Functional Analysis/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,

Weak and weak-* topologies

Let $𝒳$ be a normed space. Since $X^{\ast }$ is a Banach space, there is a canonical injection $\pi :𝒳\to {𝒳}^{\ast \ast }$ given by:

$(\pi x)f=f(x)$ for $f\in {𝒳}^{\ast }$ and $x\in 𝒳$.

One of the most important question in the study of normed spaces is when this $\pi$ is surjective; if this is the case, $𝒳$ is said to be "reflexive". For one thing, since ${𝒳}^{\ast \ast }$, as the dual of a normed space, is a Banach space even when $𝒳$ is not, a normed space that is reflexive is always a Banach space, since $\pi$ becomes an (isometrical) isomorphism. (Since $\pi (X)$ separates points in $X^{\ast }$, the weak-* topology is Hausdorff by Theorem 1.something.)

Before studying this problem, we introduce some topologies. The weak-* topology for ${𝒳}^{\ast }$ is the weakest among topologies for which every element of $\pi (𝒳)$ is continuous. In other words, the weak-* topology is precisely the topology that makes the dual of ${𝒳}^{\ast }$ $\pi (𝒳)$. (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 $𝒳$ is the weakest of topologies for which every element of ${𝒳}^{\ast }$ is continuous. (As before, the weak topology is Hausdorff.)

4 Theorem (Alaoglu) The unit ball of ${𝔅}^{\ast }$ is weak-* compact.
Proof: For every $f$, $\mathrm{ran}f$ is an element of ${𝐂}^{𝔅}$. With this identification, we have: ${𝔅}^{\ast }\subset {𝐂}^{𝔅}$. The inclusion in topology also holds; i.e., ${𝔅}^{\ast }$ is a topological subspace of ${𝐂}^{𝔅}$. The unit ball of ${𝔅}^{\ast }$ is a subset of the set

$E=\prod _{x\in 𝔅}\{\lambda ;\lambda \in 𝐂,|\lambda |\le \Vert x{\Vert }_{𝔅}\}$.

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. This is easy once we have the notion of nets, which will be introduced in the next chapter. For the sake of completeness, we give a direct argument here. (TODO) $◻$

4. Theorem Let $𝒳$ be a TVS whose dual separates points in $𝒳$. Then the weak-* topology on ${𝒳}^{\ast }$ is metrizable if and only if $𝒳$ 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,

4. Lemma Every closed convex subset $E\subset 𝒳$ is weakly closed.
Proof: Let $x$ be in the weak closure of $E$. Suppose, if possible, that $x\in ̸E$. By (the geometric form) of the Hanh-Banach theorem, we can then find $f\in {𝒳}^{\ast }$ and real number $c$ such that:

$\mathrm{Re}f(x)<c<\mathrm{Re}f(y)$ for every $y\in E$.

Set $V=\{y;\mathrm{Re}f(y)<c\}$. What we have now is: $x\in V\subset E^c$ where $V$ is weakly open (by definition). This is contradiction.$◻$

4. Corollary The closed unit ball of $𝒳$ (resp. ${𝒳}^{\ast }$) is weakly closed (resp. weak-* closed).

4 Exercise Let $B$ be the unit ball of $𝒳$. Prove $\pi (B)$ is weak-* dense in the closed unit ball of ${𝒳}^{\ast \ast }$. (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.)

Reflexive Banach spaces

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

• (i) $𝒳$ is reflexive.
• (ii) The closed unit ball of $𝒳$ 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) $\Rightarrow$ (ii) is immediate. For (iii) $\Rightarrow$ (i), we shall prove: if $𝒳$ 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) $\Rightarrow$ (iii), it suffices to prove:

4 Lemma Let $𝒳$ be a Banach space, $x_j\in X$ a sequence and $F$ be the weak closure of $x_j$. 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 $𝒳$ admits a dense countable subset $E$. Then for $u,v\in {𝒳}^{\ast }$, $u(x)=v(x)$ for every $x\in E$ implies $u=v$ by continuity. This is to say, a set of functions of the form $u\mapsto u(x)$ with $x\in E$ separates points in $X$, a fortiori, $B$, the closed unit ball of $X^{\ast }$. 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 $B^{\prime }$. It follows that $B^{\prime }$ separates points in $X$. In fact, for any $x\in X$ with $\Vert x\Vert =1$, by the Hahn-Banach theorem, we can find $f\in B$ such that $f(x)=\Vert x\Vert =1$. By denseness, there is $g\in B^{\prime }$ that is near $x$ in the sense: $|g(x)-f(x)|<2^{-1}$, and we have:

$|g(x)|\ge |f(x)|-|g(x)-f(x)|>2^{-1}$.

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

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 $𝒳$ is reflexive if and only if ${𝒳}^{\ast }$ is reflexive.'

4 Theorem Let $𝒳$ be a Banach space with a w:Schauder basis $e_j$. $𝒳$ is reflexive if and only if $e_j$ satisfies:

• (i) $\underset{n}{\sup }\Vert \sum _{j=1}^n{a}_j{e}_j\Vert <\mathrm{\infty }\Rightarrow \sum _{j=1}^{\mathrm{\infty }}{a}_j{e}_j$ converges in $𝒳$.
• (ii) For any $f\in {𝒳}^{\ast }$, $\underset{n\to \mathrm{\infty }}{\lim }\sup \{|f(x)|;x=\sum _{j\ge n}^{\mathrm{\infty }}{a}_j{e}_j,\Vert x\Vert =1\}=0$.

Proof: ($\Rightarrow$): Set $x_n=\sum _{j=1}^n{a}_j{e}_j$. By reflexivity, $x_n$ then admits a weakly convergent subsequence $x_{n_k}$ with limit $x$. By hypothesis, for any $x\in 𝒳$, we can write: $x=\sum _{j=1}^{\mathrm{\infty }}{b}_j(x)e_j$ with $b_j\in {𝒳}^{\ast }$. Thus,

$b_l(x)=\underset{k\to \mathrm{\infty }}{\lim }{b}_l(x_{n_k})=\underset{k\to \mathrm{\infty }}{\lim }\sum _{j=1}^{n_k}{a}_j{b}_l(e_j)=a_l$, and so $x=\sum _{j=1}^{\mathrm{\infty }}{a}_j{e}_j$.

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

$E_n=\{x;x\in 𝒳,\Vert x\Vert =1,b_1(x)=\mathrm{...}{b}_{n-1}(x)=0\}$.

Then (ii) means that $\underset{E_n}{\sup }|f|\to 0$ for any $f\in {𝒳}^{\ast }$. Since $E_n$ is a weakly closed subset of the closed unit ball of ${𝒳}^{\ast }$, which is weakly compact by reflexivity, $E_n$ is weakly compact. Hence, there is a sequence $x_n$ such that: $\underset{E_n}{\sup }|f|=|f(x_n)|$ for any $f\in {𝒳}^{\ast }$. It follows:

$\underset{n\to \mathrm{\infty }}{\lim }|f(x_n)|=|f(\underset{n\to \mathrm{\infty }}{\lim }{x}_n)|=|f(\sum _{j=1}^{\mathrm{\infty }}{b}_j(\underset{n\to \mathrm{\infty }}{\lim }{x}_n)e_j)|=0$

since $\underset{n\to \mathrm{\infty }}{\lim }{b}_j(x_n)=0$. (TODO: but does $\underset{n\to \mathrm{\infty }}{\lim }{x}_n$ exist?) This proves (ii).
($\Leftarrow$): Let $x_n$ be a bounded sequence. For each $j$, the set $\{b_j(x_n);n\ge 1\}$ is bounded; thus, admits a convergent sequence. By Cantor's diagonal argument, we can therefore find a subsequence $x_{n_k}$ of $x_n$ such that $b_j(x_{n_k})$ converges for every $j$. Set $a_j=\underset{n\to \mathrm{\infty }}{\lim }{b}_j(x_{n_k})$. Let $K=2\underset{n}{\sup }\Vert x_n\Vert$ and $s_n=\sup \{|f(y)|;y=\sum _{j=m+1}^{\mathrm{\infty }}{c}_j{e}_j,\Vert y\Vert \le K\}$. By (ii), $\underset{n\to \mathrm{\infty }}{\lim }{s}_n=0$. Now,

$|f(\sum _{j=1}^m{b}_j(x_{n_k})e_j)|\le |f(\sum _{j=1}^{\mathrm{\infty }}{b}_j(x_{n_k})e_j)|+|f(\sum _{j=m+1}^{\mathrm{\infty }}{b}_j(x_{n_k})e_j)|\le \Vert f\Vert \underset{n}{\sup }\Vert x_n\Vert +s_m$ for $f\in {𝒳}^{\ast }$.

Since $s_m$ is bounded, $\underset{m}{\sup }|f(\sum _{j=1}^m{a}_j{e}_j)|<\mathrm{\infty }$ for every $f$ and so $\underset{m}{\sup }\Vert \sum _{j=1}^m{a}_j{e}_j\Vert <\mathrm{\infty }$. By (i), $\sum _{j=1}^m{a}_j{e}_j$ therefore exists. Let $ϵ>0$ be given. Then there exists $m$ such that $s_m<ϵ/2$. Also, there exists $N$ such that:

$\sum _{j=1}^m(a_j-b_j(x_{n_k}))f(e_j)<ϵ/2$ for every $k\ge N$.

Hence,

$|f(x_{n_k})-f(\sum _{j=1}^{\mathrm{\infty }}{a}_j{e}_j)|\le |\sum _{j=1}^m(a_j-b_j(x_{n_k}))f(e_j)|+|f(\sum _{j=m+1}^{\mathrm{\infty }}(a_j-b_j(x_{n_k}))e_j|<ϵ$.

4 Exercise Prove that every infinite-dimensional Banach space contains a closed subspace with a Schauder basis. (Hint: construct a basis by induction.)

Compact operators on Hilbert spaces

3 Lemma Let $T\in B(ℌ)$. Then $T(\bar{B}(0,1))$ is closed.
Proof: Since $\bar{B}(0,1)$ is weakly compact and $T(\bar{B}(0,1))$ is convex, it suffices to show $T$ is weakly continuous. But if $x_n\to 0$ weakly, then $(T{x}_n|y)=(x_n|T^{\ast }y)\to 0$ for any y. This shows that T is weakly continuous on $\bar{B}(0,1)$ (since bounded sets are weakly metrizable) and thus on $ℌ$.$◻$

Since T is compact, it suffices to show that $T(\bar{B}(0,1))$ is closed. But since $T(\bar{B}(0,1))$ is weakly closed and convex, it is closed.

3 Lemma If $T\in B(ℌ)$ is self-adjoint and compact, then either $\Vert T\Vert$ or $-\Vert T\Vert$ is an eigenvalue of T.
Proof: First we prove that $\Vert T{\Vert }^2$ is an eigenvalue of $T^2$. Since $T$ is compact, by the above lemma, there is a $x_0$ in the unit ball such that $\Vert T\Vert =\Vert T{x}_0\Vert$. Since $⟨T^2{x}_0,x_0⟩=\Vert T{\Vert }^2$,

$\Vert T^{2}x-\Vert T{\Vert }^{2}x{\Vert }^2\le \Vert T{\Vert }^2-2\Vert T{\Vert }^2+\Vert T{\Vert }^2$

Thus, $T^2{x}_0=\Vert T{\Vert }^2{x}_0$. Since $(T^2-\Vert T{\Vert }^{2}I)x_0=(T+\Vert T\Vert I)(T-\Vert T\Vert I)x_0$, we see that $(T-\Vert T\Vert I)x_0$ is either zero or an eigenvector of $T$ with respect to $-\Vert T\Vert$. $◻$

3 Theorem If T is normal; that is, $T^{\ast }T=T{T}^{\ast }$, then there exists an orthonormal basis consisting of eigenvectors of T.
Proof: Since we may assume that T is self-adjoint, the theorem follows from the preceding lemma by transfinite induction. By Zorn's lemma, select U to be a maximal subset of H with the following three properties: all elements of U are eigenvectors of T, they have norm one, and any two distinct elements of U are orthogonal. Let F be the orthogonal complement of the linear span of U. If F ≠ {0}, it is a non-trivial invariant subspace of T, and by the initial claim there must exist a norm one eigenvector y of T in F. But then U ∪ {y} contradicts the maximality of U. It follows that F = {0}, hence span(U) is dense in H. This shows that U is an orthonormal basis of H consisting of eigenvectors of T.$◻$

3 Corollary (polar decomposition) Every compact operator K can be written as:

$K=R|K|$

where R is a partial isometry and $|K|$ is the square root of $K^{\ast }K$

For $T\in ℒ(ℌ)$, let $\sigma (T)$ be the set of all complex numbers $\lambda$ such that $T-\lambda I$ is not invertible. (Here, I is the identity operator on $ℌ$.)

3 Corollary Let $T\in B(ℌ)$ be a compact normal operator. Then

$\Vert T\Vert =\underset{\Vert x\Vert =1}{\max }\Vert (Tx|x)\Vert =\sup \{|\lambda ||\lambda \in \sigma (T)\}$

3 Theorem Let $T$ be a densely defined operator on $ℌ$. Then $T$ is positive (i.e., $⟨Tx,x⟩\ge 0$ for every $x\in \mathrm{dom}T$) if and only if $T=T^{\ast }$ and $\sigma (T)\subset [0,\mathrm{\infty })$.
Partial proof: $(\Rightarrow )$ We have:

$⟨Tx,x⟩=\overline{⟨T^{\ast }x,x⟩}$ for every $x\in \mathrm{dom}T$

But, by hypothesis, the right-hand side is real. That $T=T^{\ast }$ follows from Lemma 5.something. The proof of the theorem will be completed by the spectrum decomposition theorem in Chapter 5.$◻$

More materials on compact operators, especially on their spectral properties, can be found in a chapter in the appendix where we study Fredholm operators.

3 Lemma (Bessel's inequality) If $u_k$ is an orthonormal sequence in a Hilbert space $ℌ$, then

$\sum _{k=1}^{\mathrm{\infty }}|⟨x,u_k⟩{|}^2\le \Vert x{\Vert }^2$ for any $x\in ℌ$.

Proof: If $⟨x,y⟩=0$, then $\Vert x+y{\Vert }^2=\Vert x{\Vert }^2+\Vert y{\Vert }^2$. Thus,

$\Vert x-\sum _{k=1}^n⟨x,u_k⟩{\Vert }^2=\Vert x{\Vert }^2-2\mathrm{Re}\sum _{k=1}^n|⟨x,u_k⟩{|}^2+\sum _{k=1}^n|⟨x,u_k⟩{|}^2=\Vert x{\Vert }^2-\sum _{k=1}^n|⟨x,u_k⟩{|}^2$.

Letting $n\to \mathrm{\infty }$ completes the proof. $◻$.

3 Theorem (Parseval) Let $u_k$ be a orthonormal sequence in a Hilbert space $ℌ$. Then the following are equivalent:

• (i) $\mathrm{span}\{u_1,u_2,\mathrm{...}\}$ is dense in $ℌ$.
• (ii) For each $x\in ℌ$, $x=\sum _{k=1}^{\mathrm{\infty }}⟨x,u_k⟩u_k$.
• (iii) For each $x,y\in ℌ$, $⟨x,y⟩=\sum _{k=1}^{\mathrm{\infty }}⟨x,u_k⟩\overline{⟨y,u_k⟩}$.
• (iv) $\Vert x{\Vert }^2=\sum _{k=1}^{\mathrm{\infty }}|⟨x,u_k⟩{|}^2$ (the Parseval equality).

Proof: Let $ℳ=\mathrm{span}\{u_1,u_2,\mathrm{...}\}$. If $v\in ℳ$, then it has the form: $v=\sum _{k=1}^{\mathrm{\infty }}{\alpha }_k{u}_k$ for some scalars ${\alpha }_k$. Since $⟨v,u_j⟩=\sum _{k=1}^{\mathrm{\infty }}{a}_j⟨u_k,u_j⟩=a_j$ we can also write: $v=\sum _{k=1}^{\mathrm{\infty }}⟨v,u_k⟩u_k$. Let $y=\sum _{k=1}^{\mathrm{\infty }}⟨x,u_k⟩u_k$. Bessel's inequality and that $ℌ$ is complete ensure that $y$ exists. Since

$⟨y,v⟩=\sum _{k=1}^{\mathrm{\infty }}⟨x,u_k⟩⟨u_k,v⟩=\sum _{k=1}^{\mathrm{\infty }}⟨x,⟨v,u_k⟩u_k⟩=⟨x,\sum _{k=1}^{\mathrm{\infty }}⟨v,u_k⟩u_k⟩=⟨x,v⟩$

for all $v\in ℳ$, we have $x-y\in {ℳ}^{\mathrm{\perp }}=\{0\}$, proving (i) $\Rightarrow$ (ii). Now (ii) $\Rightarrow$ (iii) follows since

$|⟨x,y⟩-\sum _{k=1}^n⟨x,u_k⟩\overline{⟨y,u_k⟩}|=|⟨x,y-\sum _{k=1}^n⟨y,u_k⟩u_k⟩|\to 0$ as $n\to \mathrm{\infty }$

To get (iii) $\Rightarrow$ (iv), take $x=y$. To show (iv) $\Rightarrow$ (i), suppose that (i) is false. Then there exists a $z\in (\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\mathrm{\{}{\mathrm{u}}_{\mathrm{1}},{\mathrm{u}}_{\mathrm{2}},\mathrm{...}\mathrm{\}}{)}^{\mathrm{\perp }}$ with $z\ne 0$. Then

$\sum _{k=1}^{\mathrm{\infty }}|⟨z,u_k⟩{|}^2=0<\Vert z{\Vert }^2$.

Thus, (iv) is false.$◻$

3 Theorem Let $x_k$ be an orthogonal sequence in a Hilbert space $(ℌ,\Vert \cdot \Vert =⟨\cdot ,\cdot {⟩}^{1/2})$. Then the series $\sum _{k=1}^{\mathrm{\infty }}{x}_k$ converges if and only if the series $\sum _{k=1}^{\mathrm{\infty }}⟨x_k,y⟩$ converges for every $y\in ℌ$.
Proof: Since

$\sum _{k=1}^{\mathrm{\infty }}|⟨x_k,y⟩|\le \Vert y\Vert \sum _{k=1}^{\mathrm{\infty }}\Vert x_k\Vert$ and $\sum _{k=1}^{\mathrm{\infty }}\Vert x_k\Vert =\Vert \sum _{k=1}^{\mathrm{\infty }}{x}_k\Vert$

by orthogonality, we obtain the direct part. For the converse, let $E=\{\sum _{k=1}^n{x}_k;n\ge 1\}$. Since

$\underset{E}{\sup }|⟨\cdot ,y⟩|=\underset{n}{\sup }|\sum _{k=1}^n⟨x_k,y⟩|<\mathrm{\infty }$ for each $y$

by hypothesis, $E$ is bounded by Theorem 3.something. Hence, $\sum _{k=1}^{\mathrm{\infty }}\Vert x_k\Vert <\mathrm{\infty }$ and $\sum _{k=1}^n{x}_k$ converges by completeness.

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

4 Theorem A Hilbert space $ℌ$ 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 $𝒳$ is reflexive if and only if every element of $𝒳$ attains its maximum on the closed unit ball of $𝒳$.

4 Corollary (Krein-Smulian) Let $𝒳$ be a Banach space and $K\subset 𝒳$ a weakly compact subset of $𝒳$. then $\overline{co}(K)$ is weakly compact.
Proof: [4]

A Banach space is said to be uniformly convex if

$\Vert x_n\Vert \le 1,\Vert y_n\Vert \le 1$ and $\Vert x_n+y_n\Vert \to 0\Rightarrow \Vert x_n-y_n\Vert \to 2$

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

4 Theorem Every uniformly convex space $𝔅$ is reflexive.
Proof: Suppose, if possible, that $𝔅$ is uniformly convex but is not reflexive. $◻$

4 Theorem Every finite dimensional Banach space is reflexive.
Proof: (TODO)

4 Theorem Let ${𝔅}_1,{𝔅}_2$ be Banach spaces. If ${𝔅}_1$ has a w:Schauder basis, then the space of finite-rank operators on ${𝔅}_1$ is (operator-norm) dense in the space of compact operators on ${𝔅}_1$.

5 Theorem $L^p$ spaces with $1<p<\mathrm{\infty }$ 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