12.1 Norms
Definition 12.1.1 (Norm).label 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:
- (N)
For any $x \in E$, $\norm{x}_{E} = 0$ if and only if $x = 0$.
- (SN2)
For any $x \in E$ and $\lambda \in K$, $\norm{\lambda x}_{E} = \abs{\lambda}\norm{x}_{E}$.
- (SN3)
For any $x, y \in E$, $\norm{x + y}_{E} \le \norm{x}_{E} + \norm{y}_{E}$.
In which case, the pair $(E, \norm{\cdot}_{E})$ is a normed vector space over $K$.
Definition 12.1.2 (Banach Space).label Let $E$ be a normed vector space over $K \in \RC$, then $E$ is a Banach space if it is complete.
Lemma 12.1.3.label Let $E$ be a normed vector space, then the following are equivalent:
- (1)
$E$ is a Banach space.
- (2)
For each $\seq{x_n}\subset E$ with $\sum_{n \in \natp}\norm{x_n}_{E} < \infty$, there exists $x \in E$ such that $x = \sum_{n = 1}^{\infty} x_{n}$.
Proof. (2) $\Rightarrow$ (1): Let $\seq{x_n}\subset E$ be a Cauchy sequence, then there exists a subsequence $\seq{n_k}$ such that $\norm{x_{n_{k+1}} - x_{n_{k}}}_{E} < 2^{-k}$ for all $k \in \natp$. For each $k \in \natp$, let $y_{k} = x_{n_{k+1}}- x_{n_{k}}$, then there exists $y \in E$ such that $y = \sum_{k = 1}^{\infty} y_{k}$. Let $\eps > 0$, then there exists $K \in \natp$ such that:
- (a)
$\norm{y - \sum_{k = 1}^{K-1} y_k}_{E} < \eps$.
- (b)
For each $n \ge n_{K}$, $\norm{x_n - x_{n_K}}_{E} < \eps$.
In which case, for every $n \ge n_{K}$,
Therefore $x_{n} \to y + x_{n_1}$ as $n \to \infty$.$\square$
Proposition 12.1.4.label Let $E$ be a separated TVS over $K \in \RC$, then the following are equivalent:
- (1)
There exists $U \in \cn^{o}(0)$ bounded and convex.
- (2)
There exists a norm $\norm{\cdot}_{E}: E \to [0, \infty)$ that induces the topology on $E$.
Proof. (1) $\Rightarrow$ (2): Using Proposition 10.1.11, 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 12.1.5 (Successive Approximation).label 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:
- (a)
$\norm{x}_{E} \le C\norm{y}_{F}$.
- (b)
$\norm{y - Tx}_{F} \le \gamma \norm{y}_{F}$.
then for any $y \in F$, there exists $\seq{x_n}\subset E$ such that:
- (1)
$\sum_{n \in \natp}\norm{x_n}_{E} \le C\norm{y}_{F}/(1 - \gamma)$.
- (2)
$\sum_{n = 1}^{\infty} Tx_{n} = y$.
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 12.1.6 (Uniform Boundedness Principle).label Let $E, F$ be normed vector spaces and $\mathcal{T}\subset L(E; F)$. If
- (B1)
$E$ is a Banach space.
- (B2)
For every $x \in E$, $\sup_{T \in \mathcal{T}}\norm{Tx}_{F} < \infty$.
then $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)}< \infty$.
Proof. By the Banach-Steinhaus Theorem, $\mathcal{T}$ is equicontinuous. Therefore there exists $r > 0$ such that $\bigcup_{T \in \mathcal{T}}T[B_{E}(0, r)] \subset B_{F}(0, 1)$. In which case, $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)}\le r^{-1}$.$\square$
Proposition 12.1.7.label Let $E$ be a normed vector space, then for any $x \in E$,
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 Hahn-Banach Theorem, there exists $x \in E^{*}$ such that $\dpn{x, \phi}{E}= \norm{x}_{E}$ and $\norm{\phi}_{E^*}\le 1$.$\square$
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