# Functional Analysis/Geometry of Banach spaces

 Functional Analysis Chapter 4: Geometry of Banach spaces
 (May 27, 2008) 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 ${\displaystyle {\mathcal {X}}}$ be a normed space. Since ${\displaystyle X^{*}}$ is a Banach space, there is a canonical injection ${\displaystyle \pi :{\mathcal {X}}\to {\mathcal {X}}^{**}}$ given by:

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

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

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

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

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

Since ${\displaystyle E}$, 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) ${\displaystyle \square }$

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

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

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

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

4 Exercise Let ${\displaystyle B}$ be the unit ball of ${\displaystyle {\mathcal {X}}}$. Prove ${\displaystyle \pi (B)}$ is weak-* dense in the closed unit ball of ${\displaystyle {\mathcal {X}}^{**}}$. (Hint: similar to the proof of Lemma 4.something.)

4 Theorem A set ${\displaystyle E}$ is weak-* sequentially closed if and only if the intersection of ${\displaystyle E}$ 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 ${\displaystyle {\mathcal {X}}}$ be a Banach space. The following are equivalent:

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

4 Lemma Let ${\displaystyle {\mathcal {X}}}$ be a Banach space, ${\displaystyle x_{j}\in X}$ a sequence and ${\displaystyle F}$ be the weak closure of ${\displaystyle x_{j}}$. If ${\displaystyle F}$ is weakly compact, then ${\displaystyle F}$ is weakly sequentially compact.
Proof: By replacing ${\displaystyle X}$ with the closure of the linear span of ${\displaystyle X}$, we may assume that ${\displaystyle {\mathcal {X}}}$ admits a dense countable subset ${\displaystyle E}$. Then for ${\displaystyle u,v\in {\mathcal {X}}^{*}}$, ${\displaystyle u(x)=v(x)}$ for every ${\displaystyle x\in E}$ implies ${\displaystyle u=v}$ by continuity. This is to say, a set of functions of the form ${\displaystyle u\mapsto u(x)}$ with ${\displaystyle x\in E}$ separates points in ${\displaystyle X}$, a fortiori, ${\displaystyle B}$, the closed unit ball of ${\displaystyle X^{*}}$. The weak-* topology for ${\displaystyle B}$ is therefore metrizable by Theorem 1.something. Since a compact metric space is second countable; thus, separable, ${\displaystyle B}$ admits a countable (weak-*) dense subset ${\displaystyle B'}$. It follows that ${\displaystyle B'}$ separates points in ${\displaystyle X}$. In fact, for any ${\displaystyle x\in X}$ with ${\displaystyle \|x\|=1}$, by the Hahn-Banach theorem, we can find ${\displaystyle f\in B}$ such that ${\displaystyle f(x)=\|x\|=1}$. By denseness, there is ${\displaystyle g\in B'}$ that is near ${\displaystyle x}$ in the sense: ${\displaystyle |g(x)-f(x)|<2^{-1}}$, and we have:

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

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

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

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

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

Proof: (${\displaystyle \Rightarrow }$): Set ${\displaystyle x_{n}=\sum _{j=1}^{n}a_{j}e_{j}}$. By reflexivity, ${\displaystyle x_{n}}$ then admits a weakly convergent subsequence ${\displaystyle x_{n_{k}}}$ with limit ${\displaystyle x}$. By hypothesis, for any ${\displaystyle x\in {\mathcal {X}}}$, we can write: ${\displaystyle x=\sum _{j=1}^{\infty }b_{j}(x)e_{j}}$ with ${\displaystyle b_{j}\in {\mathcal {X}}^{*}}$. Thus,

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

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

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

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

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

since ${\displaystyle \lim _{n\to \infty }b_{j}(x_{n})=0}$. (TODO: but does ${\displaystyle \lim _{n\to \infty }x_{n}}$ exist?) This proves (ii).
(${\displaystyle \Leftarrow }$): Let ${\displaystyle x_{n}}$ be a bounded sequence. For each ${\displaystyle j}$, the set ${\displaystyle \{b_{j}(x_{n});n\geq 1\}}$ is bounded; thus, admits a convergent sequence. By Cantor's diagonal argument, we can therefore find a subsequence ${\displaystyle x_{n_{k}}}$ of ${\displaystyle x_{n}}$ such that ${\displaystyle b_{j}(x_{n_{k}})}$ converges for every ${\displaystyle j}$. Set ${\displaystyle a_{j}=\lim _{n\to \infty }b_{j}(x_{n_{k}})}$. Let ${\displaystyle K=2\sup _{n}\|x_{n}\|}$ and ${\displaystyle s_{n}=\sup\{|f(y)|;y=\sum _{j=m+1}^{\infty }c_{j}e_{j},\|y\|\leq K\}}$. By (ii), ${\displaystyle \lim _{n\to \infty }s_{n}=0}$. Now,

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

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

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

Hence,

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

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

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

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

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

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

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

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

${\displaystyle K=R|K|}$

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

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

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

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

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

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

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

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 ${\displaystyle u_{k}}$ is an orthonormal sequence in a Hilbert space ${\displaystyle {\mathfrak {H}}}$, then

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

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

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

Letting ${\displaystyle n\to \infty }$ completes the proof. ${\displaystyle \square }$.

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

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

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

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

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

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

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

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

Thus, (iv) is false.${\displaystyle \square }$

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

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

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

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

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

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

A Banach space is said to be uniformly convex if

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

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

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

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

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

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

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