\section{Norms}
\label{section:normed-banach}


\begin{definition}[Norm]
\label{definition:norm}
    Let $E$ be a vector space over $K \in \RC$ and $\norm{\cdot}_E: E \to [0, \infty)$, then $\norm{\cdot}_E$ is a \textbf{norm} if:
    \begin{enumerate}
        \item[(N)] For any $x \in E$, $\norm{x}_E = 0$ if and only if $x = 0$.
        \item[(SN2)] For any $x \in E$ and $\lambda \in K$, $\norm{\lambda x}_E = \abs{\lambda} \norm{x}_E$.
        \item[(SN3)] For any $x, y \in E$, $\norm{x + y}_E \le \norm{x}_E + \norm{y}_E$.
    \end{enumerate}
\end{definition}

\begin{proposition}
\label{proposition:norm-criterion}
    Let $E$ be a separated TVS over $K \in \RC$, then the following are equivalent:
    \begin{enumerate}
        \item There exists $U \in \cn^o(0)$ bounded and convex.
        \item There exists a norm $\norm{\cdot}_E: E \to [0, \infty)$ that induces the topology on $E$.
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1) $\Rightarrow$ (2): Using \autoref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $U$ is also circled. Let $V \in \cn^o(0)$, then there exists $\lambda \in K$ such that $\lambda V \supset U$ and $\abs{\lambda}^{-1}U \subset V$.

    Let $\norm{\cdot}_E: E \to [0, \infty)$ be the \hyperref[gauge]{definition:gauge} of $U$. For each $r > 0$, let $B_E(0, r) = \bracs{x \in E|\norm{x}_E < r}$, then
    \[
    \bracs{\lambda U|\lambda > 0} = \bracs{B_E(0, r)|r > 0}
    \]

    is a fundamental system of neighbourhoods at $0$ for $E$. Therefore $\norm{\cdot}_E$ induces the topology on $E$.
\end{proof}


\begin{theorem}[Successive Approximation]
\label{theorem:successive-approximation}
    Let $E, F$ be normed vector spaces, $T \in L(E; F)$, $C \ge 0$, and $\gamma \in (0, 1)$. If for all $y \in F$, there exists $x \in E$ such that:
    \begin{enumerate}
        \item[(a)] $\norm{x}_E \le C\norm{y}_F$.
        \item[(b)] $\norm{y - Tx}_F \le \gamma \norm{y}_F$.
    \end{enumerate}
    then for any $y \in F$, there exists $\seq{x_n} \subset E$ such that:
    \begin{enumerate}
        \item $\sum_{n \in \natp}\norm{x_n}_E \le C\norm{y}_F/(1 - \gamma)$.
        \item $\sum_{n = 1}^\infty Tx_n = y$.
    \end{enumerate}
    In particular, if $E$ is a Banach space, then for every $y \in F$, there exists $x \in E$ such that $\norm{x}_E \le C\norm{y}_F/(1 - \gamma)$ and $Tx = y$.
\end{theorem}
\begin{proof}
    Let $y_1 = y \in F$. Let $n \in \natp$ and suppose inductively that $\bracs{x_k| 0 \le k < n}$ and $\bracs{y_k| 0 \le k \le n}$ has been constructed. By (a) and (b), there exists $x_n \in E$ such that $\norm{x_k}_E \le C\norm{y_k}_F$ and $\norm{y_n - Tx_n}_F \le \gamma \norm{y_{n}}_F$. Let $y_{n+1} = y_n - Tx_n$, then $\norm{y_{n+1}}_F \le \gamma \norm{y_n}_F$.

    For each $n \in \nat$,
    \[
    \norm{y_n}_F \le \gamma^{n - 1}\norm{y_1}_F = \gamma^{n - 1}\norm{y}_F
    \]

    Since $\norm{x_n}_E \le C\norm{y_n}_F$,
    \[
    \sum_{k \in \natp}\norm{x_k}_E \le C\norm{y}_F\sum_{k \in \nat_0}\gamma^k = \frac{C\norm{y}_F}{1 - \gamma}
    \]

    In addition,
    \[
    \norm{y - \sum_{k = 1}^n Tx_k}_F = \norm{y_{n+1}}_F \le \gamma^n \norm{y}_F
    \]

    so $\sum_{n = 1}^\infty Tx_n = y$.
\end{proof}

\begin{theorem}[Uniform Boundedness Principle]
\label{theorem:uniform-boundedness}
    Let $E, F$ be normed vector spaces and $\mathcal{T} \subset L(E; F)$. If
    \begin{enumerate}
        \item For every $x \in E$, $\sup_{T \in \mathcal{T}}\norm{Tx}_F < \infty$.
        \item $E$ is a Banach space.
    \end{enumerate}
    then $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} < \infty$.
\end{theorem}
\begin{proof}
    For each $n \in \natp$, let $A_n = \bracs{x \in X|\norm{Tx}_F \le n \forall T \in \mathcal{T}}$, then each $A_n$ is closed with $\bigcup_{n \in \natp}A_n = E$. By the \hyperref[Baire Category Theorem]{theorem:baire}, there exists $n \in \natp$ and $U \subset E$ open such that $\sup_{x \in U}\sup_{T \in \mathcal{T}}\norm{Tx}_{F} < \infty$.

    Let $x \in U$ and $r > 0$ such that $\overline{B(x, r)} \subset U$, then for any $y \in E$ with $\norm{y}_E \le r$ and $T \in \mathcal{T}$,
    \[
    \norm{Ty} = \norm{Ty + Tx - Tx}_E = \normn{T\underbrace{(x + y)}_{\in U}}_E + \norm{Tx}_E \le 2n
    \]

    so $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le 2n/r$.
\end{proof}


\begin{proposition}
\label{proposition:dual-norm}
    Let $E$ be a normed vector space, then for any $x \in E$, 
    \[
    \norm{x}_E = \sup_{\substack{\phi \in E^* \\ \norm{\phi}_{E^*} = 1}}\dpn{x, \phi}{E}
    \]
\end{proposition}
\begin{proof}
    For any $\phi \in E^*$ with $\norm{\phi}_{E^*} = 1$, $\dpn{x, \phi}{E} \le \norm{x}_E \cdot \norm{\phi}_{E^*} = \norm{x}_E$. On the other hand, by the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, there exists $x \in E^*$ such that $\dpn{x, \phi}{E} = \norm{x}_E$ and $\norm{\phi}_{E^*} \le 1$. 
\end{proof}


% TODO: Replace this with a more general version involving polars in the future. 
\begin{theorem}[Alaoglu's Theorem]
\label{theorem:alaoglu}
    Let $E$ be a normed vector space over $K \in \RC$, then $B^* = \bracsn{\phi \in E^*| \norm{x}_{E^*} \le 1}$ is compact in the weak*-topology. 
\end{theorem}
\begin{proof}
    For each $x \in E$, $I_x = \bracsn{\dpn{x, \phi}{E}|\phi \in B^*}$ is compact. By \autoref{proposition:operator-space-completeness}, the closure of $B^*$ in $\prod_{x \in E}I_x$ is a subset of $\hom(E; K)$. Since $B^*$ is bounded, $I_x \subset \overline{B_K(0, 1)}$ for all $x \in E$, so the closure of $B^*$ is contained in $B^*$. By \autoref{theorem:tychonoff}, $\prod_{x \in E}I_x$ is compact. Therefore $B^*$ is compact with respect to the weak*-topology.
\end{proof}


