10.1 Norms
Definition 10.1.1 (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 norm if:
For any $x \in E$, $\norm{x}_{E} = 0$ if and only if $x = 0$.
For any $x \in E$ and $\lambda \in K$, $\norm{\lambda x}_{E} = \abs{\lambda}\norm{x}_{E}$.
For any $x, y \in E$, $\norm{x + y}_{E} \le \norm{x}_{E} + \norm{y}_{E}$.
Proposition 10.1.2. Let $E$ be a separated TVS over $K \in \RC$, then the following are equivalent:
There exists $U \in \cn^{o}(0)$ bounded and convex.
There exists a norm $\norm{\cdot}_{E}: E \to [0, \infty)$ that induces the topology on $E$.
Proof. (1) $\Rightarrow$ (2): Using Proposition 8.1.11, 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 gauge of $U$. For each $r > 0$, let $B_{E}(0, r) = \bracs{x \in E|\norm{x}_E < r}$, then
is a fundamental system of neighbourhoods at $0$ for $E$. Therefore $\norm{\cdot}_{E}$ induces the topology on $E$.$\square$
Theorem 10.1.3 (Successive Approximation). 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:
$\norm{x}_{E} \le C\norm{y}_{F}$.
$\norm{y - Tx}_{F} \le \gamma \norm{y}_{F}$.
then for any $y \in F$, there exists $\seq{x_n}\subset E$ such that:
$\sum_{n \in \natp}\norm{x_n}_{E} \le C\norm{y}_{F}/(1 - \gamma)$.
$\sum_{n = 1}^{\infty} Tx_{n} = y$.
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 $Tx = y$.
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 $n \in \nat$,
Since $\norm{x_n}_{E} \le C\norm{y_n}_{F}$,
In addition,
so $\sum_{n = 1}^{\infty} Tx_{n} = y$.$\square$
Theorem 10.1.4 (Uniform Boundedness Principle). Let $E, F$ be normed spaces and $\mathcal{T}\subset L(E; F)$. If
For every $x \in E$, $\sup_{T \in \mathcal{T}}\norm{Tx}_{F} < \infty$.
$E$ is a Banach space.
then $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)}< \infty$.
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 Baire Category Theorem, 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 $T \in \mathcal{T}$,
so $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)}\le 2n/r$.$\square$