\tit{Part III Functional Analysis 2014}
\section*{Question 1}
Pure bookwork.
\section*{Question 2}
[This solution is based off notes taken during Dr. Zsak's example class]
\textbf{(iii)} Since $Y$ has finite codimension in $X^\ast$ so there exists $\set {e_1, \ldots, e_n}$ such that $X^\ast = Y + \Span \set {e_1, \ldots, e_n}$ where $Y \cap \Span \set {e_1, \ldots, e_n} = 0$. Define:
\[\phi_j(f_i) = \begin{cases}1 & i = j \\ 0 & i \ne j\end{cases}\]
and $\phi_j(y) = 0$ for $y \in Y$. Then we have $\ker \phi_j = Y + \Span \set {e_i : i \ne j}$. So $Y = \bigcap_{j = 1}^n \ker \phi_j$.
Suppose that $x \in F \cap X$. If $x \in F$ then $Y \subseteq \ker \hat x$. But $\hat x$ is $w^\ast$--continuous, we have that $\ker \hat x$ is $w^\ast$--closed. But $Y$ is $w^\ast$--dense, which implies that $\ker \hat x = X^\ast$. So $x = 0$.
We now want to show $d = d(S_X, F) > 0$. Let $\psi \in S_F$, (which is therefore not in $X$) then $d(\psi, X) > 0$ since $X$ is closed in $X^{\ast \ast}$. The map $\psi \mapsto d(\psi, x)$ defined on $S_F$ is a continuous map on a compact space, so attains an infimum, so there exists $\delta > 0$ such that $d(\psi, X) \ge \delta$ for all $\psi \in S_F$.
Now take $\psi \in F$ general and $x \in S_X$. Then:
\begin{align*}
\norm {\psi - x} & = \norm \psi \norm {\frac {\psi} {\norm \psi} - \frac x {\norm \psi}} \ge \delta \norm \psi
\end{align*}
Hence if $\norm \psi \ge 1/2$ we have $\norm {\psi - x} \ge \delta/2$. For $\norm \psi \le 1/2$ we have $\norm {\psi - x} \ge 1/2$. So we have $d \ge \max \set {1/2, \delta} > 0$.
Let $x \in S_X$ and let $E = \Span (F \cup \set {\hat x})$. Define the map $\psi = 0$ on $F$ and $\psi(\hat x) = 1$, then extend linearly. We have $d(x, F) \ge d$, so we must have $\norm \psi \le 1/d$. Extend using Hahn--Banach to get $\psi \in X^{\ast \ast \ast}$ vanishing on $F$. Then $\norm {d \psi} \le 1$, so there exists $f \in X^\ast$ with $\norm f < 1 + \epsilon$ such that $d \psi = \hat f$ on $E$. Then $\psi = \widehat {f/d}$ with $\norm {f/d} < 1/d + \epsilon/d$. Then we have:
\[\frac {\hat f} {\lVert {\hat f}\rVert}(x) = \frac 1 {\lVert {\hat f}\rVert} \ge \paren {\frac 1 d + \frac \epsilon d}^{-1} = d \paren {1 + \epsilon}^{-1}\]
Note that $\hat f$ restricts to $0$ on $F$, so $f \in \bigcap_{i = 1}^n \ker \phi_k = Y$. Hence we obtain \underline{for $x \in S_X$} and each $\epsilon > 0$:
\[\sup \set {f(x) : f \in Y, \, \norm f \le 1} \ge d(1 + \epsilon)^{-1}\]
Hence:
\[\sup \set {f(x) : f \in Y, \, \norm f \le 1} \ge d\]
So for general non--zero $x \in X$ we have by scaling:
\[\sup \set {f(x) : f \in Y, \, \norm f \le 1} \ge d \norm x\]
The demand is then immediate.
\textbf{(iv)}: Let $f \in X^\ast \setminus \overline Z^{w^\ast}$. Then there exists a $w^\ast$--continuous map $\phi = \hat x \in X$ such that $\phi$ restricts to $0$ on $\overline Z^{w^\ast}$ and is non--zero at $f$. Then we would have:
\[0 < c \norm f \le \sup \set {g(x) : g \in S_Z} = 0\]
since $\hat x$ vanishes on $\overline Z^{w^\ast}$ hence $S_Z$.
An example for the last part is $X = \ell_1$, $Z = c_0 \subseteq \ell_\infty$. We have for each $f \in Z$ and $x \in \ell_1$:
\[f(x) = \sum_{n = 0}^\infty f_n x_n\]
for some $\sequence {f_n}_{n \in \N} \in \ell_\infty$. Notice that we can pick $f_n = \sgn(x_n)$ (clearly defining a bounded sequence) to get $f(x_n) = \norm x_1$. Hence we get the $1$--norming property.
\section*{Question 3}
[This solution is based off notes taken during Dr. Zsak's example class]
Bookwork up until last bit.
Since $F$ is finite dimensional, $S_F$ is norm compact. So there exists a cover by $\delta$--balls with centres $\set {f_1, \ldots, f_k}$. Fix $\mu \in B_{C(K)^\ast}$. Let:
\[U = \set {\nu \in B_{C(K)^\ast} : |(\nu - \mu)(f_j)| < \delta \text { for each } 1 \le j \le k}\]
Since $B_{C(K)^\ast} = \overline {\conv \set {\alpha \delta_k : k \in [0, 1], \, |\alpha| = 1}}^{w^\ast}$, $U$ intersects $\conv \set {\alpha \delta_k : k \in [0, 1], \, |\alpha| = 1}$. So there $n \in \N$, $|\alpha_j| = 1$, $w_j \in [0, 1]$ and $s_j \ge 0$ ($1 \le j \le n$) with $\sum_{j = 1}^n s_j = 1$. Then we have:
\[\abs {\int_0^1 f_i d \mu - \sum_{i = 1}^n t_i f_i(w_i)} < \delta\]
where $t_i = s_i \alpha_i$ so that $|t_i| = s_i$. We now have for $f \in S_F$:
\begin{align*}
\abs {\int_0^1 f d \mu - \sum_{i = 1}^n t_i f(w_i)} & = \abs {\int_0^1 (f - f_i) d \mu - \sum_{i = 1}^n t_i(f - f_i)(w_i)} + \abs {\int_0^1 f_i d \mu - \sum_{i = 1}^n t_i f_i(w_i)} \\ & = \delta + \sum_{i = 1}^n |t_i| + \delta \\ & = 3 \delta
\end{align*}
Since this holds for all $f \in X^\ast$, we have:
\[\abs {\int_0^1 f d \mu - \sum_{i = 1}^n t_i f(w_i)} \le 3 \delta \norm f\]
Taking $\delta = \epsilon/3$ we are done.
\section*{Question 4}
Bookwork up to invariant subspaces. Let $B(X)$ be the Banach algebra of bounded operators on $X$. Let $A$ be the maximal commutative subalgebra of $B(X)$ that contains $T$. Then $\sigma_A(T) = \sigma(T)$. Since $\sigma(T)$ is disconnected, there exists disjoint closed sets $C_1, C_2$ such that $\sigma(T) = C_1 \cup C_2$. Since $C_1, C_2$ are disjoint and closed, we have $\epsilon = d(C_1, C_2) > 0$. So let $U$ and $V$ be open neighborhoods of $C_1$ and $C_2$ that are still disjoint, say $U = C_1 + B_{\epsilon/2}(0)$ and $V = C_2 + B_{\epsilon/2}(0)$.
Then $U \cup V$ certainly contains $\sigma(T)$ and has precisely two connected components, $U$ and $V$. Define $f : U \cup V$ by $f(z) = 1$ on $U$ and $f(z) = 0$ on $V$. This is holomorphic since it is holomorphic on both connected components. Notice that $f^2 = 1$. Then consider $\Theta_T(f) = P$. This is a projection since $P^2 = \Theta_T(f^2) = \Theta_T(f) = P$. $P \not \in \set {0, I}$ since $\sigma_A(\Theta_T(f)) = f(\sigma(T)) = \set {0, 1}$ by the HFC. Since $A$ is a commutative algebra, we have $PT = TP$. Notice now that $Y = \ker(I - P) = \mathrm{im}(P)$ is a non--trivial closed subspace of $X$, and for each $y \in Y$ we have $T y = T P y = P T y \in Y$. So $Y$ is a non--trivial invariant subspace for $T$.
\section*{Question 5}
Pure bookwork.