\section*{Question 1}
Bookwork and example sheets. 

\section*{Question 2}
[This solution is based off notes taken during Dr. Zsak's example class]

\textbf{(b)} Since $x_n \xrightarrow{w} 0$ we have:
\[0 \in \overline {\conv \set {x_n : n \in \N}}^w = \overline {\conv \set {x_n : n \in \N}}\]
by Mazur. So there exists $p_1 < q_1$ and $t_{p_1}, \ldots, t_{q_1}$ such that $\sum_{i = p_1}^{q_1} t_i = 1$ such that $\norm {\sum_{i = p_1}^{q_1} t_i x_i} < 1$. Then since $0 \in \overline {\conv \set {x_n : n \ge q_1}}^w = \overline {\conv \set {x_n : n \ge q_1}}$, and so on, we can continue this construction to find $p_1 < q_1 < p_2 < q_2 < \ldots$ such that $\sum_{i = p_n}^{q_n} = 1$ and $\norm {\sum_{i = p_n}^{q_n} t_i x_i} < 1/n$. Then $\sum_{i = p_n}^{q_n} t_i x_i \to 0$ as desired. 

\textbf{(c)} Since $X^\ast$ is separable, $B_{X^{\ast \ast}}$ is $w^\ast$--metrizable, hence $B_{X^{\ast \ast}}$ is $w^\ast$--sequentially compact. Since $\sequence {x_n}_{n \in \N}$ is norm bounded in $X$, it is norm bounded (hence $w^\ast$--bounded) in $X^{\ast \ast}$. So there exists some subsequence $\sequence {\widehat {y_n}}_{n \in \N}$ converging $w^\ast$ to $\phi \in X^{\ast \ast}$. That is, for $f \in X^\ast$ we have $\widehat {y_n}(f) = f(y_n) \to \phi(f)$. So $\sequence {f(y_n)}_{n \in \N}$ is Cauchy for each $f \in X^\ast$. 

Let $u_n$ be a convex block of the $\sequence {y_n}_{n \in \N}$, say $u_n = \sum_{i = p_n}^{q_n} t_i y_i$. Then for $f \in X^\ast$ we have: 
\begin{align*}
|f(u_n - y_n)| & \le \sum_{i = p_n}^{q_n} t_i(f(y_i) - f(y_n)) \\ & \le \sup_{p_n \le i \le q_n} |f(y_i) - f(y_n)|
\end{align*}
Since $p_n, q_n \to \infty$, we have $|f(u_n - y_n)| \to 0$ for each $f \in X^\ast$. That is, $u_n - y_n \xrightarrow{w} 0$. 

\textbf{(d)} Let $\sequence {z_n}_{n \in \N}$ be a sequence in $B_Z$ with $z_n \xrightarrow{w} 0$. Pick $\sequence {w_n}_{n \in \N} \subseteq (4/3) B_X$ such that $q(w_n) = z_n$. Passing to a subsequence, we can assume that $\sequence {w_n}_{n \in \N}$ $w^\ast$--converges to some $\psi \in X^{\ast \ast}$. Take a convex block $\sequence {u_n}_{n \in \N}$ of $\sequence {z_n}_{n \in \N}$ such that $u_n \to 0$. Take the corresponding convex block of $\sequence {w_n}_{n \in \N}$, $\sequence {x_n}_{n \in \N}$ so that $q(x_n) = u_n$ for each $n \in \N$. Then by the previous part, we have $x_n - w_n \xrightarrow{w} 0$. We can now conclude, since we have $\norm {x_n - w_n} < 3$ and $\norm {q(x_n - w_n) - z_n} = \norm {q(x_n)} \to 0$. 

\section*{Question 3}
Mostly bookwork. Let $\phi \in L_p^\ast$. Then there exists $g \in L_q$ such that: 
\[\phi(f) = \int fg d \mu\]
Then we have $|f_n g| \le |fg|$. Since $|fg|$ is integrable by Holder's inequality and $f_n g \to 0$ pointwise almost everywhere, we have: 
\[\int f_n g d \mu \to 0\]
by DCT, and hence $\phi(f_n) \to 0$. So $f_n \xrightarrow{w} 0$. 

\section*{Question 4}
First, any unital $\ast$--homomorphism $f : A \to B$ between $C^\ast$--algebras is continuous: for each $x \in A$ we have $\sigma_B(f(x)) \subseteq \sigma_A(x)$. (since if $x$ is invertible in $A$, then $f(X)$ is invertible in $B$) Then $\norm {f(x)}^2 = \norm {(f(x))^\ast (f(x))} = r_B(f(x^\ast x)) \le r_A(x^\ast x) = \norm {x^\ast x} = \norm x^2$, where we have used that if $A$ is normal then $r_A(x) = \norm x$. 

Let $A$ be the maximal commutative $C^\ast$ subalgebra generated by $T$. Then $\sigma_A(T) = \sigma(T)$. Then there is a unital $\ast$--isomorphism $S : \Phi_A \to \sigma(T)$ sending $\phi \in \Phi_A$ to $\phi(T)$. Then the map $f \mapsto f \circ S$ is a unital $\ast$--isomorphism $C(\sigma(T)) \to C(\Phi_A)$. Then the inverse Gelfand map $G^{-1} : C(\Phi_A) \to A$ is a unital $\ast$--isomorphism. Finally defining a map $\Gamma : C(\sigma(T)) \to A$ by $\Gamma(f) = G^{-1}(f \circ S)$ we have a unital $\ast$--isomorphism. First, we have for $\phi \in \Phi_A$: $(G \circ \Gamma)(\mathrm{id})(\phi) = S\phi = \phi(T) = \hat T(\phi)$. So $(G \circ \Gamma)(\mathrm{id}) = \hat T = G(T)$. Since $G$ is an isomorphism it follows that $T = \Gamma(\mathrm{id})$. Now for $\lambda \in K$, we have $(G \circ \Gamma)(\lambda)(\phi) = G^{-1}(\lambda)(\phi) = \lambda = \widehat {\lambda I}(\phi) = (G(\lambda I))\phi$. So $\Gamma(\lambda) = \lambda I$. 

For uniqueness, note that the conditions $\Gamma(\mathrm{id}) = T$ and $\Gamma(\lambda) = \lambda$ together imply that for any polynomial $p \in \C[X, Y]$ we have $\Gamma(p(z, \overline z)) = p(T, T^\ast)$. Since the polynomials in $z, \overline z$ are dense in $C(\sigma(T))$, we have that $\Gamma$ is uniquely determined on a dense subset of $C(\sigma(T))$, hence uniquely determined.

Now suppose that $\sigma(T) = K_1 \cup K_2$ is disconnected. Let $U_1$ and $U_2$ be open neighborhoods of $K_1$ and $K_2$ with positive distance between them. Define $f(z) = 1$ on $U_1$ and $f(z) = 0$ on $U_2$. Then $f$ is continuous since it is continuous on both connected components. We have $\Gamma(f)^2 = \Gamma(f^2) = \Gamma(f)$, so $\Gamma(f)$ is a projection, and commutes with $A$ since $\Gamma(f) \in A$. We lastly need to verify that $\Gamma(f)$ is non--trivial. We do this by noting that since $\Gamma$ is a $\ast$--isomorphism onto $A$, it is injective. Since $f$ is not identically $0$ or the $1$, we have $\Gamma(f) \not \in \set {0, I}$. Hence $\Gamma(f)$ is a non--trivial projection commuting with $T$ and we are done. 

\section*{Question 5}

$(1) \implies (2)$: Since $T^{\ast \ast}$ extends $T$ we have $T^{\ast \ast} x \in T B_X \subseteq \overline {T B_X}$ for each $x \in X$. Hence: 
\begin{equation}
B_X \subseteq (T^{\ast \ast})^{-1} (\overline {T B_X}) \tag{*}
\end{equation}
We have that $\overline {T B_X}$ is $w$--compact. So $\overline {T B_X}$ is $w^\ast$--compact, hence $w^\ast$--closed. So $(T^{\ast \ast})^{-1} (\overline {T B_X})$ is $w^\ast$--closed. Taking closures on both sides of the inclusion $(*)$ and then applying $T^{\ast \ast}$ gives $T^{\ast \ast}(B_{X^{\ast \ast}}) \subseteq \overline {T B_X} \subseteq Y$. So $T^{\ast \ast}(X^{\ast \ast}) = \bigcup_{r = 1}^\infty T^{\ast \ast}(r B_{X^{\ast \ast}}) \subseteq Y$. So we have proved $(2)$. 

$(2) \implies (3)$: Suppose that $T^{\ast \ast}(X^{\ast \ast}) \subseteq Y$. Then for each $g \in X^{\ast \ast}$, we have $g \circ T^{\ast \ast} = T^{\ast \ast} g \in Y = (Y^\ast, w^\ast)^\ast$. So by the universal property, $T^\ast$ is $w^\ast$--to--$w$ continuous. 

$(3) \implies (4)$: Suppose that $T^\ast : Y^\ast \to X^\ast$ is $w^\ast$--to--$w$ continuous. Since $B_{X^\ast}$ is $w^\ast$--compact by Banach--Alaoglu, we have that $T^\ast(B_{X^\ast})$ is $w$--compact, in particular $w$--closed. So we immediately get that $T^\ast$ is weakly compact. 

$(4) \implies (1)$: Suppose that $T^\ast$ is compact. Using $(1) \implies (4)$, we have that $T^{\ast \ast}$ is compact. Since $T^{\ast \ast}$ extends $T$, we have $T(B_X) \subseteq T^{\ast \ast}(B_{X^{\ast \ast}})$. $B_{X^{\ast \ast}}$ is $w^\ast$--compact by Banach--Alaoglu. Since $T^{\ast \ast} : X^{\ast \ast} \to Y^{\ast \ast}$ is $w^\ast$--to--$w$ continuous by $(1) \implies (3)$, we have that $T^{\ast \ast}(B_{X^{\ast \ast}})$ is $w$--compact in $Y^{\ast \ast}$. So $\overline {T(B_X)}^w$ is a $w$--closed subset of the $w$--compact set $T^{\ast \ast}(B_{X^{\ast \ast}})$, so is $w$--compact. 

If $X$ is reflexive, then $X = X^{\ast \ast}$. Then $T^{\ast \ast}(X) = T(X) \subseteq Y$, so $T$ is weakly compact. If $Y$ is reflexive, then so is $Y^\ast$. Consider $T^{\ast} : Y^\ast \to X^\ast$ and $T^{\ast \ast \ast} : Y^\ast \to X^\ast$. Then $T^{\ast \ast \ast}(Y^\ast) = T^\ast(Y^\ast) \subseteq X^\ast$, so $T^\ast$ is weakly compact. So $T$ is weakly compact. 