
\section*{Question 1}
\textbf{(i)} First part bookwork. 

\textbf{(ii)} Let $T x = \sequence {f_\gamma(x)}_{\gamma \in \Gamma}$. If $T$ is bounded, then there exists $M$ such that $\sup_{\gamma \in \Gamma} |f_\gamma(x)| \le M \norm x$ for each $x \in X$. In particular $|f_\gamma(x)| \le M$ for each $x \in X$. So $\set {f_\gamma : \gamma \in \Gamma} \subseteq X^\ast$ and further $\norm {f_\gamma}_\infty \le M$ for each $\gamma \in \Gamma$, so this set is bounded. 

Take $\alpha = \sup_{\gamma \in \Gamma} \norm {f_\gamma}_\infty$. It is clear that: 
\[\norm {T x} = \sup_{\gamma \in \Gamma} |f_\gamma(x)| \le \alpha \norm x\]
Now note that for each $\epsilon > 0$ there exists $\gamma \in \Gamma$ such that $\alpha - \epsilon/2 < \norm {f_\gamma}_\infty \le \alpha$. Then there exists $\norm {e_\gamma} = 1$ in $X$ such that $\alpha \ge |f_\gamma e_\gamma| > \alpha - \epsilon$. So $(\alpha - \epsilon) \norm {e_\gamma} < \norm {T e_\gamma} \le \alpha \norm {e_\gamma}$. Since $\epsilon$ was arbitrary we have $\norm T = \alpha$. 

Conversely, if $\set {f_\gamma : \gamma \in \Gamma}$ is bounded and we have $\norm {T x} = \sup_{\gamma \in \Gamma} |f_\gamma(x)| \le \alpha \norm x$ as in the previous case. So $T$ is bounded. 

\textbf{(iii)} Pick $\Gamma = X$ and take $\sequence {f_x}_{x \in X}$ to be the support functional at $x$. Then $\set {f_x : x \in X} \subseteq X^\ast$ is bounded and so $T y = \sequence {f_x(y)}_{n \in \N}$ is continuous. For each $y \in X$ we have $|f_x(y)| \le \norm y$ with equality if $y = x$. Then $\norm {T y} = \sup_{x \in X} |f_x(y)| = \norm y$. So $T$ is an isometric embedding.

If $X$ is separable, take $\Gamma = \N$, a dense subset $\sequence {e_n}_{n \in \N}$ and the corresponding support functionals $\sequence {f_n}_{n \in \N}$. We have $\norm {T x} = \sup_{n \in \N} |f_n(x)|$. We have $|f_n(e_n)| \le \norm {e_n}$ for each $e_n$, while $|f_n(x)| \le \norm x$ for all $x \in X$. So we have $\norm {T e_n} = \norm {e_n}$ for each $n$. Since $\sequence {e_n}_{n \in \N}$ is dense, we have $\norm {T x} = \norm x$ for all $x \in X$ by continuity. 

If $X = Y^\ast$ for separable $Y$, take $\sequence {x_n}_{n \in \N}$ a dense subset of $B_Y$ and consider the bounded subset $\set {\widehat {x_n} : n \in \N} \subseteq Y^{\ast \ast} = X^\ast$. Then define $T f = \sequence {f(x_n)}_{n \in \N}$. For each $f \in X$ and $\epsilon > 0$ there exists $y \in S_Y$ such that $\norm f_\infty - \epsilon/2 < |f(y)| \le \norm f_\infty$. In particular, by approximating $y$ there exists some $n$ such that $\norm f_\infty - \epsilon < |f(x_n)| \le \norm f_\infty$. It follows that $\norm {T f} = \sup_{n \in \N} |f(x_n)| = \norm f_\infty$. 

\textbf{(iv)} Take $T \in B(Y, \ell_\infty(\Gamma))$. Then $T y = \sequence {f_\gamma(y)}_{\gamma \in \Gamma}$ for each $y \in Y$. By part ii, each $f_\gamma : Y \to \C$ is in $Y^\ast$, and since $\norm T = \sup_{\gamma \in \Gamma} \norm {f_\gamma}_\infty$, we have $\norm {f_\gamma}_\infty \le \norm T$ for each $\gamma \in \Gamma$. Applying Hahn--Banach to each $f_\gamma$, there exists $\widetilde {f_\gamma} : Z \to \C$ such that $\lVert {\widetilde {f_\gamma}}\rVert \le \norm T$. Then $\widetilde T = \langle {\widetilde {f_\gamma}\rangle}_{\gamma \in \Gamma}$ works and satisfies $\lVert \widetilde T\rVert \le \norm T$. 

\textbf{(v)} \textit{If}: If the condition holds, then there exists an isometric embedding $T : X \to \ell_\infty(\Gamma)$ for a set $\Gamma$. Then there exists a bounded linear map $P : \ell_\infty(\Gamma) \to X$ such that $\norm P \le \lambda$ and $P \circ T$ is the identity on $X$. Now let $Y$ be a subspace of a normed space $Z$ and let $S : Y \to X$ be a bounded linear map. Then $T \circ S : Y \to \ell_\infty(\Gamma)$ is bounded. We have shown that $\ell_\infty(\Gamma)$ is $1$--injective, so there exists $\widetilde {T \circ S} : Z \to \ell_\infty(\Gamma)$ such that $\widetilde {T \circ S}$ restricts to $T \circ S$ and $\norm {\widetilde {T \circ S}} \le \norm {T \circ S}$. Then $P \circ \widetilde {T \circ S}$ restricts to $P \circ (T \circ S) = S$. Finally we have, since $T$ is an isometry: 
\[\norm {P \circ \widetilde {T \circ S}} \le \norm {\widetilde {T \circ S}} \le \lambda \norm {T \circ S} = \lambda \norm S\]
So $X$ is $\lambda$--injective. 

\emph{Only if}: Suppose that $X$ is $\lambda$--injective and let $T : X \to Z$ be an isometric embedding. Then take $Y = T(x)$. We can take the bounded inverse $T^{-1} : T(X) \to X$. Then there exists $\widetilde {T^{-1}} \in B(Z, X)$ such that $\widetilde {T^{-1}}$ restricts to $T^{-1}$ with $\norm {\widetilde {T^{-1}}} \le \lambda \norm {T^{-1}}$. Take $P = \widetilde {T^{-1}}$. Then $P \circ T$ is the identity since $P$ restricts to $T^{-1}$ on $Y$. Further, $\norm {\widetilde {T^{-1}}} \le \lambda \norm {T^{-1}} = \lambda$. 

\section{Question 2}
\subsection*{Part a}
Bookwork up to last part, which is also partly bookwork. 

Suppose $\set {f_n : n \in \N} \subseteq B_{X^\ast}$ separates the points of $X$. Let $\sigma$ be the initial topology induced by the $\set {f_n : n \in \N} \subseteq B_{X^\ast}$ and let $K$ be a $w$--compact set. It is bookwork to prove that $\sigma$ is metrizable. Then the inclusion $(K, w) \to (K, \sigma)$ is a continuous bijection from a compact space to a Hausdorff space, and hence $(K, w) = (K, \sigma)$. That is, $(K, w)$ is metrizable. 

\subsection*{Part b}
\subsubsection*{Part bi}
Let $\sequence {x_n}_{n \in \N}$ be a sequence in a Banach space $X$ that is weakly convergent to $x$. Then $\set {x_n : n \in \N} \cup \set x$ is weakly compact, and hence is norm bounded by the bookwork in part a. We deduce that any weakly convergent sequence is norm bounded. 

Since $\sequence {f_n}_{n \in \N}$ converges weakly to $0$, it is a norm--bounded sequence, say $\norm {f_n}_\infty \le M$. Since point evaluations are bounded linear functionals, we also have that $f_n \to 0$ pointwise. Then applying the dominated convergence theorem we have $\norm {f_n}_1 \to 0$. 

\subsubsection*{Part bii}
Let $\alpha = \inf_{y \in C} \norm y$. For each $n \in \N$, pick $x_n \in C$ such that $\alpha < \norm {x_n} < \alpha + \frac 1 n$. Let $Y = \overline {\Span \set {x_n : n \in \N}}$. Since $Y$ is a closed subspace of a reflexive space, it is reflexive. That is, $(B_Y, w)$ is compact. 

$Y$ is also separable as the closed linear span of a countable set. Take a dense countable subset $\sequence {e_n}_{n \in \N}$. Then the countable subset $\sequence {\widehat {e_n}}_{n \in \N} \subseteq B_{Y^{\ast \ast}}$ separates the points of $Y^\ast$. So the topology on any weakly compact subset of $Y^\ast$ is metrizable. In particular, $(B_{Y^\ast}, w)$ is metrizable. Since $Y$ is reflexive we have $(B_{Y^\ast}, w) = (B_{Y^\ast}, w^\ast)$. So $(B_{Y^\ast}, w)$ is compact and metrizable, and hence must be weakly separable. 

Note that if $\mathcal S = \set {z_n : n \in \N}$ is $w^\ast$--dense in $B_{Y^\ast}$, it must separate the points of $Y$: if $f, g \in B_{Y^\ast}$ had $f(z_n) = g(z_n)$ for each $n$, we would obtain that $f = g$ from the continuity of $f$ and $g$. It follows that every weakly compact subset of $Y$ is metrizable, in particular $(B_Y, w)$ is metrizable, hence sequentially compact. (since we already know it to be compact)

By construction, $\sequence {x_n}_{n \in \N}$ is norm (hence weakly) bounded and hence is contained in some ball $M B_Y \cap C$. This set is a $w$--closed subset of $B_Y$, and so is metrizable and compact, hence sequentially compact. So some subsequence $\langle x_{n_j}\rangle_{j \in \N}$ converges weakly to $x \in M B_Y \cap C$. Then:
\[\norm x \le \liminf_{n \to \infty} \norm {x_n} = \alpha\]
But since $\alpha = \inf_{y \in C} \norm y$, we must therefore have $\norm x = \inf_{y \in C} \norm y$.

\subsection*{Part biii}
Let $K$ be a $w$--compact subset of $\ell_\infty$. Then the coordinate functionals sending $\sequence {x_n}_{n \in \N}$ to $x_n$ separate the points of $\ell_\infty$. So $K$ is metrizable. Then $K$ is $w$--separable, with countable dense subset $\set {z_n : n \in \N}$. Then $\overline {\Span K} = \overline {\Span \set {z_n : n \in \N}}^w = \overline {\Span \set {z_n : n \in \N}}$ by Mazur. So $\Span K$ is norm--separable, and hence so is $K$ as a metric subspace. 

\section*{Question 3}
\subsection*{Part a}
Example sheets

\subsection*{Part b}
Bookwork

\subsection*{Part c}
Bookwork: prove that if $X$ is separable then $(B_{X^\ast}, w^\ast)$ is metrizable. 

Since $X^\ast$ is separable, $(B_{X^{\ast \ast}}, w^\ast)$ is metrizable and hence sequentially compact by Banach--Alaoglu. A norm--bounded sequence $\sequence {x_n}_{n \in \N}$ in $X$ remains norm (hence $w^\ast$) bounded in $X^{\ast \ast}$. So there exists $\phi \in X^{\ast \ast}$ such that $\widehat {x_n} \xrightarrow{w^\ast} \phi$. That is, $\phi(f) = \lim_{n \to \infty} \widehat {x_n}(f) = \lim_{n \to \infty} f(x_n)$ for each $f \in X^\ast$. 

If $X$ is not reflexive, there exists $\phi \in B_{X^{\ast \ast}} \setminus B_X$. Since $(B_{X^{\ast \ast}}, w^\ast)$ is metrizable and closed, it is sequentially closed. Since $B_X$ is $w^\ast$--dense in $B_{X^{\ast \ast}}$ by Goldstine, there exists a sequence $\sequence {x_n}_{n \in \N}$ in $B_X$ (hence norm bonuded) with $\widehat {x_n} \xrightarrow{w^\ast} \phi$. If some subsequence of $\sequence {x_n}_{n \in \N}$, say $\langle x_{n_j}\rangle_{j \in \N}$ converged  weakly to $x$ in $X$, we would have $\widehat {x_{n_j}} \xrightarrow{w^\ast} \widehat x$ in $X^{\ast \ast}$, (since the embedding is $w$--to--$w^\ast$ continuous) hence $\widehat x = \phi$, contrary to our assumption that $\phi \not \in X$. So $\sequence {x_n}_{n \in \N}$ is our desired sequence. 

\section*{Question 4}
\subsection*{Part a}
Bookwork.

\subsection*{Part b}
Bookwork up to $\exp(x) = \cdots$. WLOG take $A$ to be commutative, by taking the maximal commutative subalgebra containing $x$. Then define $\exp : U \to \C$ by $f(z) = \sum_{n = 0}^\infty \frac {z^n} {n!}$, a holomorphic function that is the uniform limit of the polynomials $P_N(z) = \sum_{n = 0}^N \frac {z^n} {n!}$.  Then we have: 
\begin{align*}
\exp(x) & := \Theta_x(f) \\ & = \Theta_x(\lim_{n \to \infty} P_N) \\ & = \lim_{N \to \infty} \Theta_x(P_N) \\ & = \lim_{N \to \infty} \sum_{n = 0}^N \frac {x^n} {n!}
\end{align*}
where we have used that $\Theta_x$ is continuous and sends polynomials to the analogy in the Banach algebra. So we get $\exp(x) = \sum_{n = 0}^\infty \frac {x^n} {n!}$. 

Bookwork proving form of spectrum.

From the previous part we have $\sigma_A(f(x)) = \set {f(\lambda) : \lambda \in \sigma_A(x)} \subseteq V$ since $\sigma_A(x) \subseteq U$ and $f(U) \subseteq V$. Consider the HFCs $\Theta_x : \mathcal O(V) \to A$ and $\Theta_{f(x)} : \mathcal O(V) \to A$. We aim to show that $g \mapsto \Theta_x(g \circ f)$ and $g \mapsto \Theta_{f(x)}(g)$ both satisfy the conditions for the HFC and so must be equal. 

They are certainly both homomorphisms and $g \mapsto \Theta_x(g \circ f)$ is continuous as the composition of continuous functions. Further, both maps are unital since $\Theta_x(1 \circ f) = \Theta_x(1) = 1 = \Theta_{f(x)}(1)$. Finally we can see that $\Theta_x(\mathrm{id} \circ f) = \Theta_x(f) = f(x) = \Theta_{f(x)} (\mathrm{id})$. Hence we deduce that $\Theta_x(\bullet \circ f) = \Theta_{f(x)}$ by the uniqueness part of the HFC. So for each holomorphic $g$ we have $\Theta_x(g \circ f) = (g \circ f)(x) = \Theta_{f(x)}(g) = g(f(x))$. 

Let $U = B_{\norm x + \epsilon}(0)$ with $\epsilon$ picked so that $\norm x + \epsilon < 1$ and let $V$ be an open subset containing $U$. Define a logarithm $\log$ on $1 - B_{\norm x + \epsilon}(0)$ (note that this is still a positive distance from the origin) and let $f : U \to \C$ be defined by $f(z) = \log(1 - z)$. Define $g : f(U) \to \C$ to be the exponential $g(z) = \exp(z)$. Then we have $(g \circ f)(x) = (1_U - \mathrm{id})(x) = 1 - x = g(f(x)) = \exp(f(x))$. Letting $y = f(x)$ gives the result $\exp(y) = 1 - x$. 


\section*{Question 5}
Bookwork up to Invariant Subspace Problem. 

Let $\lambda_1$ and $\lambda_2$ be distinct points in $\sigma(T)$ and fix disjoint open neighborhoods $U_1, U_2$ thereof in $K$. Consider the projections $P_1 = P(U_1)$ and $P_2 = P(U_2)$. $P_1$ is certainly non--zero since $P(U)$ is non--zero for open $U$. We have $P(U_1)P(U_2) = P(U_1 \cap U_2) = 0$, which means that $P_1 \ne I$. $P_1$ clearly commutes with every projection as a projection itself, hence we have $TP_1 = P_1T$. Then $V = \ker(I - P_1) = \mathrm{im}(P_1)$ is a non--trivial invariant subspace. 
