Functional Analysis/Preliminaries

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Functional Analysis Chapter 1: Preliminaries

Chapter 2: Banach spaces→

This chapter gathers some standard results that will be used in sequel. In particular, we prove the Hahn-Banach theorem, which is really a result in linear algebra. The proofs of these theorems will be found in the Topology and Linear Algebra books.

Set theory

The axiom of choice states that given a collection of sets $S_{i},i\in I$, there exists a function

$f:I\to \prod _{i\in I}S_{i}$.

Exercise. Use the axiom of choice to prove that any surjection is right-invertible.

In this book the axiom of choice is almost always invoked in the form of Zorn's Lemma.

Theorem 1.1 (Zorn's Lemma]). Let $\left(X,\prec \right)$ be a poset such that for every chain, $T\subset X$, which is linearly ordered by $\prec$ there is a maximal element, $t\in T$. Then $\left(X,\prec \right)$ has a maximal element $x\in X$. That is, for any $y\in X,x\not \prec y$.

Topology

Theorem 1.2. Let $X$ be a metric space. The following are equivalent.

• $X$ is a compact space.
• $X$ is totally bounded and complete. (Heine-Borel)
• $X$ is sequentially compact; i.e., every sequence in X has a convergent subsequence.
  

Exercise. Prove that $[0,1]\cap \mathbf {Q}$ is not compact by exhibiting an open cover that does not admit a finite subcover.

Exercise. Let $X$ be a compact metric space, and $f:X\to X$ be an isometry: i.e., $d(f(x),f(y))=d(x,y)$. Then f is a bijection.

Theorem 1.3 (Tychonoff). Every product space of a nonempty collection of compact spaces is compact.

Exercise. Prove Tychonoff's theorem for finite product without appeal to Axiom of Choice (or any of its equivalences).

By definition, a compact space is Hausdorff.

Theorem 1.4 (metrization theorem). If $X$ is a second-countable compact space, then $X$ is metrizable.
Proof. Define $d:X\to \mathbf {R}$ by

$d(x,y)=\sum _{j=1}^{\infty }2^{-j}|f_{j}(x)-f_{j}(y)|$

Then $d(x,y)=0$ implies $f_{j}(x)=f_{j}(y)$ for every $j$, which in turn implies $x=y$. The converse holds too. Since $d(x,y)=d(y,x)$, $d$ is a metric then. Let $\tau _{d}$ be the topology for $X$ that is induced by $d$. We claim $\tau _{d}$ coincides with the topology originally given to $K$. In light of:

Lemma. Let $X$ be a set. If $\tau _{1}\subset \tau _{2}$ are a pair of topologies for $X$ and if $\tau _{1}$ is Hausdorff and $\tau _{2}$ is compact, then $\tau _{1}=\tau _{2}$.

it suffices to show that $\tau _{d}$ is contained in the original topology. But for any $x\in X$, since $d(\cdot ,x)$ is the limit of a sequence of continuous functions on a compact set, we see $d(\cdot ,x)$ is continuous. Consequently, an $\tau _{d}$-open ball in $d$ with center at $x$ is open (in the original topology.) $\square$

Proposition 1.5. (i) Every second-countable space is separable. (ii) Every separable metric space is second-countable.
Proof. To be written. $\square$

In particular, a compact metric space is separable.

Exercise. The w:lower limit topology on the real line is separable but not second-countable.

Theorem 1.6 (Baire). A complete metric space is not a countable union of closed subsets with dense complement.
Proof. See w:Baire category theorem. $\square$

We remark that the theorem is also true for a locally compact space, though this version will not be needed in the sequel.

Exercise. Use the theorem to prove the set of real numbers is uncountable.

Theorem 1.7 (Ascoli). Let X be a compact space. A subset of $C(X)$ is compact if and only if it is bounded, closed and equicontinuous.
Proof. See w:Ascoli's theorem. $\square$

The next exercise gives a typical application of the theorem.

Exercise. Prove Peano's existence theorem for ordinal differential equations: Let $f$ be a real-valued continuous function on some open subset of $\mathbf {R} ^{n}$. Then the initial value problem

${\dot {x}}=f(x),x(t_{0})=x_{0}$

has a solution in some open interval containing $t_{0}$. (Hint: Use w:Euler's method to construct a sequence of approximate solutions. The sequence probably does not converge but it contains a convergent subsequence according to Ascoli's theorem. The limit is then a desired solution.)

Exercise. Deduce w:Picard–Lindelöf theorem from Peano's existence theorem: Let $f$ be a real-valued locally Lipschitz function on some open subset of $\mathbf {R} ^{n}$. Then the initial value problem

${\dot {x}}=f(x),x(t_{0})=x_{0}$

has a "unique" solution in some open interval containing $t_{0}$. (Hint: the existence is clear. For the uniqueness, use w:Gronwall's inequality.)

Theorem 1.8. Given a metric space X, there exists a complete metric space ${\tilde {X}}$ such that $X$ is a dense subset of ${\tilde {X}}$.
Proof. w:Completion (metric space)#Completion $\square$

Linear algebra

Theorem 1.9. Let V be a vector space. Then every (possibly empty) linearly independent set is contained in some basis of V.
Proof. Let $\Omega$ be the set of all linearly independent set containing the given linearly independent set. $\Omega$ is nonempty. Moreover, if $\omega$ is a chain in $\Omega$ (i.e., a totally ordered subset), then $\cup \omega$ is linearly independent, since if

$a_{1}x_{1}+...+a_{n}x_{n}=0$

where $x_{i}$ are in the union, then $x_{i}$ all belong to some member of $\omega$. Thus, by Zorn's Lemma, it has a maximal element, say, E. It spans V. Indeed, if not, there exists an $x\in V\backslash E$ such that $E\cup {x}$ is a member of $\Omega$, contradicting the maximality of E. $\square$

The theorem means in particular that every vector space has a basis. Such a basis is called a Hamel basis to contrast other bases that will be discussed later.

Theorem 1.10 (Hahn-Banach). Let ${\mathcal {X}}$ be a real vector space and $p$ be a function on ${\mathcal {X}}$ such that

$p(x+y)\leq p(x)+p(y)$ and $p(tx)=tp(x)$

for any $x,y\in {\mathcal {X}}$ and any $t>0$. If ${\mathcal {M}}\subset {\mathcal {X}}$ is a closed subspace and $f$ is a linear functional on ${\mathcal {M}}$ such that $f\leq p$, then $f$ admits a linear extension $F$ defined in ${\mathcal {X}}$ such that $F\leq p$.
Proof. First suppose that ${\mathcal {X}}=\{x+tz;x\in {\mathcal {M}},t\in \mathbb {R} \}$ for some $z\not \in {\mathcal {M}}$. By hypothesis we have:

$f(x)+f(y)\leq p(x-z)+p(y+z)$ for all $x,y\in {\mathcal {M}}$,

which is equivalent to:

$\sup\{f(x)-p(x-z);x\in M\}\leq \inf\{p(y+z)-f(y);y\in M\}$.

Let $c$ be some number in between the sup and the inf. Define $F(x+tz)=f(x)+tc$ for $x\in {\mathcal {M}},t>0$. It follows that $F$ is an desired extension. Indeed, $f=F$ on ${\mathcal {M}}$ being clear, we also have:

$F(x+tz)\leq tp({x \over t}+z)$ if $t>0$

and

$F(x+tz)\leq -tp({x \over -t}-z)$ if $t<0$.

Let $\Omega$ be the collection of pairs $(H,g_{H})$ where $H$ is linear space with ${\mathcal {M}}\subset {\mathcal {H}}\subset X$ and $g_{H}$ is a linear function on ${\mathcal {M}}$ that extends $f$ and is dominated by $p$. It can be shown that $\Omega$ is partially ordered and the union of every totally ordered sub-collection of $\Omega$ is in $\Omega$ (TODO: need more details). Hence, by Zorn's Lemma, we can find the maximal element $(L,g_{L})$ and by the early part of the proof we can show that ${\mathcal {L}}={\mathcal {X}}$. $\square$

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.

Exercise State the analog of the theorem for complex vector spaces and prove that this version can be reduced to the real version. (Hint: $Ref(ix)=Imf(x)$)

Note the theorem can be formulated in the following equivalent way.

Theorem 1.11 (Geometric Hahn-Banach). Let V be a vector space, and $E\subset V$ be a convex subset. If x is not in E, then there exists a hyperplane that contains E but doesn't contain x.
Proof. We prove the statement is equivalent to the Hahn-Banach theorem above. We first show that there is a one-to-one corresponding between the set of sublinear functional and convex sets. Given a convex set $E$, define $p(x)=\inf\{t>0|tx\in E\}$. $p$ (called a w:Minkowski functional) is then sublinear. In fact, clearly we have $p(ax)=|a|p(x)$. Also, if $tx\in E$ and $sy\in E$, then, by convexity, ${t \over t+s}x+{s \over t+s}y\in E$ and so $p(x+y)\leq t+s$. Taking inf over t and s (separately) we conclude $p(x+y)\leq p(x)+p(y)$. Now, note that: $E=\{x\in V|p(x)\leq 1\}$. This suggests that we can define a set $E$ for a given sublinear functional $p$. In fact, if $p$ is sublinear, then for $x,y\in E$ we have: $p(tx+sy)\leq tp(x)+sp(y)\leq 1$ when $t,s\leq 0,t+s=1$ and this means $tx+sy\in E$. Hence, $E$ is convex. $\square$

Corollary 1.12. Every convex subset of a vector space is the intersection of all hyperplanes containing it (called convex hull).

Exercise. Prove Carathéodory's theorem.

(TODO: mention moment problem.)

Theorem 1.13. Let $V\supset W$ be linear vector spaces, and $\pi :V\to V/W$ a canonical surjection. If $T:V\to X$ (where X is some vector space) is a linear map, then there exists $F:V/W\to X$ such that $F\circ \pi =f$ if and only if $W\subset \operatorname {ker} f$. Moreover,

• (i) If $F$ exists, then $F$ is unique.
• (ii) $F$ is injective if and only if $\operatorname {ker} (f)=W$.
• (iii) $F$ is surjective if and only if $f$ is surjective.
  

Proof. If $F$ exists, then $W=\operatorname {ker} (\pi )\subset \operatorname {ker} (F\circ \pi )=\operatorname {ker} f$. Conversely, suppose $W\subset \operatorname {ker} f$, and define $F$ by:

$F(x+W)=f(x)$

for $x\in V$. $F$ is well-defined. In fact, if $x+W=y+W$, then $x-y\in W\subset \operatorname {ker} f$. Thus,

$(F\circ \pi )(x)=f(x)=f(y)=(F\circ \pi )(y)$.

By this definition, (i) is now clear. (ii) holds since $F(x+W)=(F\circ \pi )(x)=f(x)=0$ implies $x\in W$ if and only if $\operatorname {ker} f=W$. (iii) is also clear; we have a set-theoretic fact: $g\circ f$ is surjective if and only if $g$ is surjective. $\square$

Corollary 1.14. If $T:V_{1}\to V_{2}$ induces a map $T:W_{1}\to W_{2}$ where $W_{j}\subset V_{j}$ are subspaces, then we can induce

$T:V_{1}/W_{1}\to V_{2}/W_{2}$.

Proof. Obvious. $\square$

Corollary 1.15. If $f:V\to W$ is a linear map, then $V/\operatorname {ker} (f)\simeq \operatorname {ran} (f)$.
Proof. Obvious. $\square$

Exercise. Given an exact sequence

$0\to V_{1}\to V_{2}\to ...\to V_{n}\to 0$,

we have: $(-1)^{k}\operatorname {dim} V_{k}=0$