Functional Analysis 4 -- Frechet Space

This is course note from Siran Li-Shanghai Jiao Tong University.

Let \(\Omega\) be an open set. One problem is how can we understand the topology of \(C(\Omega).\)

Example:

Let \(\Omega = ]0,1[\), then there are a lot of \(f \in C(]0,1[) \ s.t. \ \underset{x \in ]0,1[}{\sup} |f(x)| = \infty.\) The problem here is the "open" set.

Now we consider instead for every \(0<a<b<1\), the "norm" \(P_{a,b}(f) := \underset{x \in [a, b]}{\sup} |f(x)|.\) However, this is not a norm on \(C(]0,1[)\) because \(\forall [a,b] \subset ]0,1[, \exists f \in C(]0,1[)\backslash\{0\}, \ s.t. f(x) = 0 \forall x \in [a,b].\)

One can check that \(\forall 0<a<b<1,\) the "norm" \(P_{a,b}: C(]0,1[) \rightarrow \mathbb{R}^+\) satisfies all requirements for norm except for \(P_{a,b}(f) = 0 \Rightarrow f = 0.\) We call this kind of norm as seminorm.

Lemma 4.1

Let \((X, || \cdot ||_o)\) be as seminorm vector space. Let \(N = \{x \in X, ||x|| = 0\}.\) Then \(N\) is a vector subspace. Then \((X\backslash N, || \cdot ||)\) defined by \(|| x+N|| := ||\tilde{X}||_o, \forall \tilde{x} \in x+N\) is a NVS.

The idea here is that we quotient the degenerate part of seminorm vector space to get a norm vector space.

Remark: It is not enough to only consider one seminorm to describe the properties of \(C(]0,1[).\)

Definition: Separating Family of Seminorm [[TODO]]

For \(C(]0,1[),\) consider a family of seminorm \(\{P_{a,b}\}_{0<a<b<1, (a,b) \in \mathbb{Q}^2}.\) Then, if \(\forall f \in C(]0,1[)\backslash\{0\}, \exists 0<a<b<1, s.t. P_{a,b}(f) \neq 0,\) we say the family of seminorm is seperating.

Note: Since the rational number is dense in \(\mathbb{R},\) we only need to consider \(a, b \in \mathbb{Q}.\) Here \((a,b)\) represents a pair of number.

Proposition 4.2

Let \(X\) be a VS with a countable family \(\{p_n\}\) of separating seminorms. Then \[ d(x,y) = \sum_{n=1}^{\infty} \frac{p_n(x,y)}{1+p_n(x,y)}\] is a metric.

Definition: Frechet Space

A complete metric space \(X\) with metric generated by a countable family of separating seminorms is a Frechet space.

Examples:

1. \(C(]0,1[)\) is a Frechet space.

2. \(C(\Omega)\) is a Frechet space, \(\forall \Omega \subset \mathbb{R}^d, \Omega\) is open.

Note: Actually, one could construct a group of compact exhaustion \(K_j := \{x \in \mathbb{R}^d \mid |x| \le 100j, dist(x, \partial \Omega) \le \frac{1}{j}\}.\) That means we can exhaust an open set from inside using compact set.

Then it is clear to say that {Banach Space} is not a subset of {Frechet Space}.