Functional Analysis/Preliminaries

Functional Analysis
Chapter 1: Preliminaries

50% developed  as of Oct, 2009 (Oct, 2009)

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 theoryEdit

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  .


TopologyEdit

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 algebraEdit

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: