2.2). = 0.. Whence Tx1 — Tx2 = 0, which implies that

Tx1 = Hence T is well defined.

Let ~ be a natural ~napping en ~ onto. )C. Let s T(X)

su.ch that •y~ converges to y0.

We want to show that y0 e T(X).

By definItion, T(~) = T(X), there exists an element e ~X

such t~riat y = T(± ). Then there exists. x of X such thatfl n n

= ~n and 1~r~11 ~l/2 xH. for n 1, 2,

We show that .tb~ sequence ~x } •is bounded. Let c =n n

I I.’~I . If the sequence is not bounded, then there exists

a subsequence which will converge •to infinity. Let {~c}

denote the sequence without loss in generality.. Hence en =

I Ix I will converge to infinity.n

We have from the norm of ~ thatn

IIxnII>j~ II2n11,2. en

or

• I I.~I I•.. .I I•x~I I...~II~2II0nII

9

Hence

~ I..—~-~t2forn 1.

n

Therefore H~’rit t• is a bounded sequence.

Since U is a co~p.1eteI~ ~continuouS operator, there e~dsts~

a sub~e~ueflCe say Pnl such that U(~n) converges to z.t.,cn.s en

Then,- .Xn U sn.,

en en

Therefore,

• .T (~x~) ~ u (Xn)

= — + Uj\ __-__-~ o + •~ =‘Cn

Rence T ~ II~f~ T(z)..Cn

Therefore,.

T(z.) = •lim T(~n\ = 1i~n U ~ z — z =

~ \~cn) n—°° \~c~ en)

Hence T(z.) = 0 ii~p1ies that z e

This ii~p1ies that ~ = = 0 e ~. Hence n I1~i> o..Cn

ThIs is impossible because

10

lix ii lix iir—~=— ~—= =1,ten lixnil

which is a contradiction. Therefore {x~} is bounded. But

iIxnIj ~• 2I~nil which implies that iix~I [ is bounded.

Since U is a completel~r continuous operator, there. exists

a subsequence Ixni} such that Uxni —~ ‘x. This implies that

= T(~x ) + U(x ) y + U(x~ )1 1

= y0 + x =

Whence

= urn y = 1i~n T(x ) = T(x ) s T(D~).n.n+vo 1 ~ 1

Therefore T(X) is closed.

Definition 2.I.~4:. If there is some integer n~ 0 such

that N(Tn) N(T~~), then the smallest suc~ integer is

called the ascent of T.

Lemma 2.~I.5: Let U be completely continuous and T = I — U.

Then th.e sequence of sets

N(T), N(T2), ... , N(Tn),

is ir.creasing and contains only a finite number of different

sets i.e.,

(i.) N(~) C N(T2)C” CN(T~)

(b) There exists p such that N(T~) = N(T~~), i.e. the

ascent of T is finite.

.11 ~.~

II

Prdof. (a): N•(T~) c N(Tfl.+l), becau.se Tk(~) = 0 implies

T(TkCx)) = 0, and from this it follows that Tk+lCx) = 0

Let N(Tn) = N(Tn+l)c N(Tfl+2). Let ~ (.~) 0.. This

implies that Tfl~+i(TC:x:)) =0,. whence T.(:x.) e N(Tr~+]~) = NCT.n).

Therefore Tn+l~±) = 0 and ~c e N•(T).

Therefore N(T~2) c N(T~) and consequently = N(T~2).

cb): We want to show that there is •a number p

N(TP) = N(TP~-). Let it be false. Let X~ =

Xn+l for all n. Then is a closed proper subspace

• B~r Riesz’a. Le:mrna .tftere.. exists x e ~X au:ch thatn+l n+i

1 and ~~n+l’ D~) >. 1/2.

Let iu~n. p~ ;x ) >1/2 for n~.l.n÷i n

Consider the ecuation

- V(~) = - T~) - ~n + TCx~)

= - x, where ~ = ~n + -

We show that ~c e X . We havein—i

Tin~Cx) = T~1-J~(±) + T~’(T(~X~)) - Tinl~T(x~))

= Tm_l(~ ) + T~’(x ) - Tm(~ )n in n

We look at x which is an element •of X , i.e.,n n

= N(T~’) = N(Tn )~N(Tinl), because n~m-l. Therefore

in-i • IllT (x~) = 0. Also T (D~) = 0 by definition. As before, it

follows that T~C~) = 0.. Eehce T~”(~) = 0,. and ~ e N(T~1) =

such that

Let X +n

n+ 1

I I~~1I I

II !d! l.~j~!:’ &~

12

Xmi~

But S X~, so p(x~, Xmi) > 1/2. This implies that

I IU~x~) — U~~)I I I I~c~ — 3E1 > 1/2 for all n.

Therefore U(x~) is not cauchy.

Since j = 1, by complete continuity of U, there

exists a subsequence ~ such that Ux1~ converges.

Therefore U~nj) is cauchy, which is a contradiction.

Defii~itioii 2.1.6: If there is some integer p ~O such

that TP+lO() = TPc~Ø, then the. smallest such integer is

called the descent of T. If such a p axist, then the descent

of T is finite.

T = I — U where U is •a completely continuous operator.

Lenma: The sequence of sets

T(DO, T2(X), ... , T~(X),

are decreasing and has only finitely many different sets, i.e.

(a) T(X) ~ T2()O .. ~Tn(X) ~ •

(b) when Tnoc) T’~’~-CX), then T”(X) = T’~(X) for It n.

If T = I — U and U is completely continuous, then there

ejcists a finite descent n.

Proof (a): Let y s T12~~(X). Then y = T’~(ic) =

• A ~I ~ ~, ~

13

Tn(T.Cx)) C T~cx), which shows that T’~~ (X) C T”(X). A ssume

that Tr~lGX) = T~GX). Let y e Tn(~) Tn+l(]O, which implies

y = T~~(~) = T(9~.(i~.)) where Tn(~) s T~X) = Tn+l(X). Thus

T~x) T1”~+l(,z), where z ~ X. Therefore

y = T(T~(~)) = T(~~1(~)) = T~~2 (~). c T~2(~).

Therefore T~1(X) c T’~2CX).

n+1(h): Assume now that •T CX) + .T ~X) for any n. This

n+1implies that T (X.) is a proper subset of T”X). By RieszTs

lemma, there axists a sequence’ {x~} e Tnl(X) such that •1 I~I I = 1

and ~ T~~-(X))~l/2 for all n.

Let m > n. Using the equation U = 1 — T we .considei~

U(X~) - U(~) = ~n - T(~) - ~m -

n - Tcx) - + T• (ij~)

= x -• x, where jc = + TCx.) - T(x ).

fl mWe show that xe

But T(~x) e Tr~(X), because x was chosen in T~(X);

because r~ ~ n+ 1;

T(±) e T~1(X) cT~’CX), because m + I > ~ ÷ 1.

Therefore ~ s T11+1CX).

Considering U = 1 — T and Riesz’s lemma, we have.

I IU~) — U(x1~•) = II~n — > Z/2. icr all n.

But ~ I

Therefore by the. comple.ta continuity of the Dperator II,.

there ecists a subsequence ~ } such that ‘~J~x~~) converges.ni 1

This is a contradiction.

Défii~ItIbh. 2.~i.8: Let r denote the descent n such. that

Tr1(~) = Tfl+lcx.). Whenever T(X) = T(X), we have ±~ = 0..

And we also let V = TrcX) and X” = N(T??).

We get a description of the operator T from the following

theorem.

• Theoi~e~r 2.:I.~9.Ca~): The operator T maps the bspace D~

onto itself.

(b):, The operator T maps X” into itself, wher.e X”

is finite—dimensional.

Cc).: For e~ei~.y ~c ~ ~, it can be writteh as ~c = ‘1’ c~”

where x’ ~ ~ and x’T E DC”,. i. e •, ]C = ~C’ ~ DC”. Th.e~e sxi st s.

a constant X > 0 such that

liii ‘~‘II~II and ~(d:): we state the ope~ato~ i~ as: ~rj ~j? ~ ~TJ?I, .wh~e

~J’ and V~t map the apace DC into DC’ and DC” respectively. TJ’~

J~4~ ~ ~ II ~

15

and U” are completely continuouS, operators, and lJ’U” =

U”U’ = 0. The operator T’ = I — U’.. has •a lineai’ invers:e.

ProOf (a).: Since .X’ = Tr(X), we have

TçX’) = T(Trcx)) = Tr+lcx) Tr(x) V..

Let Ta.) = C where X £ V = TrcX.). Let’ the’ ascent n ~ r,

the’n N(T~’) = N(~n+~~) = N(Tn+2) = ... Therefore~c ~ T~C) =

= T~~-cX). This. implies that ~c = T~(’z)’., z ‘e X 0 =

= T(T~C~.)), implies N(T1~+]~) = N(T~). Hence’ DC = Trl(Z) = 0.

I~t tha ascen~ n ~ r. Let ~ & Tr(I)~Tn(~). The±efore,

it follows obviously. Hence ‘T is 1—1..’

‘(b~): We have Tr = (I. - U)r, and X” = N,(Tr). Hence,

Tr, = CI _U)r = 1 — —

= 1 — IJO, whePe U0 = c~1U~-’ + . .. +

=T0.

Also ‘~~X”) = T(N(Tr)). We show that .T : X” —‘-

Let’ ~y’ £ T’. ThIs implies that T’~(y) = 0. Let i~ > 0. The’n

0 = Trc~) = Tr—l(T(y)) which implies T(y) ~ N(?~). Hence

= X”. If r =0, :x” =‘{o} and the’

inculsion T~”) C~’ is trivial.

‘Let’ T0 = TID~’. T0 is’ ‘a one—to—one Tilapping froin

X’ onto D~’, and T1 e±ists;. The set V is. closed and therefore

-lr.a Banach-space. Let x e DC. We ‘put DC’ = T T’ (DC), .and

= i— DC~ which implies’ DC = x’ + DC”, where DC’ £ V and ~x’ e

l.~IJM~fl I~h4~I,~ fl •~ L,~thuh! b~jflI~ ~Ih ~

16

By definition Trcx~ ~x’.

Since T is one—to—one, there e~cists z e X’ such that

T(TrcJ~)) = z.. Therefore x’ z which is an element of X’.

Then

Trcxu) T~’c~ - = T~(~) -

Tr(~) - Tr—1TcT;lTrc~))

= Tr(I) - Tr-l(Trc~))

= Tr(~) - ~2r_1(~)

- =0

We have ac” E = N(Tt’), which proves ~ = xt + xi?.

Now we show that ~c = xt + x” is unique.

Let ~c = + = x’ ÷ DC11, where xj .s V and ~“ ~ X’~. Then

— DC’ = — = z, where z e ~‘fl X”. fLence z T’~(x)

and Tr (z) = 0 w~ch i~p lies that z ~ N~ Al s:o

Tr (4 = Tr C~). + T’~ = Tr + 0 Tr cx~x.. But since

e X’, we have Tr (xj) = T~ Cx~i~ and therefore

= T~lTrcDC) ~DC’

Therefore the representation of DC is unique.

(d): Since U = .- T, for DC E V, we have Ucx). =

c we have obtained this from the proof of Cal, that

[h ~ UA~h~ (I ~!H~[h[[k

‘7

U.:~ Si~iiaa~1~U ]~—~“. The space

= ]~“ and ~c ~C is. expres~sed in the fDrrn ~ = ~‘ + ~“,

w1ie~.e• x’ .e IX’. and x” s X’. We define

= ~X’ and U”(x) 1J(c”) e X”, ~ ~

Then

Ut.c~~)÷iJl? (~) = ii•~±’ ) + nCx!’) = U(x’ + ~“) = U(x.).

If e IX’,. thei~ x= ~ + 0 whIch. shows that

mci) = = rjcx), and Utt(~) = 0..

If ~ ~ ~ then ~ .0 .+ ~ whe~e

= .0,. and U”Cx) = = U(x).

~EPurther we have

(1J.”1J~.).(~x.) U”.(IJ.’.cx)) 0,. because u’c~.) e

Si~ui1~1y’

.(jJ’iT:”.) (~x.) = jJ’.C”:cx)) + 0,. because U”.(~.) s X”.

~Frcuu this. it follows t~at ~TJ’~J” 0..

•Ii~ yi~~ :of. the o.peratoa’ IT”, we have.

IT’~ : ~-~---÷ ~X”, and t.h.e~.efore. Roi~’.) = ~“. ~ NCT~) i~

fi.ni.te~dimiensional. Re.r~ce IT”. is co~plete1y~ continuous;. . And

also Th. is; co~ip.le.teIy .cont~nuous since Ti’ .= U .— U”..

We show that T’ I -. IT’ has; an inverse. For this. we

have. .to show, that T~C~.) = 0 ii~plies~ x 0.

We have

0 + 0 = 0 = .T’.C~Y = :CI —: 1J.?)(~).

~. .i i. ~

18

= Jc? + 3:!? — Ut (3:) = 3:? — U(xt) + x~’

= T(3:’) + 3:”.

Since each element has unique representation, T(x’)

and ~“ = 0. But T : X’ -~- X’ is one—to—one which implies

that ~c’ = 0.. Since x = x’, this implies that ~c = 0.

We show that T’ is onto X.

Let y s X, then y = yT + ~ where y’ e V an~. y’t ~

Let T0 = TID~ and T0 : V ~ IX’.onto

Therefore there exists ~c’ e ]~‘ such that x’ T~(y)

We define ~c = :~‘ + y”.

Then we have

=x — U’(~x) =~‘ + y” — U’(~c)

= x’ — U(x’) + y” = T(~x’) + y”

= y’ + y”

Therefore T’ is onto.

Theoi~eii1 2.1.10:. I~ the ascent of the cperator T is

ni and the descent is r, then iii = r.

F~oof: Let 3: ~ N(Tr+l). We write :3: in the fo~m ~

= 3:’ + 3:” where x’ s ~X’ and x” e IX”. We have

= T~C~i Tr+ic~t).;+ Tr~(itt) = T~’(3:’) + 0,

he c au s e

~h~h L,k~J~lI~

19

~TT ~ ~fl = .N(Tr) c~.N(T~1).

T : X’ X. Therefore Tr+1: X’onto onto

Let Tr(J~) = 0; this iiuplies T(~.) 0 and hence ~ = 0. By

induction Tr+1c~) = 0 implies ~‘ = 0.. This shows: Tr+l is,

one—to—one. By definition, T’(~C) = T(~), we have

T2(~Z’) =. T•(~(]~)) = T(~’) = xr.

Therefore

Tr+lcx) = ~ which implies that T’~ is •onto.

Th.erefor e

0 = T’~’c~X) = Tr+l(~?) which implies that ~ = 0.

Thus x = e N(T~) which implies that N(T-) CN(Tr).

It follows that N(T~) = N(T’~). Hence m < r.

Therefore

• = N(Tr).

Let y ~ nI~(.X), then y = ~ where ~ = ~‘ + x”, x’ e

arid ~ ~ ~TT

Th en

y = Tm(x) = ~pflt(~~? ) + T~’(x~’) = T~’(x’). + .0 ~T~m(*t).

Because ~c” e = N(T~) =

But also T(T1.(±~)) •

Therefore

y = Tm(It) = T~(.T(T~l(±.’))) = pm+l(Tl(~?)) S

20

Tm+~Cx).

Since ~ Tm~~()G this implies that Tm(X) = T~11+lCx)

H.énce r < m. Therefore r = rn.

Theor~2 .1.11: The equation x — U(x) = y or T(2) =

y or T(x) = y is soluble for y e ~C if T(x) = 0 has a unique

solution x = 0.

proof: Let the. equation have a solution for each y e

i.e., there. exists ~c e ~ such that T(x) = y. Therefore T

is onto, or = T°(X) , which implies that r = 0 and

hence m = 0. Then NET) = N(T’) = N(T2)

Therefore N (T) = N (I) = { 0 }.

Therefore TC~) = 0 has a unique solution, x = 0.

Conversely, suppose that T(x) = 0 has a unique solution,0

i.e., x = 0 implies that NET) = {0} = NEI) = NET ) . Then

0 which implies that r = 0.. We have T(X) TEX) X.

Thus T is onto.

Th.~refore for each y e D~ there exists x e X such that

TCx) = y.

• T:-ie’orex 2.1.12: The sets NET) and N(T~) have the same

finite dimension.

Eroof: Let ~ ~2’•••’ xn} be a basis for NET) ~and

21

~ro.of: Let.{~1, ~2’•••’ x~} be a basis for N(T) and

{g1, g2,..., g} be a basis for N(T*). There e~cists

f]~. ~‘2’”’ ~Lfl eX’ such that

Ii if j kf;c.xk). =~< (j,k. = 1,2,....., n).

Similarly there e~dsts. y1, y2,••~, ~m e X such that

(iifj=kg.(.Y~) =K (j,1~ = 1,2,..., n).

~9ifi+k

Let iii > n. Let •W: ~ -~ ~ be defined by.

w(~) =Z . f~c~.)Yk~ D~.

Let Y = ~U + W in X. DenQte ~ = .({y1, y2,...., y~}). Note.

that W.(~c) e ~ Hence W : D~—* Xo, where DC0 is finite—

dimensional. The±~efore W. is :comp.le:tely continuous. ThIs

implies. that V is also completely continuous.

Let T = I - V. Tharefore T(~.) - V.(~c) = - U(~) -

n=T.(x) — ~ f (.x)y

k=1 k kn

Consider the equation T.(~) = T.(~.) —E f (~)y~ = 0.k=1 . k

Let. .~ be one solution so that

= TC±~) -E. f~~o)y~: 0.

1 ~ ~ L ~

22

Th.~r e~ ore

g3CT.C~.01 k=l ~ ,0

• ~ CT C~0) ~ k=1 f~C~.01 ~~ 0

• g5CTCx0l)— f5(x01 ‘= 0, which ‘implies T*’ g5~co) —

Cs.Ex01) ‘~ 0.. But g5 .s N(T*) ‘which implies T*(g5) = 0, and

hence f5~01 = 0,. Cs 1,2,...., n) .

~rom ~c, we have

TC~0 TC’D~0) 0 whIch ‘implies that ~ ~ NCT*)

n • nTherefore x0 Z ~ and ~

k~l • k~l

hence c~ 0 f~or all s which ‘implies x0 = 0.

Thus Tcxi 0 has •a unique soThution ~c 0.

Th~er.e~’ore TCXJ. = y will have a solution y e X.

Let

T~) -

Therel’ore T ~ •~n+l has ‘a solution, say ~ , i.. e •, T~’)

=

And consider.

g~~1CTC~*~

k~I~ ~ ~ ~k ~ ~

23

g~~1(T(~) k=1 = g~÷1(.T.(~c*).) k=1 ~k~fl+1~k~

— 0

=0

Therefore = 0

But = = 1, which is a contradiction.

Therefore ñ > rn.

Also if we assume that ~ < n we arrive at a contradiction

in the s52Ue fashion. Hence n = rn. In this we consider the

eqiiat ion

k=l ~k~k

• ~ II ~A b ILUU~

CH~T.ER 1E1

SPEC .T1~

3.1 ‘Spe.’c~t~uiu ‘~‘~ ‘~ E’oüi~de’d’ L’i~’~a~’ ‘Op”e~ator’:

In this ‘secti:cn one sees .ho~t the equation

(~.l.i) T)~) ~ =

or the equivalent equation,

(3..’1:.~). T~:(~) = =

~eac.ts~ with ~espect to .th~...co~ple±~par~et’er ~ or ~i. ~U_

d.eno.tes alinear operator in a .co~nplex Banach~s,pa.ce X. Th.~

cou’np.le,~ plane is. divi:ded~ into two sets;’, pcti): and X(IJ )~,

depending on the so1ubility~. o~. eq~uat.i..on (3.1.1).

~e~ji~it’j’on 3.1.3’:’ .• The set p(tJ). is. called th~ ~eso1yent

set. o~ U or regular ~a.lues. o~ the operator 11.,

p~tJ). =‘{~ : T~~- e~ists~ and~

The s~et ~(U.) is called the characteristic set o,~ U.

= (~j):)C

The s;et ~ CU) •contaThs the .ch~.acteristi’c ~al~es~,’ the ~‘ s.

3~ilarly,. for the equation ~ .we~ have. the. set.s ~

and ~

~e’finj’t’j’~n 3.”l.~’;: The set ~U)~ IS c~ile’~ the non—~

24

,~L~4á~Jh ~F~h~U&,~i 1 ~ II ~!~{I~ Ir~lwdjri

25

singular values of U where

= {ii : T’~-(i.i) exists and is bounded}.

The set cY(U) is called the spectrum of U where

~U) = (Ir(U))C.

We consider the equations

(3.1.5) Tx(x) = AU(x),

and

(3.l.6~ T’(~x) = px — U(.x) = 0

and we arrive at the followings concepts.

Defjni~ions 3.1.7: The characteristic ~talue of U is a

scalar A su.~h that T~(x) = 0 has a non—zero solution. The

solution x is called the ~igeneIernent or characteristic

element of J. The set of all eigenelements corresponding to

the characteristic value A, given by ~ N(T~) is called theA=l

eigen—subspace. The dimension (finite or infinite) of these

eigen—subspace is called the multiplicity o~ A. The rank of A

is the ascent of TA.

If we consider (3.1.6) we arrive at th~ concepts of

eigenvalue, elgenvector, and eigen—subspace.

Leunna 3.1.8: We observe that if U is a self—adjoint

operator in aHilbert space and j~i is an eigenvalue, then its

rank r 1, i.e.,

N(TJ~) ~(T~2) = ... = N(T~3~) =

26

proof.: We have to show that N(T~j) N(T~?). Obviously

N(T3~)cN(T3~2). Hence we show that N(T~j2)cN(T~). Let

x e N(T~2). This implies that T~2(~c) = 0.. Hence

(~, T~2~) = ~x, ~). 0.

(T~., T~) = 0 which implies that 0..

Thus it follows that N(T~2)CN(T)..

Therefore the rank. r = 1.

There is. a connection het.w.~e.en the .s.~.ect~n and .th~

characteristic set of the same operator Ti.

If~ CEt~), ~hen~

Concerning the above. concept we have the following

propositions.

Propos~ition 3.l.~9: We have 2~ e p(U) if and only if

there e.xists a bounded linear inverse B~ = T~’ = (I — ATJY3-.

Proof: Let T = T~ - T~, and T0 = T~. Then,

IITIl = :lIT~ - T~Il =

Conaider NG~0, e) {2L:I~ — 2~cj

1 1Lets =————

‘) TTIc~.

[~A~Ia ~i~h!ã*ib~ ~hI ~~~Ib!~LIb!III& ~~

27

Then we. ha~ve

______________ I.

[ui I. ~lI~JI.l .I.I:T~1~I. [l~l.J 1

Thus

1~

lI~[f .iIT~h1[. :1 1 JT~J

Re~ace ~ +. e~st~ and is ho~inded.

.Ther.ef~e CT,~ ÷ T~ - T~ )~ = ~ e~d~st:s;: and .~a.bounded.

~opQs~itio~n c~.I.TQ:):: The set pCU.) iscpen~, .s~o that

the s;et ~1J) is. closed.

• P~of:: Let 2~ s pELT). This implies that T~]~exists:

and i.s~ hounded.

We look at

• 1 lT~ -, T.f I ••=•1i~.i ~~v- ~ - ~~yj J. j~. -~j :j ~

Cons~de~ Nc~Q, c). ~ueh~ that c = 1 ____________

2 ,f J~TJ~ f 1 IT~ I .1

~or. each 2~ ~ NC~0, ~), ‘T~. e±±~t~ and ishounj~ed.

Eence~ ~ e p.C ...wh~cfr ~oi~e~ that NE~.0 , ~

Therefo~’e. pELT), is oper~.

28.

Pro~positidi~3.I.l1: The circular domain N(o., ~ ) =

Hij I I

1 ~ lies in the set p(U), and hence the spectrum

• HulL

a(U) belongs to {]1 : Ipi. < I IUI [}.

P~oof: Let A be such that ~ . T~ I — AU.HUH

We have I.I~uH = JM huH <1:

Therefore TA’ exists and is buonded. This inplies A e p(U).

If p s o(U), this implies ~ ~ ,xCiJ), whIch sh~ws e_j.1 3j U

- . .~ .. - ..l •1 •, •1•. IThis ruplies triat ~— 4 and hence —--— _____

hii~ I huh I hI! I lullTherefore Ij’~ -< ‘I Ii; 1•.I••.•

•~3.1.2’:. The seta ~(V). and ]c(ift.) are situated

a~mnetrically with ~es.pect to the real axis.

‘of:’ By definition TA = I -. AU and .T~ I - A1J*

Rut .T~ (I — AU ).~ = I —

Since (T~)1 exists if and only if T’ exists, and

(T~ ~ -1 = (T1 ) *.

Therefore TA’ and (T~ ~ -1 exist s imu1taneo~isly.

~ropdsit’i~d~ 3.1.13’:’ If U is co~np1ete.1y continuous,

then T1 is bounded if it exists.

29.

Proof: The range of T~ is a clos:ed sub•s~pace of a

Banach-space. Therefore RCT~) is a Banach-sp.ace. Consider

T : ~ ÷ R(T~) is an onto ~napping. Hence if ~ •e~i:sts.,

it has to be bounded by the opeh m~apping theorem.

Theorem (3.]~.ILl): If ~[J is a completel~r continuous;

operation,

(a) the. ciaracteris:tic set contains only chs~.ac.te~iati.c

values, i.e., xLTJ) = ~(u.), each characte~istic. value

having finite ~nultipli.cit~r;

(b) for eyery~ r > 0, the circle J2~ ≤ r contains.

at most only a finite number of characteristic. ~Taluea;

(~) if .e ~(~•) and ~2 ~ U~)., .wh~e. ~ =.

and i.f D~ is the. eLgeheIemuent .correspDnding to

and g2 e X~ the eigenelement corresponding to A2, then

= 0.

Proof (a): If ?~ 4; p(U), the homogehous equation

T~~) = - A U(~) = 0

has a non-zero solution or T~1 not bounded by above pro

position, T~l emdsts implies T~1 bounded. HerFDe 2~ is a

characteristi.c value, each characteristic value ~ ~

has finite multiplicity. The dimension of eigen—subspace

B = U N(T~) is called the multiplicity of A. Since IT isn=l

~IL ~ ~~

30

completely. continuous, T~ has a finite ascent, say no.

Hénce

2 noN(Tj) ~ ). c c N(T~ ) =

We now have B = V~9• N(T~). But each N(T~) is finiten=1

dimensional. Wénce B has finite dimension.

Eb) Let us assume the converse, i.e., that there exist

infinitely many characteristic values in the circle B(0,r)

12L1 -< r}. Let ~l ~ B(0,r) . Therefore there exists

e B(o,r) such that ~2 + ~• There by induction there exists

s BCO,r) such that ~ + 2k., if i + j, and we may assume

~ + 0 for all i. Let be the corresponding eigenelement

which are non—zero, i.e., Xn — ~~nU(xn~ 0, for all n.

We claim that for all n, {x1, x2, ••~ ~ xn} are

linearly independent. For n = 1 we have xl such that

+ 0. Iiénce{x1} is linearly independent. Let {x1,..., x}

be linearly independent. We show that

.“ ‘ 1ri’ ~n+1~ j~ linearly~ independent.

]f not, then

n.= E c~x...

n+l 2. 1~

Al so

,~ ~ .1 . 1 ~,1 1

31

= U~~) , and JCfi+l = U(]~+i) ~ ~ ~i JC~.

An ?ln+1 i=l i=l A1

Upon subsitution of ~n+1’ we have

n~ n n,

1=1 1 Z °~ix~ and E •~.ci~1 —. ~i)x~ = 0.An+i 1=1 i=1k~~ . 2H1

But x~ + 0, then 0~i _~i = 0, which implies that.An+1 A1

1 _1~= 0. If cx~ + 0, then 1 — 1 = 0. This• A1) .

implies that V ~1 and hence An+i = A1. This is a..~n+1 •~i

contradiction. Therefore t~~n} V is linearly independent.n=1

Let ~n = ({JC1, ~2’”’ ~n11 and ~n+1 ({.~c1, ~2’”’ ~n’

~n+1~ Therefore ~n + ~n+1 and further X~ is a proper

closed supspace of X~. Therefore by Ries’z lenma there

axists ~n-i-1 ~ .X~1 such that j Iy~~j~I I = 1, and P(Yn-f-l, ~

> 1/2 for all n. Let ~ ~ ~n so that

n.= . E B~X1. Then we have

i=l

•th ~

32

lJ(x1 E R1lJGx1~ ~ c X~.i=l 1=1 ~j

Al son ns.~

T .(~x) = - ~ 1J(x) E S~x. - E —~ ~n 1=1 ‘~ 1 n 1=1 A1

= —.

n-i= E ~.CI. - ÷ S (1 ~-~~1i

1 1 fl n

n-i= S SCi—_~l)x. e~Xn-i

We no~ let n~ > n and consi:de~ the e~xpressicn

EIJC~y~1 ~C2~nY~I - T~Cy~~i - ly — T~Ci1

~ Cy.1~ —~ +T~(y)n

.y. — y y~ +. (yl ~ T~nCYnl•

But what has been proved, T~(y~) ~ an~ A~U ~n1 ~

~• Conaecjuently y~ — T~Cy.) ~ 6 which

i~iplies that ~ e Wen~ce

I[UC2~y ~ :_ 1J(~ ~ = 1I~ - ~I1211 ~n 2

33

Therefore {TJ(2~.nYn)} is not caiich.y.

But J I I = 2~ I I I I I = I I < r.

Therefore by the camplete continuity of U the~’e .ejciats: a

subsequence {2~.yrn. } such. that TJ(?~m.ym.). ~co.nverges.. This1 1 1 j

is a contradiction.

(c) We have ~1UE~11. = ~ and A.2U*.(~2). g2.

ifénce = 2~21J*(g2) C~)

2~

]t follows that ~‘ —• ‘~a1. ~ =

~ g2~1) + 0, then ~2 ~l’• which is a contradiction.

Cdr’äl”Iax~/ 3.1.15:’ The set ~0(U), the set. of all eigen—

•e.Ieiten~s, contains at Tnost countably many •eleiiients.

‘~ro’o~: The co~p1e~ R2 can be represented as

B2 = B (0, n) where B.(0 , n3 = { 2s~ : I 2~. ~. < n }n=l

But fo~ each n 3(0,nl contains only finitely many’ chärac—

teristi.c values. Thus B2 contains, only countably many

charac~eristic values.

II ~~~IjJU~~

3L~

3.2~ We are still involved with. the equation

(3.2.11 — AU~1

oni~ -aow ~e are inte~es.ted in th~ case when it is. uniquely

so lub i.e.

:e~f’ihitidr1 3.22:. Let ~ s p.OJ) , ~ + 0, that

.)~IJ1.~~- exists; and is~ continuous;. Let B’~ .be. defined

as (3~ 2.31 B~ = 1(i_ I)..

Then B~ is called the res:olyent ope~ator of U.

~.init.ion 3..’2.L[: For non—singular. value•s, we define

~j..2:.51 = T~= c~Then is called the resolvent. o.pei~ator of. U..

i~a 3.2.6: (~) ~f j~ + :0,. :t1~n R11 ~:. ~ ~2 B

(bi. If 2~. + 0, ther~ B~ =

_________~ .We have..

H =(~~~:.l.33. 33. .1_i i~p

Considering the equation .B~, we have

3’

B1 ~(T~ -

:13. 33

fLénce

.11 AJ~~~

35

=

p p

and

I ..1 I

Ther.e~’ ci’ e

2.1ii

T:~e .s.ei’ies ~. will c onverge under, radius of conn=l

vei’.gen3e r, where. r = ._, .

liin sup

.Le~’a3:.”2.~7’: The IlEit :j1ii.~1 I~ e±ist’s;.

• ~o’of: Let urn mr I[11~I.I~ ‘ •a and li~ sup~ IIU~H~

= b. ~or e > 0 .thei~e. e~is:t’s ~ such thä~ ‘j IL~I.I~ < a + s,

which ruplies that ‘I JI~°I I’.. ~≤ (a. ÷ s) .

Let ~ =r d~i~{1, HiJ1l, .II~;2:Il,,,..., lIU~hII}. Let’ n be

an arbitrary nurnber, then by Algorith~ theore~u, we have

n = ~ + 3, where 3 e { 0,. 1, 2,..., rn—l}.

• lrrn ~ kILé.nce’:J =IJ =~ •~u3.~ II’iJ II • 1111311.’.

ii ~ th i~i ~ ‘~

36

Thepef or e

~11~ii ~ •j~1~ ~.k/nj iv~.i il/n ~ i iu~i ~

(a ÷ s~

Taking the unit, we have Ujn (a + )n-J/nNi/n = a + s.

rt follows that a < I

Thus li~ ~ ~ 1/n ~ lin ~ (a ÷ E~ n_j/nNl/n

= a ÷ ~, for all ~ > 0.

mence liiu sup ~ 1u’~’l 11/n ~ a.

~ i/nBut a ~c lim sup ) ~ a.

Ee.nce lim sup :J jij~1 [~fl = •lim j~f: ‘1~Jflj [~/n.

Therefore the 1 imit e±i at a.

‘Pheo’~e~u 3.2.8’: The resoivent B~ can .be~ e±panded as’ a

series:

(3.2.9) B~ = U ÷~2 ~ + ~fl~n~’ = ~

n=0

The radius of convergence is given by

• 1 ‘I’(3.2.10) r = ______

• urn ~f~I’u~I ‘I uim

n‘~roof’: We. consi:der.,tlie seri~ea ‘ ~ ~ ~ series,

n~ 0,

converges :jfl tile’ apace of ope~ators.. i-I H— .XJ,~ ~he~r’.e 0~ i~: a

37

Banach space. We

2~U~ ~ E ~‘~9 ~ [.TJ.”J f. conver~es if I~\~’I I l.UnIn=o

The. sum of th€ se~ies is ~ :— ~J) ~-. H:é.nce,

z Cr = ~l.

Therefore,

- = - I):~ 2~ 2L

= ~ 2~ ~TJ~ = fl

~ n=l n~l

= En=

This series, will .conve~ge within the radius of convergence

oo~nfl

of Z. 2~ U . The radi-us of convergence is given byn~o

.1• n l/nlim sup~ J.u •IJ..n ÷

1f r. < 1, then both se~i.es: converges, and if’ ~ > 1 they

diverge. Lf = 0, we do not know.

Coro1iaz~j ‘3.2.11: The res,olyent R” = ‘~ , where

~ ~n+l.

1 ______________

r ii~. 1~n.1 1.l/nfl-’- c~. -

ft ~th ~.ft ~

38

Prool’: We have ±~ram lemma (3.2.6),

1R =—I+—B~,.

~1 j~

~ l~n+1:1’ ~2 ~ ~n

=~I: + i; ~••~

n=O

= :+ + + • + 1 ~ +

.The~refore the aer~es converges when I~.I < r.ii

This implies that I]~I >

Le~a 3.2.12:: ~‘or all 2~,ji ep(U), we have

B — B (2~ —

A 31

Proor: We consider the equation

(r :+ RB). = (r ~U) —l

We arrive at

Cl + 2~B~) (1 :- =

- ~ =~

39

ABA(I AU) AU

BA — AU) ~

Similarly B~j = (I:

W~.nce BA — B~ = (~ :—)~B — ‘— ~U)~U.

~ ~niu’ti.ply this equati.on on thE ~i~ght by~ CE~ AD1 and on

the lest by CE. ~pIJ]..

We haye.

CE. ~- .Pm.CBA - B~I CE :- ~ -

- CE. ~U) (I :— pU) ~1U (I —AU)

= CE’ — pU) U — U (I—AU)

= U .— pU2 — U ÷ AU.

= (A —‘

And hence .w~ have

BA - B~ = (A - ~(I*- pU) ‘U2(I.- •AUr’

= (A - ~) CE-

= -

‘Co’o’.ll~i~f 3.2.13:’ The reso:lvents: BA and B~j are

oerinutab.I.e, i.e., B B’~ = B’BA~ ~

h ~ ~1Lb~ 0 ~0~~ . k4 hiJo ~iI~

140

Proof: If 2~ + p, then

B~B~=

Wénce it follows that

-

B2 B~ =

Co~.olIäry 3.2.a~[:: The resol~veht is a co.htinuous

function of the pa~raine.ter 2~. at every .point cf the set pGJ1.,

i.e.., ~f I I ÷ ~ where £ p(~)., then

il-IB~ .

n •0

~rdof: If I - I + 0,. then: I .1 B~ I JVJ V n.• n o

We try to show that is a continuous function.

I IB~i I

.We Consider

HIB~II.- .JJ.B~ IIH IIB~ -B~ :IJ=II(~ -~~IB~B~II.n 0 n 0. ‘nO

~ ~n ~Qj I I~I l!B~I I.

If B-~ ,B~ is not eq~ai to zero,. thei~n o

~ilb~ li IUlJU! ftd~~l.:~ ~

~1i

• .1.

— .≤—-— .-~-o.!I•~I[ IIB~II •~n •~o

H~énc e• .. •1•.•.

I IB~I I ‘I’IB~0I I

The~e~c~e

• I1B~ II -~ ‘IIB~ I I. which ‘liuplies: I.I.B~ I I. ~ I~ V n.n . 0

Co.nsi.üering lemma (3.2.12). ,. we. have

- •~II ~‘I[.(~ .~0H

~ ~ -. :I1.B.~ II: IIB~ I•••0 fl

~IIB ‘j[..I~ - .~ I,..2~c 0

~hi~ch~ converges to ëro.

The~.e.~ore’ is continuous.

Theor~e.mu 3 ‘2.15:’ • The ‘radius of conver.ge~ce r of the

series • is~ co.nve~gerit inside the irc.le: •~

BA=~ ~n=O

radius r,. ~‘or 2~ such ‘th&t •~ ~ ~ •I~~ ~ e ~ this i~up1iea

that ~‘AJ •> r, for otherwise I2~j.~≤ r implies. A s pOll.’ This

is a contradiction.

Therefore r0 = inf LAI’ > r.

~42

Let ~x e D~ and. ~ e D(*.. Define a complex function by ~(2i)

= ~(B~ (~) ):. . If ~, -p e p(TJ1, ~e have

- f C~1.) - fCB Cx))

]i_~

f(B. Cx1 - B (~x1)

fCC~. -

C]-’ -. ~iC~xll

=

iJpon taking the Ii~m~t, we have:

urn C~:-i =. ~•1i-:m fC~B~CxflJj÷~ jj_~

• Therefore the continuous derivative: e~i:sts, •~‘ ~

• ~e e2pand ~ .aa a Tay’ior s:erIea in th~ neighborhood of the

point = 0, i.e.,

= ~(~) ÷~i(~) ~\ • ______ A~ +...

1! n!

a~D~th~a JW IEthll!Ih

L{3

Therefore this expansion is valid in any circle which does

not ccntain any singular point of ~.

Let ~ ro which implies ~A e p(U)

Wênce B~(x) is defined and ~OJ. f(B~(~)) +°~

Therefore for I2~I < r0, the Taylor series is valid. But

= f(B.~(x))= f( ~~nUn+l(~))n=O

= 5’ ~fl~’ (u~”- (x1 ), where •< r.n= 0

We alEc consider the expansions

~C2~) = ~CO) -i-~ ‘CO). ~ +. . .+ ~)(0)~ +.. ., where J2~j ~ r0.1!

f~~1). +~ f(U2~i)+... + ~ f~U~~c)) +...,

where 21 -≤ ~•

But an analytic function has a unique Taylor series.

Therefore ~ is conve~geht fo~ ~ ~ r0.

Let 0 ~≤ ≤ ro which implies. Z converges..n= Q

Rénce urn ~r. ~ cun+l ~) = 0.. For 2~ = we haven ~

fClirn 1TJfl~(xui 0,. for all f e Xx..n

Thus ~ ~nUn+l 0, by the I~áhn-Banach theorem.

~~I~6,J-•~ *01,,!, -, !**I!! II IhI,~U~ It I,!k[~ ,-!,,I!l!,L~ ..]~t~j!I!!!.

1414

Therefore I I I ~ N~ for all n, and

supi ~-~U1(~) [I ~ < N1. ≤ °~ for all n,~.

By the uniform boundedness principle

sup I .1 ~fl11n+l I = N ‘<°~

Thus 11n+11 I- -< N for all n, and I2~I I I•u”~•I ~.< N. Raising

it to ~he power of 1, we haven

~I.I-1i~I I~ N~ ~

and

-~ ~-j~ 1.j~+l~ I-~ ~≤ •Ii~

~t follows that ~ ~- ~ 1 w~ch i~plies ~ .< r. for all such

that 0 .< < r0.

LettIng 2~ r0, we ha~ve r0. < r.

Therefore r

Theor-e~n 3.2.16:. - I~t ~ be a regular value of the

o.perat~on ~, then B~ B~0 + (~ — ~) B~+. ..+ .(~ ~)~

This. expansion holds ~ the circle IA — -A0I < p0, where

p0 = ___________________ = p (2~ ~, ~x CU)):1Lrn j IBfl+1Ij.in-~[I2Lo I[n

Proof-: From the definition of B~, we häye

B-k - 2~U1 ~tJ

~hI~M~ ~

145

= u + + A2u3 +...+ +

Then

— = U ÷ (~ ~U2 +...+ (A —. A0)flUfl+l + ...

The radius of convergence p0 is given by

p0= 1lint IIIJnIIi.n+°~I II~

Let F(A) = B2. Then by the Theorem of the riean,

(A) — FQAo) =B~ B-)~o = B~BA.A—0

H énc e

?‘ (A) = urn ~C2~) F(7~0) = lint B B~ B~A /~0 0A —2~

Also

F’CA) -. F?~o) = B~ B~.0 ~ + B~0)CB~ —.

A -: - -,

= (B~ + BA ) BABA + 2B30 0 A0

Further

_____________ 2:(,B~. —. B~.) 2(B~ .B~~0.)(B~ +. B~B~ .+. .B~)

A - A0 - A - A0

= 2 BAB~0(B~ + BABA + B~0) -~ 3! B~.

Considering F(A) we have

F(~) + (~ —A0). ~(A0) + (A - A0) ?Fht~0) ~

(~ — A0) ~F(n) (A0) +...

I~6~1 ~ ~ ~ ~ ~~

146

I{ence

= B~ ÷ (~. — A0)B~0 + (2k.— A0)~B~ +...+

(A —

A0

And the radius of convergence is p0 1liin IunII~.n~o~’~ II~

Coi~oIIä~y3 .2.17:: The expansion

= I r + 1 U +.. .+ 1 Un ~Ii 2. ~n+l

holds for Jp~ > 1, where 1 is the radius of the least circler r

with center at the origin that entirely contains the spectrum.

Proof: We look at R~ where = 11 + lB1p. p.2

From corollary (3.2.11). we consider B1 U = B0.

1~1

Multiplying ~l by 1, we havep.2

B1 = ~2B0 + B~ + ~jn+1 B~ +

Therefore

R ~~r+~U+” + 1 Un÷...p. pp.2 p.n+l

II ~áJI~

147

Lemma 3.2.18: If U is a seif—adjoint operator in a

FLi1ber~ space where ~. ItJ~ , then lull = lim~r

Proof: We show that for the seif—adjoint operator

11u211 hurl2. We ha~re

rj (~) 2 = (Ux ,Ux) I = (U2~c ,x) I I lu2ic2I I ~ )Ju211 1!~2fl.

fLén.ce

I I ~ I I~I I i~p1ies I [UI~,uaring both sides we have hUh2 < hIu2hI.

But I u~ I < uII2.

Therefore I lu21 I •~ I lull.?.

1 ~

CHAPTER IV

THE FREDHOLN ALTERNATIVE

The Fredhoim alternative holds for T = I - U, if for

some power U~’ of a linear operation U is a completely con—

tinuous. operator. We consider the equation

(~4.l.1) T(±) = y and its adjoint

(14.1.2) T*(g) =

We also consider the ho~ogeneous. equations

(~4.l.3) T(i~) = 0 . and

(14.1.14) T*(g) = .0.

That the Fredholm alternative holds. for operation T

implies, that,

(1) either (14.1,1) and (14.1.2) are soluble, whatever their

right—hand sides, in which. case th.e solutions, are unique;

(2) or (14.1.3) and (14.1.14> have the same rm~ber of linearly

independent solutions ~l’ ~2’•••’ x~ and g1,..., g~ re—

s:pectivel~r. Then th.e neceaaary conditions for th.e .solubility

of (1) and (2). are

= 0 and = 0, where k = 1,..., n.

The general solution of’ (14.1.1) is given by’

148

l~ ~ I~d~illI1~ ~I

149

n nx* ,+ Z and that of C~.. 1:.~) ~y• g ~ ~g* + S

k~1 k.~1

.T~e haye ~ and g~ aa any giyen .soi-ut~ons: of (14.1.1) and

C14.1.2)., where c1,..., Cn are arbitrary constants..

Th’~o~e~ 4~I.5:’ Each of the, following two conditions:

ia .ne.ces:s:ax?~r and sufficient for t.he Fredho’ln alternative

to .h~o1d~ for the operator T:

(.à..) T can be written in the’, form

T = W~ + Y,

Where W is. an operator. ha~i.ng a .bo~.nded linear inyers;e and

Y is a co~p1etely~ continuous operato~;

• T can be written in the foi~a

where W1 ~ an ‘operator haiing a bounded linea~ inverse and

~l 1~s~ a finite:—di~uer~si’onal operator.

‘~roof ‘(~J’; Let .T ‘~ W~ + Y, where W’ has. a bounded’ lineai’

inyea~se W1 and V is c~ple.t..ei~ continuous:. Theu equation

(14.1.11 is equiyalent to W~1T(~ = W~-(~r ). Also W~l =

CW.:]~1*. e~is:ts so that •(14.l.2)’. is eq~uiva1ent to.

C~4.l’.6) T*W*~l(~) =

If g0 is ‘a solution ‘of (14.1.6), W*~l(g~) is a solution of

(14.1.2), while if g~ iS: a solution of ‘(~4.l.~)’., then .W*(~)

is ‘a solution of (14.1.6).

• ~ ~ ~ •I ~ JI~ ~ di I,

50

Let U = W~-Y. Then IJ~. _Y~.CW~~i* ~ ~ Also C4..1.5)

and (14.1.6) can be written as;

(14.1.7) ~-U (i) = .w1 (v),• (14.1.8) g_TJ*(g) = f.

Since U is completely: co±itinuoiis;, (i4..l.7) and ~(14.1.8y he.c’omues

(.l.9).x — Ucx.) =0,.

(14.1.10) g - U*(~) =‘ 0..

Thus they have the same finite. number of linearly indepeh-.

dent solutions., ~ ~2’••~ ~ and g{,..g~,...., g~. Also

(.14.1.3) will haye th~ s~ue. co~up.1ete s.y~s..te~is of 1inear1~

independent~~

We show that th,~ functionals

(}4..1.1i) • = w*~lC~). (k. = 1,..~, hi,

form a comple:te. system of li.nearl5r inde.pendei~t solutions;. of

(14.1.14).. The functionals are linearly independent •.since it

n • n • nfollows from E ~ •= :0 •that ~ ~kgt — E ~W* C~,) = 0..

K1 • k=]. • . k • k~l

This, is only poss:ible ‘~‘ ~1 = = = ~n 0. If Q1...1.14X

had a solution g0 which is: not a linearly c~nbinati.on of

(14.1.11), then W~C~) would be. a solution of ‘(14.1.101. But

it would not be a linaarly~ combination of the functionals

~li~d~AI~ iU~~

51

g?, which is impossible.

Therefore (4.1.3) and (~4.1.4) have the sa~ne finite niraber

of linearly independent solutions. Also (~1.l.5) and (4.1.1)

have solutions i±~ ~(w~-Cy)) = 0,. k = 1,..., n.

Therefore by (4.1.11) we have

(W~)*(g~)Cy) = w*l(g~)(y) =g~(~) = c (~. = 1,..., n).

Cb.): Let x1,....~ x~ and g1,..., g~ be co~plete syste~ns

of 1inearly~ independent solutions of (~4..l.3) and (14.1.11).

respectively. By the blorthogonality th.eor~m the~e exists

~unctiona1s. ~l~•••~ ~ ~ and •eleiiieñts x1,...~ x~ ~ such

that[0j+k

(4.1.13) f~cXk) =~ Cj.,k, = 1,...., ri),S%~~ ~k

(4.1.14) ~~(Y1) =~ (j.,k = 1,..., h).

Let ‘~“ T(x), ~“ ~ ~ Let y e ~X be expressed

(4.1.15) y = :Y~ + yt’, y’ e Y~ y” e YT~.

~1e put ~ = k~l ~~Cy.) ~k.’ and yt. = y — ytt.

Hence b.~ (4.1.14) we have

h~ I ~ II ~~

52

g~Cy~) ~g3~ - k~l~ Ci. - 1,..., h).

Thus. TC~.) = y’ has a solution yT e ~ Alsc~ the uniqueness:

nof C~.]~.l5) rollows ~‘ram y” = E

k.~l

The ecuation Tc~.) ~ y” h~s a solution and hence

g(y~) = = Q, 3 = 1,..., fl.

Let ~ .NCf-,..., ~) and ~“ ~ ~~1~•••~ ~ Let ~ C

X iid-ieu’e ~x = x~ ± x!?, x’ e IX”. Denote W~ .b~i

~ C~ I T Ci.) +. ~ ~)k~l

1—1We .ShDW that ~ .; ~ -?X, and has linear inverse.

onto.

Let y’ .~ ~X, i~liere y = ~y” ÷ y~”, ~? c X’,. ~ .eZX~r.. .No~ ~ .s

~ NX1, and

R klkk c ~“, .i..e.~, T(~ ~ ~ ~ a .s~:lut.~.on ~ c

nput x” ~ ~ ~Y~1CYk and ~ ~. z~ .+ x~.

W~e consider TE~) 9 and. f3~.tt) = .0 to~geth~ w~ith~ (14 113).

53

W~ obtain

nw1Ci*) = T•C±’ ~. + TC~:T~. + ~ ~

T(±’) + ~ ~ .~..k.j)ykk=l 3=1

n= + ~ =

k=l

Now we show that there are ±~~o .s~oiiition~ o~’ y~ other

than ~ . I~’ so, the~ there e~ists~ ~ + .0. such, that W1(~0) = 0,

1. e •, T(~) ÷ k=lk = 0.

nThen TCx.0) E y? and E fk(x0)yk e Y”. SInce (~1..l.1•5) is

k=1

IJni;ue we. have.

T~0) = 0 and k=lk k = ~ = 0,. Ck = 1,..., n).

Therefore x~ s .K’ X”.

Therefore .~ = 0.. This is •a contradiction.

‘Le~a~ 4.1.16: Let A and B be a linear operator

mapping the formed space ]~ into itself. If th.e operator

are per~utab1e, and the operator C = AB has a bounded in

verse. A and B also have inverses.

Proof: First we show that A and C’ are permutable.

~~iL ~

5L~

We have

A = C~1CA = C1AB~ = C~’AcA~) = C~’AC.

Mutiplying on tha right by C~1, •we get AC;’ C~1A.

From the permutabi1ity~ o~’ A and C-1 ~re have

B (p~ —1) = BAC —1 = ~ ~• —.1 =

and

(AC—1)B = C~]-AB = cia-c I.

Therefore 31 AC1.

Now we show that B and C’ are pe~mutab.l.e.

We have

B = C1CB = C1ABR = P~-BCAB) = C1BC

~ulti.p1y~ing on the right hy~ C~, w:e ~et ~ C~’B.

~ro~n the perwntab.i1it~r of B and C1 we have

AEB~.1~ = ABC~~- = cc.—1 =

and

CBC—1)A = C—~-AB. .= C]-C .= I.

The~e.fore A~1 = BC 1.

Le~ua 4.1.17: Let Ii be. a 1inea~ operator in thE space

X. The characteristic set. ~.CU) of th.e o.perEtor U and the

xO~T.Y” of U’~’ are connected by the relationship ]x(U)]m ~(um)

i.e.,A E.xE11), then )~ffl 6x(.Tim).

Froo~: Let e =. e2’Tti/In. We have

I —. 2~T~J~ = Ci — 2~ e U)...Ci~ — 2~. e~1U).

L~ ~ ~

55

If Am ~ •xOJm), on setting

A = I - AU, B = (I — A s u)~..(I — A 5m1 U), c = I —. XmUm,

We see that C~- exists and is linear.

Therefore by (~4.l.l6) A’ e~cists and hence is linear, i.e.,

A EU).

~ [IJ,h,~I~i ;~II]±~NII~! liii b,U,~II**hL~.!.~ 11± -~-

CELAPTER V

APPLICATIONS TO INTEGRAL EQUATIONS

An equation of the form

(5.1.1) x(s) ~ K(s.,t) ~(t)dt = y(s)

is- called ar integral equation. This particular equation is

known as an equation of the second kind. The kernel K(s,t)

is assirmed to be continuous on the square .fO,l;O,l~ of

functions.

It i.s possible to consider an integral equation of a

inore general type than (5.1.1), i.e.,

(5.1. 2). ~ Cs) —. ~ K (5, t) c (t)- dt = y (s.~

where T is any closed bounded set in ~—dimensjona1 Euclidean

space. In this case s and t denote points of v—dimensional

space.

We consider the integral operator U where

z = UE~), z(s) K(s,t) ~(t)dt,

i_s regarded as an operator C into C. The norm of U is

given b~r

I ~ I = maxSl ~K(s,t) Idt

56

~nh~IiI ~ .~

57

and is completely continuous.

We write equation (5.1.1) in the form

(5.1.3) ~c — A Ucx) = y.

Let ~ be a solution of this. equation. If x~ is expressed

in terms of y by x* = y + AB~(y), then by theorem (3.2.8) it

can be expanded as a power series, i.e.,

(5.1.~4) K* = y + 2~U(y) +. . .+ A~U~(~) +. ~•

It converges whenevei~ 2~ ~ 1 = r where d = lim 1~Jn1 Iid n •-~-~‘.

and r is the distance fro~i the point 2~. = 0 to the charac

teristic set of U. The series (5.l.LD will also converge for

i2~I~<’ 1’ =‘‘‘ 1’”I U~ max lIk(s,t) Idt

0

The powers of U are also integral opera~ors, i.e.,

(5.1.5) z = un(~) , z(s~ c’1 K~(s,t)~(t~dt ~n =

where K~(s,t) is the iterated kernel.

We can substitute (5.l.~1} in (5.1.3) which will give us

an expansion ~f the solution of th.e integral equation (5.1.1)

as a power series in the parameter “A:

~ (s~ = y (sY + A S1 K (s , t) y Ct) dt ‘+. . . + A~ Kfl Cs , t) y (t) dt0

This series is unU’ormly’ co.nvergent for s s ~

JLIl~ d~ ~I ~ ~ ~IL~ ~h ~~

58

Since the series

(5.1.5) ~ = ~ + ~ ~n~1~n ~

is convergent in the space of operations from C into. C,

we have

1..j~1+p ___

= ~iax $ ~J~K~(s,t). dt ~ 0.s 0 3=in+1

Thus the series

(5.1.7) E ~j =1

is uniform1~ convergent in the space L for each fixed a e

,io,.11.

The function r(s,t;~), the’ su~m of the sei~ies, is te~med

the reso:lyent of the integral equation (5.1.1). It fo1io~rs

that

B~ (yl •Cs~ = F (~ , t ;~). Ct~. dt.

If J~I •< r, then the pro~es.s of successive approxi~ations

for. equation (5.1.3), is convergent. If this is applied to.

equation (5.1.11, its solution can be obtained as a limit

of a iniformly convergent sequence of continuous functions

• 1~x~ Cal }. It is given by’ the recurrence formula

~~1Cs) ~ j~C(s,tl. ~n(ti dt .+ y’(s) (n =‘ 0,1.,...).

By this fornula ~co is an ai~bitrar~ co~tinuous function.

.~h ~H1 i~,~ ~

59

Equation (5.1.1): is said to be of Fredholm type, of

the second kind. A special situation results if we assume

that K(s,t) = 0 when s < t. The equation (5.1.1) then

becomes~*5

(5.1.8) x(s) —. A 3 K(s,t) ~(t) = y(s)0

This equation is known as Volterra integral equation. The

iterated kernels for Volterra equations also vanish for

S ~< t.

Lemma 5.1.9: If the kernel K(s,t) is continuous for

0 < t ~< s ~ 1, then th~e axpansion

+ A U(y) +.. .+ .2~n~Jn(~) +.

holds for all couple~x A, i.e., that r ~

Proof: Let 1K(s,t) I ~ M. Then for K~(s,t) we have

the following inequality

(5.1.10). K~(s.,tiI ~ ~~—1 Mn (n = .1,2,...).

(n — 1)

If n = 1 the inequality is trivial. We shcw it to be true

for n > 1. We have

II~+1(~,t)~ $ IK.(s ,u) K~(u,t) Idt N ~ ~ ~0 0(n—l1~

•n

From the inequality (5.1.10) we have

~JI~~

60

I IU~I I ~ ç0~’t) Idt Mn m~c ç dt = ~fl~)

Eënce

<N.7~T1)~!,Therefore r =°~. This implies that an integral equation

of Volterra type has no charactei~istic values.

i~~IdI

BIBLIOGRAPEY

Edward, Robert E. Functional Analysis; theory and~pplication. New York: Eblt, Rinehart and Winston,1965.

Gelbaum, Bernard R. Conference on Functional Analysis,University of California, Irvine, 1966. Washington:Thompson Book Co., 1967.

Goffman, Casper. First Course in Functional AnaI~sis.New Jersey: Prentice—hall, 1950.

Godstein, Allen A. Constructive Real Analysis. New York:harper and Row Co., 1967.

Kantorovich, Leonid Vital’evich. Functional Anal~sis inNormed Spaces. Edited by Ap. P. Robertson. Oxford,New Y~ork: Pergamon Press, 19611.

Riesz, Frigyes, and Sz. Nagy, Bela. Functicnal AnalysisTranslated from the second French edition by Leo F.Boron. New York: Ungar, 1955.

Taylor, Angus B. Introduction to Funct~ona Analysis.New York: John Wiley and Sons, Inc., 1967.

Von Neumann, John. Functional Operators. Princeton:Princeton University Press, 1950.

Wilansky, Albert. FunctIonal Analysis. New York:Blaisdell Publishing Co., 19614.

Wilansky, Albert. TopIcs •in Functional Analysis. Berlin:Springer—Verlag, 1967.

Y~oshida, Kosaku. Functional Analysis. Berlin: Springer—Verlag, 1965.

61