> [!definition] > > Let $(V, ||\cdot||)$ be a [[Normed Vector Space|normed vector space]] over $\mathbb{C}$. $(V, ||\cdot||)$ is a **Banach** space if the induced [[Metric Space|metric]] yields a [[Complete Metric Space|complete]] metric space. > [!theorem] > > Let $X$ be a vector space over $\complex$ and $\norm{\cdot}_1, \norm{\cdot}_2$ be norms on $X$ such that $\norm{\cdot}_1 \le \norm{\cdot}_2$. If $X$ is complete with respect to both of them, then they are equivalent. > > *Proof*. Denote $(X, \norm{\cdot}_2)$ with $\cx$ and $(X, \norm{\cdot}_1)$ with $\cy$, and let $T: \cx \to \cy$ by $x \mapsto x$. Then $T$ is bijective and continuous as > $ > \norm{Tx}_1 \le \norm{x}_2 \quad \forall x \in X > $ > By the [[Open Mapping Theorem]], $T$ is an isomorphism, meaning that there exists $C > 0$ such that > $ > \norm{T^{-1}(x)}_2 \le C\norm{x}_1 \quad \forall x \in X > $ > Chaining the inequality yields that $\norm{\cdot}_1 \sim \norm{\cdot}_2$. # Separable Banach Space > [!theorem] > > Let $E$ be a [[Separable Topological Space|separable]] Banach space, then $E^*$ is separable with respect to the [[Weak Topology|weak* topology]]. > > *Proof*. It is sufficient to show that the unit ball $S = \bracs{x^* \in E^*: \norm{x^*}_{E^*} \le 1}$ with respect to the operator norm is separable. > > Let $\seq{x_n} \subset E$ be a [[Dense|dense]] subset and $S = \bracs{x^* \in E^*: \norm{x^*}_{E^*} \le 1}$, then for each $N \in \nat$, there exists a mapping > $ > \varphi_N: S \to \ell^2([N]) \quad x^* \mapsto (\angles{x_1, x^*}_E, \cdots, \angles{x_N, x^*}_E) > $ > Since $\ell^2([N])$ is separable, there exists $\bracsn{x_{N, k}^*}_{k = 1}^\infty \subset S$ such that $\bracsn{\varphi_N(x^*_{N, k})}_{k = 1}^\infty$ is dense in $\varphi_N(S)$. > > Let $x^* \in E^*$ with $\norm{x^*}_{E^*} \le 1$, then for each $n \in \nat$, there exists $k_n \in \nat$ such that $|\anglesn{x_j, x^*_{n, k_n} - x^*}_E| < 1/n$ for all $1 \le j \le n$. Therefore for each $N \in \nat$, $\anglesn{x_N, x_{n, k_n}^*}_E \to \angles{x_N, x^*}_E$ as $n \to \infty$. As $\bracsn{x_{n, k_n}^*}_{n = 1}^\infty$ is bounded in $E^*$ and converges pointwise to $x^*$ on a dense subset, $x_{n, k_n}^* \to x^*$ in the weak* topology. > [!definition] > > Let $E$ be a separable Banach space and $R \ge 0$, then every bounded subset of $E^*$ equipped with the weak* topology is metrisable. > [!theorem] > > Let $E$ be a separable Banach space, then the [[Borel Sigma Algebra|Borel]] $\sigma$-algebra on $E$ is generated by: > 1. Open sets in $E$ with respect to the strong topology. > 2. $\bracs{B(x, r): x \in E, r > 0}$. > 3. $\bracsn{\overline{B(x, r)}: x \in E, r > 0}$. > 4. Open sets in $E$ with respect to the [[Weak Topology|weak topology]]. > > *Proof*. $(1) \supset (2), (3), (4)$: $(2)$, $(3)$, and $(4)$ consists of open or closed sets. > > $(2) = (3)$: Via countable unions and intersections. > > $(2) \supset (1)$: Let $C \subset E$ be [[Closed Set|closed]] and $\seq{x_n} \subset C$ be a [[Dense|dense]] subset, and > $ > C' = \bigcap_{k \in \nat}\bigcup_{n \in \nat}B(x_n, 1/k) > $ > then $C \subset C'$ by density of $\seq{x_n}$. On the other hand, for any $x \in C'$ and $k \in \nat$, there exists $n_k \in \nat$ such that $x \in B(x_{n_k}, 1/k)$ and $x = \limv{k}x_{n_k}$. As $C$ is closed, $x \in C$. Thus $C = C'$ and $C \in \sigma(\bracs{B(x, r): x \in E, r > 0})$. > > $(4) \supset (3)$: Since the topology on $E$ is invariant under translation or scaling, it is sufficient to show that for every $r > 0$, $\ol{B(0, r)}$ is in the $\sigma$-algebra generated by the open sets in the weak topology. To this end, let $\seq{x_n^*} \subset \bracs{x^* \in E^*: \norm{x^*}_E \le 1}$ be dense with respect to the weak* topology, then for any $x \in E$, > $ > \norm{x}_{E} = \sup_{\norm{x^*}_{E^*}} \abs{\angles{x, x^*}_E} = \sup_{n \in \nat}\abs{\angles{x, x_n^*}_E} > $ > thus > $ > \overline{B(0, r)} = \bigcap_{n \in \nat}\bracs{y \in E: \abs{\angles{x, x_n^*}_E} \le r} > $