Functional Analysis Lecture 3

[[lecture-data]]

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Lecture Notes: Rodriguez, page 14

1. Normed Spaces and Banach Spaces

Last Time

We introduced

space of bounded linear operators
[[bounded linear operator space is banach]]

subspace and quotients

Recall

Subspace

Let $V$ be a normed vector space and $W \subseteq V$ . $W$ is a subspace if for all $λ_{1}, λ_{2} \in K$ and $w_{1}, w_{2} \in W$ then

λ_{1} w_{1} + λ_{2} w_{2} \in W

ie closed under linear combinations

see [[subspace]]

Theorem

A [[subspace]] $W \subset V$ of a [[Banach space]] is Banach if and only if $W \subset V$ is closed.

Proof

Exercise

The Proof is left as an exercise

see [[closed subspaces of banach spaces are banach]]

Quotient

Let $W \subseteq V$ be a [[subspace]]. We define an [[equivalence relation]] on $V$ by

v \sim v^{'} ⟺ v - v^{'} \in W

Define $[v] = {v^{'} \in V : v^{'} \sim v}$ the equivalence class of $v$ . Then

V | W = {[v] : v \in V}

is called the quotient space which we typically call " $V$ mod $W$ " and notate as

[v] = v + W

$V | W$ is a [[vector space]] such that for all $v_{1}, v_{2} \in V$ and $λ \in K$

$(v_{1} + W) + (v_{2} + W) := (v_{1} + v_{2}) + W$
$λ (v + W) = λ v + W$

Note

$W = 0 + W = w + W \forall w \in W$

see [[quotient of a vector space]]

Theorem

Let $| | \cdot | |$ be a [[semi-norm]] on $V$ . Then

E := {v \in V : | | v | | = 0}

is a [[subspace]] of $V$ and

| | v + E | |_{v | E} := | | v | |, v + E \in V | E

defines a [[norm]] on the space $V | E$

Proof

$E$ is a [[subspace]] since $\forall v_{1}, v_{2} \in E$ and $λ_{1}, λ_{2} \in K$ we have

\begin{aligned} | | λ_{1} v_{1} + λ_{2} v_{2} | | & \leq | λ_{1} | \cdot | | v_{1} | | + | λ_{2} | \cdot | | v_{2} | | \\ = 0 \\ ⟹ | | λ_{1} v_{1} + λ_{2} v_{2} | | & = 0 \end{aligned}

We first show $| | \cdot | |_{V | E}$ is well-defined, ie

v + E = v^{'} + E ⟹ | | v | | = | | v^{'} | |

Suppose $v + E = v^{'} + E$ . ie, there exists $e \in E$ such that $v + v^{'} + e$ . Then

\begin{aligned} | | v | | & = | | v^{'} + e | | \\ \leq | | v^{'} | | + | | e | | \\ = | | v^{'} | | \\ ⟹ | | v | | \leq | | v^{'} | | & and also | | v^{'} | | \leq | | v | | \\ ⟹ | | v | | & = | | v^{'} | | \end{aligned}

Exercise

To complete this proof, all that is left is verification that this function is a norm.

recall that since this is a semi-norm, homogeneity and the [[triangle inequality]] are already satisfied
definiteness can be shown by noting that everything that evaluates to $0$ in the norm is in the equivalence class of $0 + E$

Exercise

Let $V$ be a normed vector space and $W \subseteq V$ a closed W is a [[norm.

Baire Category Theorem

Let $M$ be a complete [[metric space]] and let ${C_{n}}_{n}$ be a collection of closed subsets of $M$ such that $$M = \bigcup_{n \in N}C_{n}$$Then at least one of the $C_{n}$ contain an open ball

B (x, r) = {y \in M : d (x, y) < r}

ie, at least one $C_{n}$ has an interior point.

Note

A set that does not contain an interior point is called nowhere dense. Sometimes when we apply this theorem, we do not need $C_{n}$ to be closed. Then the result is that their closures must contain and open ball. ie, we cannot have all $C_{n}$ be all nowhere dense.

Note

We can use this theorem to prove that there exists a continuous function that is nowhere differentiable

Proof (by contradiction)

Suppose BWOC that there is some collection of closed subsets of $M$ such that $⋃_{n} C_{n} = M$ and all $C_{n}$ are nowhere dense.

Note

we'll show that there is a sequence in $M$ that does not converge

Since $M$ contains an open ball but $C_{1}$ does not, we have $M \neq C_{1}$ . Thus there is some $p_{1} \in M ∖ C_{1}$ . Since $C_{1}$ is closed but $M ∖ C_{1}$ is open, this implies that $\exists ϵ_{1} > 0$ s.t. $B (p_{1}, ϵ_{1}) \cap C_{1} = \emptyset$ .

Now, $B (p_{1}, \frac{ϵ_{1}}{3}) ⊄ C_{2}$ means there exists some $p_{2} \in M ∖ C_{2}$ . And since $C_{2}$ is closed, there exists $0 < ϵ_{2} < \frac{ϵ_{1}}{3}$ such that $B (p_{2}, ϵ_{2}) \cap C_{2} = \emptyset$

Now, suppose there we have found $k$ such points. Then for each $j = 1, \dots, k$ we have $p_{j} ⊄ C_{j}$ with $ϵ_{j}$ such that $B (p_{j}, ϵ_{j}) \cap C_{j} = \emptyset$

Since $B (p_{k}, \frac{ϵ_{k}}{3}) ⊄ C_{k + 1}$ , there exists some $p_{k + 1} \in B (p_{k}, \frac{ϵ_{k}}{3})$ such that $p_{k + 1} \notin C_{k + 1}$

Then there exists $ϵ_{k + 1} < \frac{ϵ_{k}}{3}$ such that $B (p_{k + 1}, ϵ_{k + 1}) \cap C_{k + 1} = \emptyset$

Thus, by indiction, we have found a sequence of points ${p_{k}}_{k} \subset M$ and $ϵ_{k} \in (0, ϵ_{1})$ such that for all $k$ ,

$p_{j} \in B (p_{j - 1}, \frac{ϵ_{j - 1}}{3})$
$B (p_{j}, ϵ_{j}) \cap C_{j} = \emptyset$

This sequence ${p_{k}}_{k}$ is Cauchy because for all $k, ℓ \in N$ , we have

\begin{aligned} d (p_{k + 1}, p_{k + ℓ}) & \leq d (p_{k}, p_{k + 1}) + d (p_{k + 1}, p_{k + 2}) + \dots + d (p_{k + ℓ - 1}, p_{k + ℓ}) \\ < \frac{ϵ_{k}}{3} + \frac{ϵ_{k + 1}}{3} + \dots + \frac{ϵ_{k + ℓ - 1}}{3} \\ < \frac{ϵ_{1}}{3^{k}} + \frac{ϵ_{1}}{3^{k + 1}} + \dots + \frac{ϵ_{1}}{3^{k + ℓ}} \\ < ϵ_{1} \sum_{n = k}^{\infty} \frac{1}{3^{k}} \\ = \frac{ϵ_{1}}{3^{k}} (\frac{1}{1 - \frac{1}{3}}) \\ = \frac{ϵ_{1}}{2} 3^{- k + 1} \end{aligned}

Since $M$ is complete, there exists some $p \in M$ such that $p_{k} \to p$ as $k \to \infty$ .

Now, for all $k \in N$ , we have

\begin{aligned} d (p_{k + 1}, p_{k + ℓ + 1}) & < ϵ_{k + 1} [\frac{1}{3} + \frac{1}{3^{2}} + \dots + \frac{1}{3^{ℓ}}] \\ < ϵ_{k + 1} \sum_{n = 1}^{\infty} 3^{- n} \\ = \frac{ϵ_{k + 1}}{2} \\ ⟹ lim_{ℓ \to \infty} d (p_{k + 1}, p_{l + ℓ + 1}) & = d (p_{k + 1}, p) \\ \leq \frac{ϵ_{k + 1}}{2} \\ < \frac{ϵ_{k}}{6} \end{aligned}

So as $ℓ \to \infty$ , we have

d (p_{k + 1}, p)

⟹ d (p_{k}, p) \leq d (p_{k}, p_{k + 1}) + d (p_{k + 1}, p) \leq \frac{1}{3} ϵ + \frac{1}{6} ϵ_{k} < ϵ_{k}

ie, $p \in B (p_{k}, ϵ_{k}) = B_{k}$ for all $k$ . And each of these balls has $B_{k} \cap C_{k} = \emptyset$ . Thus $p$ is not in any of the $C_{k}$ ie $p \notin ⋃_{k} C_{k} = M$

⟹ ⟸

\begin{matrix} ◼ \end{matrix}

see [[Baire Category Theorem]]

Uniform Boundedness Theorem

Let $V$ be a [[vector space]], $B$ be a [[Banach space]], and ${T_{n}}$ a sequence in $B (B, V)$ . Then if for all $b \in B$ we have

sup_{n} | | T_{n} b | | < \infty

(ie the sequence is pointwise bounded) then we have

sup_{n} | | T_{n} | | < \infty

(ie the operator norms are bounded)

Proof

For all $k \in N$ , define

C_{k} = {b \in B : | | b | | \leq 1, sup_{n} | | T_{n} b | | \leq k}

Then each set is closed because if ${b_{n}} \subset C_{k}$ and $b_{n} \to b$ then we have $| | b | | = lim_{n \to \infty} | | b_{n} | | = 1$ . And for all $m \in N$ we have

| | T_{m} b | | = lim_{n \to \infty} | | T_{m} b_{n} | |

since each of the operators $T_{m}$ are continuous. And $| | T_{m} b_{n} | | \leq k$ because $b_{n} \in C_{k}$ , so $C_{k}$ must be closed.

Now, we also have

{b \in B : | | b | | \leq 1} = ⋃_{k \leq n} C_{k}

because for any $b \in B$ , there is no $k$ such that $sup_{m} | | T_{m} b | | \leq k$ .

So LHS is complete because it is a closed subset of $M$ , and so by Baire's Theorem, one of the $C_{k}$ must contain an open ball $B (b_{0}, δ_{0})$ .

Thus for any $b \in B (0, δ_{0})$ , we have that $b_{0} + b \in B (b_{0}, δ_{0}) \subset C_{k}$ so

sup_{n} | | T_{n} (b_{0} + b) | | \leq k

Then since both $b_{0}, b_{0} + b \in B (b_{0}, δ_{0})$ , we have

\begin{aligned} sup_{n} | | T_{n} b | | & = sup_{n} | | - T_{n} b_{0} + T_{n} (b_{0} + b) | | \\ \leq sup_{n} | | T_{n} b_{0} | | + sup_{n} | | T_{n} (b_{0} + b) | | \\ \leq k + k \end{aligned}

Thus, rescaling, we have for any $n \in N$ and all $b \in B$ with $| | b | | = 1$ , we have

| | T_{n} (\frac{δ_{0}}{2} b) | | \leq 2 k ⟹ | | T_{n} b | | \leq 4 k

ie the [[operator norm]] of $T_{n}$ is bounded for all $n$ , and therefore its $sup$ is bounded as well.

\begin{matrix} ◼ \end{matrix}

see [[uniform boundedness theorem]]

Review

#flashcards/math/functional

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