Title: Analysis of Functions

Q: Limit of measurable functions
A: Let $(X,\mathcal{A}), ((Y,d),\mathcal{B}(Y))$ be $\sigma$-algebras. If $(f_k)\to f$ and $(f_k)$ are msble, then $f$ is msble.
\begin{proof}
Enough to check $f^{-1}(U)\in\mathcal{A}$ for $U$ open.
Let
\begin{align*}
U_n&=\left\{y\in Y: d(y,Y\setminus U) > \frac{1}{n} \right\}\\
F_n&=\left\{ y\in Y: d(y,Y\setminus U) \geq \frac{1}{n} \right\}.
\end{align*}
Then $U_n\subseteq F_n\subseteq U_{n+1} \cdots \subseteq U=\bigcup_n U_n = \bigcup_n F_n$ and
\begin{align*}
f^{-1}(U) &= \bigcup_n f^{-1} (U_n) \\
&\subseteq \bigcup_n \lim\inf_k f_k^{-1}(U_n) \quad (\text{$U_n$ open}) \\
&\subseteq \bigcup_n \lim\inf_k f_k^{-1}(F_n) \quad (U_n\subseteq F_n) \\
&\subseteq \bigcup_n f^{-1}(F_n) \quad (\text{$F_n$ closed})  \\
&= f^{-1}(U)
\end{align*}
so there is equality throughout. Hence $f^{-1}(U)\in\mathcal{A}$.
\end{proof}

Q: Limit of measurable functions by inf/sup
A: If $(f_k):E\to\R$ are measurable, then $f=\lim f_k$ is measurable.
\begin{proof}
We can write
\begin{align*}
(\lim\inf f_k)^{-1}(-\infty, a] &= \bigcap_{j\geq 1}\bigcap_{n\geq 1}\bigcup_{k\geq n} f_k^{-1}(-\infty, a+\tfrac{1}{j}] \\
(\lim\sup f_k)^{-1}(-\infty, a] &= \bigcap_{j\geq 1}\bigcup_{n\geq 1}\bigcap_{k\geq n} f_k^{-1}(-\infty, a+\tfrac{1}{j}].
\end{align*}
So if $\lim\inf f_k=\lim\sup f_k =f$, then $f$ is measurable.
\end{proof}

Q: Abstract density result
A: The simple functions are dense as a subset of $L^p$ for $p\in [1,\infty]$.
\begin{proof}
Taking real/imaginary and positive/negative parts, wlog consider $f\geq 0 \in \mathcal{L}^p$. Now define
\[
f_n = n \wedge \frac{\lfloor 2^n f \rfloor}{2^n}.
\]
Case $p<\infty$. By MCT $f_n\nearrow f$ wrt $\mathcal{L}^p$.
Case $p=\infty$. For $n>\|f\|_\infty$, $\|f_n-f \|_\infty \leq \frac{1}{2^n}\xrightarrow{a.e.} 0$.
\end{proof}

Q: Density-separability of $L^p$
A: For $U\subseteq\R^d$ open, $L^p(U)$ is separable and $C_c^\infty(U)\subseteq L^p(U)$ is dense.
\begin{proof}
We cover $U$ using rational boxes with disjoint interiors: on the grid $\Z^d$ take the cubes inside $U$, bisect the borderline cubes to form $2^d$ half-length cubes and repeat.
Enough to apx $\sum_{i=1}^n\alpha_i \mathbbm{1}_{A_i}$ simple. Apx $\alpha_i$ using $\Q$ and $A_i$ by $U_i\supseteq A_i$ open. Then cover $U_i$ using rational boxes $(C_{i,k})$ with $\mu(U_i) = \sum \mu(C_{i,k})<\infty$.
For the second part, apx the indicator functions using $e^{-1/x^2}$.
\end{proof}

Q: Regularity of Lebesgue Measure
A: Every msble set can be apx from above (below) by open (compact) msble sets.
\begin{proof}
From construction.
\end{proof}

Q: Lebesgue point
A: Let $f:\R^n\to\C$ be measurable. We say $x\in\R^n$ is a Leb pt if
\[
\frac{1}{\mu(B(x,r))} \int_{B(x,r)} |f(y) - f(x) | \; \d y \to 0
\]
as $r\to 0$.

Q: Continuity $\implies$ Lebesgue point
A: Let $f:\R^n\to\C$ be msble. If $f$ is cts at $x\in\R^n$, then $x$ is a Leb pt of $f$.
\begin{proof}
Given $\varepsilon>0$, $\exists\delta >0$ such that $\forall y\in B(x, \delta)$:
\[
|f(x) - f(y) | < \varepsilon.
\]
Hence $\forall r < \delta$:
\[
\frac{1}{\mu(B(x,r)} \int_{B(x,r)} |f(y) - f(x) | \; \d y \leq \varepsilon.
\]
\end{proof}

Q: Lebesgue density theorem
A: For any $E\in\mathcal{B}(\R^n)$ and a.e. $x\in\R^n$:
\[
\frac{\mu(E\cap B(x,r))}{\mu(B(x,r))} \to \mathbbm{1}_E(x) \tag{$*$}.
\]
\begin{proof}
Consider $x\in\R^n$ with $\|x\|\leq M$, $r<1$. Then
\[
\frac{\mu(E \cap B(x,r))}{\mu(B(x,r))} = \frac{\mu(E \cap B(0, M+1) \cap B(x,r))}{\mu(B(x,r))}.
\]
Now apply Lebesgue's Differentiation Theorem to $\mathbbm{1}_{E \cap B(0, M+1)} \in L^1$.
This gives $(*)$ for $x\in B(0, M)\setminus N_m$ where $\mu(N_m)=0$. Hence $N=\bigcup_{M\in\N} N_M$ is a null set and $(*)$ holds on $N^C$.
\end{proof}

Q: Vitali's covering lemma
A: For $X\subseteq \bigcup_{i=1}^N B(x_i, r_i) \subseteq \R^d$, $\exists J \subseteq \{1, \cdots N\}$ such that
\[
X\subseteq \bigcup_{i\in J} B(x_i, 3r_i)
\]
where $(B(x_i,r_i))_{i\in J}$ are pairwise disjoint.
\begin{proof}
Basis $N=0$ is trivial.
Wlog $r_1 \geq r_2 \geq \cdots \geq r_n$. Then any ball which intersects $B(x_1, r_1)$ is contained in $B(x_1, 3r_1)$. Remove all such balls.
By inductive hypothesis the remaining balls can be covered as required.
\end{proof}

Q: Continuous functions are dense in $L^1$
A: Given $f\in L^1(\R^n)$ and $\varepsilon>0$, $\exists g\in C^0(\R^n)$ such that $f=g+h$ with $\|h\|_1 < \varepsilon$.
\begin{proof}
\end{proof}

Q: Hardy-Littlewood maximal operator
A: Define
\begin{align*}
Mf(x) &= \sup_{r>0} mf(x,r) \\
mf(x,r) &=  \frac{1}{\mu(B(x,r))} \int_{B(x,r)} |f| \; \d\mu \\
E_a &= \{ x\in \R^d: Mf(x) > a \}.
\end{align*}
Then $E_a$ is open (and thus msble) for $a>0$.
\begin{proof}
Given $x\in E_a$, $Mf(x)>a$ so $\exists r>0$ such that $mf(x,r)>a$, i.e.
\[
1 > \frac{a}{mf(x,r)} .
\]
Pick $\delta>0$ small enough that
\[
\left( \frac{r}{r+\delta} \right)^d > \frac{a}{mf(x,r)}.
\]
For $ y \in B(x,\delta)$, by $\Delta$-inequality $B(x,r) \subseteq B(y, r+\delta)$. Hence
\begin{align*}
mf(y, r+\delta) &= \frac{1}{\mu(B(y, r+\delta))} \int_{B(y, r+\delta)} |f| \; \d\mu \\
&\geq \left[ \left( \frac{r}{r+\delta}\right)^d \frac{1}{\mu(B(x,r))} \right] \int_{B(x,r)} |f| \; \d\mu \\
& = \left(\frac{r}{r+\delta}\right)^d mf(x,r) \\
&> a.
\end{align*}
Thus $y\in E_a$. So $B(x,\delta) \subseteq E_a$.
\end{proof}

Q: Hardy-Littlewood operator is weak $L^1$
A: For all $a>0$,
\[
\mu(E_a) \leq \frac{3^d}{a} \|f\|_{L^1}
\]
\begin{proof}
Given $K\subseteq E_a$ compact, for any $x\in K$, $\exists r_x$ such that
\[
mf(x,r) > a.
\]
By compactness, $\exists I$ finite such that
\begin{align*}
K &\subseteq \bigcup_{x\in K} B(x,r_x) \\
&\subseteq \bigcup_{i \in I} B(x_i,r_{x_i}). \\
\intertext{By Vitali's covering lemma, $\exists J\subseteq I$ such that}
K &\subseteq \bigcup_{j\in J} B(x_j,3r_{x_j}) \\
\intertext{and $(B(x_j,r_{x_j}))_{j\in J}$ pairwise disjoint. Hence}
\mu(K) & \leq \sum_{j\in J} \mu(B(x_j,3r_{x_j})) \\
& \leq 3^d \sum_{j\in J} \mu(B(x_j,r_{x_j})) \\
& \leq \frac{3^d}{a} \sum_{j\in J} \int_{B(x_j,r_{x_j})} |f| \; \d\mu\\
& \leq \frac{3^d}{a} \int_{\R^d} |f| \; \d\mu.
\end{align*}
Done by inner regularity.
\end{proof}

Q: Lebesgue differentiation theorem
A: For $f\in L^1(\R^n)$, a.e. point is Lebesgue.
\begin{proof}
Define $tf(x,r)=m(f(x)-f)(x,r)$ and $Tf(x)=\lim\sup tf(x,r)$. By density of $C^0$ in $L^1$, $\exists g\in C^0$ such that $f=g+h$ with $\|h\|_1$ arbitrarily small. Then
\[
tf(x,r) \leq tg(x,r) + \frac{1}{\mu(B(x,r))} \int_{B(x,r)} |h| \; \d\mu + |h(x)|.
\]
Taking $\lim\sup_{r\to 0}$ and using that $g$ is everywhere Lebesgue:
\[
Tf(x) \leq Mh(x) + |h(x)|.
\]
Taking measures, using that H-L is weak $L^1$ and Chebyshev:
\begin{align*}
&\mu\left( \left\{ x: Tf(x)> \frac{1}{k} \right\} \right) \\
\leq &\mu\left( \left\{ x: Mh(x)> \frac{1}{2k} \right\} \right) + \mu\left( \left\{ x: |h(x)|> \frac{1}{2k} \right\} \right) \\
\leq & 2k(3^n + 1) \|h\|_{L^1} \\
\to & 0 \text{ as $\|h\|_1\to 0$}. \\
\intertext{Taking unions over $k\in\N$}
0=&\mu\left( \left\{ x: Tf(x)\neq 0 \right\} \right).
\end{align*}
\end{proof}

Q: Lusin's Theorem V1: long proof
A: If $f:\R\to\C$ is msble and $\varepsilon>0$, $\exists E\in\mathcal{L}(\R)$ with $\mu(E)<\varepsilon$ and $f|_{\R\setminus E}$ is cts.
\begin{proof}
For each interval $I_\ell = [\ell, \ell+1)$, $\ell\in\Z$, pick step functions $S_n^\ell \to f$. By Egorov's Theorem, $\exists A_\ell\subseteq I_\ell$ such that $\mu(A_\ell)\leq \frac{\varepsilon}{3\cdot 2^{|\ell |}}$ and $S_n^\ell \to f$ uniformly on $I_\ell\setminus A_\ell$. Then $A=\bigcup A_\ell$ has $\mu(A)<\varepsilon$.
The set $D$ of points of discontinuity of $S_n^\ell$ is countable so $E:=\Z \cup D \cup A$ has $\mu(E)\leq\varepsilon$.
Concatenate $S_n^\ell: I_\ell\to\C$ into step functions $S_n:\R\to \C$. On $\R\setminus E$, these are cts and $S_n\to f$ uniformly.
\end{proof}

Q: Density of step functions
A: If $f:E\to\C$ is msble, $\mu(E)<\infty$, then $\exists S_n$ simple such that $S_n\to f$ a.e.
\begin{proof}
Pick $s_n$ simple such that $s_n\to f$ a.e. Pick $S_n$ step functions such that $S_n=s_n$ except on $D_n$ with $\mu(D_n)<2^{-n}$. Then
\[
D=\lim\sup D_n = \bigcap_{n\geq 1} \bigcup_{m\geq n} D_n
\]
is a null set with $S_n\to f$ on $E\setminus D$.
\end{proof}

Q: Step approximation of simple functions
A: Given a simple function $s:E\to \C$ with $\mu(E)<\infty$ and $\varepsilon>0$, $\exists S$ a step function such that
\[
\mu\left(\left\{ x\in\R: S(x)\neq s(x) \right\} \right) < \varepsilon.
\]
\begin{proof}
Write $s=\sum_{k=1}^n \alpha_k \mathbbm{1}_{A_k}$. Pick finite unions of intervals $(K_k)_{k=1}^n$ with $\mu(A_k\Delta K_k)<\frac{\varepsilon}{n}$. Let $S=\sum_{k=1}^n \alpha_k \mathbbm{1}_{K_k}$.
Then $S(x)=s(x)$ on $E\setminus \bigcup_{k=1}^n A_k\Delta K_k$ and
\[
\mu\left(\bigcup_{k=1}^n A_k\Delta K_k \right) < \varepsilon.
\]
\end{proof}

Q: Lusin's Theorem V1: short proof
A: If $f:\R\to\C$ is msble and $\varepsilon>0$, $\exists E\in\mathcal{L}(\R)$ with $\mu(E)<\varepsilon$ and $f|_{\R\setminus E}$ is cts.
\begin{proof}
Wlog $f:E\to\R$ with $\mu(E)<\infty$. Let $(V_n)$ be the rational open intervals.
By inner regularity, there exist compact sets $K_n\subseteq f^{-1}(V_n)$ and $K_n'\subseteq E \setminus f^{-1}(V_n)$ such that
\[
\mu\left( E \setminus (K_n\cup K_n')\right) < \frac{\varepsilon}{2^n}.
\]
By compactness $\exists U_n$ open such that $K_n\subseteq U_n$ and $U_n\cap K_n'=\emptyset$. Now let
\[
K:= \bigcap (K_n \cup K_n')
\]
such that $\mu(E\setminus K)<\varepsilon$.
Given $x\in K$, $n\in\N$ such that $f(x)\in V_n$, then $x\in K_n\subseteq U_n$. Similarly, we have the converse $f(U_n\cap K)\subseteq V_n$. So $f|_K^{-1}(V_n)=U_n$.
\end{proof}

Q: Lusin's Theorem V2
A: If $f:\R\to\C$ is msble and $\varepsilon>0$, $\exists G$ with $\mu(G)<\varepsilon$ and $g:\R\to\C$ cts such that $f=g$ on $\R\setminus G$.
\begin{proof}
By Lusin V1, $\exists E\subseteq\R$ such that $\mu(E)<\frac{\varepsilon}{2}$ and $f|_{\R\setminus E}$ is cts.
By outer regularity, $\exists G\supseteq E$ open such that $\mu(G)<\varepsilon$. Write $G=\bigcup_{k\in\N} (a_k, b_k)$.
Define
\[
g(x) = \begin{cases} f(x) \quad &x\in\R\setminus G \\
f(a_k) + \frac{x-a_k}{b_k-a_k} \left( f(b_k) - f(a_k) \right) \quad &x\in (a_k,b_k).
\end{cases}
\]
\end{proof}

Q: Open Hahn-Banach
A: Let $E$ be a nsp and $A, B\subseteq E$ be convex non-empty disjoint where $A$ is open. Then there is a closed weakly separating hyperplane, i.e. $\exists f\in E^*, \alpha\in\R$ such that
\[
\begin{cases}
f \leq \alpha \quad & \text{ on } A \\
f > \alpha \quad & \text{ on } B. \\
\end{cases}
\]
\begin{proof}
First we separate convex open non-empty set $C$ from a point $v\not\in C$. Let $g:\spn\{v\}\to\R$ be given by $g(\lambda v)=\lambda$. By convexity, wlog $\lambda v\in C$ only if $\lambda < 1$. So $\alpha^{-1}(\lambda v)\in C$ $\implies$ $g(\lambda v)=\lambda< \alpha$. That is $g\leq \mu_C$. Extend to $\tilde{g}:E\to\R$ with $\tilde{g}\leq \mu_C$. Then $\tilde{g}(v)=1\geq \mu_C(C)\geq g(C)$.
Finally $C=A-B$ is convex open and does not contain zero (disjointness). We can separate $C$ and $\{0\}$ by say $f$ such that $f(A)-f(B)\leq 0$.
\end{proof}

Q: Closed hyperplane
A: Let $E$ be a nvsp, $ f: E\to\R$ linear, $\alpha\in\R$. Then the affine hyperplane $H:=\{ f = \alpha\}$ is closed $\iff f$ is cts.
\begin{proof}
($\Leftarrow$) immediate.
($\Rightarrow$) For $v$ such that $f(v)<\alpha$, by openness $\exists \varepsilon>0$ such that $B_\varepsilon(v)\subseteq H^C$. By convexity $f(B_\varepsilon(v))<\alpha$ $(*)$.
For any $w\in B_1(0)$:
\[
f(v+\varepsilon w) = f(v)+\varepsilon f(w) < \alpha.
\]
So $\|f\| \leq \frac{\alpha - f(v)}{\varepsilon}$.
\end{proof}
($*$) If $w\in B_\varepsilon(v)$ such that $f(w)>\alpha$, then
\[
t\mapsto (1-t)f(v)+tf(w), \quad[0,1]\to\R
\]
is cts so hits $\alpha$ by IVT.

Q: Closed Hahn-Banach
A: Let $E$ be a nvsp and $A,B\subseteq E$ be convex non-empty disjoint. Assume $A$ closed and $B$ compact (e.g. a point). Then there is a closed hyperplane strictly separating $A,B$, i.e. $\exists f\in E^*, \alpha\in\R, \varepsilon>0$ such that
\[
\begin{cases}
f \leq \alpha - \varepsilon \quad &\text{ on } A \\
f \geq \alpha + \varepsilon \quad &\text{ on } B.
\end{cases}
\]
\begin{proof}
Claim: we can pick $\varepsilon>0$ such that $A_\varepsilon = A + B(0, \varepsilon)$, $B_\varepsilon = B + B(0,\varepsilon)$ are convex open disjoint non-empty.
Indeed, otherwise $\exists (v_n)\subseteq A, (w_n)\subseteq B$ such that $\|v_n - w_n \|\to 0$. Then $\exists \varphi(n)$ a subsequence such that $w_{\varphi(n)}\to w\in B$. Then $v_{\varphi(n)} \to w \in A \cap B$, contradicting disjointness.
Now apply open H-B to $A_\varepsilon, B_\varepsilon$.
\end{proof}

Q: Initial topology
A: Let $X, (Y_i)_{i\in I}$ be tsps and $\varphi_i: X\to Y_i$. Then the \emph{initial topology} is the weakest topology $X$ such that every $\varphi_i$ is cts.
Basis:
\[
\left\{\bigcap_{\text{finite}}f^{-1}(V): V\subseteq Y_i \text{ open}, i \in\ I \right\}.
\]
Convergence: $(x_n)\to x \iff \forall i\in I: (\varphi_i(x_n))\to \varphi_i(x)$.
Continuity: $\psi: Z\to X$ is cts $\iff$ $\forall i\in I: \varphi_i\circ\psi: Z\to Y_i$ cts.

Q: Final topology
A: Let $(X_i)_{i\in I}, Y$ be tsps and $\varphi_i: X_i \to Y$. Then the \emph{final topology} is the finest topology on $Y$ such that every $\varphi_i$ is cts.

Q: Neighbourhood system
A: $\mathcal{V}(x)$ is the set of all nbhds of $x$.

Q: Neighbourhood basis
A: $\mathcal{B}(x)$ is a set such that any $V\in\mathcal{V}(X)$ contains some $B\in \mathcal{B}(x)$.

Q: Weak topology
A: Let $E$ be a nvsp. The weak topology $\sigma (E, E^*)$ is the initial topology for $E^*$.
The strong topology is the topology induced by the norm.

Q: Weak topology is Hausdorff
A: For $E$ a nvsp, $\sigma(E,E^*)$ is Hausdorff.
\begin{proof}
Let $v\neq w\in E$. Apply H-B to $\{v\}, \{w\}$ closed compact convex non-empty. Then $\exists f\in E^*, \alpha\in\R$ such that
\[
\begin{cases}
v \in &\{ f< \alpha\}=:U_v \\
w \in &\{g > \alpha\}=:U_w
\end{cases}
\]
where $U_v, U_w$ are disjoint open nbhds.
\end{proof}

Q: Weak-$*$ topology
A: Let $E$ be a nvsp. The weak-$*$ topology on $E^*$, written $\sigma(E^*, E)$, is the initial topology generated by
\[
\{ \ev(v): v\in E\}.
\]

Q: Norm in terms of dual space
A: For $v\in E$, we have
\[
\|v\|_E = \max \left\{ |f(v)| : f \in \bar{B}_{E^*}(0,1) \right\}.
\]
\begin{proof}
RHS $\leq$ LHS by definition of operator norm.
Extend the map $g:\spn\{v\}\to\R$, $t v\mapsto t \|v\|$ to a map $\tilde{g}\in \bar{B}_{E^*}(0,1)$ which saturates the inequality.
\end{proof}

Q: Weak-$*$ topology is Hausdorff
A: $(E^*, \sigma(E^*, E))$ is Hausdorff.
\begin{proof}
Let $F_1\neq F_2\in E^*$. Then wlog $\exists v\in E, \alpha\in\R$ such that $F_1(v)<\alpha< F_2(v)$. Then separate by
\begin{align*}
U_1&= \ev(v)^{-1}((-\infty, \alpha)) \\
U_2&= \ev(v)^{-1}((\alpha, \infty)).
\end{align*}
\end{proof}

Q: Banach-Alaoglu-Bourbaki Theorem
A: $\bar{B}_{E^*}(0,1)$ is weak-$*$ compact.
\begin{proof}
Follows from Tychonoff's theorem.
\end{proof}

Q: Kakutani Theorem
A: A Bsp $E$ is reflexive $\iff$ $B_E:=\bar{B}_{E}(0,1)$ is weakly compact.
\begin{proof}
($\Rightarrow$) As $\ev$ is isometric $B_{E}=\ev^{-1}(B_{E^{**}})$, where $B_{E^{**}}$ is weak-$*$ compact by BAB. Hence ets
\[
\ev^{-1}: (E^{**}, \sigma(E^{**}, E^*) ) \to (E, \sigma(E, E^*))
\]
is continuous. Using the generators of the weak-topology, ets $\forall F\in E^*$:
\[
F\circ \ev^{-1}: (E^{**}, \sigma(E^{**}, E^*) ) \to \R
\]
is continuous. Indeed, by reflexivity $\forall \varphi = \ev(v)\in E^{**}$:
\[
F \circ \ev^{-1}(\varphi) =  F (v) = \ev(v)(F) = \varphi(F).
\]
So $F \circ \ev^{-1}$ is a generator of the weak-$*$ topology.
($\Leftarrow$) Firstly
\[
\ev: \sigma(E, E^*) \to \sigma(E^{**}, E^{*}).
\]
is cts as it is cts wrt the strong topologies.
Hence $\ev(B_E)\subseteq B_{E^{**}}$ is compact (in particular closed) and dense wrt $\sigma(E^{**}, E^{*})$. So $\ev(B_E)= B_{E^{**}}$.
\end{proof}

Q: Goldstein's Lemma
A: If $E$ is a Bsp, then $\ev(\bar{B}_E)$ is dense in $\bar{B}_{E^{**}}$ wrt the weak-$*$ topology $\sigma(E^{**}, E^*))$.
\begin{proof}
Given $\varphi\in \bar{B}_{E''}$ with open nbhd $V$ in $\sigma(E'', E')$, using the nbhd basis wlog
\[
V = \{ \psi\in E'': |\psi - \varphi|(F_i) < \varepsilon\}
\]
for some $\varepsilon>0, F_1, \cdots F_k\in E'$. Write .
Rtp $\ev(\bar{B}_E) \cap V \neq \emptyset$ so ets $\exists f\in \bar{B}_E$ such that
\[
|\ev(f) - \varphi|(F_i)  <\varepsilon.
\]
In vector form, we want
\[
\|H(f) - \boldsymbol\alpha\| \tag{$*$}
\]
small, where $\alpha_i := \varphi(F_i)$ and $H: E\to\R^k, f\mapsto (F_i(f))$ is linear cts.
Observe that $\forall\boldsymbol\beta\in\R^k$, as $\|\varphi\|_{E'}\leq 1$:
\[
|\boldsymbol\alpha \cdot \boldsymbol\beta| = | \sum_i \alpha_i \beta_i | = |\varphi(\beta_i F_i) | \leq \| \sum_i \beta_i F_i\|_{E'}. \tag{$\dagger$}
\]
Suppose $(*)$ is not satisfied, i.e. $\boldsymbol\alpha\not\in \overline{H(\bar{B}_E)}$.
Separating a subspace and point in $\R^k$, $\exists\boldsymbol\beta\in\R^k, \gamma\in\R$ such that $\forall f\in \bar{B}_E$:
\[
H(f) \cdot\boldsymbol\beta < \gamma < \boldsymbol\alpha \cdot \boldsymbol\beta.
\]
But taking sups
\[
\sup_{f\in \bar{B}_E} (H(f) \cdot\boldsymbol\beta) = \| \sum_i \beta_i F_i\|_{E'} \leq \gamma < \boldsymbol\alpha \cdot \boldsymbol\beta
\]
contradicting $(\dagger)$.
\end{proof}

Q: Weak to weak identity is continuous
A: Since weak-$*$ $\subseteq$ weak, i.e. $\sigma(E', E) \subseteq \sigma(E',E'')$, we have that
\[
\id: \quad E', \sigma(E', E'') \to E', \sigma(E', E)
\]
is continuous.

Q: Evaluation of the ball is strongly closed
A: If $E$ is a Bsp, then $\ev(\bar{B}_E)$ is closed in $E'', \|\ph\|_{E''}$. In particular, if $\ev(\bar{B}_E)$ is strongly dense in $B_{E''}$, then $E$ is reflexive.
\begin{proof}
\begin{align*}
(\ev(f_n)) \text{ convergent } &\implies (\ev(f_n)) \text{ Cauchy } \\
&\implies (f_n) \text{ Cauchy} \quad \text{(isometry)} \\
&\implies (f_n) \text{ convergent} \quad \text{(completeness)} \\
&\implies (\ev(f_n)) \to \ev(f) \quad \text{(continuity)}.
\end{align*}
\end{proof}

Q: Continuity carries from strong to weak
A: Let $\Phi: E_1 \to E_2$ be linear and cts for $E_1, E_2$ nvsps. Then
\[
\Phi: \quad E_1, \sigma (E_1, E_1') \to E_2, \sigma (E_2, E_2')
\]
is cts. In particular, as weak-$*$ $\subseteq$ weak, we have cty into weak-$*$.
\begin{proof}
Ets $\forall F \in E_2'$:
\[
F \circ \Phi: \sigma(E, E') \to \R
\]
is continuous.
But $F \circ \Phi\in E_1'$ so is indeed cts wrt weak top.
\end{proof}

Q: Closed subspace is reflexive
A: Let $E$ be a reflexive Bsp and $M\leq E$ closed. Then $M$ is reflexive.
\begin{proof}
We can check that
\[
\sigma(E, E') \cap M = \sigma(M, M')
\]
Indeed, $\subseteq$ follows by restriction and $\supseteq$ by extension using H-B.
Now $\bar{B}_M$ strongly closed $\implies$ $\sigma(E, E')$ closed.
But by Kakutani, $\bar{B}_E$ is $\sigma(E, E')$ compact. So $\bar{B}_M$ is $\sigma(E, E')$ compact. By restriction, $\bar{B}_M$ is $\sigma(M, M')$ compact.
\end{proof}

Q: Strongly closed implies weakly closed
A: Let $E$ be a nvsp and $M\leq E$ strongly closed. Then $M$ is $\sigma(E, E')$ closed.
\begin{proof}
Given $x\not\in M$, apply closed H-B to get $F\in E^*$ such that $F|_M<\alpha<F(x)$ for fixed .
Then $U_x:=F^{-1}(\alpha, \infty)$ is a weakly open set disjoint from $M$.
Hence $M=E\setminus U$ for $U=\bigcup_{x\not\in M} U_x$ weakly open.
\end{proof}

Q: Reflexivity of the dual space
A: A Bsp $E$ is reflexive $\iff$ $E'$ is reflexive.
\begin{proof}
($\Rightarrow$) If $E$ is reflexive, then since both are generated by the evaluations
\[
\sigma(E', E) = \sigma(E', E'') \tag{$*$}.
\]
Therefore
\begin{align*}
\bar{B}_{E'} \text{ is } \sigma(E', E) \text{ compact} \qquad & \text{ by BAB} \\
\bar{B}_{E'} \text{ is } \sigma(E', E'') \text{ compact} \qquad & \text{ by }(*) \\
E' \text{ is reflexive} \qquad  & \text{ by Kakutani}.
\end{align*}
($\Leftarrow$) If $E'$ is reflexive, then by first part $E''$ is reflexive. But $E$ is isometric to $\ev(E)\leq E''$ a closed subspace. So $E$ is reflexive.
\end{proof}

Q: Uniformly convex
A: A Bsp $E$ is uniformly convex if $\forall\varepsilon>0, \exists\delta>0$ such that $\forall f,g \in \bar{B}_E$:
\[
\|f - g\|_E > \varepsilon \implies \left\|\frac{f+g}{2} \right\|_E < 1 - \delta.
\]
\begin{eg}
$\mathcal{L}^p$ for $p\in (1, \infty)$.
\end{eg}

Q: Millman-Pettis Theorem
A: If $E$ is a uniformly convex Bsp, then $E$ is reflexive.
\begin{proof}
Since $\ev(\bar{B}_E)\leq \bar{B}_{E''}$ is strongly closed, ets strongly dense.
Given $\|\varphi\|_{E''}=1$ and $\varepsilon>0$, by UC $\exists\delta>0$ such that
\[
\|f-g\|_E \geq \varepsilon \implies \left\| \frac{f+g}{2} \right\|_E < 1- \delta. \quad (\forall f,g\in \bar{B}_E).
\]
By operator norm $\exists F\in \bar{B}_{E'}$ such that $\varphi(F)>1-\frac{\delta}{2}$. Consider the $\sigma(E'', E')$ open nbhd of $\varphi$
\[
V = \left\{ \psi\in E'': |\psi-\varphi|(F)<\frac{\delta}{2}\right\}.
\]
By Goldstine, $\ev(f)\in V$ for some $f\in \bar{B}_{E}$.
If $\|\ev(f)-\varphi \|_{E''}\leq \varepsilon$, then done. Otherwise
\[
\varphi \in W:= (\ev(f) + \varepsilon \bar{B}_{E''})^C = \bigcup_{F\in \bar{B}_{E'}} \{\psi: |\psi-\ev(f)|(F)>\varepsilon \}
\]
where $W$ is another $\sigma(E'', E)$ open nbhd of $\varphi$. By Goldstine $\ev(g)\in V\cap W$ for some $g\in B_E$. Therefore
\begin{align*}
 |\ev(f)-\varphi|(F)&<\frac{\delta}{2} \qquad (f\in V)\\
|\ev(g)-\varphi|(F)&<\frac{\delta}{2} \qquad (g\in V) \\
2\varphi(F) &\leq F(f+g) + \delta \qquad \text{(summing above)} \\
1 - \frac{\delta}{2} < \varphi(F) &\leq \left\|\frac{f+g}{2}\right\|_E + \frac{\delta}{2} \qquad (F\in \bar{B}_{E'}).
\end{align*}
By UC $\|f-g\|_E\leq \varepsilon$. By isometry $\|\ev(f)-\ev(g)\|_{E''}\leq \varepsilon$, contradicting $\ev(g)\in W$.
\end{proof}

Q: Convergence for uniformly convex Banach space
A: Let $E$ be a UC Bsp.  Then
\[
(f_n)\to f \iff (f_n)\wto f, \|f_n\|\to \|f\|.
\]
\begin{proof}
($\Rightarrow$) Immediate.
($\Leftarrow$) If $f=0$, then done. Otherwise, $\|f_n\|\neq 0$ ev. Now define
\[
g_n = \frac{f_n}{\|f_n\|}, \quad g = \frac{f}{\|f\|} \in \bar{B}_{E}.
\]
Then $\frac{g_n+g}{2}\in\bar{B}_E \wto g$. By Fatou
\[
1 = \|g\| =\left\| \lim_{n\to\infty} \left(\frac{g_n+g}{2}\right)\right\| \leq \lim\inf \left\| \frac{g_n+g}{2} \right\| \leq \lim\sup \left\| \frac{g_n+g}{2} \right\|\leq 1.
\]
So $\| \frac{g_n+g}{2} \| \to 1$. By UC $\|g_n-g\| \to 0$.
\end{proof}

Q: Separability of the dual
A: Let $E$ be a Bsp and $\{F_n: n\in\N \}\subseteq E'$ is dense. Then $E$ is separable.
\begin{proof}
By operator norm can pick $(f_n)\subseteq\bar{B}_E$ such that $F_n(f_n)\geq \frac{1}{2]}\|F_n\|_{E'}$.
Then $L=\spn\{f_n\}$ is separable (e.g. rational LCs). Ets $L\leq E$ is dense. Suppose not. Then $\exists f_0\in E\setminus\bar{L}$. By closed H-B $\exists G\in E'$ separating $\bar{L}$ and $\{f_0\}$, i.e. $\exists\alpha\in\R$ such that
\[
G|_L < \alpha < G(f_0).
\]
But $\im G|_L$ is a subspace so $G|_L=0$. By lemma $G=0$, contradicting $G(f_0)>0$.
\emph{Annhiliator lemma}. Suppose $F\in E'$ and $F|_L=0$. We claim $F=0$. Given $\varepsilon>0$, $\exists n_0\in\N$ such that
\begin{align*}
\|F-F_{n_0}\| &\leq \varepsilon \\
\|F_{n_0}\| &\leq 2F_{n_0}(f_{n_0}) \qquad (\text{dfn of $f_n$}) \\
&= 2(F_{n_0}-F)(f_{n_0})  \\
&\leq 2 \varepsilon \\
\|F\| &\leq 3\varepsilon.
\end{align*}
\end{proof}

Q: Reflexive and separable dual
A: A Bsp $E$ is reflexive and separable iff $E'$ is reflexive and separable.

Q: Separability via weak-$*$ ball
A: A Bsp $E$ is separable $\iff$ $(\bar{B}_{E'}, \sigma(E', E))$ is metrisable.
\begin{proof}
TBC
\end{proof}

Q: Countable BAB for separable spaces
A: Let $E$ be a \emph{separable} Bsp. Then closed bounded subsets of $E', \sigma(E', E)$ are sequentially \emph{weak-$*$} compact.
\begin{proof}
Let $(f_k)\subseteq E$ be dense, $(F_n)\subseteq \bar{B}_{E'}$. Then $\forall k\in\N$:
\[
(F_n(f_k))_n \subseteq [-\|f_k\|, \|f_k\|] \subseteq\R.
\]
Apply B-W. By diagonalisation $\exists (F_{\theta(n)}(f_k))_n$ converging for every $k$.
Given $f\in E, \varepsilon>0$, by density $\exists f_{k_0}$ such that $\|f-f_{k_0}\|<\varepsilon / 3$. Moreover by Cauchy convergence $\exists N\in\N$ such that $\forall n_1,n_2\geq N$:
\[
|F_{\theta(n_1)}(f_{k_0}) - F_{\theta(n_2)}(f_{k_0}) | < \frac{\varepsilon}{3} .
\]
By $3\varepsilon$ argument
\[
|F_{\theta(n_1)}(f) - F_{\theta(n_2)}(f) | < \varepsilon .
\]
So $(F_{\theta(n)}(f))$ is (Cauchy) convergent and thus $F_{\theta(n)}$ is weak-$*$ convergent.
\end{proof}

Q: Countable BAB for reflexive spaces
A: Let $E$ be a \emph{reflexive} Bsp. Then closed bounded subsets of $E$ are sequentially \emph{weak} compact.
\begin{proof}
Let $(f_n)\subseteq \bar{B}_E$ and define $L=\overline{\spn\{f_n\}} \subseteq E$.
\emph{Separability}. $L$ is separable (e.g. by $\Q$-LCs). In particular $L''$ is separable.
\emph{Reflexivity}. By restriction $\bar{B}_L, \sigma(L, L')$ is weakly compact. By Kakutani $L$ is reflexive.
By reflexivity $\bar{B}_L, \sigma(L,L')$ identifies with $\bar{B}_{L''}, \sigma(L'',L')$ which is sequentially compact by separable BAB.
\end{proof}

Q: Ball criterion for separability
A: A Bsp $E$ is separable $\iff$ $\bar{B}_{E'}, \sigma(E', E)$ is metrisable.
\begin{proof}
($\Rightarrow$) Let $\{f_n\}\subseteq \bar{B}_E$ be dense. Define the metric on $\bar{B}_{E'}$
\[
d(F,G) = \sum_{n\in\N} \frac{|F(f_n) - G(f_n)|}{2^n} \leq \|F-G\|_{E'}.
\]
Positive definite: $d(F,G)=0 \iff F=G$ on $\{f_n\}$ $\iff F=G$ on $E$.
$\Delta$-$\leq$: clear by truncating. Given a $d$-nbhd
\begin{align*}
B(F,\varepsilon) &= \{G\in \bar{B}_{E'}: d(F,G)<\varepsilon \} \\
&\supseteq \{G \in \bar{B}_{E'}: |G(f_i) - F(f_i)| < \frac{\varepsilon}{2}, i\leq n_0, \sum_{n\geq n_0} \frac{1}{2^n} \leq \frac{\varepsilon}{2}\}. \\
\intertext{Conversely, given a $\sigma(E, E')$ nbhd}
U&=\{G\in \bar{B}_{E'}: |G(g_i) - F(g_i)|<\varepsilon, i\leq k \}, \\
\intertext{then $\exists (n_i)_1^k$ such that $\|f_{n_i}-g_{n_i}\|<\frac{\varepsilon}{3}$. Define the $d$-nbhd}
V &=\{G\in\bar{B}_{E'}: d(F,G)<\frac{\varepsilon}{3}2^{-\max(n_i)} \}.
\end{align*}
Ets $V\subseteq U$. Indeed $\forall G\in V$, $d(F,G) < \frac{\varepsilon}{3}2^{-\max(n_i)}$, so $\forall i\leq k$:
\begin{align*}
|F(f_{n_i}) - G(f_{n_i}) | &< \frac{\varepsilon}{3} \\
|F(g_i) - G(g_i)| &\leq |F(g_i) - F(f_{n_i})| + |F(f_{n_i}) - G(f_{n_i}) | + |G(f_{n_i}) - G(g_i)| <\varepsilon.
\end{align*}
($\Leftarrow$) For each $n\in\N$, write $U_n = B(0, \frac{1}{n})$, i.e.
\begin{align*}
U_n &= \{G\in \bar{B}_{E'}: d(G,0)<\tfrac{1}{n} \}. \\
\intertext{By weak-$*$ openness, $\exists V_n\subseteq U_n$ such that}
V_n &= \{G\in \bar{B}_{E'}: |G(f)|<\varepsilon_n, f\in C_n \} \quad (|C_n|<\infty, \varepsilon_n>0).
\end{align*}
Then $C_\infty = \bigcup_n C_n \subseteq E$ is ctble and $\bigcap_n V_n\subseteq \bigcap U_n = \{0\}$.
Now if $F\in E'$ such that $F|_{C^\infty}\equiv 0$, then $F\in \bigcap_n V_n=\{0\}$. By closed H-B, $C_\infty$ is dense.
\end{proof}

Q: Clarkson inequality
A: For $p\in [2,\infty), f,g\in\mathcal{L}^p(\R)$:
\[
\left\|\frac{f+g}{2}\right\|_{p}^p + \left\|\frac{f-g}{2}\right\|_{p}^p \leq \frac{\|f\|_p^p +\|g\|_p^p}{2}.
\]
\begin{proof}
For $\theta\in[0,1]$, since $p/2\geq 1$:
\[
\theta^{p/2} + (1-\theta)^{p/2} \leq \theta + (1-\theta) = 1.
\]
Substituting $\theta = \frac{a^2}{a^2+b^2}$ for $a,b\geq 0$ gives
\[
a^p + b^p \leq (a^2+b^2)^{p/2}.
\]
Substituting $a=\frac{|x+y|}{2}$, $b=\frac{|x-y|}{2}$ and using convexity of $(\ph)^{p/2}$:
\begin{align*}
\left| \frac{x+y}{2}\right| ^p + \left| \frac{x-y}{2}\right| ^p &\leq \left( \left| \frac{x+y}{2}\right| ^2 + \left| \frac{x-y}{2}\right| ^2\right)^{p/2} \\
&= \left(\frac{x^2+y^2}{2}\right)^{p/2} \\
&\leq \frac{|x|^p+|y|^p}{2}.
\end{align*}
\end{proof}

Q: Reflexivity of $\mathcal{L}^p$
A: $\mathcal{L}^p(\R)$ is reflexive for $p\in(1,\infty)$.
\begin{proof}
\textbf{Case} $p\in[2,\infty)$. By Millman-Pettis, ets UC. Given $\varepsilon>0$ and $f,g\in\bar{B}_{\mathcal{L}^p}$ with $\|f-g\|_p<\varepsilon$, by Clarkson's inequality
\[
\left\|\frac{f+g}{2}\right\|_p^p + \left(\frac{\varepsilon}{2}\right)^p \leq \frac{\|f\|_p^p + \|g\|_p^p}{2} \leq 1.
\]
\textbf{Case} $p\in (1,2]$. Define the canonical isometric embedding $\mathcal{F}:\mathcal{L}^p\to (\mathcal{L}^q)'$ by $u\mapsto F_u$ where
\[
F_u(f)=\int_\R u f.
\]
Therefore $\mathcal{L}^p\sim S\leq (\mathcal{L}^q)'$. By isometry and completeness, $S$ is closed. By the first case $S$ is reflexive.
\end{proof}

Q: Riesz representation for $L^p$
A: $\mathcal{L}^p(\R)$ with $p\in(1,\infty)$ has $(\mathcal{L}^p)'=\mathcal{L}^q$ and $\forall F\in(\mathcal{L}^p)'$, $\exists g\in \mathcal{L}^q$ such that
\[
F(f)= \int_\R fg.
\]
\begin{proof}
Define $\mathcal{F}:\mathcal{L}^p\to (\mathcal{L}^q)'$ by $u\mapsto F_u$ given by
\[
F_u(f) = \int_\R fg.
\]
\emph{Linearity} of $\int$ gives linearity of $\mathcal{G}$.
\emph{Isometry}. By H\"older's inequality
\[
|F_u(f)| \leq \|u\|_p \|f_u\|_q.
\]
So $\|F_g\| \leq \|u\|_p$. Define $f_u=|u|^{p-2}u\in \mathcal{L}^q$ with
\[
\|f_u\|_q = \left(\int |u|^p\right)^{1/q} =\| u\|_p^{p-1}
\]
and
\[
F_u(f_u) = \int_\R |u|^p = \|u\|_p^p = \|u\|_p \|f_u\|_q.
\]
Hence the inequality is saturated: $\|F_u\|= \|u\|_p$.
\emph{Surjectivity}. Suppose $\varphi\in (\mathcal{L}^q)''$ such that $\varphi|_{\mathcal{G}(\mathcal{L}^p)}=0$. By reflexivity $\varphi=\ev(h)$ for some $h\in \mathcal{L}^q$. Therefore
\[
\ev(h)(\mathcal{F}(u)) = F_u(h) = \int_\R hu.
\]
In particular for $u=|h|^{q-2}h\in \mathcal{L}^p$:
\[
\varphi(\mathcal{F}(u)) = \int_\R |h|^q = 0.
\]
So $h=0$ and thus $\varphi=0$. By H-B $\mathcal{G}(\mathcal{L}^p)$ is dense.
\end{proof}

Q: Riesz representation for $L^1$
A: $L^1(\R)'=L^\infty(\R)$ and $\forall F\in L^1, \exists g \in L^\infty$ such that
\[
F(f) = \int_\R fg.
\]
\begin{proof}
\emph{Isometry}. Given $g\in L^\infty$, by H\"older $F_g: f\mapsto \int fg \in (L^1)'$ and $\|F_g\|\leq \|g\|_\infty$.
Suppose $\|g\|_\infty > \|F_g\|$. Then $\exists C>\|F_g\|$ such that $S=\{|g|>C\}$ is non-null. Hence $\exists A\subseteq S$ with $\mu(A)\in (0,\infty)$ (by intersecting with balls).
Define $f_0=\sgn(g)\mathbbm{1}_A\in L^1$. Then
\[
C\mu(A) \leq F_g(f_0) = \int_A |g| \leq \|F_g\| \|f_0\|_1 = \|F_g\| \mu(A)
\]
contradicting $C> \|F_g\| $.
\emph{Density}. Conversely, given $F\in (L^1)'$, define weight $w(x) = 1/ (1+|x|)\in L^2$ and $G(h)=F(wh)$. Then by C-S $wL^2\subseteq L^1$. So $G\in (L^2)'$. By Riesz $\exists v\in L^2$ such that $G(h)=\int vh $.
Define $g=vw^{-1}$. If $f\in wL^2$, then $F(f)=G(fw^{-1})=\int v (fw^{-1}) = \int gf$. By above $\|g\|_\infty = \|F\|$.
Finally $C_c^0\subseteq wL^2$ is dense in $L^1$ so $F=F_g$ on the whole of $L^1$.
\end{proof}

Q: Reflexivity of $L^1$
A: $L^1(\R)$ is not reflexive.
\begin{proof}
Define unit apxs
\[
f_n = \frac{1}{\mu(B(0,\frac{1}{n}))} \mathbbm{1}_{B(0,\frac{1}{n})}.
\]
Supposing $L^1$ reflexive, by Kakutani $\exists \theta(n)$ such that $f_{\theta(n)}\wto f\in \bar{B}_{L^1}(0,1)$. That is, $\forall g\in L^\infty$
\[
\int f_{\theta(n)} g \to \int fg.
\]
For $g\in C_c^0(\R\setminus\{0\})$, LHS $\to 0$ so $\supp(f)\subseteq\{0\}$. Thus $f=0$ a.e.
For $g=1$, LHS $=1$ but RHS $=0$!
\end{proof}

Q: Reflexivity/separability/compactness of $L^\infty$
A: $L^\infty = (L^1)'$.
$L^1$ separable $\implies \bar{B}_{L^\infty}$ metrisable and weak-$*$ (sequentially) compact (BAB).
$L^1$ not reflexive $\implies$ $L^\infty$ not reflexive.
$L^\infty$ not separable, e.g. $\{\mathbbm{1}_{[-r,r]}): r>0\}$ is uncountable and the members are all distance 1 apart.

Q: Dual of $L^\infty$
A: By H\"older $L^1\subset (L^\infty)'$.
To see strictness, take unit apxs and use weak-* sequential compactness. Alternatively extend $C_c^0\to\R, f\mapsto f(0)$ to $L^\infty$ by H-B.
In fact $(L^\infty)'$ is the finitely additive signed measures absly cts wrt LM.

Q: Continuous functions in $L^\infty$
A: The closure of $C_c^0\subseteq L^\infty$ is $C_0(\R)$, the cts fns which go to zero at infinity.

Q: Riesz-Markov Theorem
A: $C_0(\R)'=\mathcal{M}(\R)$, the real-valued radon measures. Specifically, $\mathcal{T}: \mathcal{M} \to C_0'$ defined by $\mu\mapsto F_\mu$ where
\[
F_\mu(f) = \int f \; \d \mu
\]
and
\[
\|F _\mu \|=\|\mu\|_{TV}:= \sup\left\{ \sum|\mu(A_j)|: \R=\bigsqcup A_j \right\}
\]
is an isometric isomorphism.

Q: Weak vs. strong boundedness
A: Let $E$ be a Bsp. Then $B\subseteq E$ is bounded $\iff$ $\forall F\in E': F(B)\subseteq \R$ is bounded.
\begin{proof}
$(\Rightarrow)$ Clear.
$(\Leftarrow)$ Apply Banach-Steinhaus to $(\ev(f))_{f\in B}$.
\end{proof}

Q: Weak-$*$ vs. strong boundedness
A: Let $E$ be a Bsp. Then $B'\subseteq E'$ is bounded $\iff \forall f\in E: \ev(f)(B')\subseteq\R$ is bounded.
\begin{proof}
($\Rightarrow$) Clear.
($\Leftarrow$) Apply Banach-Steinhaus to $(F)_{F\in B'}$.
\end{proof}