\section*{Question 1}
Pure bookwork. 

\section*{Question 2}
Bookwork omitted.

\textbf{(d)} Suppose $\overline C$ is weakly compact in $X$. Then it is $w^\ast$ compact, hence $w^\ast$ closed, in $X^{\ast \ast}$. Since $C \subseteq \overline C$, it follows that $\overline C^{w^\ast} \subseteq \overline C \subseteq X$. (since the $w^\ast$ topology is smaller than the norm topology, we obtain $\overline C^{w^\ast} = \overline C$ in this case) 

Now suppose that $\overline C^{w^\ast} \subseteq X$. Since $C$ is bounded, we have that $C \subseteq M B_X$ for some $M > 0$, so $\overline C^{w^\ast}$ is a $w^\ast$--closed subset of the $w^\ast$--compact set $M B_{X^{\ast \ast}}$, and hence is $w^\ast$--compact itself. But $\overline C^{w^\ast} \subseteq X$, so it is $w$--compact in $X$. (take the preimage under the embedding $X \to X \subseteq X^{\ast \ast}$, which is a $w$--to--$w^\ast$ homomorphism) Note that since $\overline C$ is norm closed and $C$ is convex, $\overline C$ is also weakly closed, so $\overline C \subseteq \overline C^{w^\ast}$. Since $\overline C$ is then a weakly closed subset of a weakly compact set, it is itself weakly compact. 

\textbf{(e)} That dual operators are $w^\ast$--to--$w^\ast$ continuous is on the example sheets. Note from this that $T^{\ast \ast}$ extends $T$, since it sends $\widehat x$ to $\widehat {T x}$. Suppose that $T^{\ast \ast}(X^{\ast \ast}) \subseteq Y$. Then in particular $T^{\ast \ast}(B_{X^{\ast \ast}}) \subseteq Y$, with $T^{\ast \ast}(B_{X^{\ast \ast}})$ being $w^\ast$--compact since $B_{X^{\ast \ast}}$ is. Since $T^{\ast \ast}$ extends $T$ we have $T B_X \subseteq T^{\ast \ast}(B_{X^{\ast \ast}})$. Then $\overline {T B_X}^{w^\ast} \subseteq T^{\ast \ast}(B_{X^{\ast \ast}}) \subseteq Y$. So $\overline {T B_X}$ is weakly compact. 

Now suppose that $\overline {T B_X}$ is weakly compact then $\overline {T B_X}^{w^\ast} \subseteq Y$. Recalling again that $T^{\ast \ast}$ extends $T$, we can see that $(T^{\ast \ast})^{-1}(\overline {T B_X}^{w^\ast}) \supseteq T^{-1}(\overline {T B_X}^{w^\ast}) \supseteq B_X$. Since $\overline {T B_X}^{w^\ast}$ is $w^\ast$--closed, so is $(T^{\ast \ast})^{-1}(\overline {T B_X}^{w^\ast})$. So by taking $w^\ast$--closures we have $B_{X^{\ast \ast}} \subseteq (T^{\ast \ast})^{-1}(\overline {T B_X}^{w^\ast})$. So $T^{\ast \ast}(B_{X^{\ast \ast}}) \subseteq \overline {T B_X}^{w^\ast} \subseteq Y$. By scaling and taking unions we get $T^{\ast \ast}(X^{\ast \ast}) \subseteq Y$. 

\section*{Question 3}
Bookwork omitted. 

Suppose that there existed a Banach space $X$ with $X^\ast = c_0$ or $X^\ast = L_1[0,1]$. Then $B_{X^\ast} = \overline {\conv(\Ext(B_{X^\ast}))}^{w^\ast}$. It is therefore enough to show that the closed unit balls of $c_0$ and $L_1[0, 1]$ have no extreme points. 

Let $x \in B_{c_0}$. Since $x_n \to 0$, there exists $n$ such that $|x_n| < 1/4$. Then setting $u = x + \frac 1 4 e_n \ne x$ and $v = x - \frac 1 4 e_n \ne x$ we have $\frac 1 2 u + \frac 1 2 v = x$, so $x$ is not extreme. 

Let $f \in C_{L_1[0, 1]}$. Then there exists $x$ such that $\int_0^x |f| = \int_x^1 |f| = 1/2$. Then we have:
\[\frac 1 2 \underbrace {f 1_{[0, x]}}_{\ne f} + \frac 1 2 \underbrace{f 1_{[x, 1]}}_{\ne f} = f\]
since $f$ cannot be identically zero on either $[0, x]$ and $[x, 1]$.

By Banach--Stone, $C[0, 1]$ being isometrically isomorphic to $C([0, 1] \cup [2, 3])$ is an equivalent claim to $[0, 1]$ being homeomorphic to $[0, 1] \cup [2, 3]$. But the former space is connected and the latter disconnected, so they cannot be homeomorphic. 

\section*{Question 4}
Pure bookwork. 

\section*{Question 5}
[I proved this in a different way to how Dr. Zsak proved in the revision class]

Consider:
\[p(x) = \limsup_{n \to \infty} \frac {x_1 + \ldots + x_n} n\]
First note that:
\[\frac {x_1 + \ldots + x_n} n \le \norm x_\infty\]
for each $n$, so we have $p(x) \le \norm x$.  Then, note that: 
\[\sup_{k \ge n} \frac {(x_1 + y_1) + \ldots + (x_n + y_n)} n \le \sup_{k \ge n} \frac {x_1 + \ldots + x_n} n + \sup_{k \ge n} \frac {y_1 + \ldots + y_n} n\]
Taking $n \to \infty$ we get $p(x + y) \le p(x) + p(y)$. Clearly we have positive homogeneity. Extending the zero functional we obtain $L$ such that $L x \le p(x) \le \norm x$. Swapping $x$ for $-x$ we have $|Lx| \le \norm x$. Hence $L \in \ell_\infty^\ast$. We prove each property. 

$(b)$: Let $x$ be a convergent sequence. Then we have: 
\begin{align*}
L x & \le \limsup_{n \to \infty} \paren {\frac 1 n \sum_{k = 1}^n x_k} \\ & \le \limsup_{n \to \infty} \paren {\frac 1 n \sum_{k = 1}^n \sup_{j \ge k} x_j} \\ & =  \lim_{n \to \infty} \sup_{j \ge n} x_j \\ & = \lim_{n \to \infty} x_n
\end{align*}
Similarly:
\begin{align*}
Lx & = -L(-x) \\ & \ge -\limsup_{n \to \infty} \paren {\frac 1 n \sum_{k = 1}^n (-x_k)} \\ & = \liminf_{n \to \infty} \paren {\frac 1 n \sum_{k = 1}^n x_k} \\ & \ge \liminf_{n \to \infty} \paren {\frac 1 n \sum_{k = 1}^n \inf_{j \ge k} x_j} \\ & = \lim_{n \to \infty} \inf_{j \ge n} x_j \\ & = \lim_{n \to \infty} x_n
\end{align*}
So we have $L x = \lim_{n \to \infty} x_n$. 

$(a)$: We already have $\norm L \le 1$, and since $L(1, 1, \ldots, 1) = \lim_{n \to \infty} 1 = 1$ while $\norm {(1, 1, \ldots, 1)}_\infty = 1$, so $\norm L = 1$. 

$(c)$: Let $S$ be the shift operator on $\ell_1$. We have:
\[L(x - S x) \le \limsup_{n \to \infty} \frac {(x_1 - x_2) + (x_2 - x_3) + \ldots + (x_{n - 1} - x_n)} n = \limsup_{n \to \infty} \frac {x_1 - x_n} n = 0
\]
since $x$ is bounded. Similarly:
\begin{align*}
L(x - S x) & = -L(S x - x) \\ & \ge -\limsup_{n \to \infty} \frac {(x_2 - x_1) + (x_3 - x_2) + \ldots + (x_n - x_{n - 1})} n \\ & = \liminf_{n \to \infty} \frac {x_1 - x_n} n \\ & = 0
\end{align*}
So we have $Lx = L(S x)$, proving $(a)$. 