# Functional Analysis/Preliminaries

←Chapter 0: Preface | 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[edit | edit source]

The axiom of choice states that given a collection of sets , there exists a *function*

- .

**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 be a poset such that for every chain, , which is linearly ordered by there is a maximal element, . Then has a maximal element . That is, for any .*

## Topology[edit | edit source]

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

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

**Exercise.** *Prove that is not compact by exhibiting an open cover that does not admit a finite subcover.*

**Exercise.** *Let be a compact metric space, and be an isometry: i.e., . 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 is a second-countable compact space, then is metrizable.*

*Proof.* Define by

Then implies for every , which in turn implies . The converse holds too. Since , is a metric then. Let be the topology for that is induced by . We claim coincides with the topology originally given to . In light of:

**Lemma.** *Let be a set. If are a pair of topologies for and if is Hausdorff and is compact, then .*

it suffices to show that is contained in the original topology. But for any , since is the limit of a sequence of continuous functions on a compact set, we see is continuous. Consequently, an -open ball in with center at is open (in the original topology.)

**Proposition 1.5.** *(i) Every second-countable space is separable. (ii) Every separable metric space is second-countable.*

*Proof.* To be written.

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.

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 is compact if and only if it is bounded, closed and equicontinuous.*

*Proof.* See w:Ascoli's theorem.

The next exercise gives a typical application of the theorem.

**Exercise.** *Prove Peano's existence theorem for ordinal differential equations: Let be a real-valued continuous function on some open subset of . Then the initial value problem*

has a solution in some open interval containing . (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 be a real-valued locally Lipschitz function on some open subset of . Then the initial value problem*

has a "unique" solution in some open interval containing . (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 such that is a dense subset of .*

*Proof.* w:Completion (metric space)#Completion

## Linear algebra[edit | edit source]

**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 be the set of all linearly independent set containing the given linearly independent set. is nonempty. Moreover, if is a chain in (i.e., a totally ordered subset), then is linearly independent, since if

where are in the union, then all belong to some member of . Thus, by Zorn's Lemma, it has a maximal element, say, *E*. It spans *V*. Indeed, if not, there exists an such that is a member of , contradicting the maximality of *E*.

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 be a real vector space and be a function on such that*

- and

for any and any . If is a closed subspace and is a linear functional on such that , then admits a linear extension defined in such that .*Proof.* First suppose that for some . By hypothesis we have:

- for all ,

which is equivalent to:

- .

Let be some number in between the sup and the inf. Define for . It follows that is an desired extension. Indeed, on being clear, we also have:

- if

and

- if .

Let be the collection of pairs where is linear space with and is a linear function on that extends and is dominated by . It can be shown that is partially ordered and the union of every totally ordered sub-collection of is in (TODO: need more details). Hence, by Zorn's Lemma, we can find the maximal element and by the early part of the proof we can show that .

We remark that a different choice of 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: )*

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

**Theorem 1.11 (Geometric Hahn-Banach).** *Let *V* be a vector space, and 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 , define . (called a w:Minkowski functional) is then sublinear. In fact, clearly we have . Also, if and , then, by convexity, and so . Taking inf over *t* and *s* (separately) we conclude . Now, note that: . This suggests that we can define a set for a given sublinear functional . In fact, if is sublinear, then for we have: when and this means . Hence, is convex.

**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 be linear vector spaces, and a canonical surjection. If (where X is some vector space) is a linear map, then there exists such that if and only if . Moreover,*

- (i) If exists, then is unique.
- (ii) is injective if and only if .
- (iii) is surjective if and only if is surjective.

*Proof.* If exists, then . Conversely, suppose , and define by:

for . is well-defined. In fact, if , then . Thus,

- .

By this definition, (i) is now clear. (ii) holds since implies if and only if . (iii) is also clear; we have a set-theoretic fact: is surjective if and only if is surjective.

**Corollary 1.14.** *If induces a map where are subspaces, then we can induce*

- .

*Proof.* Obvious.

**Corollary 1.15.** *If is a linear map, then .*

*Proof.* Obvious.

**Exercise.** *Given an exact sequence*

- ,

*we have: *