Wikibooks Functional Analysis

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Functional Analysis A Brief Overview Collected by Jyoti Swaroop Repaka 12/2/2008

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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 the subject, 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 unintegrated part 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 theorems will be found in the Topology and Linear Algebra books.

Contents       

1 Theorem (Heine-Borel) 2 Theorem (Tychonoff Theorem) 3 Theorem (Baire Category Theorem) 4 Theorem (Urysohn Metrization Theorem) 5 Theorem (Arzelà–Ascoli)  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. Exercise Show by contradiction that the set of real numbers is uncountable, using the theorem. In particular, a dense set may be meager. 1. Exercise Give 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 every j x = y. The converse of this also holds. Since d(x,y) = d(y,x), d is a metric. Let ηd be the topology for K that is induced by d. We claim ηd coincides with the topology originally given to K. In light of: 1 Lemma Let X be a set. If and τ2 is compact, then τ1 = τ2.

are a pair of topologies for X and if τ1 is Hausdorff

it suffices to show that ηd 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 ηd-open ball in d with center at x is open (in the original topology.)

Theorem (Arzelà–Ascoli) Definitions  

A set of functions F defined on [a,b] is uniformly bounded if there exists an M such that for any function f within F, f(x)<M for all x within [a,b]. A set of functions F defined on [a,b] is equicontinuous if for all ε > 0, there exists a δ > 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 (Arzelà–Ascoli) 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 at An 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 Hölder'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 | by f and g, respectively. First we have:

Then Hölder's inequality followed by division gives:

where .

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

By letting μ 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

that

and p(tx) = tp(x)for any

and any t > 0.

If is a closed subspace and f is a linear functional on then f admits a linear extension F defined in such that . Proof: First suppose that hypothesis we have:

which is equivalent to:

such that

for some

for all

such , . By

,

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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 F is an desired extension. Indeed, f = F on clear, we also have:

being

if t > 0 and

if t < 0. Let Ω be the collection of pairs (H,gH) where H is linear space with and gH is a linear function on that extends f and is dominated by p. It can be shown that Ω is partially ordered and the union of every totally ordered sub-collection of Ω is in Ω (TODO: need more details). 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 of c in the proof results in a different extension. Thus, an extension given by the Hahn-Banach theorem in general is not unique. 1. Exercise State 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 Lemma Let 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 is y such that f1(y) = 1. For any x, since

,

g(x − f1(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 may suppose that there is some y such that fn(y) = 1 while f1(y) = f2(y) = ... = fn − 1(y) = 0. For any , since fk(x − fn(x)y) = 0 for every k = 1,2,...,n, g(x − fn(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 Γ of real (or complex)-valued functions defined on a set X is said to separate points in X if and f(x) = f(y) for every

implies x = y

With regard to this notion, there are two important facts that will be used in Chapter 4. 1. Theorem Let X be a topology generated by a set Γ of real(or complex)-valued functions. Then X is Hausdorff if and only if Γseparates 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 function f on

notation 

(i)



(ii)



(iii)

, with the

, such that (w:triangular inequality) for any scalar λ implies x = 0.

(ii) implies that

. This and (ii) then

implies negative. Note that (i) implies that:

for all x; that is, norms are always non-

and and so:

. (So, the map

(ii) hold, then

is continuous.) If only (i) and

is called a seminorm.

A linear space with a norm is called a normed space. With the metric normed space is a metric space. We define theoperator norm of a continuous linear operator f between normed spaces

and

, denoted by

a

, 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 from w: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 by yn. We claim:

as Let ε > 0 be given. Since xk is Cauchy, there is N such 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) T is continuous if and only if there is a constant C > 0 such that all



(ii) element.

any C as in (i) }, and

. for

if

has nonzero

Proof: See w:bounded operator and see w: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 a Banach 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 function f defined on a set E is 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 of f. 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 Since

and Tfj is defined for every j; i.e., the sequence fj is in the domain of T.

, Tfj is Cauchy. It follows that (fj,Tfj) is Cauchy and, by completeness, has limit (g,Tg). Since f = g, Tf is defined; i.e., f is in the domain of T. 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 true for 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 T is

surjective, some B(k)contains an interior point; thus, it is a neighborhood of 0. A linear operator from a normed space set

to

, is closed in

. Then by Baire's Theorem,

is said to be closed if its graph, that is the .

2 Corollary If 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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Let I be a subspace of a normed space norm:

. The quotient space

where is a canonical projection. That triangular inequality. But since

for all

is a normed space with

is a norm is obvious except for the

. Taking inf over i,j separately we get:

Suppose, further, that is also a commutative algebra and I is an ideal. Then quotient algebra. In fact, as above, we have:

becomes a

, since π is a homomorphism. Taking inf completes the proof.

for all

So, the only nontrivial part is the completeness. It turns out that algebra) if I is closed. (TODO: a proof of this.) 2 Corollary If and

and

is a Banach space (or

are Banach spaces, then the norms

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

Proof: Let

be the identity map. Then we have: .

This is to say, I is continuous. Since Cauchy sequences apparently converge in the norm , the open mapping theorem says that the inverse of I is also continuous, which means explicitly: . By the same argument we can show that

is dominated by

2 Corollary Any 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 operator. The following are equivalent.

be Banach spaces, and

a linear

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 in X. , andTxj − Tx is convergent. Thus, if (ii)

Then xj converges to some x0 or holds, gives the inequality:

. Finally, to prove (iii)

(i), we note that Corollary 2.something

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 Banach space), 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 Corollary A linear functional u on a normed space is continuous if and only if it has closed kernel. Proof:

is closed if u 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 is z such that u(z) = 1. Then the sequence xj − u(xj)z has a limit in the kernel of u, 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 operators from to . 2 Theorem If

we denote the space of all continuous linear

is complete, then every Cauchy sequence Tn in

limit T and . Proof: Let Tn be a Cauchy sequence in operator norm. For each

converges to a , since

and is complete, there is a limit y to which Tn(x) converges. Define T(x) = y. T is 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 functions meager and that:

. be a family of continuos

where Y is a normed linear space. Suppose that

is non-

for each It then follows: there is some

open and such that

(a) If we assume in addition that each member of space, then

is a linear operator and X is a normed linear

(b) Proof: Let

be a sequence. By

hypothesis, and each Ej is closed since is open by continuity. It then follows that some EN has an interior point y; otherwise, M fails to be nonmeager. Hence, (a) holds. To show (b), making additional assumptions, we can find an open ball with

. It then follows: for any

and any

, .

A family Γ of linear operators is said to be equicontinuous if given any neighborhood W of 0 we can find a neighborhood V of 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 equicontinuous. That is, there is K > 0 such that:

shows the family is

for every The theorem now follows since

and every

.

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 dominating z. Then:

as a sublinear functional

; that is,

.

2. Corollary Let be a subspace of a normed linear space if and only if z(x) = 0 for any that vanishes on . Proof: By continuity

. Thus, if

, then

.

. Then there is a δ > 0 such that

Conversely, suppose

for

. Define a linear functional z(y + λx) = λ for

every any

. Then x is in the closure of

, since

and scalars λ. For

, .

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

while we still have z = 0 on

and

.

Here is a classic application. 2 Theorem Let

be Banach spaces,

implies that Proof: Suppose continuity of z,

for every and

.

be a linear operator. If , then T is continuous. . For every , by hypothesis and the

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Now, by the preceding corollary y = 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



(ii) Given

, if f(x) = 0 for every

for every , then x = 0.

Proof: (i) By continuity, . This proves the direct part. For the converse, define Txf = f(x) for hypothesis for every

. By

.

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, f is continuous since its domain is finite-dimensional, and so by the Hahn-Banach theorem we could extend the domain of f in such a way we have

.

2. Corollary Let Then

be Banach, for every

and

if and only if

dense and linear. and

for

every . Proof: Since fj is Cauchy, it is bounded. This shows the direct part. To show the converse, let

. If

, then

By denseness, we can take yj so that

.

2 Theorem Let 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 each x in the range of T. Proof: For

, we have:

. Since the series is geometric by hypothesis, the right-hand side is finite. Let . By the above, each time x 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 each x boundedness that:

, it follows from the principle of uniform

. 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 Corollary The space of invertible continuous linear operators of

is an open subspace

.

Proof: If

and

, then S is invertible.

If is a scalar field and is a normed space, then is called a dual of denoted by . In view of Theorem 2.something, it is a Banach space.

and is

A linear operator T is said to be a compact operator if the image of the open unit ball under T is 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 operator

be a reflexive Banach space and be a Banach space. Then a linear is a compact operator if and only if T sends weakly convergent

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sequence to norm convergent ones. Proof: [1] Let xn converges weakly to 0, and suppose Txn is not convergent. That is, there is an ε > 0 such that for infinitely many n. Denote this subsequence by yn. 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, let E be a bounded set. Then since is reflexive every countable subset of E contains a sequence xn 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 Lemma Let 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, using w:Riesz's_lemma, we can construct a sequence xj such that:

for any sequence of scalars ak. Thus, in particular,

for all j,k. (TODO: fill gaps)

2 Corollary  

(i) Every finite-rank linear operator T (i.e., a linear operator with finite-dimensional range) 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 Theorem The 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 T be a linear operator and ω be the open unit ball in the domain of T. If T is compact, then is bounded (try scalar multiplication); thus, T is continuous. Since the sum of two compacts sets is again compact, the sum of two compact operators is again compact. For the similar reason, αT is compact for any scalar α. 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 of E. Let ε > 0 be given. Then there is some compact operator T such that . Also, since T is a compact operator, we can cover T(ω) by a finite number of open balls of radius ε / 2 centered at z1,z2,...zn, respectively. It then follows: for

, we can find some j so that

and

so . This is to say, S(ω) is totally bounded and since the completeness its closure is compact. 2 Corollary If Tn is a sequence of compact operators which converges in operator norm, then its limit is a compact operator.

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2 Theorem (transpose) Let

be Banach spaces, and

be a continuous linear

operator. Define

by the identity

. Then

continuous both in operator norm and the weak-* topology, and Proof: For any

Thus,

and

is

.

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

> 0 be given. Then there is

with

. Using the Hahn-Banach

theorem we can also find

and z0(u(x0)) = | u(x0) | . Hence,

. We conclude

. To show weak-* continuity let V be a neighborhood of 0 in

that is, some

;

for . If we let yj = u(xj), then

since

. This is to say,

is weak-* continuous.

2 Theorem Let , , and p q If is closed in both L and L , then the topology for

a linear subspace. inherited from Lp and Lq coincide.

Proof: The identity map is continuous. Since I is a bijection between Banach spaces, the open mapping theorem says I is a homeomorphism. 2 Theorem (lifting property of l1) Let linear surjection. If

be Banach spaces, and

be a

is a a continuous linear operator, there exists a continuous

linear operator such that . 1 Proof: Let en be a canonical basis for l . Using Corollary 2.something to the open mapping theorem we obtain a sequence xn and constant K so that: Qxn = Ten, and

.

Then by

. Define . Then

References

, and

is continuous since

.

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1. ↑ This proof and a few more related results appear in [1]

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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 a pre-Hilbert space if for each pair (x,y) of elements in the space there is a unique complex (or real) number called an inner product of x and y, denoted by subject to the following conditions: 

(i) The functional



(ii)



(iii)

,

is linear. for every nonzero x

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

for scalars α. We define

and this becomes a norm.

Indeed, it is clear that and (iii) is the reason that 0. Finally, the triangular inequality follows from the next lemma. 3.1 Lemma (Schwarz's inequality) if we can write x = λy for some scalar λ.

, implies that x =

where the equality holds if and only

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

since

for any complex number α

Proof of Lemma: First suppose

. If

, it then follows:

where the equation becomes 0 if and only if x = λy. Since we may suppose that general case follows easily.

, the

3.2 Theorem A 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 Moreover, since the calculation:

,

and

.

, we have: . If α is a real scalar and αj is a sequence of rational numbers converging to α, then by continuity and the above, we get: 3.3 Lemma Let

be a pre-Hilbert. Then

and Proof: The direct part holds since:

in norm if and only if for any as

.

as

.

Conversely, we have: as

3.4 Lemma Let D be a non-empty convex closed subset of a Hilbert space. Then D admits a unique element z such that . Proof: By δ denote the right-hand side. Since D is nonempty, δ > 0. For each n = 1,2,..., there is some convex,

such that

. That is,

and so

.

It follows:

as

. Since D is

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This is to say, xn is Cauchy. Since D is a closed subset of a complete metric space, whence it is complete, there is a limit we have

where the right side is

with

. The uniqueness follows since if

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, Finally, if

, then

3.5 Lemma Let

is a closed (linear) subspace of

.

andx = 0.

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 if z and z + w are linear dependent. quite well written.) 3.6 Theorem (orthogonal decomposition) Let subspace. For every we can write

(TODO: the proof isn't

be a Hilbert space and

be a closed

x=y+z where Proof: Clearly

and

, and y and z are uniquely determined by x. is convex, and it is also closed since a translation of closed set is again

closed. Lemma 3.4 now gives a uniqueelement that the uniqueness, suppose we have written:

such . Let z = x − y. By Lemma 3.5,

. For

x = y' + z' where and . By Lemma 3.5, But, as noted early, such y' must be unique; i.e., y' = y. 3.7 Corollary Let 

(i)

be a subspace of a Hilbert space if and only if

is dense in

.

. Then .

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

(ii)

.

Proof: By continuity,

. (Here,

set E under the map

denotes the image of the

.) 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 . Since f is continuous, is closed. If 0. If not, by Corollary 3.6, there is a nonzero orthogonal to . By replacing z with we may suppose that in the kernel of f and thus is orthogonal to z, we have:

. For any

, then take y = , since zf(x) − f(z)x is

and so:

The uniqueness follows since that

for all

means

. Finally, we have the identity:

where the last inequality is Schwarz's inequality. 3.9 Exercise Using 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 each y, which is uniquely determined by x, by π(x). The function π then turns out to be a linear operator. Indeed, for given , we write:

25

x1 = y1 + z1,x2 = y2 + z2 and x1 + x2 = y3 + z3 where

and

for j = 1,2,3. By the uniqueness of decomposition

π(x1) + π(x2) = y1 + y2 = y3 = π(x1 + x2). The similar reasoning shows that π commutes with scalars. Now, for (where

and

), we have:

That is, π is continuous with

. In particular, when

is with π(x0) = x0 and an orthogonal projection (onto

is a nonzero space, there . Such π is called

and consequently ).

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 extend f so 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 the uniqueness. For this, let h be another extension with the desired properties. Since the kernel of f − h is closed and thus contain , f = h on . Hence, for any , . The extension g is thus unique. 3 Theorem Let of

be an increasing sequence of closed subspaces, and be the closure . If is an orthogonal projection onto , then for every .

Proof: Let if

and so Let follows: let

. Then and

. Since

is closed. Indeed,

, then

, the proof is complete. be Hilbert spaces. The direct sum of and define

is defined as

26

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

. Let

is a

. Then for every v.

That is to say, For the converse, suppose

, which is a graph of a linear operator by assumption. Thus, u = 0. . Then (j = 1,2)

and so for every v in the domain of T, dense. Thus, u1 = u2, and graph of a function, say, S. The linear of S can be checked in the similar manner.

is a

Remark: In the proof of the lemma, the linear of T was 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 * f is defined if and only if

is continuous for every . The operator T * is called the Hilbert adjoint (or just adjoint) of T. If T is closed in addition to having dense domain, then

27

Here, . By the above lemma, T * is densely defined. More generally, if a densely defined operator T has a closed ), then S and S * are both densely defined. It

extension S (i.e., follows: from the next theorem.

. That is, T * is densely defined and T * * exists. That S = T * * follows

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

3 Theorem Let then

for any closed extension S of T. Proof: As above,

Here, the left-hand side is a graph of T * * . For the second identity, since space, it suffices to show 3.something.

is a Hilbert

. But this follows from Lemma

The next corollary is obvious but is important in application. 3 Corollary Let linear operator. Then

be Hilbert spaces, and a closed densely defined if and only if there is some K > 0 such that: for every

3 Lemma Let

be a densely defined linear operator.

Then Proof: f is 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

.

28

Proof: Suppose T is surjective. Since T has closed range, it suffices to show the estimate for

. Let

of T restricted to

with Tu= f. Denoting by G the inverse

, 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 functional L by

for

., for every

.

Thus, L is continuous on the range of T * . It follows from the Hahn-Banach theorem that we may assume that L is defined and continuous on . Thus, by Theorem 3.something, we can write

in

with some u. Since L(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 T has closed range (or equivalently S is surjective.) Suppose S * fj is convergent. The preceding theorem gives: as

.

Thus, is Cauchy in the graph of S * , which is closed. * Hence, S fj converges within the range of S * . The converse holds since T * * = T. We shall now consider some concrete examples of densely defined linear operators. 3 Theorem If and

is continuous, then T * is defined everywhere and continuous,

. Proof: It is clear that T * is defined everywhere, and its continuity is then the consequence of the closed graph theorem. Now, for every f. Thus, T * is continuous with

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

29

for every f. . Applying this result to T * in place

That is to say, of T completes the proof.

(Recall that a isometric linear surjection is called unitary.) is unitary if and only if U * U and UU * are

3 Corollary A linear operator 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 Exercise Construct 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 T is called "symmetric" if . If the equality in the above holds, then T is called "self-adjoint". In light of Theorem 3.something, every selfadjoint is closed and densely defined. If T is symmetric, then since T * * is an extension of T, . 3 Theorem Let

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

Then

where the equality holds if

1,2) and

(j =

is closed and densely defined.

Proof: Let

. Then for

every But, by definition, of

. denotes

. Hence,

is an extension

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

30

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

Let

, there is some *

such that:

. It follows from the closedness

*

of T that T f = v. Since closed, T * Tu = T * f = v.

and Tis

3 Theorem Let 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: If Tu = 0, and u = 0 if T has a dense range (for example, it is surjective). Thus, T is injective. Since T − 1 is closed (by Lemma 2.something) and operator. Finally, we have:

,

is a continuous linear

. Here, , and the equality holds since the domains * −1 of T and T coincide. Hence, T 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 If

for

over

to itself.

, then S = T.

Proof: Let R = T − S. We have and get: 0.

. Summing the two we for

. Taking y = Rx gives

for all

3 Exercise Show that the above lemma is false if the underlying field is consider a finite-dimensional Hilbert space.)

or R =

. (Hint: it suffices to

For , let ζ(T) be the the set of all complex numbers λ such that T − λI is not invertible. (Here, I is the identity operator on .) 3 Theorem Let T be a densely defined operator on every

) if and only if T = T * and

Partial proof:

We have:

. Then T is positive (i.e., .

for

31

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 projection onto if and only if π = π * = π2 and the range of π is Proof: The direct part is clear except for π = π * . But we have:

. Then π is an orthogonal .

since π(x) and x − π(x) are orthogonal. Thus, π is positive and so self-adjoint then. For the for every x. But we have: π(x − π(x)) =

converse, we only have to verify 0 and

.

3 Lemma (Bessel's inequality) If uk is an orthonormal sequence in a Hilbert space

for any Proof: If

, then

, then

. . Thus,

. Letting

completes the proof.

2 Lemma A normed space

if

is complete (in the sense of metric spaces) if and only

implies

Proof: If

as n > m and

.

exists.

, then since

,

32

exists by completeness. Conversely, suppose xj is a Cauchy sequence. Thus, for each j = 1,2,..., there exists an index kj such that

Let

for any

. Then

.

. Hence, by assumption we can get the

limit

, and since

as

,

we conclude that xj has a subsequence converging to x; thus, it converges to x. 3 Theorem (Parseval) Let uk be a orthonormal sequence in a Hilbert space following are equivalent: 

(i)



(ii) For each



(iii) For each



(iv)

is dense in ,

. .

,

. (the Parseval equality).

Proof: Let

. If

some scalars αk. Since write: . Let complete ensure that y exists. Since

for all follows since

. Then the

, we have

, then it has the form:

for

we can also . Bessel's inequality and that

, proving (i)

(ii). Now (ii)

is

(iii)

33

as To get (iii)

(iv), take x = y. To show (iv)

a

with

(i), suppose that (i) is false. Then there exists . Then

. Thus, (iv) is false. 3 Theorem Let xk be an orthogonal sequence in a Hilbert space

the series Proof: Since

converges if and only if the series

. Then

converges for every

.

and

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

.

for each y

by hypothesis, E is bounded by Theorem 3.something. Hence, converges by completeness.

and

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

34

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 in Banach spaces and (iv) complemented (and uncomplemented) subspaces of Banach spaces. It turns out those are a geometric property, Let be a normed space. Since X * is a Banach space, there is a canonical injection given by: π(x)f = f(x) for

and

.

One of the most important question in the study of normed spaces is when this π is surjective; if this is the case, is said to be "reflexive". For one thing, since is a Banach space even when is not a Banach, a normed space that is reflexive is always a Banach space. (Since π(X) separates points in X * , the weak-* topology is Hausdorff by Theorem 1.something.) Before studying this problem, we introduce some topologies. The weak-* topology for weakest among topologies for which every element of

is the

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 continuous. (As before, the weak topology is Hausdorff.) 4 Theorem (Alaoglu) The unit ball of Proof: For every f,

is an element of

The inclusion in topology also holds; i.e., of is a subset of the set

is

is weak-* compact. . With this identification, we have: is a topological subspace of

. The unit ball

. Since E, a product of disks, is weak-* compact by w: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,

.

35

4. Lemma Every closed convex subset

is weakly closed.

Proof: Let x be in the weak closure of E. Suppose, if possible, that form) of the Hanh-Banach theorem, we can then find for every

and real number c such that: .

Set . What we have now is: (by definition). This is contradiction. 4. Corollary The closed unit ball of

(resp.

. By (the geometric

where V is weakly open

) is weakly closed (resp. weak-* closed).

4 Exercise Let B be the unit ball of . Prove π(B) is weak-* dense in the closed unit ball of . (Hint: similar to the proof of Lemma 4.something.) 4 Theorem A 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 can find 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 replacing X with the closure of the linear span of X, we may assume that admits a dense countable subset E. 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 in X, a fortiori, B, the closed unit ball of X . The weak-* topology for B is therefore metrizable by Theorem 1.something. Since a compact metric space is second countable; thus, separable, B admits a countable (weak-*) dense subset B'. It follows that B' separates points in X. In fact, for any

with

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

, by the Hahn-Banach theorem, we can find . By denseness, there is

that is near x in the sense: | g(x)

. Again by theorem 1.something, F is now metrizable.

36

Remark: Lemma 4.something is a special case of w:Eberlein–Šmulian theorem, which states that every 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 Exercise Give 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 a w:Schauder basis ej. Prove that and only if ej satisfies:



(i)



(ii) For any

converges in

is reflexive if

.

,

.

Proof: ( ): Set . By reflexivity, xn then admits a weakly convergent subsequence with limit x. By hypothesis, for any , we can

write:

with

. Thus,

, and so

.

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

Then (ii) means that closed unit ball of

for any . Since En is a weakly closed subset of the , which is weakly compact by reflexivity, En is weakly compact. Hence,

there is a sequence xn such that:

for any

. It follows:

37

since

. (TODO: but does

exist?) This proves (ii).

( ): Let xn be a bounded sequence. For each j, the set is bounded; thus, admits a convergent sequence. By Cantor's diagonal argument, we can therefore find a subsequence

of xn such that

Let

converges for every j. Set

and

(ii),

.

. By

. Now,

for

.

Since sm is bounded,

for every f and so

.

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

for every

.

Hence,

.

4 Exercise Prove that every infinite-dimensional Banach space contains a closed subspace with a Schauder basis. (Hint: construct a basis by induction.) 4 Theorem A 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 maximum on the closed unit ball of .

attains its

38

4 Corollary (Krein-Smulian) Let subset of Proof: [4]

. then

be a Banach space and

a weakly compact

is weakly compact.

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 Theorem Every finite dimensional Banach space is reflexive. Proof: (TODO) 4 Theorem Let finite-rank operators on

be Banach spaces. If has a w:Schauder basis, then the space of is (operator-norm) dense in the space of compact operators on .

5 Theorem Lp spaces are uniformly convex (thus, reflexive). Proof: (TODO) 5 Theorem (M. Riesz extension theorem) (see w:M. Riesz extension theorem)

References  

SEPARABLE BANACH SPACE THEORY NEEDS STRONG SET EXISTENCE AXIOMS Functional Analysis and Infinite-dimensional Geometry

39

Chapter 5: Topological vector spaces A vector space endowed by a topology that makes translations (i.e., x + y) and dilations (i.e., αx) continuous is called a topological vector space or TVS for short. A subset E of a TVS is said to be: 

bounded if for every neighborhood V of 0 there exist s > 0 such that every t > s



balanced if convex if 1.



for every scalar λ with for any

for

with λ1x + λ2y =

and any

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, if E is convex,

, or Since

for any

.

holds in general, the proof is complete.

Define f(λ,x) = λx for scalars λ, vectors x. If E is a balanced set, for any

, by continuity,

. Hence, the closure of a balanced set is again balanced. In the similar manner, if E is convex, for s,t > 0 , meaning the closure of a convex set is again convex. Here the first equality holds since is injective if for

. Moreover, the interior of E, denoted by with λ1 + λ2 = 1

, is also convex. Indeed,

, 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, .

40

Hence,

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

Let V be a neighborhood of 0. By continuity there exists a δ > 0 and a neighborhood W of 0 such that:

It follows that the set {λ; | λ | < δ}W is a union of open sets, contained in V and is balanced. In other 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 V such that for every n = 1,2,.... That is, there is a unbounded sequence x1,x2,... in E. Finally, to show that (iii) implies (i), let U be a neighborhood of 0, and V be a balanced open set with . Choose t so that , using the hypothesis. Then for any s > t, we have:

1 Corollary Every 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 Lemma Let f be a linear operator between TVSs. If f(V) is bounded for some neighborhood V of 0, then f is continuous.

6 Theorem Let 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 of f is not closed. That means: there is a y which is in the closure of but to say, every element of Thus,

. For any , is in the kernel of f. This is is a linear combination of y and some other element in .

is dense. (ii) If f is continuous,

is closed. Conversely,

suppose is closed. Since f is continuous when f is identically zero, suppose there is a point y with f(y) = 1. Then there is a balanced neighborhood V of 0 such

41

that Then

. It then follows that

. Indeed, suppose

if

.

, which is a contradiction.

The continuity of f now follows from the lemma. 6 Theorem Let

be a TVS and

is dense

its subspace. Suppose: in

implies z = 0 in

.

(Note this is the conclusion of Corollary 2.something) Then every continuous linear function f on a subspace of extends to an element of . Proof: We essentially repeat the proof of Theorem 3.8. So, let be the kernel of f, which is closed, and we may assume 0 in M, but

. Thus, by hypothesis, we can find

such that:g =

for some point p outside M. By Lemma 1.6, g = λf for some scalarλ.

Since both f and g do not vanish at p,

.

Lemma Let 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 holds since for every n. Thus, assume that the lemma has been proven until k − 1. First, suppose n1,...,nk are not all distinct. By permutation, we may then assume that n1 = n2. It then follows:

and

.

The inductive hypothesis now gives: . Next, suppose n1,...,nk are 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, metric d such that d(x,y) = d(x + z,y + z) and

is metrizable with the

for every

42 

(ii) For every neighborhood

of 0, there is a continuous function g such that

g(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, such n1,...,nk we obtain:

. Taking inf over all

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,...,nk such that and Thus, 0 since

by the lemma. In particular, if is Hausdorff. Since Vn are balanced, if

, .

for "every" m, then x = ,

for every n1,...,nk with

.

That means , and in particular . Defining d(x,y) = f(x − y) will complete the proof of (i). In fact, the properties of f we have collected shows the function d is a metric with the desired properties. The lemma then shows that given any m, for some 0 forms a local base for the original topology.

. That is, the sets {x;f(x) < δ} over δ >

The second property of d in (i) implies that open ball about the origin in terms of this d is balanced, and when has a countable local base consisting of convex sets it can be strengthened to: Indeed, if

, which implies open balls about the origin are convex. , and if and with λ1 + λ2, then

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

43

and by iteration and continuity it can be shown that

for every

.

Corollary For every neighborhood V of some point x, there is a neighborhood of x with Proof: Since we may assume that x = 0, take W = {x;g(x) < 2 − 1}. Corollary If every finite set of a TVS

is closed,

is Hausdorff.

Proof: Let x,y be given. By the preceding corollary we find an open set containing x. 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 every finite subset of every locally convex is closed, hence Hausdorff in view of Theorem something. Lemma Let

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

Given a sequence pn of semi-norms, define:

. d then becomes a metric. In fact, Since

for any

seminorm p,

References 

lp with 0 < p < 1 is not locally convex

.

44

Chapter 6: C*-algebras A Banach space

over

is called a Banach algebra if it is an algebra and satisfies .

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

as

, the

is continuous. For

, let ζ(x) be the the set of all complex numbers λ such that x − λ1 is not invertible.

5 Theorem For every

, σ(x) is nonempty and closed and .

Moreover,

(r(x) is called the spectral radius of x) Proof: Let

be the group of units. Define

the proof x is fixed.) If Similarly, we have:

by f(λ) = λ1 − x. (Throughout

, then, by definition, . Thus,

or . Since f is clearly

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

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

Since ζ(x) is compact, there is a induction to see this),

.

such that r(x) = a. Since

for

. By

. This

(use

45

Next, we claim that the sequence

is bounded for | s | > r(x). In view of the uniform

boundedness principle, it suffices to show that since

is bounded for every

. But

, this is in fact the case. Hence, there is a constant c such that follows:

for every n. It

. Taking inf over | s | > r(x) completes the proof of the spectral radius formula. Finally, suppose, on the contrary, that ζ(x) is empty. Then for every

, the map

is analytic in . Since , by Liouville's theorem, we must −1 −1 have: g((x − s) ) = 0. Hence, (x − s) = 0 for every , a contradiction. 5 Corollary (Gelfand-Mazur theorem) If every nonzero element of is invertible, then isomorphic to . Proof: Let be a nonzero element. Since ζ(x) is non-empty, we can then find such that λ1 − x is not invertible. But, by hypothesis, λ1 − x is invertible, unless λ1 = x.

is

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 homeomorphism 5 Theorem 5 Lemma Let

be the set of all nonzero . (The members of

are calledcharacters.)

is bijective to the set of all maximal ideals of . Then x is invertible if and only if

. for every

46

5 Theorem An involution is an anti-linear map such that x * * = 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 for x * in place of x. In particular,

Furthermore, the C * -identity is equivalent to the condition: implies

(if 1 exists). , for this and

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 Theorem If C0(X) is isomorphic to C0(Y), then it follows that X and Y are homeomorphic. 3 Lemma Let T be a continuous linear operator on a Hilbert space only if

for all

. Then TT * = T * T if and

.

Continuous linear operators with the above equivalent conditions are said to be normal. For example, an orthogonal projection is normal. See w:normal operator for additional examples and the proof of the above lemma. 3 Lemma Let N be a normal operator. If α and β are distinct eigenvalues of N, then the respective eigenspaces of α and β are orthogonal to each other.

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Proof: Let I be the identity operator, and x,y be arbitrary eigenvectors for α,β, respectively. Since the adjoint of αI is , we have: . That is,

, and we thus have:

is nonzero, we must have α = β.

If

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

a sequence

such that

is bounded, by Cantor's diagonal argument, we can find 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: (see w:Von Neumann bicommutant theorem)

References  

Lecture Notes on C ∗-Algebras and K-Theory A Good Spectral Theorem

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