Wikibooks Functional Analysis

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WIKIBOOKS AND WIKIPEDIA ARTICLES

Functional Analysis

A Brief Overview

Collected by Jyoti Swaroop Repaka

12/2/2008

From Wikibooks, the open-content textbooks collection


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Functional Analysis

From Wikibooks, the open-content textbooks collection

Functional Analysis can mean different things, depending on who you ask. The core of thesubject, however, is to study linear spaces with some topology which allows us to do analysis;

ones like spaces of functions, spaces of operators acting on the space of functions, etc. Our

interest in those spaces is twofold: those linear spaces with topology (i) often exhibit interesting

properties that are worth investigating for their own sake, and (ii) have important application in

other areas of mathematics (e.g., partial differential equations) as well as theoretical physics; in

particular, quantum mechanics. (ii) was what initially motivated the development of the field;

Functional Analysis has its historical roots in linear algebra and the mathematical formulation of

quantum mechanics in the early 20 century. (Seew:Mathematical formulation of quantum

mechanics)

The aim of the book is to cover those two interests simultaneously. The book consists of two

parts. The first part covers the basics of Banach spaces theory with the emphasis on its

applications. The second part covers topological vector spaces, especially locally convex ones,

generalization of Banach spaces. In both parts, we give principal results e.g., the closed graph

theorem, resulting in some repetition. One reason for doing this organization is that one often

only needs a Banach-version of such results. Another reason is that this approach seems more

pedagogically sound; the statement of the results in their full generality may obscure its

simplicity. Exercises are meant to be unintegratedpart of the book. They can be skipped

altogethFrom Wikibooks, the open-content textbooks collectioner, and the book should be fully

read and understood. Some alternative proofs and additional results are delegated as exercises

when their inclusion may disrupt the flow of the exposition.

Knowledge of measure theory will not be needed except for Chapter 5, where measures play vital

roles in the formulation of the spectrum theorem, a key machinery in Functional Analysis. On the

other hand, solid knowledge in general topology is mandatory, for topologies that are not metric

and (topological notions such as compactness) play important roles. (For that, read, for

example,Topology, which contains more than you need to know.)
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Contents

Part 1:

Chapter 1:Preliminaries (May 28, 2008) Chapter 2:Banach spaces (Sep 1, 2007) Chapter 3:Hilbert spaces (June 4, 2008) Chapter 4:Geometry of Banach spaces (May 27, 2008)

Part 2:

Chapter 5:Topological vector spaces- (May 28, 2008) Chapter 6:C*-algebras (October 30, 2008)

Part 3:

Chapter 7:Special topics (June 6, 2008)Fredholm theory
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Chapter 1: Preliminaries

In this chapter we gather some standard results to primarily to fix language and formulation,

though some of them may not belong to functional analysis proper. In particular, we prove the

Hahn-Banach theorem, which is really a result in linear algebra. These proofs of these theoremswill be found in theTopologyandLinear Algebrabooks.

Contents

1 Theorem (Heine-Borel) 2 Theorem (Tychonoff Theorem) 3 Theorem (Baire Category Theorem) 4 Theorem (Urysohn Metrization Theorem) 5 Theorem (ArzelAscoli)

5.1 Definitions 6 Measure theory 7 Theorem (Hahn-Banach)

Theorem (Heine-Borel)

1.Theorem (Heine-Borel)A metric space is compact if and only if it is totally bounded and

complete.

Theorem (Tychonoff Theorem)

1. Theorem (Tychonoff)Every product space of a nonempty collection of compact spaces is

compact.

An linear operator from a linear space to a scalar field is called a linear functional.

Theorem (Baire Category Theorem)

1.Theorem (Baire)Every complete metric space is not the union of nowhere dense subsets.

By modifying the proof so to use compact sets instead of balls we can show that every locally

compact (Hausdorff?) space is also non-meager, though we will not be needing this.

1. ExerciseShow by contradiction that the set of real numbers is uncountable, using the theorem.

In particular, a dense set may be meager.

1. ExerciseGive an example of a dense but meager set. (Hint: you can find such an example in

the book.)
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Theorem (Urysohn Metrization Theorem)

1.Theorem (metrization theorem)If K is normal and second-countable, then K is metrizable.

Proof: Define by

Then for everyj x =y. The converse of this also holds.

Since d(x,y) = d(y,x), dis a metric. Let dbe the topology for Kthat is induced by d. Weclaim dcoincides with the topology originally given to K. In light of:

1 Lemma Let X be a set. If are a pair of topologies for X and if1 is Hausdorff

and2 is compact, then 1= 2.

it suffices to show that d is contained in the original topology. But for any ,

since is the limit of a sequence of continuous functions on a compact set, is

continuous. Consequently, an d-open ball in dwith center atx is open (in the original topology.)

Theorem (ArzelAscoli)

Definitions

A set of functions F defined on [a,b] is uniformly bounded if there exists an M such thatfor any function f within F, f(x) 0, there exists a >0 such that for all and for all

, .

Now, the following is the statement of the theorem:

A set of continuous functions F defined on [a,b] is relatively compact if and only if it is

equicontinuous and uniformly bounded.

1.Theorem (ArzelAscoli) Let be a family of continuous real-valued functions on a

compact space K. If is both equicontinuous and uniformly bounded, then is relatively

compact (or totally bounded) in the metric .

(A proof can be found atAn Introduction to Analysis/Continuous functions on a compact space.)

Measure theory

2 Theorem (Minkowski's inequality) If and p > 1, then:
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.

Proof: The inequality is a consequence of Hlder's inequality:

(where we let q by 1 /p + 1 /q =

1)

which is then a simple consequence of the following inequality:

(where a, b > 0)

To simplify notation we denote |f| , | g | byfand g, respectively. First we have:

Then Hlder's inequality followed by division gives:

where . The desired inequality follows after noting (p 1)q =p and.

By letting be a counting measure we also obtain the analog for the series:

2 Corollary

Theorem (Hahn-Banach)

1. Theorem (Hahn-Banach) Let be a real vector space and p be a function on such

that and p(tx) = tp(x)for any and any t > 0.

If is a closed subspace and f is a linear functional on such that ,

then f admits a linear extension F defined in such that .

Proof: First suppose that for some . By

hypothesis we have:

for all ,

which is equivalent to:


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.

Let c be some number in between the sup and the inf. Define F(x + tz) =f(x)

+ tc for . It follows that Fis an desired extension. Indeed,f= Fon being

clear, we also have:

ift> 0

and

ift< 0.

Let be the collection of pairs (H,gH) whereHis linear space with and gHis alinear function on that extendsfand is dominated byp. It can be shown that is partiallyordered and the union of every totally ordered sub-collection of is in (TODO: need moredetails). Hence, by Zorn's Lemma, we can find the maximal element (L,gL) and by the early part

of the proof we can show that .

We remark that a different choice ofc in the proof results in a different extension. Thus, an

extension given by the Hahn-Banach theorem in general is not unique.

1. ExerciseState the analog of the theorem for complex vector spaces and prove that this version

can be reduced to the real version. (Hint: Ref(ix) = Imf(x))

(TODO: mention moment problem.)

1 LemmaLet g,f1,...,fn be linear functionals on the same linear space. g is a linear combination

of f1,...,fn if and only if .

Proof: The direct part is clear. We shall show the converse by induction. Suppose n = 1. We may

suppose that there isy such thatf1(y) = 1. For anyx, since ,

g(xf1(x)y) = 0 or g(x) = g(y)f1(x).

The basic case is thus proven. Now, suppose the lemma holds for some n - 1. As before, we maysuppose that there is somey such that

fn(y) = 1 whilef1(y) =f2(y) = ... =fn 1(y) = 0.

For any , sincefk(xfn(x)y) = 0 for every k= 1,2,...,n, g(xfn(x)y) = 0.Hence, the application of the inductive hypothesis to the linear

functional gives:


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for some scalars a1,...,an. (TODO: give a simpler proof that does not use induction.)

A set of real (or complex)-valued functions defined on a setXis said to separate pointsin Xif and

f(x) =f(y) for every impliesx =y

With regard to this notion, there are two important facts that will be used in Chapter 4.

1. TheoremLet X be a topology generated by a setof real(or complex)-valued functions.

Then X is Hausdorff if and only ifseparates points in X.

Proof: Left as an exercise.


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Chapter 2: Banach spaces

Let be a linear space. A norm is a real-valued functionfon , with the

notation , such that

(i) (w:triangular inequality) (ii) for any scalar (iii) impliesx = 0.

(ii) implies that . This and (ii) then

implies for allx; that is, norms are always non-

negative. Note that (i) implies that:

and

and so: . (So, the map is continuous.) If only (i) and

(ii) hold, then is called a seminorm.

A linear space with a norm is called a normed space. With the metric a

normed space is a metric space. We define theoperator norm of a continuous linear

operatorfbetween normed spaces and , denoted by , by

We thus obtained an example of a normed space. Another example, which is more historical but

we will be using recurrently throughout the book, is the lp space; that is, the space of convergent

series. It is clear that lp is a linear space. That the lp norm is in fact a norm follows

fromw:Minkowski's inequality. (See Chapter 1) It remains to show that it is complete. For that,

let be a Cauchy sequence. This means explicitly that

as

For each n, by completeness, exists and we denote it byyn. We claim:

as

Let > 0 be given. Sincexkis Cauchy, there isNsuch that
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for k,j >N

Fix k>N. Then, for any m,

Thus,

.

Hence, with . (Note

since .) lp is also separable; i.e., it has a countable

dense subset. This follows from the fact that lp can be written as a union of subspaces with

dimensions 1, 2, ..., which are separable. (TODO: need more details.)

2 Theorem Let T be a linear operator from a normed space to a normed space .

(i) Tis continuous if and only if there is a constant C> 0 such that forall

(ii) any Cas in (i) }, and if has nonzeroelement.

Proof: Seew:bounded operatorand seew:operator norm.

It is clear that an addition and a scalar multiplication are both continuous. (Use a sequence to

check this.) Since the inverse of an addition is again addition, an addition is also an open

mapping. Ditto to nonzero-scalar multiplications. In other words, translations and dilations of

open (resp. closed) sets are again open (resp. closed).

A complete normed space is called aBanach space. While there is seemingly no prototypical

example of a Banach space, we still give one example of a Banach space: , the space of all

continuous functions on a compact space , can be identified with a Banach space by

introducing the norm:

It is a routine exercise to check that this is indeed a norm. The completeness holds since, from

real analysis, we know that a uniform limit of a sequence of continuous functions is continuous.

In concrete spaces like this one, one can directly show the completeness. More often than that,
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however, we will see that the completeness is a necessary condition for some results (especially,

reflexivity), and thus the space has to be complete. The matter will be picked up in the later

chapter.

A graph of any functionfdefined on a setEis the set . A continuous

function between metric spaces has closed graph. In fact, suppose . By

continuity, ; in other words,y =f(x) and so (x,y) is in the graph off. It follows

(in the next theorem) that a continuous linear operator with closed graph has closed domain.

(Note the continuity here is a key; we will shortly study a linear operator that has closed graph

but has non-closed domain.)

2 Theorem Let be a continuous densely defined linear operator between Banach

spaces. Then its domain is closed; i.e., T is defined everywhere.

Proof: Suppose and Tfj is defined for everyj; i.e., the sequencefj is in the domain ofT.

Since

,

Tfj is Cauchy. It follows that (fj,Tfj) is Cauchy and, by completeness, has limit (g,Tg).

Sincef= g, Tfis defined; i.e.,fis in the domain ofT.

The theorem is frequently useful in application. Suppose we wish to prove some linear formula.

We first show it holds for a function with compact support and of varying smoothness, which is

usually easy to do because the function vanishes on the boundary, where much of complications

reside. Because of th linear nature in the formula, the theorem then tells that the formula is truefor the space where the above functions are dense.

We shall now turn our attention to the consequences of the fact that a complete metric space is a

Baire space. They tend to be more significant than results obtained by directly appealing to the

completeness. Note that not every normed space that is a Baire space is a Banach space.

2 Theorem (open mapping theorem) Let be Banach spaces. If is a

continuous linear surjection, then it is an open mapping; i.e., it maps open sets to open sets.

Proof: Let . Since Tis

surjective, . Then by Baire's Theorem,someB(k)contains an interior point; thus, it is a neighborhood of 0.

A linear operator from a normed space to is said to be closedif its graph, that is the

set , is closed in .

2 CorollaryIf T is a continuous linear operator between Banach spaces with closed range, then

there exists a K > 0 such that if then for some x with Tx = y.

Proof: This is immediate once we have the notion of a quotient map, which we now define as

follows.


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LetIbe a subspace of a normed space . The quotient space is a normed space with

norm:

where is a canonical projection. That is a norm is obvious except for the

triangular inequality. But since

for all . Taking inf over i,j separately we get:

Suppose, further, that is also a commutative algebra andIis an ideal. Then becomes aquotient algebra. In fact, as above, we have:

,

for all since is a homomorphism. Taking inf completes the proof.

So, the only nontrivial part is the completeness. It turns out that is a Banach space (or

algebra) ifIis closed. (TODO: a proof of this.)

2 Corollary If and are Banach spaces, then the norms

and are equivalent; i.e., each norm is dominated by the other.

Proof: Let be the identity map. Then we have:

.

This is to say,Iis continuous. Since Cauchy sequences apparently converge in the

norm , the open mapping theorem says that the inverse ofIis also continuous,

which means explicitly:

.

By the same argument we can show that is dominated by

2 CorollaryAny finite-dimensional normed spaces are equivalent.

It is easy to show that any continuous closed linear operator has a closed domain. The next result

is arguably the most important theorem in the theory of Banach spaces.

2 Theorem (closed graph theorem) Let be Banach spaces, and a linearoperator. The following are equivalent.


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13

(i) T is continuos. (ii) If and Txj is convergent, then . (iii) The graph of T is closed.

Proof: That (i) implies (ii) is clear. To show (iii), suppose (xj,Txj) is convergent inX.

Thenxj converges to somex0 or , andTxjTx is convergent. Thus, if (ii)

holds, . Finally, to prove (iii) (i), we note that Corollary 2.something

gives the inequality:

since by hypothesis the norm in the left-hand side is complete. Hence, if ,

then .

Note that when the domain of a linear operator is not a Banach space (e.g., just dense in a Banachspace), the condition (ii) is not sufficient for the graph of the operator to be closed. (It is not hard

to find an example of this in other fields, but the reader might want to construct one himself as an

exercise.)

2 CorollaryA linear functional u on a normed space is continuous if and only if it has closed

kernel.

Proof: is closed ifu is continuous. To show the converse,

suppose and u(xj) is convergent. Since the corollary is obvious when u is identically

zero, we may suppose that there isz such that u(z) = 1. Then the sequencexju(xj)z has a limit in

the kernel ofu, since the kernel is closed. It follows:

.

Finally, note that an injective linear operator has closed graph if and only if its inverse is closed,

since the map sends closed sets to closed sets.

When are normed spaces, by we denote the space of all continuous linear

operators from to .

2 Theorem If is complete, then every Cauchy sequence Tn in converges to a

limit T and .

Proof: Let Tn be a Cauchy sequence in operator norm. For each , since

and is complete, there is a limity to which Tn(x) converges. Define T(x) =y. Tis linear since

the limit operations are linear. It is also continuous

since .


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Finally,

and as .

2 Theorem (uniform boundedness principle) Let be a family of continuos

functions where Y is a normed linear space. Suppose that is non-

meager and that:

for each

It then follows: there is some open and such that

(a)

If we assume in addition that each member of is a linear operator andXis a normed linearspace, then

(b)

Proof: Let be a sequence. By

hypothesis, and eachEj is closed since is open by

continuity. It then follows that someENhas an interior pointy; otherwise,Mfails to be non-

meager. Hence, (a) holds. To show (b), making additional assumptions, we can find an open

ball . It then follows: for any and any

with ,

.

A family of linear operators is said to be equicontinuous if given any neighborhood Wof 0 wecan find a neighborhood Vof 0 such that:

for every

The conclusion of the theorem, therefore, means that the family satisfying the hypothesis of the

theorem is equicontinuous.

2 Corollary Let be normed spaces. Let be a bilinear or sesquilinear

operator. If T is separately continuous (i.e., the function is continuous when all but one variables

are fixed) and is complete, then T is continuous.

Proof: For each ,


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where the right-hand side is finite by continuity. Hence, the application of the principle of

uniform boundedness to the family shows the family is

equicontinuous. That is, there is K> 0 such that:

for every and every .

The theorem now follows since is a metric space.

Since scalar multiplication is a continuous operation in normed spaces, the corollary says, in

particular, that every linear operator on finite dimensional normed spaces is continuous. The next

is one more example of the techniques discussed so far.

2. Theorem (Hahn-Banach) Let be normed space and be a linear

subspace. If z is a linear functional continuous on , then there exists a continuos linear

functional w on such that z = w on and .Proof: Apply the Hahn-Banach stated in Chapter 1 with as a sublinear functional

dominatingz. Then:

;

that is, .

2. Corollary Let be a subspace of a normed linear space . Then x is in the closure of

if and only if z(x) = 0 for any that vanishes on .Proof: By continuity . Thus, if , then .

Conversely, suppose . Then there is a > 0 such that for

every . Define a linear functionalz(y+ x) = for and scalars . For

any , since ,

.

Since the inequality holds for = 0 as well,z is continuous. Hence, in view of the Hahn-Banach

theorem, while we still havez = 0 on and .

Here is a classic application.

2 Theorem Let be Banach spaces, be a linear operator. If

implies that for every , then T is continuous.

Proof: Suppose and . For every , by hypothesis and the

continuity ofz,

.


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Now, by the preceding corollaryy = 0 and the continuity follows from the closed graph

theorem.

2 Theorem Let be a Banach space.

(i) Given , E is bounded if and only if for every (ii) Given , if f(x) = 0 for every , then x = 0.

Proof: (i) By continuity,

.

This proves the direct part. For the converse, define Txf=f(x) for . By

hypothesis

for every .

Thus, by the principle of uniform boundedness, there is K> 0 such that:

for every

Hence, in view of Theorem 2.something, for ,

.

(ii) Suppose . Define for scalars s. Now,fis continuous since its

domain is finite-dimensional, and so by the Hahn-Banach theorem we could extend the domain

offin such a way we have .

2. Corollary Let be Banach, and dense and linear.

Then for every if and only if and for

every .

Proof: Sincefj is Cauchy, it is bounded. This shows the direct part. To show the converse,

let . If , then

By denseness, we can takeyj so that .

2 TheoremLet T be a continuous linear operator into a Banach space. If

where I is the identity operator, then the inverseT 1 exists, is continuous and can be written by:


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for eachx in the range ofT.

Proof: For , we have:

.

Since the series is geometric by hypothesis, the right-hand side is finite.

Let . By the above, each timex is fixed, Sn(x) is a Cauchy sequence and

the assumed completeness implies that the sequence converges to the limit, which we denote

by S(x). Since for eachx , it follows from the principle of uniformboundedness that:

.

Thus, by the continuity of norms,

.

This shows that S is a continuous linear operator since the linearity is easily checked. Finally,

.

Hence, S is the inverse to T.

2 CorollaryThe space of invertible continuous linear operators is an open subspace

of .

Proof: If and , then S is invertible.

If is a scalar field and is a normed space, then is called a dual of and is

denoted by . In view of Theorem 2.something, it is a Banach space.

A linear operator Tis said to be a compact operatorif the image of the open unit ball under Tis

relatively compact. We recall that if a linear operator between normed spaces maps bounded sets

to bounded sets, then it is continuous. Thus, every compact operator is continuous.

2 Theorem Let be a reflexive Banach space and be a Banach space. Then a linearoperator is a compact operator if and only if T sends weakly convergent


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sequence to norm convergent ones.

Proof:[1]Letxn converges weakly to 0, and suppose Txn is not convergent. That is, there is an >

0 such that for infinitely many n. Denote this subsequence byyn. By hypothesis we

can then show (TODO: do this indeed) that it contains a subsequence such that

converges in norm, which is a contradiction. To show the converse, letEbe a bounded set. Then

since is reflexive every countable subset ofEcontains a sequencexn that is Cauchy in the

weak topology and so by the hypothesis Txn is a Cauchy sequence in norm. Thus, T(E) is

contained in a compact subset of .

2 LemmaLet r > 0. A normed space is finite-dimensional if and only if its closed ball of

radius r is compact.

Proof: If is not finite dimensional, usingw:Riesz's_lemma, we can construct a

sequencexj such that:

for any sequence of scalars ak.

Thus, in particular, for allj,k. (TODO: fill gaps)

2 Corollary

(i) Every finite-rank linear operator T (i.e., a linear operator with finite-dimensionalrange) is a compact operator.

(ii) Every linear operator T with the finite-dimensional domain is continuous.Proof: (i) is clear, and (ii) follows from (i) since the range of a linear operator has dimension less

than that of the domain.

2 TheoremThe set of all compact operators into a Banach space forms a closed subspace of the

set of all continuous linear operators in operator norm.

Proof: Let Tbe a linear operator and be the open unit ball in the domain ofT. IfTis compact,

then is bounded (try scalar multiplication); thus, Tis continuous. Since the sum of two

compacts sets is again compact, the sum of two compact operators is again compact. For the

similar reason, Tis compact for any scalar . We conclude that the set of all compact operators,

which we denote byE

, forms a subspace of continuous linear operators. To show the closedness,suppose S is in the closure ofE. Let > 0 be given. Then there is some compact operator Tsuch

that . Also, since Tis a compact operator, we can cover T() by a finitenumber of open balls of radius / 2 centered atz1,z2,...zn, respectively. It then follows:

for , we can find somej so that and

so . This is to say, S() is totallybounded and since the completeness its closure is compact.

2 CorollaryIf Tn is a sequence of compact operators which converges in operator norm, then its

limit is a compact operator.
http://en.wikibooks.org/w/index.php?title=Functional_Analysis/Banach_spaces&printable=yes#cite_note-0http://en.wikibooks.org/w/index.php?title=Functional_Analysis/Banach_spaces&printable=yes#cite_note-0http://en.wikibooks.org/w/index.php?title=Functional_Analysis/Banach_spaces&printable=yes#cite_note-0http://en.wikipedia.org/wiki/Riesz%27s_lemmahttp://en.wikipedia.org/wiki/Riesz%27s_lemmahttp://en.wikipedia.org/wiki/Riesz%27s_lemmahttp://en.wikipedia.org/wiki/Riesz%27s_lemmahttp://en.wikibooks.org/w/index.php?title=Functional_Analysis/Banach_spaces&printable=yes#cite_note-0


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2 Theorem (transpose) Let be Banach spaces, and be a continuous linear

operator. Define by the identity . Then is

continuous both in operator norm and the weak-* topology, and .

Proof: For any

Thus, and is continuous in operator norm. To show the opposite inequality, let

> 0 be given. Then there is with . Using the Hahn-Banach

theorem we can also find andz0(u(x0)) = | u(x0) | . Hence,

.

We conclude . To show weak-* continuity let Vbe a neighborhood of 0 in ;

that is, for

some . If we letyj = u(xj), then

since . This is to say, is weak-* continuous.

2 Theorem Let , , and a linear subspace.

If is closed in both Lp and Lq , then the topology for inherited from Lp and Lq coincide.

Proof: The identity map is continuous. SinceIis a

bijection between Banach spaces, the open mapping theorem saysIis a homeomorphism.

2 Theorem (lifting property ofl1) Let be Banach spaces, and be a

linear surjection. If is a a continuous linear operator, there exists a continuous

linear operator such that .

Proof: Let en be a canonical basis for l1. Using Corollary 2.something to the open mapping

theorem we obtain a sequencexn and constant Kso that:

Qxn = Ten, and .

Then . Define

by . Then , and is continuous since .

References


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1. This proof and a few more related results appear in[1]
http://en.wikibooks.org/w/index.php?title=Functional_Analysis/Banach_spaces&printable=yes#cite_ref-0http://en.wikibooks.org/w/index.php?title=Functional_Analysis/Banach_spaces&printable=yes#cite_ref-0http://www.math.gatech.edu/~loss/03falltea/6580/compact.pdfhttp://www.math.gatech.edu/~loss/03falltea/6580/compact.pdfhttp://www.math.gatech.edu/~loss/03falltea/6580/compact.pdfhttp://www.math.gatech.edu/~loss/03falltea/6580/compact.pdfhttp://en.wikibooks.org/w/index.php?title=Functional_Analysis/Banach_spaces&printable=yes#cite_ref-0


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Chapter 3: Hilbert spaces

The chapter is almost done, but there are still some errors in the proofs that have to be

rectified. (Also, we could add a discussion of the polar decomposition of unbounded

operators.)

A normed space is called apre-Hilbert space if for each pair (x,y) of elements in the space there

is a unique complex (or real) number called an inner productofx andy, denoted by ,

subject to the following conditions:

(i) The functional is linear. (ii) (iii) for every nonzerox

The inner product in its second variable is not linear but antilinear: i.e., if ,

then for scalars . We define and this becomes a norm.

Indeed, it is clear that and (iii) is the reason that implies thatx =

0. Finally, the triangular inequality follows from the next lemma.

3.1 Lemma (Schwarz's inequality) where the equality holds if and only

if we can write x = y for some scalar.

If we assume the lemma for a moment, it follows:

since for any complex number

Proof of Lemma: First suppose . If , it then follows:

where the equation becomes 0 if and only ifx= y. Since we may suppose that , thegeneral case follows easily.

3.2 TheoremA normed linear space is a pre-Hilbert space if and only

if .

Proof: The direct part is clear. To show the converse, we define

.


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It is then immediate that , and .

Moreover, since the calculation:

,

we have: . If is a real scalar and j is a sequence ofrational numbers converging to , then by continuity and the above, we

get:

3.3 Lemma Let be a pre-Hilbert. Then in norm if and only if for any

and as .

Proof: The direct part holds since:

as .

Conversely, we have:

as

3.4 LemmaLet D be a non-empty convex closed subset of a Hilbert space. Then D admits a

unique element z such that

.

Proof: By denote the right-hand side. SinceD is nonempty, > 0. For each n = 1,2,..., there is

some such that . That is, . SinceD is

convex,

and so .

It follows:

as


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This is to say,xn is Cauchy. SinceD is a closed subset of a complete metric space, whence it is

complete, there is a limit with . The uniqueness follows since if

we have

where the right side is for the same reason as before.

The lemma may hold for a certain Banach space that is not a Hilbert space; this question will be

investigated in the next chapter.

For a nonempty subset , define to be the intersection of the kernel of the linear

functional taken all over . (In other words, is the set of all

that is orthogonal to every .) Since the kernel of a continuos function is closed and the

intersection of linear spaces is again a linear space, is a closed (linear) subspace of .

Finally, if , then andx = 0.

3.5 Lemma Let be a linear subspace of a pre-Hilbert space. Then if and only

if .

Proof: The Schwarz inequality says the inequality

is actually equality if and only ifz andz + w are linear dependent. (TODO: the proof isn't

quite well written.)

3.6 Theorem (orthogonal decomposition) Let be a Hilbert space and be a closed

subspace. For every we can write

x =y +z

where and , andy andz are uniquely determined byx.

Proof: Clearly is convex, and it is also closed since a translation of closed set is again

closed. Lemma 3.4 now gives a uniqueelement such

that . Letz =xy. By Lemma 3.5, . Forthe uniqueness, suppose we have written:

x =y' +z'

where and . By Lemma 3.5, .

But, as noted early, suchy' must be unique; i.e.,y' =y.

3.7 Corollary Let be a subspace of a Hilbert space . Then

(i) if and only if is dense in .


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(ii) .

Proof: By continuity, . (Here, denotes the image of the

setEunder the map .) This gives:

and so

by the orthogonal decomposition. (i) follows. Similarly, we have:

.

Hence, (ii).

3.8 Theorem (representation theorem)Every continuous linear functional f on a Hilbert

space has the form:

with a unique and

Proof: Let . Sincefis continuous, is closed. If , then takey =

0. If not, by Corollary 3.6, there is a nonzero orthogonal to . By

replacingz with we may suppose that . For any , sincezf(x) f(z)x isin the kernel offand thus is orthogonal toz, we have:

and so:

The uniqueness follows since for all means

that . Finally, we have the identity:

where the last inequality is Schwarz's inequality.

3.9 ExerciseUsing Lemma 1.6 give an alternative proof of the preceding theorem.

In view of Theorem 3.5, for each , we can write:x =y +z where , a closed

subspace of , and . Denote eachy, which is uniquely determined byx, by (x). The

function then turns out to be a linear operator. Indeed, for given , we write:


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x1 =y1 +z1,x2 =y2 +z2 andx1 +x2 =y3 +z3

where and forj = 1,2,3. By the uniqueness of decomposition

(x1) + (x2) =y1 +y2 =y3= (x1 +x2).

The similar reasoning shows that commutes with scalars. Now, for

(where and ), we have:

That is, is continuous with . In particular, when is a nonzero space, there

is with (x0) =x0 and and consequently . Such is calledan orthogonal projection (onto ).

The next theorem gives an alternative proof of the Hahn-Banach theorem.

3 Theorem Let be a linear (not necessarily closed) subspace of a Hilbert space. Every

continuous linear functional on can be extended to a unique continuous linear functional

on that has the same norm and vanishes on .

Proof: Since is a dense subset of a Banach space , by Theorem 2.something, we

can uniquely extendfso that it is continuous on . Define . By the same

argument used in the proof of Theorem 2.something (Hahn-Banach) and the fact

that , we obtain . Since g = 0 on , it remains to show theuniqueness. For this, let h be another extension with the desired properties. Since the kernel

offh is closed and thus contain ,f= h on . Hence, for any ,

.

The extension g is thus unique.

3 Theorem Let be an increasing sequence of closed subspaces, and be the closure

of . If is an orthogonal projection onto , then for every

.

Proof: Let . Then is closed. Indeed,

if and , then

and so . Since , the proof is complete.

Let be Hilbert spaces. The direct sum of is defined as

follows: let and define


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.

It is then easy to verify that is a Hilbert space. It is also clear that this

definition generalizes to a finite direct sum of Hilbert spaces. (For an infinite direct sum of

Hilbert spaces, see Chapter 5.)

Recall from the previous chapter that an isometric surjection between Banach spaces is called

"unitary".

3 Lemma (Hilbert adjoint)Define

by . (Clearly, V is a unitary operator.) Then is a

graph (of some linear operator) if and only if T is densely defined.

Proof: Set . Let . Then

for every v.

That is to say, , which is a graph of a linearoperator by assumption. Thus, u = 0.

For the converse, suppose . Then

(j = 1,2)

and so for every v in the domain ofT, dense. Thus, u1 = u2, and is a

graph of a function, say, S. The linear ofS can be checked in the similar manner.

Remark: In the proof of the lemma, the linear ofTwas never used.

For a densely defined T, we thus obtained a linear operator which we call T* . It is characterized

uniquely by:

for every u,

or, more commonly,

for every u.

Furthermore, T*fis defined if and only if

is continuous for every . The operator T*

is called theHilbert adjoint(or just

adjoint) ofT. IfTis closed in addition to having dense domain, then


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Here, . By the above lemma, T*

is densely defined. More

generally, if a densely defined operator Thas a closed

extension S (i.e., ), then S and S*

are both densely defined. It

follows: . That is, T*

is densely defined and T* *

exists. That S = T* *

follows

from the next theorem.

3 Theorem Let be a densely defined operator. If T* is also densely defined,

then

for any closed extension S ofT.

Proof: As above,

Here, the left-hand side is a graph ofT* * . For the second identity, since is a Hilbert

space, it suffices to show . But this follows from Lemma

3.something.

The next corollary is obvious but is important in application.

3 Corollary Let be Hilbert spaces, and a closed densely defined

linear operator. Then if and only if there is some K > 0 such that:

for every

3 Lemma Let be a densely defined linear operator.

Then

Proof:fis in either the left-hand side or the right-hand side if and only if:

for every u.

(Note that for every u implies .)

In particular, a closed densely defined operator has closed kernel. As an application we shall

prove the next theorem.

3 Theorem Let be a closed densely defined linear operator. Then T is

surjective if and only if there is a K > 0 such that

for every .


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Proof: Suppose Tis surjective. Since Thas closed range, it suffices to show the estimate

for . Let with Tu=f. Denoting by G the inverse

ofTrestricted to , we have:

The last inequality holds since G is continuous by the closed graph theorem. To show the

converse, let be given. Since T*

is injective, we can define a linear

functionalL by for .,

for every .

Thus,L is continuous on the range ofT* . It follows from the Hahn-Banach theorem that we may

assume thatL is defined and continuous on . Thus, by Theorem 3.something, we canwrite in with some u. SinceL(T

*f) is continuous for ,

for every .

Hence, Tu = T* *

u = g.

3 Corollary Let be as given in the preceding theorem. Then is closed if and

only if is closed.

Proof: Define by S = T. It thus suffices to show S* is surjective when Thas

closed range (or equivalently S is surjective.) Suppose S*fj is convergent. The preceding theoremgives:

as .

Thus, is Cauchy in the graph ofS*

, which is closed.

Hence, S*fj converges within the range ofS

*. The converse holds since T

* *= T.

We shall now consider some concrete examples of densely defined linear operators.

3 Theorem If is continuous, then T* is defined everywhere and continuous,

and

.

Proof: It is clear that T* is defined everywhere, and its continuity is then the consequence of the

closed graph theorem. Now,

for everyf.

Thus, T*

is continuous with . In particular, T*Tis continuous, and so:


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for everyf.

That is to say, . Applying this result to T* in place

ofTcompletes the proof.

(Recall that a isometric linear surjection is called unitary.)

3 Corollary A linear operator is unitary if and only if U* U and UU* are

identities.

Proof: The direct part follows from

and the converse from the identity

.

(Note that: .)

Actually, this was how unitary operators were defined historically. We give a much stronger

characterization of unitary operators in Chapter 5, where we have the spectrum decomposition

theorem.

3 ExerciseConstruct an example so as to show that an isometric operator (i.e., a linear operator

that preserves norm) need not be unitary. (Hint: a shift operator.)

A densely defined linear operator Tis called "symmetric" if . If the equality in

the above holds, then Tis called "self-adjoint". In light of Theorem 3.something, every self-

adjoint is closed and densely defined. IfTis symmetric, then since T* * is an extension ofT,

.

3 Theorem Let be densely defined linear operators for j = 1,2.

Then where the equality holds if (j =1,2) and is closed and densely defined.

Proof: Let . Then

for

every .

But, by definition, denotes . Hence, is an extension

of . For the second part, the fact we have just proved gives:

.


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3 Theorem Let be a Hilbert spaces. If is a closed densely

defined operator, then T* T is a self-adjoint operator (in particular, densely defined and closed.)

Proof: In light of the preceding theorem, it suffices to show that T*Tis closed.

Let be a sequence such that (uj,T*Tuj)converges to limit (u,v). Since

,

there is some such that: . It follows from the closedness

ofT*

that T*f= v. Since and Tis

closed, T*Tu = T*f= v.

3 TheoremLet T be a symmetric densely defined operator. If T is surjective, then T is self-adjoint

and injective and T 1 is self-adjoint and bounded.

Proof: IfTu = 0,

and u = 0

ifThas a dense range (for example, it is surjective). Thus, Tis injective. Since T 1 is closed (by

Lemma 2.something) and , is a continuous linear

operator. Finally, we have:

.

Here, , and the equality holds since the domainsofTand T

*coincide. Hence, T 1 is self-adjoint. Since we have just proved that the inverse of a

self-adjoint is self-adjoint, we have: (T 1) 1 is self-adjoint.

3 Lemma Let S,T be linear operators from a pre-Hilbert space over to itself.

If for , then S = T.

Proof: LetR = TS. We have

and . Summing the two we

get: for . Takingy =Rx gives for all orR =

0.

3 ExerciseShow that the above lemma is false if the underlying field is . (Hint: it suffices to

consider a finite-dimensional Hilbert space.)

For , let (T) be the the set of all complex numbers such that T Iis notinvertible. (Here, I is the identity operator on .)

3 Theorem Let T be a densely defined operator on . Then T is positive (i.e., for

every ) if and only if T = T* and .

Partial proof: We have:


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for every

But, by hypothesis, the right-hand side is real. That T= T* follows from Lemma 5.something.

The proof of the theorem will be completed by the spectrum decomposition theorem in Chapter 5.

3 Theorem Let be a closed linear subspace of a Hilbert space . Thenis an orthogonal

projection onto if and only if = *= 2 and the range ofis .

Proof: The direct part is clear except for = * . But we have:

since (x) andx (x) are orthogonal. Thus, is positive and so self-adjoint then. For the

converse, we only have to verify for everyx. But we have: (x (x)) =

0 and .

3 Lemma (Bessel's inequality)If ukis an orthonormal sequence in a Hilbert space , then

for any .

Proof: If , then . Thus,

.

Letting completes the proof. .

2 Lemma A normed space is complete (in the sense of metric spaces) if and only

if implies exists.

Proof: If , then since

as n > m and ,


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exists by completeness. Conversely, supposexj is a Cauchy sequence. Thus, for eachj =

1,2,..., there exists an index kj such that for any .

Let . Then . Hence, by assumption we can get the

limit , and since

as ,

we conclude thatxj has a subsequence converging tox; thus, it converges tox.

3 Theorem (Parseval)Let ukbe a orthonormal sequence in a Hilbert space . Then the

following are equivalent:

(i) is dense in .

(ii) For each , .

(iii) For each , .

(iv) (the Parseval equality).

Proof: Let . If , then it has the form: for

some scalars k. Since we can also

write: . Let . Bessel's inequality and that is

complete ensure thaty exists. Since

for all , we have , proving (i) (ii). Now (ii) (iii)

follows since


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as

To get (iii) (iv), takex =y. To show (iv) (i), suppose that (i) is false. Then there exists

a with . Then

.

Thus, (iv) is false.

3 TheoremLet xkbe an orthogonal sequence in a Hilbert space . Then

the series converges if and only if the series converges for every .

Proof: Since

and

by orthogonality, we obtain the direct part. For the converse, let .

Since

for eachy

by hypothesis,Eis bounded by Theorem 3.something. Hence, and

converges by completeness.

The theorem is meant to give an example. An analogous issue in the Banach space will bediscussed in the next chapter.


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Chapter 4: Geometry of Banach spaces

In the previous chapter we studied a Banach space having a special geometric property: that is, a

Hilbert space. This chapter continues this line of the study. The main topics of the chapter are (i)

the notion of reflexibility of Banach spaces (ii) weak-* compactness, (iii) the study of a basis inBanach spaces and (iv) complemented (and uncomplemented) subspaces of Banach spaces. It

turns out those are a geometric property,

Let be a normed space. SinceX* is a Banach space, there is a canonical

injection given by:

(x)f=f(x) for and .

One of the most important question in the study of normed spaces is when this is surjective; if

this is the case, is said to be "reflexive". For one thing, since is a Banach space evenwhen is not a Banach, a normed space that is reflexive is always a Banach space.

(Since (X) separates points inX* , the weak-* topology is Hausdorff by Theorem 1.something.)

Before studying this problem, we introduce some topologies. The weak-* topology for is the

weakest among topologies for which every element of is continuous. In other words, the

weak-* topology is precisely the topology that makes the dual of . (Recall that it

becomes easier for a function to be continuous when there are more open sets in the domain of

the function.)

The weak topology for is the weakest of topologies for which every element of is

continuous. (As before, the weak topology is Hausdorff.)

4 Theorem (Alaoglu)The unit ball of is weak-* compact.

Proof: For everyf, is an element of . With this identification, we have: .

The inclusion in topology also holds; i.e., is a topological subspace of . The unit ball

of is a subset of the set

.

SinceE, a product of disks, is weak-* compact byw:Tychonoff's theorem(see Chapter 1), it

suffices to show that the closed unit ball is weak-* closed. (TODO: complete the proof.)

4. Theorem Let be a TVS whose dual separates points in . Then the weak-* topology

on is metrizable if and only if has a at most countable Hamel basis.

Obviously, all weakly closed sets and weak-* closed sets are closed (in their respective spaces.)

The converse in general does not hold. On the other hand,
http://en.wikipedia.org/wiki/Tychonoff%27s_theoremhttp://en.wikipedia.org/wiki/Tychonoff%27s_theoremhttp://en.wikipedia.org/wiki/Tychonoff%27s_theoremhttp://en.wikipedia.org/wiki/Tychonoff%27s_theorem


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4. Lemma Every closed convex subset is weakly closed.

Proof: Letx be in the weak closure ofE. Suppose, if possible, that . By (the geometric

form) of the Hanh-Banach theorem, we can then find and real number c such that:

for every .

Set . What we have now is: where Vis weakly open

(by definition). This is contradiction.

4. CorollaryThe closed unit ball of (resp. ) is weakly closed (resp. weak-* closed).

4 Exercise Let B be the unit ball of . Prove(B) is weak-* dense in the closed unit ball

of . (Hint: similar to the proof of Lemma 4.something.)

4 TheoremA set E is weak-* sequentially closed if and only if the intersection of E and the(closed?) ball of arbitrary radius is weak-* sequentially closed.

Proof: (TODO: write a proof using PUB.)

4 Theorem (Kakutani) Let be a Banach space. The following are equivalent:

(i) is reflexive. (ii) The closed unit ball of is weakly compact. (iii)Every bounded set admits a weakly convergent subsequence. (thus, the unit ball in

(ii) is actually weakly sequentially compact.)

Proof: (i) (ii) is immediate. For (iii) (i), we shall prove: if is not reflexive, then we canfind a normalized sequence that falsifies (iii). For that, see[1], which shows how to do this.

Finally, for (ii) (iii), it suffices to prove:

4 Lemma Let be a Banach space, a sequence and F be the weak closure of xj.

If F is weakly compact, then F is weakly sequentially compact.

Proof: By replacingXwith the closure of the linear span ofX, we may assume that admits a

dense countable subsetE. Then for , u(x) = v(x) for every implies u = v by

continuity. This is to say, a set of functions of the form with separates

points inX, a fortiori,B, the closed unit ball ofX* . The weak-* topology forB is therefore

metrizable by Theorem 1.something. Since a compact metric space is second countable; thus,

separable,B admits a countable (weak-*) dense subsetB'. It follows thatB' separates points inX.

In fact, for any with , by the Hahn-Banach theorem, we can find

such that . By denseness, there is that is nearx in the sense: | g(x)

f(x) | < 2 1, and we have:

.

Again by theorem 1.something, Fis now metrizable.
http://www.ams.org/proc/1998-126-08/S0002-9939-98-04691-7/S0002-9939-98-04691-7.pdfhttp://www.ams.org/proc/1998-126-08/S0002-9939-98-04691-7/S0002-9939-98-04691-7.pdfhttp://www.ams.org/proc/1998-126-08/S0002-9939-98-04691-7/S0002-9939-98-04691-7.pdfhttp://www.ams.org/proc/1998-126-08/S0002-9939-98-04691-7/S0002-9939-98-04691-7.pdf


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Remark: Lemma 4.something is a special case ofw:Eberleinmulian theorem, which states thatevery subset of a Banach space is weakly compact if and only if it is weakly sequentially

compact. (See[2],[3])

In particular, since every Hilbert space is reflexive, either (ii) or (iii) in the theorem always holds

for all Hilbert spaces. But for (iii) we could have used alternatively:

4 ExerciseGive a direct proof that (iii) of the theorem holds for a separable Hilbert space. (Hint:

use an orthonormal basis to directly construct a subsequence.)

4 Corollary A Banach space is reflexive if and only if is reflexive.'

4 Theorem Let be a Banach space with aw:Schauder basisej. Prove that is reflexive if

and only if ej satisfies:

(i) converges in .

(ii) For any , .

Proof: ( ): Set . By reflexivity,xn then admits a weakly convergent

subsequence with limitx. By hypothesis, for any , we can

write: with . Thus,

, and so .

This proves (i). For (ii), set

.

Then (ii) means that for any . SinceEn is a weakly closed subset of the

closed unit ball of , which is weakly compact by reflexivity,En is weakly compact. Hence,

there is a sequencexn such that: for any . It follows:
http://en.wikipedia.org/wiki/Eberlein%E2%80%93%C5%A0mulian_theoremhttp://en.wikipedia.org/wiki/Eberlein%E2%80%93%C5%A0mulian_theoremhttp://en.wikipedia.org/wiki/Eberlein%E2%80%93%C5%A0mulian_theoremhttp://en.wikipedia.org/wiki/Eberlein%E2%80%93%C5%A0mulian_theoremhttp://www.ams.org/proc/2003-131-10/S0002-9939-03-06894-1/S0002-9939-03-06894-1.pdfhttp://www.ams.org/proc/2003-131-10/S0002-9939-03-06894-1/S0002-9939-03-06894-1.pdfhttp://www.ams.org/proc/2003-131-10/S0002-9939-03-06894-1/S0002-9939-03-06894-1.pdfhttp://misuma.um.es/beca/Investigacion/22_Version%20Proceddings%20Charrla%20Huelva.pdfhttp://misuma.um.es/beca/Investigacion/22_Version%20Proceddings%20Charrla%20Huelva.pdfhttp://misuma.um.es/beca/Investigacion/22_Version%20Proceddings%20Charrla%20Huelva.pdfhttp://en.wikipedia.org/wiki/Schauder_basishttp://en.wikipedia.org/wiki/Schauder_basishttp://en.wikipedia.org/wiki/Schauder_basishttp://en.wikipedia.org/wiki/Schauder_basishttp://misuma.um.es/beca/Investigacion/22_Version%20Proceddings%20Charrla%20Huelva.pdfhttp://www.ams.org/proc/2003-131-10/S0002-9939-03-06894-1/S0002-9939-03-06894-1.pdfhttp://en.wikipedia.org/wiki/Eberlein%E2%80%93%C5%A0mulian_theorem


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since . (TODO: but does exist?) This proves (ii).

( ): Letxn be a bounded sequence. For eachj, the set is bounded; thus,

admits a convergent sequence. By Cantor's diagonal argument, we can therefore find a

subsequence ofxn such that converges for everyj. Set .

Let and . By

(ii), . Now,

for .

Since sm is bounded, for everyfand so .

By (i), therefore exists. Let > 0 be given. Then there exists m such that sm< / 2.Also, there existsNsuch that:

for every .

Hence,

.

4 ExerciseProve that every infinite-dimensional Banach space contains a closed subspace with a

Schauder basis. (Hint: construct a basis by induction.)

4 TheoremA Hilbert space is separable if and only if it has an (countable) orthonormal basis.

It is plain that a Banach space is separable if it has a Schauder basis. Unfortunately, the converse

is false.

4 Theorem (James) A Banach space is reflexive if and only if every element of attains its

maximum on the closed unit ball of .


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4 Corollary (Krein-Smulian) Let be a Banach space and a weakly compact

subset of . then is weakly compact.

Proof:[4]

A Banach space is said to be uniformly convex if

and

Clearly, Hilbert spaces are uniformly convex. The point of this notion is the next result.

4 Theorem Every uniformly convex space is reflexive.

Proof: Suppose, if possible, that is uniformly convex but is not reflexive.

4 TheoremEvery finite dimensional Banach space is reflexive.

Proof: (TODO)

4 Theorem Let be Banach spaces. If has aw:Schauder basis, then the space of

finite-rank operators on is (operator-norm) dense in the space of compact operators on .

5 TheoremLp spaces are uniformly convex (thus, reflexive).

Proof: (TODO)

5 Theorem (M. Riesz extension theorem) (seew:M. Riesz extension theorem)

References

SEPARABLE BANACH SPACE THEORY NEEDS STRONG SET EXISTENCEAXIOMS

Functional Analysis and Infinite-dimensional Geometry
http://misuma.um.es/beca/Investigacion/03_BanachEncy_I_Final.pdfhttp://misuma.um.es/beca/Investigacion/03_BanachEncy_I_Final.pdfhttp://misuma.um.es/beca/Investigacion/03_BanachEncy_I_Final.pdfhttp://en.wikipedia.org/wiki/Schauder_basishttp://en.wikipedia.org/wiki/Schauder_basishttp://en.wikipedia.org/wiki/Schauder_basishttp://en.wikipedia.org/wiki/M._Riesz_extension_theoremhttp://en.wikipedia.org/wiki/M._Riesz_extension_theoremhttp://en.wikipedia.org/wiki/M._Riesz_extension_theoremhttp://www.ams.org/tran/1996-348-10/S0002-9947-96-01725-4/S0002-9947-96-01725-4.pshttp://www.ams.org/tran/1996-348-10/S0002-9947-96-01725-4/S0002-9947-96-01725-4.pshttp://www.ams.org/tran/1996-348-10/S0002-9947-96-01725-4/S0002-9947-96-01725-4.pshttp://www.ams.org/tran/1996-348-10/S0002-9947-96-01725-4/S0002-9947-96-01725-4.pshttp://www.ams.org/tran/1996-348-10/S0002-9947-96-01725-4/S0002-9947-96-01725-4.pshttp://books.google.com/books?id=NFClZiX2hMkC&printsec=frontcover&dq=Auerbach+basishttp://books.google.com/books?id=NFClZiX2hMkC&printsec=frontcover&dq=Auerbach+basishttp://books.google.com/books?id=NFClZiX2hMkC&printsec=frontcover&dq=Auerbach+basishttp://www.ams.org/tran/1996-348-10/S0002-9947-96-01725-4/S0002-9947-96-01725-4.pshttp://www.ams.org/tran/1996-348-10/S0002-9947-96-01725-4/S0002-9947-96-01725-4.pshttp://en.wikipedia.org/wiki/M._Riesz_extension_theoremhttp://en.wikipedia.org/wiki/Schauder_basishttp://misuma.um.es/beca/Investigacion/03_BanachEncy_I_Final.pdf


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Chapter 5: Topological vector spaces

A vector space endowed by a topology that makes translations (i.e.,x +y) and dilations (i.e., x)continuous is called a topological vector space or TVS for short.

A subsetEof a TVS is said to be:

boundedif for every neighborhood Vof 0 there exist s > 0 such that forevery t> s

balancedif for every scalar with convex if for any and any with 1x+ 2y =

1.

1 Corollary(s + t)E = sE + tE for any s,t > 0 if and only if E is convex.

Proof: Supposing s + t= 1 we obtain for all . Conversely, ifEisconvex,

, or for any .

Since holds in general, the proof is complete.

Definef(,x) = x for scalars , vectorsx. IfEis a balanced set, for any , by continuity,

.

Hence, the closure of a balanced set is again balanced. In the similar manner, ifEis convex,

for s,t> 0

,

meaning the closure of a convex set is again convex. Here the first equality holds since

is injective if . Moreover, the interior ofE, denoted by , is also convex. Indeed,

for with 1 + 2 = 1

,

and since the left-hand side is open it is contained in . Finally, a subspace of a TVS is a

subset that is simultaneously a linear subspace and a topological subspace. Let be a subspace

of a TVS. Then is a topological subspace, and it is stable under scalar multiplication, as

shown by the argument similar to the above. Let g(x,y) =x +y. If is a subspace of a TVS, by

continuity and linearity,

.


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Hence, is a linear subspace. We conclude that the closure of a subspace is a subspace.

Let Vbe a neighborhood of 0. By continuity there exists a > 0 and a neighborhood Wof 0 suchthat:

It follows that the set {; | | < }Wis a union of open sets, contained in Vand is balanced. Inother words, every TVS admits a local base consisting of balanced sets.

1 Theorem Let be a TVS, and . The following are equivalent.

(i) E is bounded. (ii) Every countable subset of E is bounded. (iii) for every balanced neighborhood V of 0 there exists a t > 0 such that .

Proof: That (i) implies (ii) is clear. If (iii) is false, there exists a balanced neighborhood Vsuch

that for every n = 1,2,.... That is, there is a unbounded sequencex1,x2,... inE. Finally,

to show that (iii) implies (i), let Ube a neighborhood of 0, and Vbe a balanced open set

with . Choose tso that , using the hypothesis. Then for any s > t, we

have:

1 CorollaryEvery Cauchy sequence and every compact set in a TVS are bounded.

Proof: If the set is not bounded, it contains a sequence that is not Cauchy and does not have a

convergent subsequence.

1 LemmaLet f be a linear operator between TVSs. If f(V) is bounded for some

neighborhood V of 0, then f is continuous.

6 TheoremLet f be a linear functional on a TVS .

(i) f has either closed or dense kernel. (ii) f is continuous if and only if is closed.

Proof: To show (i), suppose the kernel offis not closed. That means: there is ay which is in the

closure of but . For any , is in the kernel off. This is

to say, every element of is a linear combination ofy and some other element in .

Thus, is dense. (ii) Iffis continuous, is closed. Conversely,

suppose is closed. Sincefis continuous whenfis identically zero, suppose there is apointy withf(y) = 1. Then there is a balanced neighborhood Vof 0 such


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that . It then follows that . Indeed, suppose .

Then

if , which is a contradiction.

The continuity offnow follows from the lemma.

6 Theorem Let be a TVS and its subspace. Suppose:

is dense in impliesz = 0 in .

(Note this is the conclusion of Corollary 2.something) Then every continuous linear functionfon

a subspace of extends to an element of .

Proof: We essentially repeat the proof of Theorem 3.8. So, let be the kernel off, which isclosed, and we may assume . Thus, by hypothesis, we can find such that:g =

0 inM, but for some pointp outsideM. By Lemma 1.6, g= ffor some scalar.

Since bothfand g do not vanish atp, .

LemmaLet V0,V1,... be a sequence of subsets of a a linear space containing 0 such

that for every . If

and , then .

Proof: We shall prove the lemma by induction over k. The basic case k= 1 holdssince for every n. Thus, assume that the lemma has been proven until k 1.First, suppose n1,...,nkare not all distinct. By permutation, we may then assume that n1 = n2. It

then follows:

and .

The inductive hypothesis now gives: . Next, suppose n1,...,nkare all distinct. Again by

permutation, we may assume that n1 < n2< ...nk. Since no carry-over occurs then and m < n1, m +

1 < n2 and so:

.

Hence, by inductive hypothesis, .

1 Theorem Let be a TVS.

(i) If is Hausdorff and has a countable local base, is metrizable with themetric d such that

d(x,y) = d(x + z,y + z) and for every


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(ii) For every neighborhood of 0, there is a continuous function g such thatg(0) = 0, g = 1 on Vc and for any x,y.

Proof: To show (ii), let V0,V1,... be a sequence of neighborhoods of 0 satisfying the condition in

the lemma and V= V0. Define g =

1on Vc and for every .

To show the triangular inequality, we may assume that g(x) andg(y) are both < 1, and thus

suppose and . Then

Thus, . Taking inf over all

such n1,...,nkwe obtain:

and do the same for the rest we conclude . This proves (ii)

since g is continuous at 0 and it is then continuous everywhere by the triangular inequality. Now,

to show (i), choose a sequence of balanced sets V0,V1,... that is a local base, satisfies the condition

in the lemma and is such that . As above,

define for each . For

the same reason as before, the triangular inequality holds. Clearly,f(0) = 0. If ,

then there are n1,...,nksuch that and .

Thus, by the lemma. In particular, if for "every" m, thenx =

0 since is Hausdorff. Since Vn are balanced, if ,

for every n1,...,nkwith .

That means , and in particular .

Defining d(x,y) =f(xy) will complete the proof of (i). In fact, the properties offwe havecollected shows the function dis a metric with the desired properties. The lemma then shows that

given any m, for some . That is, the sets {x;f(x) < } over >

0 forms a local base for the original topology.

The second property ofdin (i) implies that open ball about the origin in terms of this dis

balanced, and when has a countable local base consisting of convex sets it can be

strengthened to: , which implies open balls about the origin are convex.

Indeed, if , and if and with 1+ 2, then

since the sum of convex sets is again convex. This is to say,


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and by iteration and continuity it can be shown that for every .

CorollaryFor every neighborhood V of some point x, there is a neighborhood

of x with

Proof: Since we may assume thatx = 0, take W= {x;g(x) < 2 1}.

Corollary If every finite set of a TVS is closed, is Hausdorff.

Proof: Letx,y be given. By the preceding corollary we find an open set

containingx.

A TVS with a local base consisting of convex sets is said to be locally convex. Since in this book

we will never study non-Hausdorff locally convex spaces, we shall assume tacitly that everyfinite subset of every locally convex is closed, hence Hausdorff in view of Theorem something.

LemmaLet be locally convex. The convex hull of a bounded set is bounded.

Given a sequencepn of semi-norms, define:

.

dthen becomes a metric. In fact,Since for any

seminormp, .

References

lp with 0 < p < 1 is not locally convex
http://www.math.umn.edu/~garrett/m/fun/ell_p_not_convex.pdfhttp://www.math.umn.edu/~garrett/m/fun/ell_p_not_convex.pdfhttp://www.math.umn.edu/~garrett/m/fun/ell_p_not_convex.pdf


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Chapter 6: C*-algebras

A Banach space over is called aBanach algebra if it is an algebra and satisfies

.

We shall assume that every Banach algebra has the unit 1 unless stated otherwise.

Since as , the

map

is continuous.

For , let (x) be the the set of all complex numbers such thatx 1 is not invertible.

5 TheoremFor every , (x) is nonempty and closed and

.

Moreover,

(r(x) is called the spectral radius ofx)

Proof: Let be the group of units. Define byf() = 1 x. (Throughout

the proofx is fixed.) If , then, by definition, or .

Similarly, we have: . Thus, . Sincefis clearly

continuous, is open and so (x) is closed. Suppose that for . Bythe geometric series (which is valid by Theorem 2.something), we have:

Thus, is invertible, which is to say, s1 x is invertible. Hence, . Thiscomplete the proof of the first assertion and gives:

Since (x) is compact, there is a such that r(x) = a. Since (useinduction to see this),


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Next, we claim that the sequence is bounded for | s | > r(x). In view of the uniform

boundedness principle, it suffices to show that is bounded for every . But

since

,

this is in fact the case. Hence, there is a constant c such that for every n. It

follows:

.

Taking inf over | s | > r(x) completes the proof of the spectral radius formula. Finally, suppose, on

the contrary, that (x) is empty. Then for every , the map

is analytic in . Since , by Liouville's theorem, we must

have: g((xs) 1

) = 0. Hence, (xs) 1

= 0 for every , a contradiction.

5 Corollary (Gelfand-Mazur theorem) If every nonzero element of is invertible, then is

isomorphic to .

Proof: Let be a nonzero element. Since (x) is non-empty, we can then findsuch that 1 x is not invertible. But, by hypothesis, 1 x is invertible, unless 1 =x.

Let be a maximal ideal of a Banach algebra. (Such exists by the usual argument involving

Zorn's Lemma in abstract algebra). Since the complement of consists of invertible

elements, is closed. In particular, is a Banach algebra with the usual quotient norm. By

the above corollary, we thus have the isomorphism:

Much more is true, actually. Let be the set of all nonzero

homeomorphism . (The members of are calledcharacters.)

5 Theorem is bijective to the set of all maximal ideals of .

5 Lemma Let . Then x is invertible if and only if for every


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5 Theorem

An involution is an anti-linear map such thatx* * =x. Prototypical

examples are the complex conjugation of functions and the operation of taking the adjoint of a

linear operator. These examples explain why we require an involution to be anti-linear.

Now, the interest of study in this chapter. A Banach algebra with an involution is called a C*-

algebra if it satisfies

(C*-identity)

From the C*-identity follows

,

for and the same forx*

in place ofx. In particular, (if 1 exists).

Furthermore, the C*

-identity is equivalent to the condition: , for this and

implies and so .

For each , let C* (x) be the linear span of . In other

words, C* (x) is the smallest C*-algebra that containsx. The crucial fact is that C* (x) is

commutative. Moreover,

Theorem Let be normal. Then

A state on C* -algebra is a positive linear functional f such that (or

equivalently f(1) = 1). Since S is convex and closed, S is weak-* closed. (This is Theorem

4.something.) Since S is contained in the unit ball of the dual of , S is weak-* compact.

5 Theorem Every C^*-algebra is *-isomorphic to C0(X) where X is the spectrum of .

5 TheoremIf C0(X) is isomorphic to C0(Y), then it follows that X and Y are homeomorphic.

3 LemmaLet T be a continuous linear operator on a Hilbert space . Then TT* = T* T if and

only if for all .

Continuous linear operators with the above equivalent conditions are said to be normal. For

example, an orthogonal projection is normal. Seew:normal operatorfor additional examples and

the proof of the above lemma.

3 LemmaLet N be a normal operator. If andare distinct eigenvalues of N, then the

respective eigenspaces of andare orthogonal to each other.
http://en.wikipedia.org/wiki/normal_operatorhttp://en.wikipedia.org/wiki/normal_operatorhttp://en.wikipedia.org/wiki/normal_operatorhttp://en.wikipedia.org/wiki/normal_operator


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Proof: LetIbe the identity operator, andx,y be arbitrary eigenvectors for ,, respectively. Sincethe adjoint ofIis , we have:

.

That is, , and we thus have:

If is nonzero, we must have = .

5 Exercise Let be a Hilbert space with orthogonal basis e1,e2,..., and xn be a sequence

with . Prove that there is a subsequence of xn that converges weakly to some x and

that . (Hint: Since is bounded, by Cantor's diagonal argument, we can find

a sequence such that is convergent for every k.)

5 Theorem (Von Neumann double commutant theorem)M is equal to its double commutant if

and only if it is closed in either weak-operator topology or strong-operator topology.

Proof: (seew:Von Neumann bicommutant theorem)

References

Lecture Notes on C -Algebras and K-Theory A Good Spectral Theorem
http://en.wikipedia.org/wiki/Von_Neumann_bicommutant_theoremhttp://en.wikipedia.org/wiki/Von_Neumann_bicommutant_theoremhttp://en.wikipedia.org/wiki/Von_Neumann_bicommutant_theoremhttp://staff.science.uva.nl/~npl/CK.pdfhttp://staff.science.uva.nl/~npl/CK.pdfhttp://staff.science.uva.nl/~npl/CK.pdfhttp://staff.science.uva.nl/~npl/CK.pdfhttp://www.math.umn.edu/~garrett/m/fun/good_spectral_thm.pdfhttp://www.math.umn.edu/~garrett/m/fun/good_spectral_thm.pdfhttp://www.math.umn.edu/~garrett/m/fun/good_spectral_thm.pdfhttp://staff.science.uva.nl/~npl/CK.pdfhttp://en.wikipedia.org/wiki/Von_Neumann_bicommutant_theorem


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