# Functional Analysis/Geometry of Banach spaces

 Functional Analysis Chapter 4: Geometry of Banach spaces
 This is not even a draft.

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

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

One of the most important question in the study of normed spaces is when this $\pi$  is surjective; if this is the case, ${\mathcal {X}}$  is said to be "reflexive". For one thing, since ${\mathcal {X}}^{**}$ , as the dual of a normed space, is a Banach space even when ${\mathcal {X}}$  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 $X^{*}$ , the weak-* topology is Hausdorff by Theorem 1.something.)

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

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

$E=\prod _{x\in {\mathfrak {B}}}\{\lambda ;\lambda \in \mathbf {C} ,|\lambda |\leq \|x\|_{\mathfrak {B}}\}$ .

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) $\square$

4. Theorem Let ${\mathcal {X}}$  be a TVS whose dual separates points in ${\mathcal {X}}$ . Then the weak-* topology on ${\mathcal {X}}^{*}$  is metrizable if and only if ${\mathcal {X}}$  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 $E\subset {\mathcal {X}}$  is weakly closed.
Proof: Let $x$  be in the weak closure of $E$ . Suppose, if possible, that $x\not \in E$ . By (the geometric form) of the Hanh-Banach theorem, we can then find $f\in {\mathcal {X}}^{*}$  and real number $c$  such that:

$\operatorname {Re} f(x)  for every $y\in E$ .

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

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

4 Exercise Let $B$  be the unit ball of ${\mathcal {X}}$ . Prove $\pi (B)$  is weak-* dense in the closed unit ball of ${\mathcal {X}}^{**}$ . (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 ${\mathcal {X}}$  be a Banach space. The following are equivalent:

• (i) ${\mathcal {X}}$  is reﬂexive.
• (ii) The closed unit ball of ${\mathcal {X}}$  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 ${\mathcal {X}}$  is not reflexive, then we can find a normalized sequence that falsifies (iii). For that, see , which shows how to do this. Finally, for (ii) $\Rightarrow$  (iii), it suffices to prove:

4 Lemma Let ${\mathcal {X}}$  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 ${\mathcal {X}}$  admits a dense countable subset $E$ . Then for $u,v\in {\mathcal {X}}^{*}$ , $u(x)=v(x)$  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 $X^{*}$ . 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 $x\in X$  with $\|x\|=1$ , by the Hahn-Banach theorem, we can find $f\in B$  such that $f(x)=\|x\|=1$ . By denseness, there is $g\in B'$  that is near $x$  in the sense: $|g(x)-f(x)|<2^{-1}$ , and we have:

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

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

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 , )

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

4 Theorem Let ${\mathcal {X}}$  be a Banach space with a w:Schauder basis $e_{j}$ . ${\mathcal {X}}$  is reflexive if and only if $e_{j}$  satisfies:

• (i) $\sup _{n}\left\|\sum _{j=1}^{n}a_{j}e_{j}\right\|<\infty \Rightarrow \sum _{j=1}^{\infty }a_{j}e_{j}$  converges in ${\mathcal {X}}$ .
• (ii) For any $f\in {\mathcal {X}}^{*}$ , $\lim _{n\to \infty }\sup\{|f(x)|;x=\sum _{j\geq n}^{\infty }a_{j}e_{j},\|x\|=1\}=0$ .

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 $x\in {\mathcal {X}}$ , we can write: $x=\sum _{j=1}^{\infty }b_{j}(x)e_{j}$  with $b_{j}\in {\mathcal {X}}^{*}$ . Thus,

$b_{l}(x)=\lim _{k\to \infty }b_{l}(x_{n_{k}})=\lim _{k\to \infty }\sum _{j=1}^{n_{k}}a_{j}b_{l}(e_{j})=a_{l}$ , and so $x=\sum _{j=1}^{\infty }a_{j}e_{j}$ .

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

$E_{n}=\{x;x\in {\mathcal {X}},\|x\|=1,b_{1}(x)=...b_{n-1}(x)=0\}$ .

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

$\lim _{n\to \infty }|f(x_{n})|=|f(\lim _{n\to \infty }x_{n})|=|f(\sum _{j=1}^{\infty }b_{j}(\lim _{n\to \infty }x_{n})e_{j})|=0$

since $\lim _{n\to \infty }b_{j}(x_{n})=0$ . (TODO: but does $\lim _{n\to \infty }x_{n}$  exist?) This proves (ii).
($\Leftarrow$ ): Let $x_{n}$  be a bounded sequence. For each $j$ , the set $\{b_{j}(x_{n});n\geq 1\}$  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 $a_{j}=\lim _{n\to \infty }b_{j}(x_{n_{k}})$ . Let $K=2\sup _{n}\|x_{n}\|$  and $s_{n}=\sup\{|f(y)|;y=\sum _{j=m+1}^{\infty }c_{j}e_{j},\|y\|\leq K\}$ . By (ii), $\lim _{n\to \infty }s_{n}=0$ . Now,

$|f(\sum _{j=1}^{m}b_{j}(x_{n_{k}})e_{j})|\leq |f(\sum _{j=1}^{\infty }b_{j}(x_{n_{k}})e_{j})|+|f(\sum _{j=m+1}^{\infty }b_{j}(x_{n_{k}})e_{j})|\leq \|f\|\sup _{n}\|x_{n}\|+s_{m}$  for $f\in {\mathcal {X}}^{*}$ .

Since $s_{m}$  is bounded, $\sup _{m}|f(\sum _{j=1}^{m}a_{j}e_{j})|<\infty$  for every $f$  and so $\sup _{m}\|\sum _{j=1}^{m}a_{j}e_{j}\|<\infty$ . By (i), $\sum _{j=1}^{m}a_{j}e_{j}$  therefore exists. Let $\epsilon >0$  be given. Then there exists $m$  such that $s_{m}<\epsilon /2$ . Also, there exists $N$  such that:

$\sum _{j=1}^{m}(a_{j}-b_{j}(x_{n_{k}}))f(e_{j})<\epsilon /2$  for every $k\geq N$ .

Hence,

$|f(x_{n_{k}})-f(\sum _{j=1}^{\infty }a_{j}e_{j})|\leq |\sum _{j=1}^{m}(a_{j}-b_{j}(x_{n_{k}}))f(e_{j})|+|f(\sum _{j=m+1}^{\infty }(a_{j}-b_{j}(x_{n_{k}}))e_{j}|<\epsilon$ .

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 $T\in B({\mathfrak {H}})$ . Then $T({\overline {B}}(0,1))$  is closed.
Proof: Since ${\overline {B}}(0,1)$  is weakly compact and $T({\overline {B}}(0,1))$  is convex, it suffices to show $T$  is weakly continuous. But if $x_{n}\to 0$  weakly, then $(Tx_{n}|y)=(x_{n}|T^{*}y)\to 0$  for any y. This shows that T is weakly continuous on ${\overline {B}}(0,1)$  (since bounded sets are weakly metrizable) and thus on ${\mathfrak {H}}$ .$\square$

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

3 Lemma If $T\in B({\mathfrak {H}})$  is self-adjoint and compact, then either $\|T\|$  or $-\|T\|$  is an eigenvalue of T.
Proof: First we prove that $\|T\|^{2}$  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 $\|T\|=\|Tx_{0}\|$ . Since $\langle T^{2}x_{0},x_{0}\rangle =\|T\|^{2}$ ,

$\|T^{2}x-\|T\|^{2}x\|^{2}\leq \|T\|^{2}-2\|T\|^{2}+\|T\|^{2}$

Thus, $T^{2}x_{0}=\|T\|^{2}x_{0}$ . Since $(T^{2}-\|T\|^{2}I)x_{0}=(T+\|T\|I)(T-\|T\|I)x_{0}$ , we see that $(T-\|T\|I)x_{0}$  is either zero or an eigenvector of $T$  with respect to $-\|T\|$ . $\square$

3 Theorem If T is normal; that is, $T^{*}T=TT^{*}$ , 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.$\square$

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 $K^{*}K$

For $T\in {\mathcal {L}}({\mathfrak {H}})$ , 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 ${\mathfrak {H}}$ .)

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

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

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

$\langle Tx,x\rangle ={\overline {\langle T^{*}x,x\rangle }}$  for every $x\in \operatorname {dom} T$

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.$\square$

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 ${\mathfrak {H}}$ , then

$\sum _{k=1}^{\infty }|\langle x,u_{k}\rangle |^{2}\leq \|x\|^{2}$  for any $x\in {\mathfrak {H}}$ .

Proof: If $\langle x,y\rangle =0$ , then $\|x+y\|^{2}=\|x\|^{2}+\|y\|^{2}$ . Thus,

$\|x-\sum _{k=1}^{n}\langle x,u_{k}\rangle \|^{2}=\|x\|^{2}-2\operatorname {Re} \sum _{k=1}^{n}|\langle x,u_{k}\rangle |^{2}+\sum _{k=1}^{n}|\langle x,u_{k}\rangle |^{2}=\|x\|^{2}-\sum _{k=1}^{n}|\langle x,u_{k}\rangle |^{2}$ .

Letting $n\to \infty$  completes the proof. $\square$ .

3 Theorem (Parseval) Let $u_{k}$  be a orthonormal sequence in a Hilbert space ${\mathfrak {H}}$ . Then the following are equivalent:

• (i) $\operatorname {span} \{u_{1},u_{2},...\}$  is dense in ${\mathfrak {H}}$ .
• (ii) For each $x\in {\mathfrak {H}}$ , $x=\sum _{k=1}^{\infty }\langle x,u_{k}\rangle u_{k}$ .
• (iii) For each $x,y\in {\mathfrak {H}}$ , $\langle x,y\rangle =\sum _{k=1}^{\infty }\langle x,u_{k}\rangle {\overline {\langle y,u_{k}\rangle }}$ .
• (iv) $\|x\|^{2}=\sum _{k=1}^{\infty }|\langle x,u_{k}\rangle |^{2}$  (the Parseval equality).

Proof: Let ${\mathcal {M}}=\operatorname {span} \{u_{1},u_{2},...\}$ . If $v\in {\mathcal {M}}$ , then it has the form: $v=\sum _{k=1}^{\infty }\alpha _{k}u_{k}$  for some scalars $\alpha _{k}$ . Since $\langle v,u_{j}\rangle =\sum _{k=1}^{\infty }a_{j}\langle u_{k},u_{j}\rangle =a_{j}$  we can also write: $v=\sum _{k=1}^{\infty }\langle v,u_{k}\rangle u_{k}$ . Let $y=\sum _{k=1}^{\infty }\langle x,u_{k}\rangle u_{k}$ . Bessel's inequality and that ${\mathfrak {H}}$  is complete ensure that $y$  exists. Since

$\langle y,v\rangle =\sum _{k=1}^{\infty }\langle x,u_{k}\rangle \langle u_{k},v\rangle =\sum _{k=1}^{\infty }\langle x,\langle v,u_{k}\rangle u_{k}\rangle =\langle x,\sum _{k=1}^{\infty }\langle v,u_{k}\rangle u_{k}\rangle =\langle x,v\rangle$

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

$|\langle x,y\rangle -\sum _{k=1}^{n}\langle x,u_{k}\rangle {\overline {\langle y,u_{k}\rangle }}|=|\langle x,y-\sum _{k=1}^{n}\langle y,u_{k}\rangle u_{k}\rangle |\to 0$  as $n\to \infty$

To get (iii) $\Rightarrow$  (iv), take $x=y$ . To show (iv) $\Rightarrow$  (i), suppose that (i) is false. Then there exists a $z\in (\operatorname {span\{u_{1},u_{2},...\}} )^{\bot }$  with $z\neq 0$ . Then

$\sum _{k=1}^{\infty }|\langle z,u_{k}\rangle |^{2}=0<\|z\|^{2}$ .

Thus, (iv) is false.$\square$

3 Theorem Let $x_{k}$  be an orthogonal sequence in a Hilbert space $({\mathfrak {H}},\|\cdot \|=\langle \cdot ,\cdot \rangle ^{1/2})$ . Then the series $\sum _{k=1}^{\infty }x_{k}$  converges if and only if the series $\sum _{k=1}^{\infty }\langle x_{k},y\rangle$  converges for every $y\in {\mathfrak {H}}$ .
Proof: Since

$\sum _{k=1}^{\infty }|\langle x_{k},y\rangle |\leq \|y\|\sum _{k=1}^{\infty }\|x_{k}\|$  and $\sum _{k=1}^{\infty }\|x_{k}\|=\left\|\sum _{k=1}^{\infty }x_{k}\right\|$

by orthogonality, we obtain the direct part. For the converse, let $E=\left\{\sum _{k=1}^{n}x_{k};n\geq 1\right\}$ . Since

$\sup _{E}|\langle \cdot ,y\rangle |=\sup _{n}|\sum _{k=1}^{n}\langle x_{k},y\rangle |<\infty$  for each $y$

by hypothesis, $E$  is bounded by Theorem 3.something. Hence, $\sum _{k=1}^{\infty }\|x_{k}\|<\infty$  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 ${\mathfrak {H}}$  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 ${\mathcal {X}}$  is reflexive if and only if every element of ${\mathcal {X}}$  attains its maximum on the closed unit ball of ${\mathcal {X}}$ .

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

A Banach space is said to be uniformly convex if

$\|x_{n}\|\leq 1,\|y_{n}\|\leq 1$  and $\|x_{n}+y_{n}\|\to 0\Rightarrow \|x_{n}-y_{n}\|\to 2$

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

4 Theorem Every uniformly convex space ${\mathfrak {B}}$  is reflexive.
Proof: Suppose, if possible, that ${\mathfrak {B}}$  is uniformly convex but is not reflexive. $\square$

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

4 Theorem Let ${\mathfrak {B}}_{1},{\mathfrak {B}}_{2}$  be Banach spaces. If ${\mathfrak {B}}_{1}$  has a w:Schauder basis, then the space of finite-rank operators on ${\mathfrak {B}}_{1}$  is (operator-norm) dense in the space of compact operators on ${\mathfrak {B}}_{1}$ .

5 Theorem $L^{p}$  spaces with $1  are uniformly convex (thus, reflexive).
Proof: (TODO)

5 Theorem (M. Riesz extension theorem) (see w:M. Riesz extension theorem)