We introduced

• space of bounded linear operators
• bounded linear operator space is banach

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

ie closed under linear combinations

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

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

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

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

$V|W$ 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. $\lambda (v+W)=\lambda v+W$

$$W=0+W=w+W\mathrm{\forall }w\in W$$

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

is a subspace of $V$ and

defines a norm on the space $V|E$

$E$ is a subspace since $\mathrm{\forall }{v}_1,v_2\in E$ and ${\lambda }_1,{\lambda }_2\in \mathbb{K}$ we have

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

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

Let $V$ be a normed vector space and $W\subseteq V$ a closed subspace. Then $V|W$ is a normed vector space.

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

ie, at least one $C_n$ has an interior point.

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.

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

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.

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 $\mathrm{\exists }{ϵ}_1>0$ s.t. $B(p_1,{ϵ}_1)\cap C_1=\varnothing$.

Now, $B(p_1,\frac{{ϵ}_1}{3})\not\subset C_2$ 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 $B(p_k,\frac{{ϵ}_k}{3})\not\subset C_{k+1}$, 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 $B(p_{k+1},{ϵ}_{k+1})\cap C_{k+1}=\varnothing$

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. $p_j\in B(p_{j-1},\frac{{ϵ}_{j-1}}{3})$
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

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

Now, for all $k\in \mathbb{N}$, we have

So as $\ell \to \mathrm{\infty }$, we have

ie, $p\in B(p_k,{ϵ}_k)=B_k$ 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 $p\notin \bigcup _k{C}_k=M$

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

(ie the sequence is pointwise bounded) then we have

(ie the operator norms are bounded)

For all $k\in \mathbb{N}$, define

Then each set is closed because if $\{b_n\}\subset C_k$ and $b_n\to b$ then we have $||b||=\underset{n\to \mathrm{\infty }}{lim}||b_n||=1$. And for all $m\in \mathbb{N}$ we have

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

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

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

Then since both $b_0,b_0+b\in B(b_0,{\delta }_0)$, we have

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

ie the operator norm of $T_n$ is bounded for all $n$, and therefore its $sup$ is bounded as well.