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/Part 3: Functional Analysis/Chapter 12: Normed Vector Spaces

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:

  1. (N)

    For any $x \in E$, $\norm{x}_{E} = 0$ if and only if $x = 0$.

  2. (SN2)

    For any $x \in E$ and $\lambda \in K$, $\norm{\lambda x}_{E} = \abs{\lambda}\norm{x}_{E}$.

  3. (SN3)

    For any $x, y \in E$, $\norm{x + y}_{E} \le \norm{x}_{E} + \norm{y}_{E}$.

Proposition 12.1.2.label Let $E$ be a separated TVS over $K \in \RC$, then the following are equivalent:

  1. (1)

    There exists $U \in \cn^{o}(0)$ bounded and convex.

  2. (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

\[\bracs{\lambda U|\lambda > 0}= \bracs{B_E(0, r)|r > 0}\]

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

Theorem 12.1.3 (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:

  1. (a)

    $\norm{x}_{E} \le C\norm{y}_{F}$.

  2. (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. (1)

    $\sum_{n \in \natp}\norm{x_n}_{E} \le C\norm{y}_{F}/(1 - \gamma)$.

  2. (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 $n \in \nat$,

\[\norm{y_n}_{F} \le \gamma^{n - 1}\norm{y_1}_{F} = \gamma^{n - 1}\norm{y}_{F}\]

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 $\sum_{n = 1}^{\infty} Tx_{n} = y$.$\square$

Theorem 12.1.4 (Uniform Boundedness Principle).label Let $E, F$ be normed vector spaces and $\mathcal{T}\subset L(E; F)$. If

  1. (1)

    For every $x \in E$, $\sup_{T \in \mathcal{T}}\norm{Tx}_{F} < \infty$.

  2. (2)

    $E$ 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 $T \in \mathcal{T}$,

\[\norm{Ty}= \norm{Ty + Tx - Tx}_{E} = \normn{T\underbrace{(x + y)}_{\in U}}_{E} + \norm{Tx}_{E} \le 2n\]

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

Proposition 12.1.5.label Let $E$ be a normed vector space, then for any $x \in E$,

\[\norm{x}_{E} = \sup_{\substack{\phi \in E^* \\ \norm{\phi}_{E^*} = 1}}\dpn{x, \phi}{E}\]

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 $\norm{\phi}_{E^*}\le 1$.$\square$

Theorem 12.1.6 (Alaoglu’s Theorem).label 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.

Proof. For each $x \in E$, $I_{x} = \bracsn{\dpn{x, \phi}{E}|\phi \in B^*}$ is compact. By Proposition 10.11.10, 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 Theorem 5.16.6, $\prod_{x \in E}I_{x}$ is compact. Therefore $B^{*}$ is compact with respect to the weak*-topology.$\square$

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