Title: Linear Analysis

Q: Continuity of addition
A: \begin{prop}
Addition $+: V\times V \to V$ is continuous.
\end{prop}
\begin{cor}
Translations $\mathbf{v}\mapsto \mathbf{v}+\mathbf{a}$ are homeomorphisms $V\to V$.
\end{cor}
\begin{proof}
Let $U\subseteq V$ be open and $(\mathbf{v}_1, \mathbf{v}_2) \in (+)^{-1}(U)$, ie. $\mathbf{v}_1 + \mathbf{v}_2 \in U$. By openness $\exists\varepsilon > 0$ such that $B_\varepsilon(\mathbf{v}_1 + \mathbf{v}_2) \subseteq U$. By the triangle inequality
\[
B_{\frac{\varepsilon}{2}}(\mathbf{v}_1) + B_{\frac{\varepsilon}{2}}(\mathbf{v}_2) \subseteq B_\varepsilon(\mathbf{v}_1 + \mathbf{v}_2) \subseteq U.
\]
Hence we have $(\mathbf{v}_1, \mathbf{v}_2) \in B_{\frac{\varepsilon}{2}} \left((\mathbf{v}_1, \mathbf{v}_2)\right) \subseteq (+)^{-1}(U)$. So $(+)^{-1}(U)$ is open.
\end{proof}

Q: Continuity of scalar multiplication
A: \begin{prop}
Scalar multiplication $(\cdot): \F \times V \to V$ is continuous.
\end{prop}
\begin{cor}
Dilations $\mathbf{v}\mapsto\lambda\mathbf{v}$ for $\lambda\neq 0$ are homeomorphisms $V\to V$.
\end{cor}
\begin{proof}
Let $\lambda_n\to \lambda, \mathbf{v}_n\to\mathbf{v}$. Then as $\mathbf{v}_n$ is bounded
\begin{align*}
\left| \lambda_n\mathbf{v}_n - \lambda\mathbf{v} \right| &= \left|  (\lambda_n - \lambda)\mathbf{v}_n + \lambda(\mathbf{v}_n - \mathbf{v}) \right| \\
&\leq |\lambda_n - \lambda | |\mathbf{v}_n| + |\lambda| |\mathbf{v}_n - \mathbf{v}| \\
& \to 0.
\end{align*}
\end{proof}

Q: Topological vector space
A: A \emph{topological vector space} $(V, \mathcal{U})$ is a vector space $V$ together with a topology $\mathcal{U}$ such that addition and scalar multiplication are continuous maps, and moreover singleton points $\{\mathbf{v}\}$ are closed sets.

Q: Topological $\implies$ Hausdorff
A: \begin{prop}
A topological vector space $X$ is Hausdorff
\end{prop}
\begin{proof}
By continuity of translations, wlog we separate 0 and $x\in X\setminus \{0\}$.
Then $W = X\setminus \{x\}$ is an open nbhd of 0. By continuity of addition, $\exists V_1, V_2$ open nbhds of 0, such that $V_1 + V_2 \subseteq W$. That is $x\notin V_1+V_2$.
So $V_1, (x - V_2)$ are disjoint open nbhds of $0, x$.
\end{proof}

Q: Absolutely convex vs. convex and balanced
A: $C\subseteq V$ is \emph{absolutely convex} (or \emph{balanced convex}) if $\forall\lambda, \mu \in \F$:
\[
|\lambda| + |\mu| \leq 1 \implies \lambda C + \mu C \subseteq C.
\]
Equivalently, $C$ is convex and balanced: $\forall |\lambda|\leq 1: \lambda C\subseteq C$.
\begin{proof}
If $C$ is absolutely convex, then it is convex (picking $\mu=1-t$) and balanced (picking $\mu=0$).
If $C$ is convex and balanced, then $\forall \lambda, \mu$ as above
\[
\lambda C + \mu C = (\lambda+\mu) \left(\frac{\lambda}{\lambda + \mu} C +  \frac{\mu}{\lambda + \mu} C\right) \subseteq \frac{\lambda}{\lambda + \mu} C +  \frac{\mu}{\lambda + \mu} C \subseteq C.
\]
\end{proof}

Q: Convexity of open balls
A: \begin{prop}
If $(V, \|\ph \|)$ is a normed vector space, then $D=B_1(\mathbf{0})$ is absolutely convex.
\end{prop}
\begin{proof}
Given $\lambda, \mu \in \F$ such that $|\lambda| + |\mu| \leq 1$, $\mathbf{v}_1, \mathbf{v}_2 \in D$, we have
\[
\left| \lambda \mathbf{v}_1 + \mu \mathbf{v}_2 \right| < |\lambda| + |\mu| \leq 1.
\]
\end{proof}

Q: Convexity is closed under vector operations
A: If $C\subseteq V$ is convex, then so is $\mathbf{v}+\lambda C$. Moreover if $C$ is absolutely convex, then so is $\lambda C$.
\begin{proof}
For any $t\in[0,1]$, $\mathbf{c}_1, \mathbf{c}_2\in C$:
\[
t(\mathbf{v}+\lambda\mathbf{c_1}) + (1-t)(\mathbf{v}+\lambda\mathbf{c}_2) = \mathbf{v} + \lambda(t\mathbf{c}_1+(1-t)\mathbf{c}_2) \in \mathbf{v}+\lambda C.
\]
\end{proof}

Q: Bounded set/space
A: $B\subseteq V$ is \emph{bounded} if for every open nbhd $U\subseteq V$ of $\mathbf{0}$, $\exists s > 0$ such that
\[
\forall t > s: B\subseteq t U.
\]
We say $V$ is \emph{locally bounded} if there exists a bounded neighbourhood of 0.

Q: Normable vector space
A: \begin{prop}
$(V, \mathcal{U})$ is normable $\iff$ $\exists C$ an absolutely convex, bounded open nbhd of $0$ (i.e. $V$ is locally convex/bounded).
\end{prop}
\begin{proof}
($\Rightarrow$) $B_1(0)$ works.
($\Leftarrow$) Define the \emph{Minkowski functional}, or gauge of $C$, $\mu_C: V \to \R$ by
\[
\mu_C(\mathbf{v}) = \inf \{t > 0: \mathbf{v}\in tC\}.
\]
By definition $\forall t < \mu_C(\mathbf{v})$, $\mathbf{v}\not\in tC$ and by convexity $\forall t > \mu_C(\mathbf{v}): \mathbf{v}\in tC$.
\emph{Pos def}. Given $\mathbf{v} \not= \mathbf{0}$, $U = V\setminus \{\mathbf{v}\}$ is an open nbhd of $0$. Hence $\exists t>0$ such that $C\subseteq tU$. That is $\frac{1}{t}C \subseteq U$, or $\mathbf{v}\not\in \frac{1}{t}C$. So $\mu_C (\mathbf{v}) \geq \frac{1}{t} > 0$.
\emph{Scalar mult}. Since $C$ is balanced
\begin{align*}
\mu_C(\lambda \mathbf{v}) &= \inf \left\{ t > 0: \lambda \mathbf{v}\in t C \right\} \\
&= \inf \left\{ t > 0: \mathbf{v}\in \frac{t}{|\lambda|} C \right\} \\
&= \inf \left\{|\lambda| \frac{t}{|\lambda|} > 0: \mathbf{v}\in \frac{t}{|\lambda|} C\right\} \\
&= |\lambda| \mu_C(\mathbf{v}).
\end{align*}
$\triangle$ \emph{inequality}.  This is equivalent to convexity:
\begin{align*}
&\mu_C (\mathbf{v} + \mathbf{w}) \leq \mu_C(\mathbf{v}) + \mu_C(\mathbf{w}) \\
\iff & \inf\{t > 0: \mathbf{v} + \mathbf{w} \in tC\} \leq \inf\{t > 0: \mathbf{v}\in tC\} + \inf\{r > 0: \mathbf{w}\in rC\} \\
\iff & \mathbf{v}\in tC, \mathbf{w}\in rC \implies (\mathbf{v} + \mathbf{w})\in (t + r)C \\
\iff & \mathbf{v}' = \mathbf{v}/t, \mathbf{w}' = \mathbf{w}/r \in C \implies  \frac{1}{(t + r)}(t \mathbf{v}' + r \mathbf{w}') \in C.
\end{align*}
\end{proof}

Q: Locally convex
A: $(X,\mathcal{U})$ is locally convex if $\mathcal{U}$ has a basis at the origin of balanced, convex, absorbing sets

Q: Finding a balanced subset
A: Every open neighborhood $U$ of 0 in a vector space $(X, \mathcal{U} )$ contains a balanced open neighborhood $W$ of 0.
\begin{proof}
By continuity of scalar multiplication, $\exists\delta>0$ and open nbhd $V$ of 0 such that $\forall |\lambda| <\delta: \lambda V \subseteq U$.
Then $W := \bigcup_{|\lambda|<\delta} \lambda V\subseteq U$. For any $|t|\leq 1, |\lambda|<\delta$, we have that $|t\lambda|<\delta$. So $t\lambda V\subseteq U$. Hence $tW\subseteq W$.
\end{proof}

Q: Banach spaces
A: A normed vector space is a \emph{Banach space} if it is complete as a metric space, ie. every Cauchy sequence converges.

Q: Fr\'echet Space
A: A locally convex complete vsp with a translation-invariant metric.
\begin{eg}
Let $U\subseteq\R^n$ be open. Let $C(U) = \{ f: U\to \R, f \text{ cts} \}$.
Let $(K_i)\subseteq B$ be compact such that $(K_i) \nearrow\bigcup_{i = 1}^\infty K_i = U$. We define the basis to include
\[
V(i,m)= \left\{f \in C(U): \sup_{x \in K_i} |f(x)| < \frac{1}{m}\right\}
\]
for $m, i \in\N$ and translations of these sets.
It follows that $C(U)$ is locally convex, $f_n \to f$ iff $f_n \to f$ uniformly in every compact set so $C(U)$ is complete but not normable. The metric is
\[
d(f,g)=\sum_{i=0}^\infty 2^{-i}\frac{\|f-g\|_{C(K_i)}}{1+\|f-g\|_{C(K_i)}}.
\]
\end{eg}

Q: Bounded linear map
A: For $T\in\mathcal{L}(X,Y)$
\[
T\in\mathcal{B}(X, Y)\iff\exists C > 0 \text{ s.t. } \forall\mathbf{x}\in X: \|T\mathbf{x}\|_Y \leq C\|\mathbf{x}\|_X.
\]
This generalises to topological spaces by
\[
T\in\mathcal{B}(X, Y)\iff \forall B\subseteq X \text{ bounded: } T(B) \text{ is bounded}.
\]

Q: Continuous vs. Bounded
A: \begin{prop}
Let $X$, $Y$ be normed vector spaces, $T: X\to Y$ a linear map. Then TFAE:
\begin{enumerate}
\item $T$ is continuous.
\item $T$ is continuous at $\mathbf{0}$.
\item $T$ is bounded.
\end{enumerate}
\end{prop}
\begin{proof}
(i) $\Rightarrow$ (ii) is immediate.
(ii) $\Rightarrow $ (iii): Consider $B_1(0) \subseteq Y$. Since $T$ is cts at $0$, $\exists\varepsilon > 0$ such that $T(B_\varepsilon(0)) \subseteq B_1(0)$. So by linearity $T(B_1(0))\subseteq B_\frac{1}{\varepsilon}(0)$.
(iii) $\Rightarrow$ (i): $T$ bounded $\implies$ $T$ Lipschitz $\implies$ $T$ continuous.
\end{proof}

Q: Continuous vs. Locally continuous
A: Let $X,Y$ be topological vector spaces and $T\in\mathcal{L}(X,Y)$. Then $T$ is continuous $\iff$ $T$ is continuous at 0.
\begin{proof}
($\Rightarrow$) Immediate.
($\Leftarrow$). Let $y=Tx$ and $W\in Y$ be an open nbhd of $y$. Then $W-y$ is an open nbhd of 0. By cty $T^{-1}(W-y)$ contains an open nbhd $V\subseteq X$ of 0. By linearity
\[
T(x+V)=Tx + T(V)\subseteq Tx + W - y = W.
\]
Since $x+V$ is an open nbhd of $x$, this means that $T$ is continuous at $x$.
\end{proof}

Q: Dual space
A: The \emph{dual space} of $V$ is the space of continuous linear maps $V\to\F$. If $V$ is normed, then
\[
V^* = \mathcal{B}(V, \F).
\]
The \emph{algebraic dual} of $V$ is
\[
V' = \mathcal{L}(V, \F),
\]
where we do not require boundedness/continuity.

Q: $\mathcal{B}$ is Banach
A: Let $V$ be normed and $W$ Banach. Then $\mathcal{B}(V,W)$ is Banach. In particular the dual space $V^*$ is Banach.
\begin{proof}
Given $(T_i) \in \mathcal{B}(V,W)$ Cauchy,  $\forall\mathbf{v}\in V$, $(T_i(\mathbf{v}))\subseteq W$ is Cauchy. Since $W$ is complete, $T_i\to T$ pointwise.
\textbf{Linear}. Using linearity of the $T_i$:
\[
T(\lambda v +\mu w) = \lim_{i\to\infty} T_i (\lambda v +\mu w) =\lambda \lim_{i\to\infty}T_iv + \mu \lim_{i\to\infty} T_iw=\lambda T(v)+\mu T(w).
\]
\textbf{Bounded}. Let $\|\mathbf{v}\| \leq 1$. Pick $N(\mathbf{v})$ such that for $i>N(\mathbf{v})$: $\|T_i\mathbf{v}-T\mathbf{v}\|\leq 1$. Then for $i>N(\mathbf{v})$, since $T_i$ is Cauchy and thus bounded:
\[
\|T\mathbf{v}\| \leq \|T_i\mathbf{v}\| + \|T_i\mathbf{v}-T\mathbf{v}\| \leq \|T_i\| + 1 \leq 1+ \sup_i \|T_i\|<\infty.
\]
\textbf{Convergence}. Given $\varepsilon > 0$, $\exists N$ such that $\forall i, j > N$, we have
\[
\|T_i - T_j\| < \frac{\varepsilon}{2}.
\]
For each $\|\mathbf{v}\| \leq 1$, $\exists M(\mathbf{v})>N$ such that for $j\geq M(\mathbf{v})$:
\[
|T_j(\mathbf{v}) - T(\mathbf{v})| < \frac{\varepsilon}{2}.
\]
Therefore for $i >N, j\geq M(\mathbf{v})$:
\[
|T_i(\mathbf{v}) - T(\mathbf{v})| \leq |T_i(\mathbf{v}) - T_j(\mathbf{v})| + |T_j(\mathbf{v}) - T(\mathbf{v})| <\varepsilon.
\]
Since this is true for any $\mathbf{v}$, we have $\forall i>N$:
\[
\|T_i - T\| < \varepsilon.
\]
\end{proof}

Q: Reflexive space
A: A nvsp $E$ is reflexive if $\ev: E\to E^{**}$ is surjective.
In this case $E\cong E^{**}$ is Banach.

Q: Equivalence of norms on $\R^n$
A: \begin{prop}
Let $V$ be an $n$-dimensional vector space. Then any norm $\|\ph \|$ on $V$ is equivalent to $\|\ph\|_1$.
\end{prop}
\begin{proof}
By triangle inequality:
\[
\|\mathbf{v}\|\leq \left(\sup_i \|\mathbf{e}_i\|\right) \|\mathbf{v}\|_{1}=: C\|\mathbf{v}\|_{1}.
\]
We show $\|\ph \|: (S^1, \|\ph \|_{1}) \to \R$ is continuous on a compact set so $\exists m,M\in\R$ such that $\forall\mathbf{v}\in V$:
\[
m\leq \left\|\frac{\mathbf{v}}{\|\mathbf{v}\|_{1}}\right\| \leq M,
\]
which gives the Lipschitz equivalence:
\[
m \|\mathbf{v}\|_{1} \leq \|\mathbf{v}\| \leq M \|\mathbf{v}\|_{1}.
\]
\emph{Continuous} By triangle inequality, $\forall \mathbf{v}, \mathbf{w}\in V$, we have
\[
\big|\|\mathbf{v}\| - \|\mathbf{w}\|\big| \leq \|\mathbf{v} - \mathbf{w}\| \leq C\|\mathbf{v} - \mathbf{w}\|_{1}.
\]
\emph{Compact} Ets $B = \{\mathbf{v} \in V: \|\mathbf{v}\|_{1}\leq 1\}$ is (s.) compact, since $S_1\subseteq B$ is closed. Let $\{\mathbf{v}^{(k)}\}_{k = 1}^\infty\subseteq B$. Since $\mathbf{v}^{(k)} \in B$, we have
\[
\sum_{i = 1}^n |v^{(k)}_i| \leq 1.
\]
By repeated B-W, $\exists \Lambda_n \subseteq \cdots \Lambda_1\subseteq \N$ such that $\mathbf{v}^{(k)}_i\to v_i$ as $k\to\infty$ with $k\in\Lambda_i$ for $1\leq i \leq n$ and $\sum_{i = 1}^n |v_i| \leq 1$ so $\mathbf{v}_i \in B$.
\end{proof}

Q: Finite dimensional $\implies$ Banach
A: \begin{prop}
Let $V$ be a fd normed vsp. Then $V$ is Banach.
\end{prop}
\begin{proof}
Let $(\mathbf{v}_i) \in V$ be a Cauchy sequence. By boundedness, $\exists R>0$ such that $\{\mathbf{v}_i\} \subseteq \bar{B}(R)$. Since $\bar{B}(R)$ is compact, $(\mathbf{v}_i)$ converges.
\end{proof}

Q: Finite dimensional subspaces are closed
A: Let $V$ be a normed vsp, $W\subseteq V$ a fd subspace. Then $W$ is closed.
\begin{proof}
Since $W$ is fd, it is Banach. By completeness, $W$ is closed.
\end{proof}

Q: Linear maps on fd vsps are bounded
A: \begin{prop}
Let $V, W$ be normed vspaces, $V$ is fd and $T: V\to W$ be a linear map. Then $T$ is bounded.
\end{prop}
\begin{proof}
Since $V$ is fd, $\im T$ is fd. So wlog $W$ is fd. Wlog take $V=(\F^n, \|\ph \|_1)$, $W=(\F^n, \|\ph \|_\infty)$. Let $T_{ij}$ represent $T$ such that $\forall \mathbf{v}\in V$:
\begin{align*}
\|T\mathbf{v}\|_\infty &= \max_{1\leq i \leq m} \left|\sum_{j=1}^n T_{ij}v_j \right| \\
&\leq \max_{i,j} |T_{ij}| \sum_{j=1}^n |v_i| \\
&= C(T)\|\mathbf{v}\|_1.
\end{align*}
\end{proof}
\begin{proof}(alternative)
Define a norm on $V$ by $\|\mathbf{v}\|' = \|\mathbf{v}\|_V + \| T \mathbf{v}\|_W$.
Since $V$ is fd, by Lipschitz equivalence $\exists C>0$ such that for all $\mathbf{v}\in V$:
\[
\|\mathbf{v}\|' \leq C\|\mathbf{v}\|_V.
\]
In particular, we have
\[
\|T\mathbf{v}\|_W \leq (C-1)\|\mathbf{v}\|_V.
\]
\end{proof}

Q: Compact unit ball vs. fd
A: \begin{prop}
Let $V$ be a normed vsp. Then $\bar{B}(1)$ is compact $\iff$ $V$ is fd.
\end{prop}
\begin{proof}
($\Leftarrow$) Immediate by Lipschitz equivalence.
($\Rightarrow$) By compactness,  $\exists \mathbf{w}_1, \cdots, \mathbf{w}_n\in V$ such that
\[
\bar{B}(1) \subseteq \bigcup_{i = 1}^n \mathbf{w}_i + B\left(\frac{1}{2}\right).
\]
Now let $W = \spn\{\mathbf{w}_1, \cdots, \mathbf{w}_n\}\subseteq V$. Clearly $\dim W\leq n<\infty$. Finally
\begin{align*}
B(1) &\subseteq W + B\left(\frac{1}{2}\right) \\
B(1) &\subseteq W + \frac{1}{2} \left(W + B\left(\frac{1}{2}\right) \right) \\
B(1) &\subseteq W + B\left(\frac{1}{4}\right) \\
B(1) &\subseteq W + B\left(\frac{1}{2^n}\right) \\
B(1) &\subseteq \bar{W}=W \\
V &=W.
\end{align*}
\end{proof}

Q: Sublinear map
A: A map $p: V \to \R$ is \emph{sublinear} if it obeys
\begin{enumerate}
\item (positive homogeneity) $\forall\mathbf{v}\in V, \lambda > 0$:
\[
p (\lambda \mathbf{v}) = \lambda p(\mathbf{v})
\]
\item (subadditivity) $\forall\mathbf{v}_i\in V$:
\[
p(\mathbf{v}_1 + \mathbf{v}_2) \leq p (\mathbf{v}_1) + p(\mathbf{v}_2).
\]
\end{enumerate}

Q: Codimension
A: Let $W\subseteq V$. Then
\[
\codim(W) = \dim(V/W) = \dim\coker(W\to V).
\]
If $\dim V<\infty$, then
\[
\codim(W) =\dim V - \dim W.
\]

Q: Hahn-Banach for codimension 1
A: \begin{prop}
Let $V$ be a normed $\R$-vsp, $\codim_V (W)=1$, $p:V\to\R$ sublinear, $g \in W^*$, $g|_W\leq p|_W$. Then there exists a linear extension $f: V\to \R$ such that $f\leq p$.
\end{prop}
\begin{proof}
Pick $\mathbf{v}_0 \in V\setminus W$. We need that
\[
f(\mathbf{w} + a \mathbf{v}_0) \leq p(\mathbf{w} + a \mathbf{v}_0) \quad (\forall\mathbf{w} \in W, a \in \R) \tag{$*$}
\]
If $a = 0$, then $(*)$ is given. If $a > 0$ then $(*)$ is equivalent to
\begin{align*}
f(a^{-1}\mathbf{w}) + f(\mathbf{v_0}) &\leq p(a^{-1}\mathbf{w} + \mathbf{v}_0) \\
f(\mathbf{v}_0) &\leq p(\mathbf{w}' + \mathbf{v}_0) - g(\mathbf{w}') \quad (\mathbf{w}' := a^{-1} \mathbf{w}).
\end{align*}
If $a < 0$, then $(*)$ is equivalent to
\begin{align*}
 f(a^{-1}\mathbf{w}) + f(\mathbf{v}_0) &\geq -(-a^{-1})p(\mathbf{w} + a\mathbf{v}_0) \\
f(\mathbf{v}_0) &\geq -p(-a^{-1} \mathbf{w} - \mathbf{v_0}) - g(a^{-1}\mathbf{w}) \\
f(\mathbf{v}_0) &\geq -p(\mathbf{w}' - \mathbf{v_0}) + g(\mathbf{w}') \quad (\mathbf{w}' = -a^{-1}\mathbf{w}).
\end{align*}
$f(\mathbf{v_0})$ can be chosen to satisfy both iff $\forall \mathbf{w}_1, \mathbf{w}_2 \in W$:
\begin{align*}
-p(\mathbf{w}_1 - \mathbf{v}_0) + g(\mathbf{w}_1) &\leq p(\mathbf{w}_2 + \mathbf{v}_0) - g(\mathbf{w}_2) \\
g(\mathbf{w}_1 + \mathbf{w}_2) &\leq p(\mathbf{w}_2 + \mathbf{v}_0) + p(\mathbf{w}_1 - \mathbf{v}_0)
\end{align*}
which follows from $g|_W\leq p|_W$ and sublinearity of $p$.
\end{proof}

Q: Poset
A: $X$ equipped with $\leq$ is a \emph{partially ordered set} if it satisfies
\begin{enumerate}
\item $x \leq x$\hfill (reflexivity)
\item $x \leq y, y \leq x \implies x = y$ \hfill (antisymmetry)
\item $x \leq y, y \leq z \implies x \leq z$ \hfill (transitivity).
\end{enumerate}

Q: Totaly ordered set / chain
A: Let $(S, \leq)$ be a partial order. $T\subseteq S$ is \emph{totally ordered} $\forall x, y\in T$:
\[
x \leq y\text{ or } y\leq x.
\]

Q: Maximal element
A: Let $(S, \leq)$ be a partial order. Then $m\in S$ is a \emph{maximal element} if
\[
x\geq m \implies x=m.
\]

Q: Zorn's lemma
A: Let $(P, \leq)$ be a non-empty poset such that every non-empty chain has an upper bound in $P$. Then $P$ has a maximal element.

Q: Extending LI sets to a basis
A: Let $V\neq\{ 0 \}$ be a vsp and $S\subseteq V$ LI. Then $V$ has a basis $B\supseteq S$.
\begin{proof}
\textbf{Poset}. Let $P=\{ T\subseteq V: T\supseteq S, T \text{ LI } \}$ form a poset under $\subseteq$.
\textbf{Non-empty}. Clearly $S\in P$ so $P\neq \emptyset$.
\textbf{UB property}. Given $L\subseteq P$ totally ordered, define $U=\bigcup_{T\in L} T$. Clearly, $S\subseteq U$. Let $\alpha_i\in\F, v_i\in U$ such that $\sum_{i=1}^m \alpha_i v_i = 0$. Since $L$ is totally ordered and $m<\infty$, $\exists T\in L$ such that $v_1, \cdots v_m \in T$. Since $T$ is LI, we must have $\alpha_1=\cdots \alpha_m=0$. So $U$ is LI. Hence $U=\sup L \in P$.
\textbf{Zorn}. $\exists B\in P$ maximal.
\textbf{Basis}. As $B\in P$, $B$ is LI. Suppose $\spn B\neq V$. Then $\exists v\in V\setminus\spn B$ and $B\cup\{v\}$ is LI, contradicting maximality of $B$.
\end{proof}

Q: Hahn-Banach theorem (real case)
A: Let $V$ be a normed $\R$-vsp, $W\subseteq V$, $p: V \to \R$ sublinear, $g\in W^*$ such that $g|_W\leq p|_W$. Then $\exists f\in V^*$ such that $f|_W=g$ and $f\leq p$.
\begin{proof}
\textbf{Poset}. Let $P=\{ (N,h): W\subseteq N\subseteq V, h\in N^*, h|_W=g, h|_N\leq p|_N \}$ form a poset under
\[
(N,h) \leq (N', h') \iff N\subseteq N' \text{ and } h'|_N=h.
\]
\textbf{Non-empty}. $(W,g)\in P$ so $P\neq \emptyset$.
\textbf{UB property}. Given $(N_i, h_i)_{i\in I} \subseteq P$ a chain, define
\[
N=\bigcup_{i\in I} N_i \subseteq V; \quad h|_{N_i}=h_i|_{N_i}.
\]
That $h$ is well-defined, linear and bounded by $p$ is inherited from the larger of $(N_i, h_i)$ containing any given point. So $(N,h)\in P$.
\textbf{Zorn}. $\exists (M,f)\in P$ maximal.
\textbf{Whole space}. Suppose $M\neq V$. Then the codimension 1 case gives $(M',f')$ with $(M,f)<(M',f')$, contradicting maximality.
\end{proof}

Q: Hahn-Banach for normed spaces
A: Let $W \subseteq V$ be normed $\R$-vsp. Given $g \in W^*$, $\exists f \in V^*$ such that $f|_W = g$ and $\|f\|_{V^*} = \|g\|_{W^*}$.
\begin{proof}
Define the norm $p:V\to \R$ by
\[
p(\mathbf{x}) = \|g\|_{W^*}\|\mathbf{x}\|_V \quad (\forall \mathbf{x}\in V).
\]
$p$ is a norm so is sublinear. By definition $g \leq p|_W$.
By Hahn-Banach theorem, $\exists f\in V^*$ such that $f|_W = g$ and $f(\mathbf{v}) \leq p(\mathbf{v}) = \|g\|_{W^*} \|\mathbf{v}\|_V$.
\end{proof}

Q: Support functional
A: \begin{prop}
Let $V$ be a normed $\R$-vsp. For every $\mathbf{v}\in V\setminus \{0\}$, $\exists f_{\mathbf{v}} \in V^*$ such that $f_{\mathbf{v}}(\mathbf{v}) = \|\mathbf{v}\|$ and $\|f_{\mathbf{v}}\| = 1$.
\end{prop}
\begin{proof}
Let $W = \spn\{\mathbf{v}\}$, $g(t\mathbf{v}) = t\|\mathbf{v}\|_V$. Then $\|g\|=1$ and $g(\mathbf{v})=\|\mathbf{v}\|$. Done by normed Hahn-Banach.
\end{proof}

Q: Functionals separate points
A: Let $V$ be a normed $\R$-vsp. Then
\[
\mathbf{v} = \mathbf{0} \iff \forall f\in V^*: f(\mathbf{v}) = 0.
\]
In particular, $V^*\neq 0$.
\begin{proof}
($\Rightarrow$) Immediate by linearity.
($\Leftarrow$) Using the support functional: $f_\mathbf{v}(\mathbf{v}) = \|\mathbf{v}\| = 0$ so $\mathbf{v} = 0$.
\end{proof}
Let $V$ be a normed $\R$-vsp, $\mathbf{v}\neq\mathbf{w}\in V$. Then $\exists f\in V^*$ such that $f(\mathbf{v}) \not= f(\mathbf{w})$.
\begin{proof}
Contrapositive of first result applied to $\mathbf{v} - \mathbf{w}$.
\end{proof}

Q: Evaluation is an isometry
A: The map $\ev: V\to V^{**}$ is an isometry. In particular $\ev$ is an injective, isometric embedding of $V$ into $V^{**}$ with $\|\ev\|=1$.
\begin{proof}
($\leq$) $\forall \mathbf{v}\neq 0$:
\begin{align*}
\|\ev(\mathbf{v})\| &= \sup_{g\in V^*}\frac{|\ev(\mathbf{v})(g)|}{\|g\|}\\
&= \sup_{g\in V^*}\frac{|g(\mathbf{v})|}{\|g\|}\\
&\leq \|\mathbf{v}\|.
\end{align*}
($\geq$) Using the support functional, $\forall \mathbf{v}\neq 0$:
\begin{align*}
\|\ev(\mathbf{v})\| &= \sup_{f\in V^*} \frac{|\ev(\mathbf{v})(f)|}{\|f\|} \\
&\geq \frac{|\ev(\mathbf{v})(f_\mathbf{v})|}{\|f_\mathbf{v}\|} \\
&= |f_\mathbf{v}(\mathbf{v})| \\
&= \|\mathbf{v}\|.
\end{align*}
\end{proof}

Q: Norm of the adjoint
A: For any $T\in\mathcal{B}(V,W)$: $\|T^*\| = \|T\|$. In particular, $T^*\in \mathcal{B}(W^*,V^*)$.
\begin{proof}
($\leq$) For any $\mathbf{v}\in V, \theta\in W^*$:
\begin{align*}
\|(T^*\theta)(\mathbf{v})\| = \|(\theta\circ T)\mathbf{v})\| &\leq \|\theta \| \|T \| \|\mathbf{v} \| \\
\|(T^*\theta)\| &\leq \|\theta \| \|T \| \\
\|T^*\| &\leq \| T \|.
\end{align*}
($\geq$) Given $\varepsilon > 0$, $\exists \mathbf{v}\in B_1(0)$ such that $\|T\mathbf{v}\| \geq \|T\| - \varepsilon$. Using support functionals:
\begin{align*}
\|T^*\| &\geq \|T^*(f_{T\mathbf{v}})\| \\
&\geq |T^*(f_{T\mathbf{v}})(\mathbf{v})| \\
&= |f_{T\mathbf{v}}(T\mathbf{v})| \\
&= \|T\mathbf{v}\| \\
&\geq \|T\| - \varepsilon.
\end{align*}
\end{proof}

Q: Nowhere dense set
A: Let $(X,\tau)$ be a topological space. A subset $E\subseteq X$ is \emph{nowhere dense} if $\Int(\bar{E})=\emptyset$. Equivalently, $E$ is not dense is any non-empty open set.
\begin{proof}
Observe that
\[
E \text{ is dense in } U \iff U\subseteq\bar E.
\]
Hence
\[
E \text{ is nowhere dense } \iff \forall U\in\tau\setminus\emptyset: U\not\subseteq\bar E \iff \Int(\bar E)=\emptyset.
\]
\end{proof}

Q: Meagre
A: Let $X$ be a topological space. We say that $Z\subseteq X$ is of \emph{first category} or \emph{meagre} if it is a countable union of nowhere dense sets.

Q: Non-meagre
A: A subset is of \emph{second category} or \emph{non-meagre} if it is not of first category.

Q: Residual
A: Let $X$ be a topological space. We say that $Z\subseteq X$ is \emph{residual} if $X\setminus Z$ is meagre.

Q: Baire category theorem (open form)
A: Let $(X,d)$ be complete. If $(U_n)$ are open dense, then $\bigcap_{n = 1}^\infty U_n$ is non-empty (in fact dense).
\begin{proof}
Pick $x_1, \varepsilon_1$ such that $\bar B_{\varepsilon_1}(x_1) \subseteq U_1$ (using density then openness).
Inductively define $x_{n }\in X, \varepsilon_{n } < \frac{1}{n}$ such that $\bar B_{\varepsilon_{n }}(x_{n }) \subseteq \bar B_{\varepsilon_{n-1}} (x_{n-1}) \cap U_{n}$ (using density then openness).
For $m\geq n: d(x_m,x_n) \leq \varepsilon_n < \frac{1}{n} \to 0$ as $n\to\infty$. So $x_n$ is Cauchy. By completeness, $x_n \to x \in X$. By definition
\[
\forall n\in\N: x\in \bar B_{\varepsilon_n}(x_n) \implies \forall n\in\N: x\in U_n.
\]
\end{proof}

Q: Baire Category Theorem (closed form)
A: Let $(X,d)$ be complete. Then $X$ is non-meagre. That is, if $X = \bigcup_{n = 1}^\infty C_n$, where $(C_n)$ are closed, then $\exists C_n$ with non-empty interior.
\begin{proof}
Let $U_n=X\setminus C_n$. Then $C_n$ closed $\implies U_n$ open and
\[
X = \bigcup_{n = 1}^\infty C_n \implies \bigcap_{n = 1}^\infty U_n = \emptyset.
\]
By BCT (open form), $\exists U_n$ which is not dense. That is, $C_n$ has non-empty interior.
\end{proof}

Q: Nowhere dense and closed
A: $C\subseteq X$ is nowhere dense and closed $\iff X\setminus C$ is open and dense.
In particular, $Y\subseteq X$ nowhere dense $\iff X\setminus \bar Y$ is dense.

Q: Residual subsets of a complete space
A: Let $(X,d)$ be complete. Then residual sets are non-meagre and dense.
\begin{proof}
Write $X=Z\cup R$, with $R$ residual and $Z=\bigcup_j A_j$ meagre, $A_j$ nowhere dense. Equivalently, $U_j:=X\setminus \bar A_j$ are open and dense. By BCT, $\bigcap_j U_j$ is dense. So $R\supseteq \bigcap_j U_j$ is dense.
Since $X=Z\cup R$ is complete so non-meagre, $R$ must be non-meagre.
\end{proof}

Q: Category of open sets
A: Let $X$ be a complete msp, $U\subseteq X$ open. Then $U=\emptyset$ or $U$ is non-meagre.
\begin{proof}
Suppose $U$ is open and meagre. Then $X\setminus U$ is closed and residual, so dense. Hence $X\setminus U = X$ and $U=\emptyset$.
\end{proof}

Q: Principle of uniform boundedness
A: \begin{thm}
Let $X$ be a complete msp. Let $f_\lambda: X\to \R$ be continuous functions for $\lambda\in\Lambda$. If $(f_\lambda)$ are pointwise bounded, then they are locally uniformly bounded, i.e.
\[
\forall x \in X: \sup_{\lambda\in\Lambda} |f_\lambda(x)|<\infty \implies \exists B_r(x_0): \sup_{\lambda\in\Lambda} \sup_{B_r(x_0)} |f_\lambda(x) | < \infty.
\]
\end{thm}
\begin{proof}
Define the closed sets
\[
A_k  = \bigcap_{\lambda\in\Lambda} \{ x \in X:  |f_\lambda(x) | \leq k\}.
\]
Since $(f_\lambda)$ are pointwise bounded
\[
\bigcup_{k\in\N} A_k = X.
\]
By the BCT, $\exists A_k$ with non-empty interior. So $A_k$ contains some a ball on which $(f_\lambda)$ are uniformly bounded by $k$.
\end{proof}

Q: Banach-Steinhaus theorem
A: \begin{thm}
Let $V$ be a Bsp and $W$ a nvsp. If $(T_\lambda)_{\lambda\in\Lambda} \subseteq \mathcal{B}(V,W)$ are pointwise bounded, then they are uniformly bounded, i.e.
\[
\sup_\lambda \|T_\lambda (v)\| < \infty \implies \sup_\lambda \|T_\lambda\| < \infty.
\]
In particular, $\mathcal{B}(V,W)$ is closed under pointwise limits.
\end{thm}
\begin{proof}
Let $f_\lambda: V\to \R$ be given by $v\mapsto\|T_\lambda v \|$. Then $(f_\lambda)$ are continuous and PB. By PoUB, $\exists B_r(v_0)\subseteq V$ such that $(f_\lambda)$ is UB, i.e.
\[
\sup_{\lambda\in\Lambda} \sup_{v\in B_r(v_0)} \|T_\lambda v \| < \infty.
\]
By linearity, $\forall v\in B_1(0), \lambda\in\Lambda$:
\begin{align*}
\| T_\lambda v \| &= \frac{1}{r} \| T_\lambda(v_0+rv) - T_\lambda v_0 \| \\
\sup_{\lambda\in\Lambda} \|T_\lambda\| & \leq \frac{1}{r} \sup_{\lambda\in\Lambda} \sup_{v \in B_r(v_0)} \| T_\lambda v \| + \frac{1}{r} \sup_{\lambda\in\Lambda} \| T_\lambda v_0 \| <\infty.
\end{align*}
\end{proof}

Q: Ball condition for open maps
A: Let $V$ and $W$ be normed vsps. Then $T:V\to W$ is open $\iff \exists r>0: B_r(0) \subseteq T(B_1(0)) $.
\begin{proof}
($\Rightarrow$) If $T$ is open, then $T(B_1(0))$ is open.
($\Leftarrow$) Let $U\subseteq V$ be open and $u\in U$. Since $U$ is open,  $\exists \delta > 0$ such that $B_\delta(u) \subseteq U$. By linearity
\begin{align*}
T(U) &\supseteq T(B_\delta(u)) \\
&= T(u + B_\delta(0)) \\
&= Tu + \delta T(B_1(0)) \\
&\supseteq Tu + \delta r B_1(0) \\
&= Tu + B_{r\delta} (0)\\
&= B_{r\delta} (Tu).
\end{align*}
\end{proof}

Q: Ball condition for open maps (with closure)
A: Let $V$ be a Bsp, $W$ a nsp, $T\in\mathcal{B}(V,W)$. Then
\[
\overline{T(B_1(0))} \supseteq B_1(0) \implies T(B_1(0)) \supseteq B_1(0).
\]
\begin{proof}
Let $w_0\in B_1(0)\subseteq W$.
Since $\exists \tilde w_0\in B_\frac{1}{2}(0)$ with $w_0-\tilde w_0< \frac{1}{2}$ and (rescaling assumption by half) $\exists v_1\in B_\frac{1}{2}(0)$ such that $\|Tv_1 - \tilde w_0\|$ is small enough that $w_1=w_0-Tv_1\in B_\frac{r}{2}(0)$.
By induction, if $w_1, \cdots w_k$ and $v_1, \cdots v_k$ are known such that
\[
v_i, w_i=w_{i-1} - Tv_i \in B_{2^{-i}}(0) \quad (1\leq i \leq k),
\]
then choose $v_{k+1}\in B_{2^{-k-1}}(0)\subseteq V$ such that
\[
w_{k+1} = w_k - Tv_{k+1} \in B_{2^{-k-1}}(0).
\]
So $\sum_{k=1}^\infty \|v_k \| < \sum_{k= 1}^\infty 2^{-k} = 1$ and $v=\sum_{k= 1}^\infty v_k \in B_1(0)$ exists since $V$ is complete. Substituting in, we can check
\begin{align*}
w_0 - Tv &= \lim_{n\to\infty} \left(w_0 - \sum_{k=1}^n Tv_k\right) \\
&=  \lim_{n\to\infty} \left(w_1 - \sum_{k=2}^n Tv_k\right) \\
\cdots &= \lim_{n\to\infty} w_n \\
&= 0.
\end{align*}
\end{proof}

Q: Open mapping theorem
A: \begin{thm}
Let $V$ and $W$ be Bsps and $T\in\mathcal{B}(V,W)$.
\begin{enumerate}
\item If $T$ is surjective, then it is open.
\item If $T$ is bijective, then $T^{-1}\in\mathcal{B}(V,W)$.
\end{enumerate}
\end{thm}
\begin{proof}
Since $T$ is surjective
\[
W = \bigcup_{n = 1}^\infty \overline{T(B_n(0))}.
\]
Since $W$ is complete, by BCT, $\exists n \geq 1$ such that $\overline{T(B_n(0))}$ has non-empty interior, i.e. $\exists \delta > 0$, $w_0=Tv_0 \in W$ such that
\[
B_\delta(w_0) \subseteq \overline{T(B_n(0))}.
\]
By linearity
\begin{align*}
B_\delta(0) &= B_\delta(w_0) - Tv_0 \\
&\subseteq \overline{T(B_n(0))} - Tv_0 \\
&= \overline{T(B_n(-v_0))} \\
&\subseteq \overline{T(B_{n+m}(0))} \qquad (m\geq \|v_0\|)\\
&= (n+m)\overline{T(B_1(0))}  \\
B_r(0) & \subseteq \overline{T(B_1(0))} \qquad \left(r:=\frac{\delta}{n+m}\right).
\end{align*}
So $T$ is an open map by the ball criterion. If $T$ is bijective, then this further implies that $T^{-1}$ is continuous and bounded.
\end{proof}

Q: Closed Graph Theorem
A: \begin{thm}[Closed graph theorem]
Let $V, W$ be Banach spaces, and $T\in\mathcal{L}(V, W)$. The graph of $T$,
\[
\Gamma = \{(\mathbf{v}, T(\mathbf{v})): \mathbf{v}\in V\} \subseteq V\times W,
\]
is closed $\iff$ $T$ is bounded.
\end{thm}
\begin{proof}
($\Rightarrow$) $\Gamma$ is a closed subset of a complete space so is Banach (with a norm). The projections $\pi_V, \pi_W$ are continuous and $\pi_V:\Gamma\to V$ is a bijection. By the OMT, $\pi_V^{-1}\in\mathcal{B}(V,\Gamma)$. So $T=\pi_W\circ\pi_V^{-1}\in\mathcal{B}(V,W)$.
($\Leftarrow$) Let $(v_k, Tv_k)\to (v, w)$. Since $v_k\to v$ and $T$ is continuous, $w=Tv$. So $\Gamma$ is closed.
\end{proof}

Q: Normal space
A: A tsp $X$ is \emph{normal} if $\forall C_1, C_2\subseteq X$ disjoint closed,  $\exists U_1, U_2 \subseteq X$ disjoint open such that $C_1 \subseteq U_1, C_2 \subseteq U_2$.

Q: Compact sets are separable by open sets
A: Let $X$ be Hausdorff and $K_1,K_2$ compact disjoint. Then $\exists U_i\supseteq K_i$ open disjoint.
\begin{proof}
Fix $x\in K_1$. Let $U_{xy}, V_{xy}$ be disjoint open nbhds of $x,y$ for any $y\in K_2$.
Since $\bigcup_{y\in K_2} V_{xy}$ covers $K_2$, it has a finite subcover $V_x:=\bigcup_{i=1}^n V_{xy_i}$. Define $U_x := \bigcap_{i=1}^n U_{xy_i}$. Then $U_x, V_x$ are disjoint open and $x\in U_x, K_2\subseteq V_x$.
Since $\bigcup_{x\in K_1} U_x$ covers $K_1$, it has a finite subcover $U_1:=\bigcup_{i=1}^m U_{x_i}$. Let $U_2:=\bigcap_{i=1}^m V_{x_i}$. Then $U_i\supseteq K_i$ are open disjoint.
\end{proof}

Q: Normality by bisecting sets
A: Let $X$ be normal. If $A\subseteq U$, with $A$ closed and $U$ open, then $\exists V$ open and $B$ closed such that
\[
A \subseteq V \subseteq B \subseteq U.
\]
\begin{proof}
Let $A'=X\setminus U$. Then $A,A'$ are disjoint closed. By normality $\exists V, V'$ disjoint open such that $V\supseteq A, V'\supseteq A'$. Take $B=X\setminus V'$.
\end{proof}

Q: Urysohn's lemma
A: Let $X$ be normal and $C_0, C_1$ be disjoint closed subsets of $X$. Then there is a $f \in C(X)$ such that $f|_{C_0} = 0$ and $f|_{C_1} = 1$, and $0 \leq f(x) \leq 1$ for all $X$.
\begin{proof}
We start by defining $U_1 = X \setminus C_1 \supseteq C_0$. By normality, $\exists C_{\frac{1}{2}}$ and $U_{\frac{1}{2}}$ such that
\[
C_0 \subseteq U_{\frac{1}{2}} \subseteq C_{\frac{1}{2}} \subseteq U_1.
\]
Then we can find $C_{\frac{1}{4}}, C_{\frac{3}{4}}, U_{\frac{1}{4}}, U_{\frac{3}{4}}$ such that
\[
C_0 \subseteq U_{\frac{1}{4}}\subseteq C_{\frac{1}{4}} \subseteq U_{\frac{1}{2}} \subseteq C_{\frac{1}{2}} \subseteq U_{\frac{3}{4}} \subseteq C_{\frac{3}{4}} \subseteq U_1.
\]
Let $D$ be the dyadic rationals in $[0,1]$. Iterating this, we have that $\forall q< q' \in D$: $\exists U_q$ open, $C_q$ closed such that $U_q \subseteq C_q$, with $C_q \subseteq U_{q'}$.
We now define $f$ by
\[
f(x) = \inf\left\{q \in D: x \in U_q\right\},
\]
where $\inf \emptyset = 1$. Then clearly $f(X)\subseteq[0,1]$, $f|_{C_0}=0$, $f_{C_1}=1$. For continuity, observe that $f^{-1}(-\infty, t) =\bigcup_{q < t} U_q$ and $f^{-1}(t,\infty)= \bigcup_{q > t} X\setminus C_q$.
\end{proof}

Q: Normality of compact Hausdorff space
A: If $X$ is compact Hausdorff, then it is normal.
\begin{proof}
If $C_1,C_2\subseteq X$ are disjoint closed, then they are compact (since $X$ is compact). Therefore $C_1,C_2$ are compact subsets of a Hausdorff space which are separable by open sets.
\end{proof}

Q: Tietze-Urysohn extension theorem
A: Let $X$ be a normal space, $C\subseteq X$ closed and $f: C \to \R$ is continuous. Then $\exists\tilde{f}: X \to \R$ continuously extending $f$ and $\|\tilde{f}\|_\infty \leq \|f\|_\infty$.
\begin{proof}
Wlog $f(X)\subseteq [0,1]$ (otherwise compose with $\frac{1}{2}+\frac{1}{\pi}\arctan$).
We define $(f_i): C \to \R$ and $(g_i): X \to \R$ as follows. Let $f_0 = f$.
By Urysohn's lemma, $\exists 0\leq g_0 \leq \frac{1}{3}$ such that
\[
g_0(x) =
\begin{cases}
0 & x \in f_0^{-1}\left(\left[0, \frac{1}{3}\right]\right)\\
\frac{1}{3} & x \in f_0^{-1}\left(\left[\frac{2}{3}, 1\right]\right).
\end{cases}.
\]
We define $0 \leq f_1 \leq \frac{2}{3}$ by
\[
f_1 = f_0 - g_0|_C.
\]
Given $0 \leq f_i \leq \left(\frac{2}{3}\right)^i$, by Ursohn's Lemma $\exists 0 \leq g_i \leq \frac{1}{3}\left(\frac{2}{3}\right)^i$.
\[
g_i(x) =
\begin{cases}
0 & x \in f_i^{-1}\left(\left[0, \frac{1}{3}\left(\frac{2}{3}\right)^i\right]\right)\\
\frac{1}{3}\left(\frac{2}{3}\right)^i & x \in f_i^{-1}\left(\left[\left(\frac{2}{3}\right)^{i + 1}, \left(\frac{2}{3}\right)^i\right]\right)\\
\end{cases},
\]
We then define $0 \leq f_{i + 1} \leq \left(\frac{2}{3}\right)^{i + 1}$ by $f_{i + 1} = f_i - g_i|_C$.
By telescoping, on $C$ we have that
\[
\sum_{i = 0}^n g_i =f_0 - f_{n + 1} \to f_0.
\]
Then
\[
g = \sum_{i = 0}^\infty g_i \in C(X)
\]
converges absolutely uniformly. Moreover $\|g\|_\infty \leq \frac{1/3}{1-2/3}=1$.
Finally, we can ensure $\|\hat f\|_\infty \leq \|f\|_\infty$ by taking
\[
\hat f(x) = \begin{cases}
\tilde f(x) \quad & | g(x)| \leq \|f\|_\infty \\
\exp(i\arg  g(x)) \|f\|_\infty \quad & \text{otherwise}.
\end{cases}
\]
\end{proof}

Q: Equicontinuous
A: Let $K$ be a tsp and $F\subseteq C(K)$. We say $F$ is \emph{equicontinuous at $x \in K$} if
\[
\forall \varepsilon>0, \exists U \text{ open nbhd of $x$ },\forall f \in F, y \in U: \; |f(y) - f(x)| < \varepsilon.
\]

Q: Pre-compact
A: $C\subseteq X$ is pre-compact (or relatively compact) if $\bar{C}$ is compact.

Q: $\varepsilon$-net
A: Let $X$ be a msp and $E\subseteq X$. For $\varepsilon > 0$, we say that $N \subseteq X$ is an \emph{$\varepsilon$-net} for $E$ iff
\[
\bigcup_{x \in N}B_\varepsilon(x) \supseteq E.
\]

Q: Totally bounded subset
A: Let $X$ be a msp and $E\subseteq X$. We say that $E$ is \emph{totally bounded} if
\[
\forall \varepsilon>0, \exists \varepsilon\text{-net finite for }E.
\]
Note that this implies $E$ is bounded by the most distant centre plus the greatest radius of the $\varepsilon$-net.
The converse holds in finite dimensions as we can cover bounded sets with a closed ball which is compact so totally bounded.

Q: Arzel\'a-Ascoli Theorem
A: Let $K$ be a compact tsp. Then $\mathcal{F}\subseteq C(K)$ is pre-compact $\iff$ $\mathcal{F}$ is bounded and equicontinuous.
\begin{proof} Since $C(K)$ is complete, $\mathcal{F}$ is pre-compact $\iff$ $\mathcal{F}$ is totally bounded.
($\Rightarrow$) Given $\varepsilon > 0$, $\exists \{f_1, \cdots, f_n\} \subseteq \mathcal{F}$ a finite $\frac{\varepsilon}{3}$-net, i.e. $\forall f \in \mathcal{F}: \min_i \|f - f_i\| < \frac{\varepsilon}{3}$.
 Fix $x\in K$. By cty $\exists$ nbhds $(U_i)$ of $x$ such that $\sup_{y\in U_i}|f_i (y) - f_i(x)| < \frac{\varepsilon}{3}$. Then $U = \bigcap_{i = 1}^n U_i$ is open and $\sup_{y\in U, 1\leq i \leq n}|f_i (y) - f_i(x)| < \frac{\varepsilon}{3}$. Hence $F$ is equicontinuous at $x$, i.e. $\forall f \in \mathcal{F}, y \in U$:
  \[
    |f(y) - f(x)| \leq \min_i \left\{ |f(y) - f_i(y)| + |f_i(y) - f_i(x)| + |f_i(x) - f(x)| \right\} < \varepsilon.
  \]
  $(\Leftarrow)$ Given $\varepsilon > 0$, by equicty, $\forall x \in K$, $\exists U_x$ an open nbhd of $x$ such that $\sup_{y \in U_x, f \in \mathcal{F}}|f(y) - f(x)| < \frac{\varepsilon}{3}$. Since $K$ is compact $\exists (x_i)$ s.t. $\bigcup_{i = 1}^n U_{x_i}\supseteq K$.
  Since $\mathcal{F}$ is uniformly bounded, evaluating at the $(x_i)$ gives a (totally) bounded subset of $\ell^n_{\infty}$. Hence $\exists\{f_1, \cdots, f_m\}$ an $\frac{\varepsilon}{3}$-net, i.e. $\forall f \in \mathcal{F}$,  $\exists j\leq m$ such that
  \[
   \max_i |(f - f_j)(x_i)| < \frac{\varepsilon}{3}.
  \]
Hence $(f_i)$ are a finite $\varepsilon$-net for $\mathcal{F}$, i.e. $\forall f \in \mathcal{F}, \exists j \leq m$ such that
  \begin{align*}
    \|f - f_j\| &= \sup_y |(f - f_j)(y)|\\
    &= \max_i \sup_{y \in U_{x_i}}|(f - f_j)(y)|\\
    &\leq \max_i \sup_{y \in U_{x_i}} \big(|f(y) - f(x_i)| + |f(x_i) - f_j(x_i)| + |f_j(x_i) - f_j(y)|\big)\\
    &< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} +\frac{\varepsilon}{3}.
  \end{align*}
\end{proof}

Q: Totally bounded: subsequence criterion
A: A metric space $X$ is totally bounded $\iff$ every sequence has a Cauchy subsequence. In particular, if $X$ is complete then $X$ is totally bounded $\iff$ it is compact.
\begin{proof}
($\Rightarrow$) Suppose $X$ not compact. Then $\exists \varepsilon>0$ with no finite $\varepsilon$-net. By induction we can find $(x_n)$ such that $\forall m,n\in\N: d(x_m, x_n)\geq \varepsilon$, with no Cauchy subsequence.
($\Leftarrow$) Let $(x_n)\in X$. Enough to find a Cauchy subsequence.
Let $M_n$ be a finite $\frac{1}{n}$-net for $X$. Pick $m_1\in M_1$ such that $B_1(m_1)$ contains infinitely many $(x_n)$. Let $n_1$ be minimal such that $x_{n_1} \in B_1(m_1)$.
Inductively, given $m_1, \ldots m_k \in M_k$ such that $B_\frac{1}{k}(m_k)$ contains infinitely many of $(x_n)\cap \bigcap_{i< k} B_\frac{1}{i} (m_i) $ and $x_{n_\ell} \in \bigcap_{i\leq \ell} B_\frac{1}{i} (m_i)$, pick $m_{k+1}$ such that $B_\frac{1}{k+1}(m_{k+1})$ contains infinitely many points from $(x_n)\cap \bigcap_{j\leq k} B_\frac{1}{j} (m_j)$ and $n_{k+1}>n_k$ such that $x_{n_{k+1}} \in \bigcap_{j\leq k+1} B_\frac{1}{j}(m_j)$.
Then $(x_n)$ is Cauchy, i.e. $\forall\ell\geq k$:
\[
d(x_{n_k}, m_k) \leq d(x_{n_k}, m_k) + d(m_k, x_{n_l}) \leq \frac{1}{k} + \frac{1}{k} \to 0.
\]
\end{proof}

Q: Pre-compact vs totally bounded
A: Let $X$ be a complete msp. Then $E\subseteq X$ is totally bounded $\iff$ $E$ is pre-compact.
\begin{proof}
$\bar E$ is a closed subset of a complete space so is complete. Hence $E$ is totally bounded $\iff$ $\bar E$ is totally bounded $\iff$ $\bar E$ is compact.
\end{proof}

Q: Totally bounded closure
A: Let $X$ be a msp and $E\subseteq X$. Then $E$ is totally bounded $\iff$ $\bar E$ is totally bounded.
\begin{proof}
($\Rightarrow$) Given $\varepsilon>0$, we can find a finite $\frac{\varepsilon}{2}$-net $\{x_1, \cdots x_m\}\in E$ such that $E\subseteq \bigcup_i B_\frac{\varepsilon}{2}(x_i)$.
Given $c\in \bar E$, $\exists x\in E, i \in \N$ such that
\[
d(c,x)<\frac{\varepsilon}{2}, \quad d(x,x_i)<\frac{\varepsilon}{2} \implies d(z,x_i)<\varepsilon.
\]
Hence $\bar E\subseteq \bigcup_i B_\varepsilon(x_i)$.
($\Leftarrow$)  Given $\varepsilon>0$, we can find a finite $\frac{\varepsilon}{2}$-net $\{c_1, \cdots c_m\}\in \bar E$ such that $\bar E\subseteq \bigcup_i B_\frac{\varepsilon}{2}(c_i)$. For each $i$, pick $x_i\in B_\frac{\varepsilon}{2}(c_i)\cap E$. Then $\forall x\in B_\frac{\varepsilon}{2}(c_i)$:
\[
d(x, x_i) \leq d(x, c_i) + d(c_i, x) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.
\]
So $B_\frac{\varepsilon}{2}(c_i) \subseteq B_\varepsilon(x_i)$. Hence $(x_i)$ is a finite $\varepsilon$-net of $E$.
\end{proof}

Q: Algebra
A: A vsp $V$ is an algebra if there exists a bilinear map $V^2\to V$ called the product which is associative.
Equivalently, $(V, +, \ph)$ is a \emph{rng} (ie. ring not necessarily with multiplicative identity) where the product commutes with scalar multiplication.

Q: Normed algebra
A: If $V$ is in an algebra with a norm such that
\[
\|\mathbf{v}\cdot \mathbf{w}\|_V \leq \|\mathbf{v}\|_V \cdot \|\mathbf{w}\|_V
\]
for all $\mathbf{v}, \mathbf{w} \in V$, then we say $V$ is a \emph{normed algebra}.

Q: Banach algebra
A: If $V$ is a complete normed algebra, we say $V$ is a \emph{Banach algebra}.

Q: Commutative Algebra
A: If $V$ is an algebra that is commutative as a rng, then we say $V$ is a \emph{commutative algebra}.

Q: Unital
A: If $V$ is an algebra with multiplicative identity, then $V$ is a \emph{unital algebra}.

Q: Weierstrass Approximation Theorem
A: If $f\in C([0,1],\R)$, then $\exists p_n\in P_n$ such that $p_n\xrightarrow{\text{unif.}} f$.
\begin{proof}
Define the Bernstein polynomials
\[
b_{k,n}(x) = {n\choose k} x^k (1-x)^{n-k} \quad (0\leq k \leq n)
\]
and then pick
\[
p_n(x)= \sum_{k=0}^n f\left(\frac{k}{n}\right) b_{k,n}(x).
\]
Pick $\delta_m>0$ such that $|x-y| < \delta_m \implies | f(x) - f(y)| < \frac{1}{m}$.
Now fix $x\in [0,1]$ and let $X\sim B(n,x)$ such that $p_n(x)=\E[f(\tfrac{X}{n})]$. Let $A=\{|f(\bar X) - f(x)|\geq \frac{1}{m} \}\subseteq \{ |\bar X - x| \geq \delta_m\}$. By Chebyshev
\[
\P(A) \leq \P(|\bar X- x| \geq \delta_m) \leq \frac{x(1-x)}{n\delta_m^2} \leq \frac{1}{4n\delta_m^2}.
\]
Then
\begin{align*}
|p_n(x) - f(x) | &= \left| \E\left[f\left(\bar X\right)- f(x)\right]\right| \\
&\leq \E\left[\left|f\left(\bar X\right)- f(x)\right| | A \right]\P(A) + \frac{1}{m}\P(A^C) \\
&\leq  2\|f\|_\infty\P(A) + \frac{1}{m} \\
&< \frac{\|f\|_\infty}{2n\delta_m^2}+\frac{1}{m} \\
&\to 0 \text{ for $m,n\in\N$ large enough}.
\end{align*}
\end{proof}

Q: Polynomials approximate the modulus function
A: $\exists P_n:[-1,1]\to [0,1]$ polynomials such that $P_n\xrightarrow{\text{unif.}}|\ph|$.
\begin{proof}
If $q:[0,1]\to [0,1]$ satisfies
\[
q(t) = \tfrac{1}{2} (t+q(t)^2), \tag{$*$}
\]
then
\begin{align*}
(1-q(t))^2 &= 1-t \\
1-q(t) &= \sqrt{1-t} \\
|t| &= 1-q(1-t^2).
\end{align*}
Define $Q_n:[0,1]\to [0,1]$ by $Q_0(t)=0$ and
\[
Q_{n+1}(t)=\tfrac{1}{2}(t+Q_n(t)^2).
\]
By subtracting consecutive terms we find inductively
\begin{align*}
Q_{n+1}(t) - Q_n(t) &= \tfrac{1}{2} (Q_n(t)^2 - Q_{n-1}(t)^2) \tag{$\dagger$}\\
&= \tfrac{1}{2} (Q_n(t) - Q_{n-1}(t))(Q_n(t) + Q_{n-1}(t)) \\
& \geq 0 \text{ and $\nearrow$ in $t$}.
\end{align*}
Hence $Q_n(t) \nearrow q(t)$ and this is uniform since $\forall m > n$:
\begin{align*}
Q_{n+1}(t) - Q_n(t) &\leq Q_{n+1}(1) - Q_n(1) \\
Q_m(t) - Q_n(t) &\leq Q_m(1) - Q_n(1) \\
1-\sqrt{1-t} -Q_n(t) &\leq 1-Q_n(1)  \xrightarrow{\text{unif.}} 0.
\end{align*}
Then $P_n(t) := 1-Q_n(1-t^2) \xrightarrow{\text{unif.}} |t|$.
\end{proof}

Q: Lattice
A: A lattice is a poset $L$ with closure under joins ($u\vee v$) and meets ($u\wedge v$).

Q: Closed subalgebra is a lattice
A: Let $\mathcal{A}\subseteq C_\R(K)$ be a closed subalgebra. Then $\mathcal{A}$ is a lattice.
\begin{proof}
First note that
\begin{align*}
\max\{f(x), g(x)\} &= \frac{1}{2}(f(x) + g(x)) + \frac{1}{2}|f(x) - g(x)|,\\
\min\{f(x), g(x)\} &= \frac{1}{2}(f(x) + g(x)) - \frac{1}{2}|f(x) - g(x)|.
\end{align*}
So ets $\mathcal{A}$ is closed under taking $| \ph |$.
Given $f\in\mathcal{A}\setminus\{0\}$, wlog $\| f\|_\infty=1$ (else replace with $f / \|f\|_\infty$).
Pick $p_n\in P_n[-1,1]$ such that $p_n\xrightarrow{\text{unif}} | \ph |$. Then
\[
\|P_n\circ f - |f|\|_\infty \leq \|P_n - |\ph| \|_\infty \to 0.
\]
So $P_n\circ f \xrightarrow{\text{unif}} | f | \in \mathcal{A}$ by closure.
\end{proof}

Q: Two point sufficiency for lattice
A: \begin{lemma}
Let $\mathcal{L} \subseteq C_\R(K)$ be a sub-lattice.
Let $g \in C_\R(K)$ such that $\forall\varepsilon > 0, x, y \in K$, $\exists f_{x, y} \in \mathcal{L}$ such that
\[
|f_{x, y}(x) - g(x)| < \varepsilon,\quad |f_{x, y}(y) - g(y)| < \varepsilon. \tag{$*$}
\]
Then $g \in \bar{\mathcal{L}}$.
\end{lemma}
\begin{proof}
Since $f_{x, y}$ is continuous, $\exists V_{xy}$ an open nbhd of $y$ such that ($*$) holds.
By compactness $\exists\{ V_{x, y_i}: 1\leq i \leq n\}$ covering $K$.
Now let
\[
f_x = \min\{f_{x, y_1}, \cdots, f_{x, y_n}\} \in \mathcal{L}.
\]
By inheritance from some $f_{x,y_i}$, we have the UB $f_x< g + \varepsilon$.
By inheritance also $|f_x(x) - g(x)| < \varepsilon$. By continuity this holds for an open nbhd $U_x$ of $x$. By compactness, $\exists \{U_{x_1}, \cdots, U_{x_m}\}$ covering $K$.
Define
\[
f = \max\{f_{x_1}, \cdots, f_{x_m}\}  \in \mathcal{L} .
\]
By inheritance from some $f_{x_i}$, we have the LB $ > g - \varepsilon$.
Moreover the original UB still holds so
\[
\|f - g\|_\infty < \varepsilon.
\]
\end{proof}

Q: Stone-Weierstrass theorem
A: Let $\mathcal{A} \subseteq C_{\R}(K)$ be a subalgebra which separates points and vanishes nowhere. Then $\bar{\mathcal{A}} = C_\R(K)$.
\begin{proof}
Given $g \in C_{\R}(K)$, enough to apx $g$ at two points $x \not= y \in K$.
Claim: $\exists h \in \mathcal{A}$ such that $h(x), h(y) \not= 0$, and $h(x) \not= h(y)$.
By our assumptions, we can find
\begin{enumerate}
\item $h_{x, y}$: $h_{x, y}(x) \not= h_{x, y}(y)$
\item $h_{x}$: $h_{x}(x) \not= 0$.
\item $h_{y}$: $h_{y}(y) \not= 0$.
\end{enumerate}
If $h_{xy}(x)\neq 0 \neq h_{xy}(y)$ then take $h=h_{xy}$. Otherwise wlog
\begin{align*}
h_{xy}(y)&=0 \\
h_{xy}(x)&=1 \\
h_y(y)&=1 \\
h_y(x)&=C
\end{align*}
so take $h = h_y + (2-C)h_{xy}$.
Since $(h(x), h(y))$ and $(h(x)^2, h(y)^2)$ are LI in $\R^2$, $\exists \alpha, \beta \in \R$ such that
\[
\alpha(h(x), h(y)) + \beta(h(x)^2, h(y)^2) = (g(x), g(y)).
\]
So $f,g$ intersect at $x$ and $y$ where $f=\alpha h + \beta h ^2 \in \mathcal{A}$.
\end{proof}

Q: Separable space
A: A vector space $V$ is separable if $\exists W\subseteq V$ countable and dense.

Q: Separability of $C(K, \R)$
A: Let $U\subseteq\R^n$ be bounded open domain. Then $C(\bar U)$ is separable.
\begin{proof}
By WAT, the countable space of polynomials over $\Q$ is dense in polynomials over $\R$ is dense in $C(\bar U)$.
\end{proof}

Q: Complex version of Stone-Weierstrass theorem
A: Let $\mathcal{A} \subseteq C(K, \C)$ be a subalgebra which separates points, vanishes nowhere and is closed under complex conjugation. Then $\bar{\mathcal{A}} = C(K, \C)$. In particular, this holds for $C^*$ subalgebras which separate points.
\begin{proof}
Consider
\[
\mathcal{A}_\R = \left\{\frac{f + \bar{f}}{2}: f \in \mathcal{A}\right\} \cup \left\{\frac{f - \bar{f}}{2i}: f \in \mathcal{A}\right\} \subseteq \mathcal{A}.
\]
By S-W/$\R$, $\bar{\mathcal{A}_\R} = C(K, \R)$. So $\mathcal{A}_\R + i \mathcal{A}_\R = \mathcal{A}$ is dense in $C(K, \C) = C(K, \R) + i C(K, \R)$.
\end{proof}

Q: $C^*$-algebra
A: A complex unital Banach algebra with antilinear involution $*$ such that
\[
(ab)^*=b^*a^*, 1^*=1, \|a^*\|=\|a\|.
\]

Q: Convergence of Fourier Series
A: For $f \in C(S^1)$: $\| f - S_N f \|_2 \to 0$ as $N\to\infty$.
\begin{proof}
Write $\mathcal{A}$ for the trigonometric polynomials. Given $\varepsilon>0$, by S-W/$\C$, $\exists P\in\mathcal{A}$ such that $\|f-P\|_\infty < \varepsilon$.
For $n>\deg P$:
\begin{align*}
|f-S_n f |^2 &\leq 2|f-P |^2 + 2|S_n P - S_n f |^2 \qquad (S_n P = P) \\
\int_{-\pi}^\pi |f-S_n f |^2 &\leq 4 \int_{-\pi}^\pi |f-P|^2 \qquad \text{(Bessel's inequality)} \\
&\leq 8 \pi \varepsilon^2.
\end{align*}
\end{proof}

Q: Uniqueness of inner product
A: \begin{prop}
Let $(E, \|\ph\|)$ be a Euclidean space. Then there is a \emph{unique} inner product $\bra \ph,\ph\ket$ such that $\|\mathbf{v}\| = \sqrt{\bra \mathbf{v}, \mathbf{v}\ket}$.
\end{prop}
\begin{proof}
Follows from polarisation identities. In the real case
\begin{align*}
\bra \mathbf{v} + \mathbf{w}, \mathbf{v} + \mathbf{w}\ket &= \|\mathbf{v}\|^2 + 2\bra \mathbf{v}, \mathbf{w}\ket + \|\mathbf{w}\|^2 \\
\bra \mathbf{v}, \mathbf{w}\ket &= \frac{1}{2} (\|\mathbf{v} + \mathbf{w}\|^2 - \|\mathbf{v}\|^2 - \|\mathbf{w}\|^2).
\end{align*}
In the complex case
\begin{align*}
\bra \mathbf{v} + \mathbf{w}, \mathbf{v} + \mathbf{w}\ket &= \|\mathbf{v}\|^2 + \|\mathbf{w}\|^2 + \bra \mathbf{v}, \mathbf{w}\ket + \bra \mathbf{w}, \mathbf{v}\ket\\
\bra \mathbf{v} - \mathbf{w}, \mathbf{v} - \mathbf{w}\ket &= \|\mathbf{v}\|^2 + \|\mathbf{w}\|^2 - \bra \mathbf{v}, \mathbf{w}\ket - \bra \mathbf{w}, \mathbf{v}\ket\\
\bra \mathbf{v} +i\mathbf{w}, \mathbf{v} +i\mathbf{w}\ket &= \|\mathbf{v}\|^2 + \|\mathbf{w}\|^2 -i\bra \mathbf{v}, \mathbf{w}\ket +i\bra \mathbf{w}, \mathbf{v}\ket\\
\bra \mathbf{v} -i\mathbf{w}, \mathbf{v} -i\mathbf{w}\ket &= \|\mathbf{v}\|^2 + \|\mathbf{w}\|^2 +i\bra \mathbf{v}, \mathbf{w}\ket -i\bra \mathbf{w},\mathbf{v}\ket \\
4\bra \mathbf{v}, \mathbf{w}\ket &= \|\mathbf{v} + \mathbf{w}\|^2 - \|\mathbf{v} - \mathbf{w}\|^2 + i\|\mathbf{v} + i\mathbf{w}\|^2 - i\|\mathbf{v} - i\mathbf{w}\|^2 .
\end{align*}
\end{proof}

Q: Completion of a metric space
A: Given a metric space $(X,d)$, let $\bar X$ be the Cauchy sequences of $X$ with equivalence $(x_n)\sim (y_n)$ if $d(x_n,y_n)\to 0$ as $n\to\infty$.
The \emph{metric} is $\bar d(\bar x, \bar y) = \lim_{n\to\infty} d(x_n,y_n)$.
There is a natural isometric (dense) \emph{embedding} $X\subseteq\bar X$ by constant (Cauchy) sequences.
$\bar X$ is \emph{complete} by diagonalisation.

Q: Orthogonal complement of span
A: Let $E$ be a Esp and $S\subseteq E$. Then $S^\perp\leq E$ is closed and
\[
S^{\perp} = (\overline{\spn S})^{\perp}.
\]
If $E$ is a Hsp, then $\overline{\spn S}$ is complete so
\[
\overline{\spn S} = ((\overline{\spn S})^{\perp})^\perp = (S^\perp)^\perp.
\]
\begin{proof}
Immediately closed under LCs. Moreover $S^\perp = \bigcap_{w\in S} f_w^{-1}(0)$ is closed since $f_w(v)=(v,w)$ is continuous. One direction is easy:
\[
\begin{alignedat}{2}\overline{\spn S} &\supseteq \spn S &&\supseteq S \\
S^\perp &\supseteq (\spn S)^\perp &&\supseteq (\overline{\spn S})^\perp.
\end{alignedat}
\]
For the converse, let $v\in S^\perp, w=\lim_{k\to\infty} w_k \in \overline{\spn S}, w_k \in \spn S$. Then
\[
(v,w) = \lim_{k\to\infty} (v,w_k)=0.
\]
So $v\in(\overline{\spn S})^\perp$.
\end{proof}

Q: Parallelogram law
A: Let $(E, \|\ph\|)$ be a Esp. Then $\forall \mathbf{v}, \mathbf{w} \in E$:
\[
\|\mathbf{v} - \mathbf{w}\|^2 + \|\mathbf{v} + \mathbf{w}\|^2 = 2\|\mathbf{v}\|^2 + 2\|\mathbf{w}\|^2,
\]
i.e. the sum of the squares of the diagonals is the sum of squares of the sides.
\begin{proof}
We compute
\begin{align*}
\|\mathbf{v} - \mathbf{w}\|^2 + \|\mathbf{v} + \mathbf{w}\|^2 &= \bra \mathbf{v} - \mathbf{w}, \mathbf{v} - \mathbf{w}\ket + \bra \mathbf{v} + \mathbf{w}, \mathbf{v} + \mathbf{w}\ket\\
&= \bra \mathbf{v}, \mathbf{v}\ket - \bra \mathbf{v}, \mathbf{w}\ket - \bra \mathbf{w}, \mathbf{v}\ket + \bra \mathbf{w}, \mathbf{w}\ket\\
&+ \bra \mathbf{v}, \mathbf{v}\ket + \bra \mathbf{v}, \mathbf{w}\ket + \bra \mathbf{w}, \mathbf{v}\ket + \bra \mathbf{w}, \mathbf{w}\ket\\
&= 2\bra \mathbf{v}, \mathbf{v}\ket + 2 \bra \mathbf{w}, \mathbf{w}\ket.
\end{align*}
\end{proof}

Q: Orthogonal projection in a Hilbert space
A: Let $(V, \|\ph\|)$ be a Hsp and $W\subseteq V$ closed convex non-empty. Then $\exists ! P: V\to W$ such that
\[
\|v-P(v) \| = \inf_{w\in W} \|v-w\|.
\]
Moreover if $W\leq V$, then $W \oplus W^\perp = V$ and $P$ is linear idempotent self-adjoint.
\begin{proof}
Let $v\in V$ and $(w_i)\subseteq W$ such that
\[
\lim_{i \to \infty} \|w_i - v\| = \inf_{w \in W} \|w - v\| = d.
\]
By the parallelogram law $\forall i,j \geq N$:
\begin{align*}
\|w_j - w_i\|^2 + \|2v - w_i - w_j\|^2 &= 2\|w_i - v\|^2 + 2\|w_j - v\|^2 \\
\|w_i - w_j\|^2 &= \underbrace{2\|w_i - v\|^2 + 2\|w_j - v\|^2}_{\to 4d^2} - \underbrace{4\left\|v - \frac{w_i - w_j}{2}\right\|^2}_{\leq -4d^2}\\
&\to 0 \text{ as } N \to\infty.
\end{align*}
By completeness, $w_i \to v_\parallel \in W$ where $ d = \|w- v_\parallel\|$.
Now let $v_\perp = v - v_\parallel$. Suppose $v_\perp\not\in W^\perp$. Then wlog $\exists\tilde{w} \in W$ such that $\bra \tilde{w}, v_\perp \ket > 0$. Then $\forall t > 0$:
\begin{align*}
\|(v_\parallel + t\tilde{w}) - v \|^2 &= \|v_\perp - t\tilde w \|^2 \\
&= d^2 - 2t \bra \tilde{w}, v_\perp\ket + t^2 \|\tilde{w}\|^2 \\
&<d^2
\end{align*}
for sufficiently small $t$, contradicting minimality. Finally $\forall v,w \in V$:
\[
\bra Pv, w \ket = \bra v_\parallel, w_\parallel \ket = \bra v, Pw \ket.
\]
\end{proof}

Q: Riesz representation for Hilbert spaces
A: Let $H$ be a Hsp. Then $\phi: H\to H^*$ given by $\mathbf{v} \mapsto \bra \ph, \mathbf{v}\ket$ is an isometric anti-isomorphism.
\begin{proof}
\textbf{Injectivity}. If $\phi_{\mathbf{v}} = \phi_{\mathbf{u}}$, then $\forall\mathbf{w}$: $\bra \mathbf{w}, \mathbf{v}\ket = \bra \mathbf{w}, \mathbf{u}\ket$ so $\bra \mathbf{w}, \mathbf{v} - \mathbf{u}\ket = 0$. In particular, $\| \mathbf{v} - \mathbf{u}\|^2 = 0$.
\textbf{Anti-isomorphism}. Immediate from antilinearity in the second argument.
\textbf{Isometric}. Let $\mathbf{v}, \mathbf{w} \in H$ and $\|\mathbf{w}\| = 1$. Then by Cauchy-Schwarz
\[
|\phi_\mathbf{v}(\mathbf{w})| = |\bra \mathbf{w}, \mathbf{v}\ket| \leq \|\mathbf{w}\| \|\mathbf{v}\| = \|\mathbf{v}\|
\]
with equality if $\mathbf{v}=\mathbf{w}$. So $\| \phi_\mathbf{v} \| = \| \mathbf{v} \|$.
\textbf{Surjectivity}. Let $\xi \in H^*$. If $\xi = 0$, then $\xi = \phi_{\mathbf{0}}$. Otherwise, $\ker \xi$ is closed so $H = (\ker \xi) \oplus (\ker \xi)^\perp$. Pick $\mathbf{v}_0\in(\ker\xi)^\perp$ with $\|\mathbf{v}_0\|=1$. Then we can decompose any $\mathbf{w}\in W$ as
\[
\mathbf{w} = \frac{\xi(\mathbf{w})}{\xi(\mathbf{v}_0)}\mathbf{v}_0 + \left(\mathbf{w} - \frac{\xi(\mathbf{w})}{\xi(\mathbf{v}_0)}\mathbf{v}_0\right).
\]
So $(\ker \xi)^\perp = \spn \{ \mathbf{v}_0\}$. Now define $\mathbf{v}_\xi = \overline{\xi(\mathbf{v}_0)}\mathbf{v}_0$.
If $\mathbf{v}\in\ker\xi$, then
\[
\bra \mathbf{v}, \mathbf{v}_\xi \ket = \xi(\mathbf{v}_0) \bra \mathbf{v}, \mathbf{v}_0 \ket = 0 = \xi(\mathbf{v}).
\]
If $\mathbf{v}= \lambda \mathbf{v}_0\in\ker(\xi)^\perp $, then
\[
\bra \mathbf{v}, \mathbf{v}_\xi \ket = \lambda \xi(\mathbf{v}_0) \bra \mathbf{v}_0, \mathbf{v}_0 \ket =  \lambda \xi(\mathbf{v}_0) = \xi(\mathbf{v}).
\]
By linearity, $\xi = \phi_{\mathbf{v}_\xi}$.
\end{proof}

Q: Projection map
A: Let $V$ be a Esp. A linear operator $P:V\to V$ is a projection if $P^2=P$. If $P$ is also self-adjoint, then it is an orthogonal projection.

Q: Norm of projections
A: Let $P:V\to V$ be an orthogonal projection. Then $\|P\|=0$ or $\|P\|=1$.
\begin{proof}
If $P=0$, then $\|P\|=0$. Otherwise $\exists v\not\in\ker P$. By Cauchy-Schwarz
\[
\|Pv\|= \frac{\|Pv\|^2}{\|Pv\|} =  \frac{\bra Pv, Pv \ket}{\|Pv\|} =  \frac{\bra v, Pv \ket}{\|Pv\|} \leq \|v \|
\]
with equality iff $v\in\im P$.
\end{proof}

Q: Fourier sums are projections
A: For $f \in C(S^1)$, define for each $k \in \Z$,
\[
\hat{f}_k = \frac{1}{2\pi} \int_{-\pi}^\pi e^{-ikx}f(x)\;\d x.
\]
Then $S_N: C(S^1) \to C(S^1)$ given by
\[
S_N(f)(x) = \sum_{k = -N}^N e^{ikx} \hat{f}_k
\]
is the orthogonal projection onto $W=\im S_N = \spn\{e^{ikx}: |k|\leq N\}$.
\begin{proof}
If $f \in W$, say $f = \sum_{k = -N}^N a_k  e^{ikx}$, then by orthogonality
\[
S_N (f) = \sum_{k = -N}^N \bra f, e^{ikx}\ket e^{ikx} = \sum_{k= -N}^N a_k e^{ikx} = f.
\]
If $f \in W^\perp$, then $\bra e^{ikx}, f \ket =0$ for $|k| < N$. So $S_N(f) = 0$.
Finally $\forall f,g \in C(S^1)$:
\begin{align*}
\bra S_N f, g\ket &= \left\bra \sum_{k=-N}^N \bra f, e^{ikx} \ket e^{ikx}, g \right\ket\\
&= \sum_{k=-N}^N \bra f, e^{ikx} \ket \overline{\bra g, e^{ikx} \ket} \\
&= \left\bra f,\sum_{k=-N}^N  e^{ikx} \bra g, e^{ikx} \ket \right\ket \\
&=  \bra  f, S_N g\ket.
\end{align*}
\end{proof}

Q: Hilbert basis
A: A maximal ONS in a Hsp is called a complete ONS or Hilbert basis.

Q: Closure of Hilbert basis
A: Let $H$ be a Hsp and $S$ an ONS. Then $S$ is a Hbs iff
\[
\overline{\spn S} = H.
\]
\begin{proof}
Write $W=\overline{\spn S}$. Then $H=W\oplus W^\perp$. Hence
\[
W \neq H \iff W^\perp=S^\perp \neq \{ 0 \} \iff S \text{ is not maximal}.
\]
\end{proof}

Q: Basis of a separable space
A: Let $H$ be a separable Hsp, i.e. $\exists (\mathbf{y}_i)_{i \in \N}$ such that
\[
\overline{\spn \{\mathbf{y}_i\}} = H.
\]
Then $H$ has a countable Hbs.
\begin{proof}
Reduce $(\mathbf{y}_i)_{i \in \N}$ a LI set and apply Gram-Schmidt.
\end{proof}

Q: Bessel's inequality
A: Let $E$ be Esp and $S=(\mathbf{e}_i)$ an ONS. Then
\[
\sum_i |\bra v, e_i \ket|^2 \leq \|v\|^2.
\]
\begin{proof}
Wlog $|S|=n<\infty$ (else take $n\to\infty$). Let $P$ be the orthogonal projection onto $\spn S$, i.e.
\[
Pv = \sum_{i=1}^n \bra v, e_i \ket e_i.
\]
Then by Pythagoras
\[
\sum_{i=1}^n |\bra v, e_i \ket|^2 = \|Pv\|^2 \leq \|v\|^2.
\]
\end{proof}

Q: Riesz-Fischer Theorem for $\ell^2$
A: Let $H$ be a separable Hsp with basis $(\mathbf{e}_i)$.
\begin{enumerate}
\item If $\mathbf{x}\in H$, then $x_i=\bra \mathbf{x}, \mathbf{e}_i \ket\in\ell^2$ and
\[
\mathbf{x} = \sum_i  x_i\mathbf{e}_i.
\]
\item If $(x_i)\in \ell^2$, then $\mathbf{x}=\sum_i  x_i\mathbf{e}_i \in H $ and $x_i=\bra \mathbf{x}, \mathbf{e}_i \ket$.
\item If $\mathbf{x}, \mathbf{y} \in H$, then
\[
\bra \mathbf{x}, \mathbf{y}\ket = \sum_i x_i \bar{y}_i
\]
where the sum converges absolutely.
\end{enumerate}
In particular $\phi: H\to\ell^2$, $\mathbf{x}\mapsto (\bra \mathbf{x}, \mathbf{e}_i \ket)$ is an isometric isomorphism.
\begin{proof}
By Bessel's inequality, $\mathbf{x}\in H \implies (\bra \mathbf{x},\mathbf{e}_i \ket )\in\ell^2$.
Conversely, given $(x_i)\in\ell^2$, define the partial sums
\[
\mathbf{s}_n = \sum_{i = 1}^n x_i \mathbf{e}_i.
\]
Then $\forall n>m$
\[
\|\mathbf{s}_n - \mathbf{s}_m\| = \sum_{i = m + 1}^n |x_i|^2 \leq \sum_{i = m + 1}^\infty |x_i|^2 \to 0
\]
as $m \to \infty$. By completeness $ \mathbf{s}_n\to  \mathbf{s}\in H$.
If $\mathbf{s} = \sum_i  \bra \mathbf{x}, \mathbf{e}_i \ket \mathbf{e}_i$, then for each $i$
\[
\bra \mathbf{x} - \mathbf{s}, \mathbf{e}_i\ket = \lim_{n\to\infty} \bra \mathbf{x} - \mathbf{s}_n, \mathbf{e}_i\ket = \bra \mathbf{x}, \mathbf{e}_i \ket - \bra \mathbf{x}, \mathbf{e}_i \ket=0.
\]
So $\mathbf{x} - \mathbf{s}\in \spn\{ \mathbf{e}_i \}^\perp \subseteq \overline{\spn\{ \mathbf{e}_i \}}^\perp = H^\perp=\{0\}$.
Continuity of the inner product gives Parseval's identity. Cauchy-Schwarz gives absolute convergence.
\end{proof}

Q: Spectrum and resolvent
A: Let $X$ be a Bsp and $T \in \mathcal{B}(X)$. The \emph{spectrum} of $T$ is
\[
\sigma(T) = \{\lambda \in \C: T - \lambda I\text{ is not invertible}\}
\]
and the \emph{resolvent set} is
\[
\rho(T) = \C \setminus \sigma(T) = \{ z\in\C: T-z \text{ bijective and } (T-z)^{-1}\in\mathcal{B}(X)\}.
\]
The \emph{resolvent} is the map $R: \rho(T) \to \mathcal{B}(X)$ given by
\[
\lambda \mapsto (T - \lambda I)^{-1}.
\]

Q: Geometric series of linear maps
A: Let $X$ be a Bsp, $T \in \mathcal{B}(X)$ and $\|T\|_{\mathcal{B}(X)} < 1$. Then $I - T$ is invertible.
\begin{proof}
$\sum_{i = 0}^\infty T^i$ is Cauchy so converges. Moreover by cty
\[
(I-T)\sum_{i = 0}^\infty T^i = \lim_{n\to\infty} [I-T^{n+1}]  = I.
\]
\end{proof}

Q: Openness of invertible linear maps
A: Let $X$ be a Bsp, $S_1 \in \mathcal{B}(X)$ be invertible. Then $\forall S_2 \in \mathcal{B}(X)$ in the disc
\[
\|S_1^{-1}\| \|S_1 - S_2\| < 1,
\]
$S_2$ is invertible.
\begin{proof}
We can write
\[
S_2 = S_1(I - S_1^{-1}(S_1 - S_2))
\]
where
\[
\|S_1^{-1}(S_1 - S_2)\| \leq \|S_1^{-1} \| \|S_1 - S_2\| < 1
\]
so
\[
S_2^{-1} = (I - S_1^{-1}(S_1 - S_2))^{-1} S_1^{-1}.
\]
\end{proof}

Q: Resolvent is open
A: Let $z_0\in\rho(T)$. Then $\rho(T)\supseteq D$ where
\[
D = \{ z\in\C: |z-z_0| |R_T(z_0)| < 1\}.
\]
In particular, $\rho(T)$ is open and $\sigma(T)$ is closed. Moreover, $R_T$ is analytic.
\begin{proof}
For any $z\in D$
\begin{align*}
T-z &= T-z_0 - (z-z_0) \\
&= (T-z_0) (\id - R_T(z_0)(z-z_0) ) \\
(\id - R_T(z_0)(z-z_0) ) ^{-1} &= \sum_{n=0}^\infty (z-z_0)^n R_T(z_0)^n \\
R_T(z) &= \sum_{n=0}^\infty (z-z_0)^n R_T(z_0)^{n+1}.
\end{align*}
\end{proof}

Q: Bounds on spectrum
A: $\sigma(T)\neq \emptyset$ and $\sigma(T)\subseteq D=\{ z\in\C: |z| \leq \|T\|\}$.
\begin{proof}
For any $|z|>\|T\|$
\begin{align*}
R_T(z) &= -\frac{1}{z} (1-\tfrac{T}{z})^{-1} \\
&= -\frac{1}{z}  \sum_{n=0}^\infty \frac{T^n}{z^n}.
\end{align*}
So $D^c\subseteq \rho(T)$, $\sigma(T)\subseteq D$ and $R_T(z)\to 0$ as $|z|\to\infty$.
If $\sigma(T)=\emptyset$, then $R_T$ is entire. By Liouville, $R_T=0$. But $-zR_T(z)\to \id$ as $|z|\to\infty$. Contradiction.
\end{proof}

Q: Liouville's theorem for Banach spaces
A: Let $X$ be a Bsp and $F: \C \to X$ be entire (i.e. $F$ is given by a locally absolutely convergent PS) and norm bounded, ie.
\[
\sup_{z \in \C} \|F(z)\| < \infty.
\]
Then $F$ is constant.
\begin{proof}
Given $f \in X^*$
\[
\sup_{z \in \C} |f\circ F(z)| \leq \sup_{ z \in \C} \|f\| \|F(z)\| < \infty.
\]
Hence $f\circ F:\C\to \C$ is bounded and entire.
By Liouville $(f \circ F)(z) = (f\circ F)(0)$. In particular $f(F(z) - F(0)) = 0$ for any $f\in X^*$.
By Hahn-Banach $F(z) - F(0) = 0$ for all $z \in \C$.
\end{proof}

Q: Invertibility on Banach spaces
A: Let $X$ be a Bsp, $Y$ a nvsp, $T\in\mathcal{B}(X,Y)$. Then $T^{-1}\in\mathcal{B}(Y,X)$ $\iff$ $\im T$ is dense in $Y$ and $T$ is bounded below.
\begin{proof}
($\Rightarrow$) Immediate.
($\Leftarrow$) $T$ is injective (via kernels). So $T:X\to\im T$ is a bijection with inverse $S$. Since $X$ is complete, we can extend $S$ to $Y$ by BLT.
\end{proof}

Q: Point spectrum
A: Let $X$ be a Bsp. The \emph{point spectrum} is
\[
\sigma_p(T) = \{\lambda\in \C: T-\lambda \text{ not injective} \} =\{\lambda \in \C: \lambda\text{ is an eigenvalue of }T\}.
\]

Q: Continuous spectrum
A: The continuous spectrum is
\[
\sigma_c(T) = \{ \lambda\in\sigma(T): T-\lambda \text{ injective and } \im(T-\lambda) \text{ is dense}\}.
\]
In particular, $T-\lambda$ is not bounded below.

Q: Residual spectrum
A: The residual spectrum is
\[
\sigma_r(T) = \{ \lambda\in \sigma(T): T-\lambda \text{ injective and } \im(T-\lambda) \text{ not dense}\}.
\]

Q: Approximate point spectrum
A: Let $X$ be a Bsp. The \emph{approximate point spectrum} is defined as
\[
\sigma_{ap}(T) = \{ \lambda \in \C: \exists \{\mathbf{x}_n\} \subseteq X: \|\mathbf{x}_n\|_X = 1\text{ and }\|(T - \lambda I)\mathbf{x}_n\|_X \to 0\}.
\]

Q: Comparison of spectra types
A: We have the inclusions
\[
\sigma_p(T) \subseteq \sigma_{ap}(T) \subseteq \sigma(T).
\]

Q: Adjoint vs dual
A: Given $T\in\mathcal{B}(H)$, by Riesz $T^*(y)$ is the representation of
\[
T^*(y)(x) = (\bra\ph,y\ket \circ T)(x) = \bra T(x), y\ket = \bra x, T^\dagger (y) \ket.
\]
So $T^*=T^\dagger$.

Q: Compact operator
A: Let $X, Y$ be Bsp. We say $T \in \mathcal{L}(X, Y)$ is \emph{compact} if $\forall E \subseteq X$ bounded, $T(E)$ is totally bounded (or equivalently pre-compact).
Write $\mathcal{B}_0(X)=\{ T \in \mathcal{B}(X): T \text{ compact}\}$.

Q: Eigenvalues of self-adjoint operators
A: \begin{prop}
Let $H$ be a Hsp and $T \in \mathcal{B}(H)$ self-adjoint. Then $\sigma_p(T) \subseteq \R$.
\end{prop}
\begin{proof}
Let $\lambda \in \sigma_p(T)$ and $\mathbf{v} \in E_\lambda$. Then
\[
\lambda = \frac{\bra T \mathbf{v}, \mathbf{v}\ket}{\|\mathbf{v}\|^2} = \frac{\bra \mathbf{v}, T \mathbf{v}\ket}{\|\mathbf{v}\|^2} = \bar{\lambda} \in\R.
\]
\end{proof}

Q: Eigenvectors of self-adjoint operators
A: \begin{prop}
Let $H$ be a Hsp and $T \in \mathcal{B}(H)$ self-adjoint. If $\lambda\neq \mu \in \sigma_p(T)$, then $E_\lambda \not \perp E_\mu$.
\end{prop}
\begin{proof}
Let $\mathbf{v} \in E_\lambda$ and $\mathbf{w} \in E_\mu$. Then
\[
\lambda \bra \mathbf{v}, \mathbf{w}\ket = \bra T\mathbf{v}, \mathbf{w}\ket = \bra \mathbf{v}, T \mathbf{w}\ket  = \mu \bra \mathbf{v}, \mathbf{w}\ket.
\]
So $\bra \mathbf{v}, \mathbf{w}\ket = 0$.
\end{proof}

Q: Norm of self-adjoint operator from inner product
A: Let $H$ be a Hsp and $T \in \mathcal{B}(H)$ self-adjoint. Then
\[
\|T\| =  \sup_{\|\mathbf{x}\| = 1} |\bra \mathbf{x}, T \mathbf{x}\ket|.
\]
\begin{proof}
($\geq$) Immediate.
($\leq$) By Riesz representation
\[
\|T\| = \sup_{\|\mathbf{x}\| = 1} \sup_{\|\mathbf{y}\| = 1} |\bra \mathbf{y}, T \mathbf{x}\ket|.
\]
Take $\mathbf{x}, \mathbf{y} \in H$ such that $\|\mathbf{x}\| = \|\mathbf{y}\| = 1$. Wlog $\bra T\mathbf{x}, \mathbf{y}\ket\in\R$ (else multiply $\mathbf{y}$ by a scalar). Then we have the polarisation identity
\begin{align*}
|\bra T \mathbf{x}, \mathbf{y}\ket| &= \frac{1}{4}\left|\bra T(\mathbf{x} + \mathbf{y}), \mathbf{x} + \mathbf{y} \ket - \bra T(\mathbf{x} - \mathbf{y}), \mathbf{x} - \mathbf{y}\ket \right|\\
&\leq \frac{1}{4} \sup_{\|\mathbf{u}\| = 1} |\bra \mathbf{u}, T \mathbf{u}\ket|( \|\mathbf{x} + \mathbf{y}\|^2 + \|\mathbf{x} - \mathbf{y}\|^2)\\
&= \frac{1}{4} \sup_{\|\mathbf{u}\| = 1} |\bra \mathbf{u}, T \mathbf{u}\ket|(2\|\mathbf{x}\|^2 + 2 \|\mathbf{y}\|^2)\\
&= \sup_{\|\mathbf{u}\| = 1} |\bra \mathbf{u}, T \mathbf{u}\ket|.
\end{align*}
\end{proof}

Q: Approximate eigenvalues of self-adjoint operators
A: Let $H$ be a Hsp and $T \in \mathcal{B}(H)$ self-adjoint. Then $\|T\|$ or $-\|T\|$ $\in \sigma_{ap}(T)$.
\begin{proof}
Let $\|\mathbf{x}_n\| = 1$ such that $\bra \mathbf{x}_n, T \mathbf{x}_n\ket \to \lambda \in \{\pm \|T\|\}$. Then
\begin{align*}
0 &\leq \|T \mathbf{x}_n - \lambda \mathbf{x}_n \|^2\\
&= \underbrace{\|T \mathbf{x}_n\|^2}_{\leq\lambda^2} - 2\lambda \underbrace{\bra T \mathbf{x}_n, \mathbf{x}_n\ket}_{\to\lambda} + \lambda^2 \|\mathbf{x}_n\|^2\\
&\to 0.
\end{align*}
\end{proof}

Q: Eigenspaces of adjoint
A: Let $T$ be normal. If $Tx=\lambda x$, then $T^*x=\lambda^*x$. That is, $\sigma_p(T)^* = \sigma_p(T^*)$ and $\sigma_{ap}(T)^*=\sigma_{ap}(T^*)$. In particular, $T$ self-adjoint $\implies$ $\sigma(T)\subseteq\R$ and $T$ unitary $\implies$ $\sigma(T)\subseteq S^1$.
\begin{proof}
$T$ normal $\implies$ $T-\lambda$ is normal and $(T-\lambda)^* = T^*-\lambda^*$.
\end{proof}

Q: Eigenvalues of compact operators
A: Let $T$ be compact and $\lambda\neq 0\in\sigma_{ap}(T)$. Then $\lambda\in\sigma_{p}(T)$.
\begin{proof}
Let $(T-\lambda)x_n \to 0$ with $\|x_n\|=1$. By compactness, wlog $(Tx_n)\to y$ as $n\to\infty$. Then $y=\lim_{n\to\infty} \lambda x_n\neq 0$ and
\[
Ty = T\left(\lim_{n \to \infty} \lambda x_n\right) =\lambda \lim_{n \to \infty} T( x_n)= \lambda y.
\]
\end{proof}

Q: Self-adjoint map on $E^\perp$
A: Let $T\in\mathcal{B}(H)$ be self-adjoint with $(\lambda_i)_{i\in I}$ non-zero eigenvalues and
\[
E= \bigoplus_{i\in I} E_{\lambda_i}.
\]
Then $T(E^\perp)\subseteq E^\perp$.
\begin{proof}
For any $x\in E_\lambda$, $y\in E^\perp$:
\[
0 = \bra x, y \ket = \frac{1}{\lambda} \bra Tx, y \ket = \frac{1}{\lambda} \bra x, Ty \ket.
\]
So $\bra x, Ty \ket=0$.
\end{proof}

Q: Eigenspaces of compact maps are fd
A: Let $T\in\mathcal{B}(H)$ be compact self-adjoint. Then $\forall\varepsilon>0$:
\[
E = \bigoplus_{\lambda\in\sigma_p(T), |\lambda|>\varepsilon} E_\lambda
\]
is fd.
\begin{proof}
Suppose not, then $\exists (x_i)$ ON eigenvectors such that
\[
\|Tx_i - Tx_j \|^2 = \|Tx_i\|^2 + \|Tx_j\|^2 \geq 2\varepsilon^2
\]
contradicting compactness of $T$.
\end{proof}

Q: Hilbert-Schmidt Theorem
A: Let $T\in\mathcal{B}(H)$ be compact self-adjoint. Then $\{\lambda_i\}$ is countable and if infinite, then $(\lambda_i)\to 0$. Moreover
\[
T=\sum_{j=1}^\infty \lambda_j P_{E_j}, \qquad H = \ker T \oplus \overline{\bigoplus_{i=1}^\infty E_{\lambda_i}}.
\]
\begin{proof}
First $\lambda_1=\pm \|T\|\neq 0$. Given $\lambda_1 \cdots \lambda_k$ the evals with $|\ph|\geq |\lambda_k|$, let
\[
H_k = \left( \bigoplus_{i=1}^k E_{\lambda_i} \right)^\perp.
\]
Then $H_k$ is closed so is a Hsp. Also $T|_{H_k}:H_k\to H_k$ has $\|T|_{H_k} \|\leq |\lambda_k|$.
Either $T|_{H_k}=0$ or $\exists \lambda_{k+1}$ with $|\lambda_{k+1}|=\|T|_{H_k} \|\leq |\lambda_k|$.
Hence $\{ \lambda_k\}$ are countable and by compactness only infinite if $(|\lambda_k|)\searrow 0$.
Since $T|_{(\bigoplus_{i=1}^\infty E_{\lambda_i})^\perp}$ does not have a non-zero eval, it must vanish. Finally
\[
\left\| T - \sum_{i\leq n} \lambda_i P_{\lambda_i} \right\| \leq |\lambda_n| \to 0.
\]
\end{proof}

Q: Basis of eigenvectors
A: Let $H$ be separable and $T\in\mathcal{B}(H)$ compact self-adjoint. Then $\exists$ON Hbs for $H$ of $T$ eigenvectors.
\begin{proof}
Apply G-S to the eigenspaces and kernel.
\end{proof}

Q: Full compact self-adjoint spectrum
A: Let $T\in\mathcal{B}(H)$ be compact self-adjoint. Then $\sigma(T)\setminus\{0\} = \sigma_p(T)\setminus\{0\}$.
\begin{proof}
Let $T_n = \sum_{i=1}^n \lambda_i P_{ \lambda_i}$ and $\mu\not\in\sigma_p(T)\cup\{0\}$. We compute
\begin{align*}
T_n - \mu &= \sum_{i=1}^n (\lambda_i-\mu) P_{\lambda_i} - \mu \sum_{i=n+1}^\infty P_{ \lambda_i} \\
(T_n - \mu)^{-1} &= \sum_{i=1}^n (\lambda_i-\mu)^{-1} P_{\lambda_i} - \mu^{-1} \sum_{i=n+1}^\infty P_{ \lambda_i} \\
\|(T_n - \mu)^{-1}\| &\leq \max_{1\leq i \leq n}\left\{|\lambda_i-\mu|^{-1}, |\mu^{-1}| \right\}=:C  \\
T- \mu &= (T_n-\mu) + (T - T_n)  \\
&= (T_n-\mu)(1+\underbrace{(T_n-\mu)^{-1}}_{\|\ph\|\leq C} \underbrace{(T-T_n)}_{\|\ph\|\to 0}) \\
(T-\mu)^{-1}&\in\mathcal{B}(H).
\end{align*}
So $\mu\not\in\sigma(T)\cup\{0\}$.
\end{proof}

Q: Green's integral operator is compact
A: Let $K\in C^0([a,b]^2, \R)$. Then
\[
f\mapsto \int_a^b K(x,y) f(y) \; \d y
\]
is bounded and compact $C[a,b], \|\ph\|_2 \to C[a,b], \|\ph\|_\infty$ (or $\to C[a,b], \|\ph\|_2$).
\begin{proof}
By Cauchy-Schwarz, $K$ is bounded
\begin{align*}
\|Kf\|_\infty & \leq \sup_x \int_a^b |K(x,y)| |f(y)| \; \d y \\
&\leq \sup_x \left[ \int_a^b |K(x,y)|^2 \; \d y \right]^{\tfrac{1}{2}} \|f\|_2.
\end{align*}
Similarly, $\{Kf: f\in C[a,b], \|f\|_2\leq 1 \}$ is equicts:
\[
|Kf(x) - Kf(y) | \leq \left[ \int_a^b |K(x,z) - K(y,z) |^2 \; \d z\right]^{\tfrac{1}{2}} \|f\|_2 \xrightarrow{\text{unif.}} 0
\]
as $x\to y$. By A-A, $\{Kf: f\in C[a,b], \|f\|_2\leq 1 \}$ is relatively compact.
\end{proof}

Q: BLT Theorem
A: Let $X,Y$ be Bsps and $D\subseteq X$ dense. Then a bounded (compact) operator $T:D\to Y$ extends uniquely to a bounded (compact) operator $T:X\to Y$.
\begin{proof}
ES4.
\end{proof}

Q: Extension of Green's operator
A: $H=L^2[a,b]$ is the completion of $C[a,b], \|\ph \|_2$. $K$ extends uniquely to a compact, self-adjoint operator $K:H\to C[a,b]$.
\begin{proof}
Compactness follows from BLT.
Self-adjointness follows from symmetry of $K$.
$K(H)\subseteq C[a,b]$ follows from boundedness of $K: C[a,b], \|\ph\|_2\to C[a,b], \|\ph\|_\infty$.
\end{proof}

Q: Spectrum of Green's operator
A: Let $K$ be the Green's function for
\[
Tu=f\in C[a,b], \quad u(a)=u(b)=0.
\]
 By the spectral theorem, $\exists (f_n)\subseteq L^2[a,b]$ an ONS, $(\mu_n)\to 0$ such that
\[
Kf = \sum_{n=1}^\infty \mu_n\bra f, f_n \ket f_n
\]
where for $\mu_n\neq 0$: $f_n=\frac{1}{\mu_n} Kf_n \in C^0$. So $f_n\in K(C^0)\subseteq C^2$.
Assuming $\ker K=0$, wlog $(f_n)$ is a Hbs of $C^2$ $K$-efns. Then the $(f_n)$  are also $T$-efns with evals $\lambda_n=1/\mu_n$:
\[
Tf_n = \lambda_n TKf_n = \lambda_n f_n.
\]

Q: Continuous function of compact operator
A: Let $T\in\mathcal{B}(H)$ be compact self-adjoint. Then $\Phi: C(\sigma(T)) \to\mathcal{B}(H)$ given by
\[
T \mapsto f(T)=\sum_{\lambda\in\sigma(T)} f(\lambda)P_{E_\lambda}
\]
is a positive $*$-hom. Moreover
\begin{enumerate}
\item $\|\Phi(f)\|\leq \|f\|$
\item $\forall x\in E_\lambda: f(T)x=f(\lambda)x$
\item $\sigma(f(T))=f(\sigma(T))$.
\end{enumerate}
\begin{proof}
By S-W, since polys are dense in $C(\sigma(T))$, $\Phi$ extends uniquely to a $*$-hom on $C(\sigma(T))$ with (i).
For (ii), let $(p_n)\to f\in C(\sigma(T))$, $Tx=\lambda x$. Then
\[
f(T)x=\lim_{n\to\infty} p_n(T)x = \lim_{n\to\infty} p_n(\lambda)x=f(\lambda)x.
\]
For (iii), $f(T)$ is normal:
\[
f(T)^*f(T)=(\bar f f) (T) = ( f \bar f) (T) = f(T)f(T)^*.
\]
If $\mu\not\in f(\sigma(T))$, define $g$ by
\[
\lambda \in\sigma(T) \mapsto \frac{1}{\mu - f(\lambda)}
\]
Then $\id = g(T)(\mu - f(T))$ so $\mu\not\in\sigma(f(T))$.
Conversely, let $(p_n)\to f$, $\lambda\in\sigma(T)$ and $(p_n(T)-p_n(\lambda))x_n \to 0$ with $\|x_n\|=1$. Then
\[
(f(T)-f(\lambda))x_n = (f(T)-p_n(T))x_n  +  (p_n(T)-p_n(\lambda))x_n + (p_n(\lambda)-f(\lambda))x_n \to 0.
\]
So $f(\lambda)\in \sigma(f(T))$.
\end{proof}

Q: Algebra homomorphism
A: A map $\phi:A\to B$ between algebras obeys $\forall x,y\in A, k\in \mathbb{F}$:
\begin{itemize}
\item $\phi (x+y)=\phi (x)+\phi (y)$
\item $\phi (xy) = \phi (x)\phi (y)$
\item $\phi (kx)=k\phi (x)$.
\end{itemize}

Q: Star-Homomorphism
A: An algebraic homomorphism $\phi:A\to B$ is a $*$-homomorphism if $\forall a \in A: \phi(a^*)=\phi(a)^*$.

Q: Spectrum of a polynomial
A: Let $P(t)=\sum_{n=0}^N a_n t^n$. Then $\sigma(P(T))=P(\sigma(T))$.
\begin{proof}
Given $\lambda\in\sigma(T)$, write $P(t)-P(\lambda) = (t-\lambda)Q(t)$. Then
\[
P(T)-P(\lambda) = (T-\lambda)Q(T) = Q(T)(T-\lambda)
\]
is either not injective or not surjective so is non-invertible. So $P(\lambda)\in\sigma(P(T))$.
Conversely, given $\mu\in\sigma(P(T))$, write
\[
P(t) - \mu = a_N(t-\lambda_1) \cdots (t-\lambda_n).
\]
If $\lambda_1, \cdots \lambda_n \not\in\sigma(T)$, then
\[
(P(T)-\mu)^{-1} = a_N^{-1} (T-\lambda_1)^{-1} \cdots (T-\lambda_n)^{-1} \in\mathcal{B}(H),
\]
i.e. $\mu\not\in\sigma(P(T))$. So $\exists\lambda_i\in\sigma(T)$ such that $\mu=P(\lambda_i)\in P(\sigma(T))$.
\end{proof}

Q: Norm of self-adjoint operator from spectrum
A: Let $T\in\mathcal{B}(H)$ be self-adjoint. Then
\[
\|T\|= \sup_{\lambda\in\sigma(T)} |\lambda|.
\]
\begin{proof}
$\pm\|T\|\in\sigma(T)$ $\implies$ $\|T\|\leq \sup_{\lambda\in\sigma(T)} |\lambda|$.
$\sigma(T)\subseteq D(0, \|T\|)$ $\implies$ $\|T\|\geq \sup_{\lambda\in\sigma(T)} |\lambda|$.
\end{proof}

Q: Norm of polynomial
A: Let $T\in\mathcal{B}(H)$ be self-adjoint, $P$ a poly. Then
\[
\|P(T)\|= \sup_{\lambda\in\sigma(T)} |P(\lambda)|.
\]
\begin{proof}
\begin{align*}
\|P(T)\|^2 &= \| P(T)^* P(T) \| \\
&= \sup_{\lambda\in\sigma(\bar P P (T))} |\lambda| \\
&= \sup_{\lambda\in\sigma(T)} |\bar P P (\lambda)| \\
&=  \sup_{\lambda\in\sigma(T)} | P (\lambda)|^2.
\end{align*}
\end{proof}

Q: $\lim_{n\to\infty} f(nx) =0 \implies \lim_{x\to\infty} f(x) = 0$
A: Let $f:\R^+\to\R^+$ be a cts function such that $\forall x \in\R^+$:
\[
\lim_{n\to\infty} f(nx) = 0.
\]
Then $f(x)\to 0$ as $x\to\infty$.
\begin{proof}
Fix some $\epsilon >0$. The sets
\[
E_N=\{x: \forall n \geq N, f(nx)\leq \epsilon\} = \bigcap_{n\geq N} \{x:f(nx)\leq\epsilon\}
\]
are closed by cty of $f$. By assumption $\R^+ = \bigcup_{N\in\N} E_N$.
By BCT, one of them, say $E_{N}$, contains an open segment $(a,b)$. That is, $\forall n\geq N$, then $f|_{(na,nb)}\leq \epsilon$. But for $M\geq \max\{N, \frac{a}{b-a}\}$:
\[
(Ma,\infty) = \bigcup_{n \geq M} (na, nb).
\]
So $f|_{(Ma, \infty)}\leq \varepsilon$.
\end{proof}

Q: Absorbent set
A: A subset $C\subseteq V$ is \emph{absorbent} or \emph{absorbing} if
\[
\bigcup_{t>0} tC \supseteq V. \qquad\qquad\qquad\qquad
\]

Q: Kernel and image of normal operators
A: Let $T\in\mathcal{B}(H)$ be normal. Then
\begin{enumerate}
\item $\forall x \in H: \|Tx\| = \|T^*x\|$,
\item $\ker T = \ker T^* = (\im T)^\perp$,
\item $\sigma_{ap}(T)=\sigma(T)=\sigma(T^*)$.
\end{enumerate}

Q: Ordering of $\ell_p$ spaces
A: For $x\in\C^n$, $1<p<q<\infty$
\[
\|x\|_q \leq \|x \|_p \leq n^{\tfrac{1}{p} - \tfrac{1}{q}} \|x\|_q.
\]
\begin{proof}
Wlog $\|x\|_q=1$. Then $|x_i| \leq 1$  so
\[
\|x\|_q = \left(\sum |x_i|^q \right)^{\tfrac{1}{q}} = \left(\sum |x_i|^q \right)^{\tfrac{1}{p}} \leq \left(\sum |x_i|^p \right)^{\tfrac{1}{p}} = \|x\|_p.
\]
By H\"older's inequality for $(\frac{q}{p}, \frac{q}{q-p})$.
\[
\sum |x_i|^p (1) \leq \left(\sum |x_i|^q \right)^{\tfrac{p}{q}} n^{\frac{q-p}{q}} .
\]
\end{proof}

Q: Completeness of $\ell_p$ spaces
A: $\ell_p(\R)$ and $\ell_p(\C)$ are complete for $1\leq p < \infty$.
\begin{proof}
Let $(x^n)\subseteq\ell_p$ be a Cauchy sequence of column vectors.
By completeness of $\R$, the pointwise row limits $x=(x_i)$ exist. Moreover $(x^n)$ Cauchy $\implies$ $(x^n)$ bounded $\implies$ $x\in\ell_p$ (truncation argument).
Given $\varepsilon>0$, pick $N\in\N$ such that $\forall m,n \geq N$: $\|x^m-x^n\|_p<\varepsilon$.
Given $n\geq N$ pick $I\in\N$ such that $\|(x_n-x)_{i>I}\|_p<\varepsilon$.
Now pick $m>N$ such that $\|(x^m-x)_{i\leq I}\|_p < \varepsilon$.
Then
\[
\|x^n - x\|_p \leq \|(x^n - x^m)_{i\leq I}\|_p + \|(x^m - x)_{i\leq I}\|_p + \|(x^n - x)_{i> I}\|_p < 3\varepsilon.
\]
\end{proof}

Q: Completeness of $\ell_\infty$
A: $\ell_\infty(\R)$ and $\ell_\infty(\C)$ are complete.
\begin{proof}
Let $(x^n)\subseteq\ell_\infty$ be a Cauchy sequence of column vectors.
By completeness of $\R$, the pointwise row limits $x=(x_i)$ exist. Moreover $(x^n)$ Cauchy $\implies$ $(x^n)$ bounded $\implies$ $x\in\ell_\infty$.
Given $\varepsilon>0$, pick $N\in\N$ such that $\forall m,n \geq N$: $\|x^m-x^n\|_\infty<\varepsilon$.
Given $n\geq N$ pick $I\in\N$ such that $\|x^n-x\|_\infty< |x^n_I - x_I| + \varepsilon$.
Now pick $m>N$ such that $|x^m_I-x_I| < \varepsilon$. Then
\[
\|x^n - x\|_\infty < \varepsilon +  |x^n_I - x^m_I | + |x^m_I - x_I| < 3\varepsilon.
\]
\end{proof}

Q: Duals of $\ell_p$ spaces
A: For H\"older conjugates $1<p, q<\infty$, $(\ell_p)^*=\ell_q$.
\begin{proof}
Define $\varphi:\ell_q\to (\ell_p)^*$ by $\varphi(y)(x)=\sum x_n y_n$.
\emph{Isometry}. By H\"older, $\|\varphi(y)\|\leq\|y\|_q$. Fix $(\lambda_n)\subseteq S^1$ such that $\lambda_n y_n=|y_n|$. Define $x_n=\lambda_n|y_n|^{q-1}$. Then $\|x\|_p^p=\|y\|_q^q$ and
\[
\varphi(y)(x) = \sum |y_n|^q =  \|y\|_q \|y\|_q^{q-1} =\|y\|_q \|x\|_p.
\]
\emph{Surjectivity}. Given $\xi\in (\ell_p)^*$, wlog $\|\xi\|=1$. Define $x \in \ell_p^n$ by truncating the above for $y=(\xi(e_i))\in \ell_p^n$. Then again
\[
\xi(x) = \|y\|_q \|x\|_p \leq \|x\|_p.
\]
Hence $\|(\xi(e_i))\|_q \leq 1$.
\end{proof}

Q: Dual space of $\ell_1$
A: $(\ell_1)^*=\ell_\infty$.
\begin{proof}
Define $\varphi:\ell_\infty\to (\ell_1)^*$ by $\varphi(y)(x)=\sum x_n y_n$.
\emph{Isometry}. Clearly $\|\varphi(y)\|\leq\|y\|_\infty$. Given $\varepsilon>0$, pick $n\in\N$ such that $\|y\|_\infty < |y_n| + \varepsilon$. Then $|\varphi(y)(e_n)| = |y_n| > \|y\|_\infty - \varepsilon$.
\emph{Surjectivity}. Given $\xi\in (\ell_1)^*$, wlog $\|\xi\|=1$. Then $|\xi(e_n)| \leq 1$ so $\|(\xi(e_n))\|_\infty\leq 1$.
\end{proof}

Q: Dual space of $c_0$
A: $c_0^*=\ell_1$.
\begin{proof}
Define $\varphi:\ell_1\to c_0^*$ by $\varphi(y)(x)=\sum x_n y_n$.
\emph{Isometry}. Clearly $\|\varphi(y)\|\leq\|y\|_1$. Let $x^n_i=\lambda_i\mathbbm{1}_{\{i\leq n\}}$ where $\lambda_iy_i=|y_i|$. Then $|\varphi(y)(x^n)| \to \|y\|_1$.
\emph{Surjectivity}. Given $\xi\in c_0^*$, wlog $\|\xi\|=1$. Then as above $1\geq |\xi(x^n)| \to \|(\xi(e_n))\|_1$.
\end{proof}

Q: Dini's Theorem
A: Let $(f_n), f\subseteq C(K)$, $K$ compact Hausdorff, $(f_n)\nearrow f \implies (f_n)\xrightarrow{\text{unif}}f$.
\begin{proof}
Define $g_n=f-f_n\searrow 0$. Given $\varepsilon>0$, define the open sets
\[
E_n = \{g_n(x) < \varepsilon \} \nearrow K.
\]
Hence there exists a finite subcover, i.e. $K=E_N$ for some $N\in\N$.
\end{proof}

Q: Stone-Weierstrass with vanishing point
A: Let $A\subseteq C(K)$ be a subalgebra which separates points but vanishes at $x_0$. Then $\bar{A}=\{f\in C(K): f(x_0)=0\}$.
\begin{proof}
Define $A'=A+\R\subseteq C(K)$ dense (by S-W).
Given $f\in C(K)$ with $f(x_0)=0$, let $(f_n+\lambda_n)\to f$. Then
\begin{align*}
\|f_n-f\|_\infty &\leq \|f_n+\lambda_n - f\|_\infty + |\lambda_n| \\
&=\|f_n+\lambda_n - f\|_\infty + |f_n(x_0) + \lambda_n - f(x_0)| \\
&\leq 2\|f_n+\lambda_n - f\|_\infty\\
&\to 0.
\end{align*}
\end{proof}

Q: Parallelogram identity $\implies$ Euclidean
A: The nvsp $E$ is Euclidean iff the paralleogram identity holds.
\begin{proof}
\emph{Real case}. Define the positive definite symmetric form
\[
\bra x, y \ket = \frac{1}{4} \left( \|x+y\|^2 - \|x-y\|^2 \right).
\]
Applying paralleogram identity for $(a\pm c, b)$ and $(b\pm c, a)$:
\begin{align}
\|a+b+c\|^2 + \|a-b+c\|^2 &= 2\|a+c\|^2 + 2\|b\|^2 \\
\|a+b-c\|^2 + \|a-b-c\|^2 &= 2\|a-c\|^2 + 2\|b\|^2 \\
\|a+b+c\|^2 + \|a-b-c\|^2 &= 2\|b+c\|^2 + 2\|a\|^2 \\
\|a+b-c\|^2 + \|a-b+c\|^2 &= 2\|b-c\|^2 + 2\|a\|^2 .
\end{align}
$ (1)-(2)+(3)-(4)$ gives $\bra a+b, c \ket = \bra a, c \ket + \bra b, c \ket$ (dividing by 8).
Linearity: $\bra \lambda  x, y \ket = \lambda \bra x, y \ket$. Checking $\lambda=-1$ and using additivity gives $\lambda\in\Q$. Hence $t\mapsto \frac{1}{t}\bra tx, y \ket $ equals $\bra x, y\ket $ on $t\in\Q\setminus\{0\}$ and is continuous on $\R\setminus\{0\}$.
\emph{Complex case}. Define
\[
\bra x, y \ket = \frac{1}{4} \sum_{k=0}^3 i^k \|x+i^ky\|^2.
\]
Observe $\bra ix, y \ket = i\bra x, y \ket$ and $\bra y, x \ket = \overline{\bra x, y \ket}$. Linearity comes from real case applied to real and imaginary parts.
\end{proof}

Q: Hahn-Banach (Complex case)
A: Let $V$ be a normed $\C$-vsp, $W\subseteq V$, $p: V \to \R$ sublinear, $g\in W^*$ such that $\Re g \leq p|_W$. Then $\exists f\in V^*$ such that $f|_W=g$ and $\Re f\leq p$. Moreover if $p$ is a seminorm, then $|f|\leq p$.
\begin{proof}
By real case, $\exists r:V\to\R$ an extension of $\Re g$. Define $f(x)=r(x)-ir(ix)$. Then
\[
f(ix)=r(ix)-ir(-x)=r(ix)+ir(x) = i(r(x) - ir(ix))=if(x)
\]
so $f$ is $\C$-linear. Moreover $g(x)=\Re g(x)-i \Re g(ix)$ since
\[
\Im g(x) = -\Re i g(x) = -\Re g(ix).
\]
For second part, observe $\forall\lambda\in S^1$
\[
\Re \lambda f(x) = \Re f(\lambda x) \leq p(\lambda x) = p(x).
\]
Pick $\lambda \in S^1$ such that $|f(x)|=\lambda f(x)$. Then
\[
|f(x)| = \Re \lambda f(x) \leq p(x).
\]
\end{proof}

Q: Closed kernel
A: $f:V\to \C$ linear is bounded $\iff$ $\ker f$ is closed.
\begin{proof}
($\Rightarrow$) $\ker f = f^{-1}(\{0\})$.
($\Leftarrow$) Let $f$ be unbounded and $v\not\in\ker f$. Pick $(v_n)\subseteq V$ with $\|v_n\|=1$ and $|f(v_n)| \to \infty$. Then
\[
k_n:= v - \frac{f(v)}{f(v_n)} v_n \in \ker f
\]
but $\|k_n - v \|\to 0$.
\end{proof}

Q: Vanishing derivatives $\implies$ polynomial
A: Let $f:\R\to\R$ be $C^\infty$ such that $\forall x \in \R, \exists n\in\N$ such that $\forall m \geq n: f^{(m)}(x)=0$. Then $f$ is a polynomial.
\begin{proof}
Suppose not. Define
\[
C_n = \bigcap_{m\geq n} \{f^{(m)}(x)=0\}
\]
and
\[
X = \{ \forall (a,b) \ni x: f \text{ is not a poly} \}.
\]
Then $C_n, X$ are closed and $\bigcup C_n\supseteq X$. By BCT, $\exists [a,b]\subseteq\R$, $n\in\N$ such that $[a,b]\cap X\subseteq C_n$.
Let $(r,s)\subseteq [a,b]\setminus X$ be a maximal interval. Then $f|_{(r,s)}$ is a poly of say degree $d$ so in particular $f^{(d)}=\text{constant}\neq 0$. Since one of $r,s \in X$, we must have $d<n$.
Hence $f^{(n)}|_{[a,b]}=0$, contradicting $[a,b]\cap X\neq \emptyset$.
\end{proof}

Q: Proving compactness of operators
A: Every finite rank operator is compact.
Arzela-Ascoli: e.g. for integral operators in $C[0,1]$
Limit of compact operators (an $\varepsilon$-net of $T_nB$ gives a $2\varepsilon$ of $TB$).