Added relevant Hahn-Banach results.
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\section{Normed and Banach Spaces}
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\section{Norms}
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\label{section:normed-banach}
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\begin{definition}[Norm]
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\label{definition:norm}
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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:
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\begin{enumerate}
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\item[(N)] For any $x \in E$, $\norm{x}_E = 0$ if and only if $x = 0$.
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\item[(SN2)] For any $x \in E$ and $\lambda \in K$, $\norm{\lambda x}_E = \abs{\lambda} \norm{x}_E$.
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\item[(SN3)] For any $x, y \in E$, $\norm{x + y}_E \le \norm{x}_E + \norm{y}_E$.
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\end{enumerate}
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\end{definition}
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\begin{proposition}
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\label{proposition:norm-criterion}
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Let $E$ be a separated TVS over $K \in \RC$, then the following are equivalent:
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\begin{enumerate}
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\item There exists $U \in \cn^o(0)$ bounded and convex.
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\item There exists a norm $\norm{\cdot}_E: E \to [0, \infty)$ that induces the topology on $E$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1) $\Rightarrow$ (2): Using \ref{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 $\abs{\lambda}^{-1}U \subset V$.
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Let $\norm{\cdot}_E: E \to [0, \infty)$ be the gauge (\ref{definition:gauge}) of $U$. For each $r > 0$, let $B_E(0, r) = \bracs{x \in E|\norm{x}_E < r}$, then
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\[
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\bracs{\lambda U|\lambda > 0} = \bracs{B_E(0, r)|r > 0}
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\]
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is a fundamental system of neighbourhoods at $0$ for $E$. Therefore $\norm{\cdot}_E$ induces the topology on $E$.
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\end{proof}
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\begin{theorem}[Successive Approximation]
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\label{theorem:successive-approximation}
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Let $E, F$ be normed 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:
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