Functional Analysis Lecture 3

[[lecture-data]]

← Lecture 2 | Lecture 4 → 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 ${\lambda }_1,{\lambda }_2\in \mathbb{K}$ and $w_1,w_2\in W$ then ${\lambda }_1{w}_1+{\lambda }_2{w}_2\in W$ 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 $v\sim v^{\prime }⟺v-v^{\prime }\in W$ Define $[v]=\{v^{\prime }\in V:v^{\prime }\sim v\}$ 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 $\lambda \in \mathbb{K}$

1. $(v_1+W)+(v_2+W):=(v_1+v_2)+W$
2. $\lambda (v+W)=\lambda v+W$

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 ${\lambda }_1,{\lambda }_2\in \mathbb{K}$ we have

\lvert \lvert \lambda_{1}v_{1} + \lambda_{2}v_{2} \rvert \rvert &\leq \lvert \lambda_{1} \rvert \cdot\lvert \lvert v_1 \rvert \rvert +\lvert \lambda_{2} \rvert \cdot \lvert \lvert v_{2} \rvert \rvert \\ &= 0 \\ \implies \lvert \lvert \lambda_{1}v_{1}+\lambda_{2}v_{2} \rvert \rvert &=0 \end{align}$$ We first show $\lvert \lvert \cdot \rvert \rvert_{V|E}$ is well-defined, ie $$v+E = v'+E \implies \lvert \lvert v \rvert \rvert =\lvert \lvert v' \rvert \rvert $$ Suppose $v+E=v'+E$. ie, there exists $e \in E$ such that $v+v' +e$. Then $$\begin{align} \lvert \lvert v \rvert \rvert &=\lvert \lvert v'+e \rvert \rvert \\ &\leq \lvert \lvert v' \rvert \rvert +\lvert \lvert e \rvert \rvert \\ &=\lvert \lvert v' \rvert \rvert \\ \implies \lvert \lvert v \rvert \rvert \leq \lvert \lvert v' \rvert \rvert &\text{ and also }\lvert \lvert v' \rvert \rvert \leq \lvert \lvert v \rvert \rvert \\ \implies \lvert \lvert v \rvert \rvert &=\lvert \lvert v' \rvert \rvert \end{align}$$ > [!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 subspace. Then $V|W$ is a normed vector space.

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 ${\bigcup }_n{C}_n=M$ and all $C_n$ are nowhere dense.

$M$ that does not converge

we’ll show that there is a sequence in

Since $M$ contains an open ball but $C_1$ does not, we have $M\ne C_1$. Thus there is some $p_1\in M\setminus C_1$. Since $C_1$ is closed but $M\setminus C_1$ is open, this implies that $\exists {ϵ}_1>0$ s.t. $B(p_1,{ϵ}_1)\cap C_1=\varnothing$.

Now, $B\left(p_1,\frac{{ϵ}_1}{3}\right)\not\subset C_2$ means there exists some $p_2\in M\setminus C_2$. And since $C_2$ is closed, there exists $0<{ϵ}_2<\frac{{ϵ}_1}{3}$ such that $B(p_2,{ϵ}_2)\cap C_2=\varnothing$

Now, suppose there we have found $k$ such points. Then for each $j=1,\dots ,k$ we have $p_j\not\subset C_j$ with${ϵ}_j$ such that $B(p_j,{ϵ}_j)\cap C_j=\varnothing$

Since $B\left(p_k,\frac{{ϵ}_k}{3}\right)\not\subset C_{k+1}$, there exists some $p_{k+1}\in B\left(p_k,\frac{{ϵ}_k}{3}\right)$ such that $p_{k+1}\notin C_{k+1}$

Then there exists ${ϵ}_{k+1}<\frac{{ϵ}_k}{3}$ such that $B(p_{k+1},{ϵ}_{k+1})\cap C_{k+1}=\varnothing$

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 $k$,

1. $p_j\in B\left(p_{j-1},\frac{{ϵ}_{j-1}}{3}\right)$
2. $B(p_j,{ϵ}_j)\cap C_j=\varnothing$

This sequence $\{p_k{\}}_k$ is Cauchy because for all $k,\ell \in \mathbb{N}$, we have

d(p_{k+1}, p_{k+\ell}) &\leq d(p_{k}, p_{k+1}) + d(p_{k+1}, p_{k+2}) + \dots + d(p_{k+\ell-1}, p_{k+\ell}) \\ &< \frac{\epsilon_{k}}{3} + \frac{\epsilon_{k+1}}{3} + \dots+\frac{\epsilon_{k+\ell-1}}{3} \\ &< \frac{\epsilon_{1}}{3^k}+\frac{\epsilon_{1}}{3^{k+1}}+\dots+\frac{\epsilon_{1}}{3^{k+\ell}} \\ &< \epsilon_{1} \sum_{n=k}^\infty \frac{1}{3^k} \\ &= \frac{\epsilon_{1}}{3^k}\left( \frac{1}{1-\frac{1}{3}} \right) \\ &= \frac{\epsilon_{1}}{2} 3^{-k+1} \end{align}$$ Since $M$ is complete, there exists some $p \in M$ such that $p_{k} \to p$ as $k \to \infty$. Now, for all $k \in \mathbb{N}$, we have $$\begin{align} d(p_{k+1}, p_{k+\ell+1}) &< \epsilon_{k+1} \left[ \frac{1}{3} + \frac{1}{3^2} + \dots+\frac{1}{3^\ell} \right] \\ &< \epsilon_{k+1} \sum_{n=1}^\infty 3^{-n} \\ &= \frac{\epsilon_{k+1}}{2} \\ \implies \lim_{ \ell \to \infty } d(p_{k+1}, p_{l+\ell+1}) &=d(p_{k+1}, p) \\ &\leq \frac{\epsilon_{k+1}}{2} \\ &< \frac{\epsilon_{k}}{6} \\ \end{align}$$ So as $\ell\to \infty$, we have $$d(p_{k+1}, p)$$ $$\implies d(p_{k}, p) \leq d(p_{k}, p_{k+1}) + d(p_{k+1}, p) \leq \frac{1}{3}\epsilon + \frac{1}{6}\epsilon_{k} < \epsilon_{k}$$ ie, $p \in B(p_{k}, \epsilon_{k}) = B_{k}$ *for all* $k$. And each of these balls has $B_{k} \cap C_{k} = \varnothing$. Thus $p$ is not in any of the $C_{k}$ ie $p \not\in\bigcup_{k}C_{k} = M$ $$\implies \impliedby$$ $$\tag*{$\blacksquare$}$$

see Baire Category Theorem

Uniform Boundedness Theorem

Let $V$ be a vector space, $B$ be a Banach space, and $\{T_n\}$ a sequence in $\mathcal{B}(B,V)$. 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 $k\in \mathbb{N}$, define $C_k=\{b\in B:||b||\le 1,{\sup }_n||T_{n}b||\le k\}$ 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 $m\in \mathbb{N}$ we have $||T_{m}b||={\lim }_{n\to \infty }||T_m{b}_n||$ since each of the operators $T_m$ are continuous. And $||T_m{b}_n||\le k$ because $b_n\in C_k$, so $C_k$ must be closed.

Now, we also have $\{b\in B:||b||\le 1\}={\bigcup }_{k\le n}{C}_k$ because for any $b\in B$, there is no $k$ such that ${\sup }_m||T_{m}b||\le k$.

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 $B(b_0,{\delta }_0)$.

Thus for any $b\in B(0,{\delta }_0)$, we have that $b_0+b\in B(b_0,{\delta }_0)\subset C_k$ so

${\sup }_n||T_n(b_0+b)||\le k$ Then since both $b_0,b_0+b\in B(b_0,{\delta }_0)$, we have

\sup_{n} \lvert \lvert T_{n}b \rvert \rvert &= \sup_{n} \lvert \lvert -T_{n}b_{0} + T_{n}(b_{0}+b)\rvert \rvert \\ &\leq \sup_{n} \lvert \lvert T_{n}b_{0} \rvert \rvert + \sup_{n} \lvert \lvert T_{n}(b_{0}+b) \rvert \rvert \\ &\leq k +k \end{align}$$ Thus, rescaling, we have for any $n \in \mathbb{N}$ and all $b \in B$ with $\lvert \lvert b \rvert \rvert = 1$, we have $$\left\lvert \left\lvert T_{n}\left( \frac{\delta_{0}}{2}b \right) \right\rvert \right\rvert \leq 2k \implies \lvert \lvert T_{n}b \rvert \rvert \leq 4k$$ ie the [[operator norm]] of $T_{n}$ is bounded for all $n$, and therefore its $\sup$ is bounded as well. $$\tag*{$\blacksquare$}$$

see uniform boundedness theorem

Review

functional

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