Housekeeping.

src/fa/lc/hahn-banach.tex

@@ -163,3 +163,20 @@
     is lower semicontinuous and Borel measurable by \autoref{proposition:semicontinuous-properties}. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:closed-subspace-induction}
+    Let $E$ be a locally convex space over $K \in \RC$, $F \subset E$ be a closed subspace, and $y \in E \setminus F$, then $F + Ky$ is also a closed subspace. 
+
+    In particular, all finite-dimensional subspaces of $E$ are closed. 
+\end{proposition}
+\begin{proof}
+    Since $F$ is closed and $E$ is locally convex, there exists a continuous seminorm $\rho: E \to [0, \infty)$ such that $\rho(F) = \bracsn{0}$ and $\rho(y) \ne 0$. By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, there exists $\phi \in E^*$ such that $F \subset \ker(\phi)$ and $\dpn{y, \phi}{E} = 1$. In which case, $\phi$ induces an isomorphism
+    \[
+    E \to \ker(\phi) \times K \quad x \mapsto (x - \dpn{x, \phi}{E}y, \dpn{x, \phi}{E})
+    \]
+
+    whose restriction to $F + Ky$ is an isomorphism onto $F \times K$.
+
+    Let $\angles{x_\alpha + \lambda_\alpha y}_{\alpha \in A} \subset F + Ky$ be net convergent in $E$. For each $\alpha \in A$, $\dpn{x_\alpha + \lambda_\alpha y, \phi}{E} = \lambda_\alpha$. By continuity of $\phi$, $\angles{\lambda_\alpha}_{\alpha \in A}$ is convergent, and $\angles{x_{\alpha}}_{\alpha \in A}$ is also convergent. Since $\angles{x_\alpha}_{\alpha \in A} \subset F$ and $F$ is closed, $\lim_{\alpha \in A}x_\alpha \in F$. Therefore $\lim_{\alpha \in A}(x_\alpha + \lambda_\alpha y) \in F + Ky$, and $F + Ky$ is also a closed subspace. 
+\end{proof}
+

src/fa/lc/quotient.tex

@@ -24,6 +24,14 @@
     As this holds for all $y \in x + M$ and $y' \in x' + M$, $\rho_M(x + x') \le \rho_M(x) + \rho_M(x')$.
 \end{proof}
 
+\begin{lemma}
+\label{lemma:quotient-norm-optimiser}
+    Let $E$ be a vector space over $K \in \RC$, $\rho: E \to [0, \infty)$ be a seminorm, $M \subset E$ be a vector subspace, $\rho_M: E/M \to [0, \infty)$ be its quotient by $M$, and $x \in E$ with $\rho_M(x) < 1$, then there exists $x_0 \in B_E(0, 1)$ such that $x + M = x_0 + M$.
+\end{lemma}
+\begin{proof}
+    Since $\inf_{y \in M}\rho(x - y) = \rho_M(x) < 1$, there exists $y \in M$ with $\rho(x - y) < 1$. Let $x_0 = x - y$, then $\rho(x_0) < 1$ as well and $x_0 + M = (x - y) + M = x + M$. 
+\end{proof}
+
 \begin{definition}[Quotient Locally Convex Space]
 \label{definition:lc-quotient}
     Let $E$ be a locally convex space over $K \in \RC$ and $M \subset E$ be a vector subspace, then there exists $(\td E, \pi)$ such that:

src/fa/norm/normed.tex

@@ -122,7 +122,7 @@
 
 \begin{proposition}
 \label{proposition:dual-norm}
-    Let $E$ be a normed vector space, then for any $x \in E$, 
+    Let $E$ be a normed vector space over $K \in \RC$, then for any $x \in E$, 
     \[
     \norm{x}_E = \sup_{\substack{\phi \in E^* \\ \norm{\phi}_{E^*} = 1}}\dpn{x, \phi}{E}
     \]
@@ -132,3 +132,19 @@
 \end{proof}
 
 
+\begin{lemma}[Riesz]
+\label{lemma:riesz-almost-orthogonal}
+    Let $E$ be an infinite-dimensional normed vector space over $K \in \RC$ and $\seq{\alpha_n} \subset (0, 1)$, then there exists $\seq{x_n} \subset \partial B_E(0, 1)$ such that
+    \[
+    d(x_n, \text{span}\bracsn{x_j|1 \le j < n}) > \alpha_n \quad \forall n \in \natp
+    \]
+\end{lemma}
+\begin{proof}
+    Let $N \in \natp$ and suppose inductively that $\bracsn{x_n|1 \le n < N}$ have been constructed to satisfy the lemma for all $1 \le n < N$. Let $F_N = \text{span}\bracsn{x_n|1 \le n < N}$, then $F_N$ is finite-dimensional, and closed by \autoref{proposition:closed-subspace-induction}. As $E$ is infinite-dimensional, $E/F_N$ is non-trivial, and there exists $y_N \in E$ such that $\norm{y_N + F_N}_{E/F_N} > \alpha_N$. By \autoref{lemma:quotient-norm-optimiser}, there exists $z_N \in B_E(0, 1)$ such that $z_N + F_N = y_N + F_N$. As $z_N + F_N \ne 0$, $z_N \ne 0$ as well. Let $x_N = z_N/\norm{z_N}_{E} \in \partial B_E(0, 1)$, then
+    \[
+        \norm{x_N + F_N}_{E/F} > \norm{z_N + F_N}_{E/F} = \norm{y_N + F_N}_{E/F} > \alpha_N
+    \]
+
+    Thus $d(x_N, \text{span}\bracsn{x_n|1 \le n < N}) = \norm{x_N + F_N}_{E/F} > \alpha_N$.
+\end{proof}
+