Functional Analysis/Geometry of Banach spaces

←Chapter 3: Hilbert spaces

Functional Analysis Chapter 4: Geometry of Banach spaces

Chapter 5: Operational calculus and representation theory→

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:

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

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

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 [1], 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 [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 ${\mathcal {X}}$ 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: [4]

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 $1<p<\infty$ 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