# Functional Analysis > [!definition] > > Let $\cx$ be a [[Normed Vector Space|normed space]], then the weak topology generated by its [[Topological Dual|dual]] $\cx^*$ is known as the **weak topology** on $\cx$, with the corresponding convergence known as **weak convergence**. > [!definition] > > Let $\cx$ be a normed space, then the weak topology on $\cx^{*}$ generated by $\hat \cx \subset \cx^{**}$ is known as the **weak-\* topology**, which corresponds to pointwise convergence. > [!theorem] > > Let $\cx$ be a normed space, then the weak topology on $\cx$ and the weak\* topology on $\cx^*$ are [[Hausdorff Space|Hausdorff]]. > > *Proof*. Since the [[Bounded Linear Functional|bounded linear functionals]] separate points on $\cx$ and $\cx^*$ ([[Hahn-Banach Theorem|Hahn-Banach theorem]]), the weak topology generated by them is Hausdorff (4.34). > [!theorem] > > Let $\cx$ be a normed space and $\seq{x_n} \subset \cx$ be a [[Sequence|sequence]]. Then > 1. $x_n \to x$ weakly if and only if $f(x_n) \to f(x) \forall f \in \cx^*$. > 2. If $x_n \to x$ strongly, then $x_n \to x$ weakly. > 3. If $x_n \to x$ weakly, then $\norm{x_n}$ is bounded, and $\norm{x} \le \liminf{\norm{x_n}}$. > 4. If $x_n \to x$ weakly and $f_n \to f$ strongly, then $f_n(x_n) \to f(x)$. > > *Proof*. For part 1, see net case in the topology section. If $x_n \to x$, then since $f \in \cx^*$ is continuous, $f(x_n) \to f(x)$ for all $f \in \cx^*$, and $x_n \to x$ weakly. > > ### Weak Convergence and Boundedness > > If $x_n \to x$ weakly, then $\norm{x_n}$ is bounded, and $\norm{x} \le \liminf{\norm{x_n}}$. > > *Proof*. Consider the collection $\bracs{\hat x_n: n \in \nat}$, then $\hat x_n(f) \to \hat x(f)$ for all $f \in \cx^*$ implies that $\sup_{n \in \nat}\abs{\hat x_n(f)} < \infty$ for all $f \in \cx^*$. By the uniform boundedness principle, $\sup_{n \in \nat}\norm{\hat x_n} = \sup_{n \in \nat}\norm{x_n} < \infty$. Taking the limit yields > $ > \begin{align*} > \abs{f(x_n)} &\le \norm{f} \norm{x_n} \\ > \abs{f(x)}&\le \norm{f}\liminf\norm{x_n} \\ > \norm{x} &= \sup_{\norm{f} = 1}\abs{f(x)} \le \liminf\norm{x_n} > \end{align*} > $ > > ### Double Limit > > If $x_n \to x$ weakly and $f_n \to f$ strongly, then $f_n(x_n) \to f(x)$. > > *Proof*. > $ > \begin{align*} > \abs{f_n(x_n) - f(x)} &\le \abs{(f_n - f)(x_n)} + \abs{f(x_n - x)} \\ > &\le \norm{f_n - f}\norm{x_n} + \abs{f(x_n - x)} \to 0 > \end{align*} > $ > > [!theorem] > > Let $\cx$ be an infinite-dimensional normed space, then > $ > S = \bracs{x \in \cx: \norm{x} = 1} > $ > is not closed in the weak topology. with > $ > \ol{S} = \bracs{x \in E: \norm{x} \le 1} > $ > *Proof*. Denote $U_{xi\varepsilon} = \bracs{y \in \cx: \abs{f_i(x - y)} < \varepsilon}$. Let $x \in E$ such that $\norm{x} = 1$, and $U \in \cn(x)^o$, then there exists $\seqf{U_{x, i_k, \varepsilon_k}}$ such that $\bigcap_{k = 1}^{n}U_{x, i_k, \varepsilon_k} \subset U$. > > Since $\cx$ is infinite-dimensional, there exists $y \in \cx$ such that $f_i(y) = 0$ for all $i \in [1, n]$. The function $\phi(t) = \norm{x + ty}$ is continuous with $g(0) < 1$ and $\limv{t}\phi(t) = \infty$. By the intermediate value theorem there exists $t_0$ such that $\norm{x + t_0y} = 1$, and $x + t_0y \in U \cap S$. Therefore $x$ cannot be separated from $S$ using open neighbourhoods, $x \in \ol{S}$, and $B_E \subset \ol{S}$. > > As $B_E$ itself is closed, $B_E = \ol S$. > [!theorem] > > Let $\cx$ be a normed space and $C$ be a [[Convexity|convex]] subset, then $C$ is [[Closed Set|closed]] in the weak topology if and only if it is closed in the strong topology. > > *Proof*. Suppose that $C$ is closed in the strong topology. Let $x \in C^c$, then by the [[Hahn-Banach Theorem]], there exists a [[Affine Hyperplane|closed hyperplane]] strictly separating $x$ and $C$, a linear functional $f \in \cx^*$, and $\alpha \in \real$ such that > $ > f(x) < \alpha < f(y) \quad \forall y \in C > $ > Let $V = \bracs{y \in \cx: f(x) < \alpha} \subset C^c$, then $x \in (C^c)^o$ for all $x \in C^c$, and $C$ is open in the weak topology. > [!theorem] > > Let $\cx$ be a normed space and $\seq{x_n} \subset \cx$ such that $x_n \to x$ weakly. Then there exists a sequence $y_n$ made up of convex combinations of $x_n$s that converges strongly to $x$. > > *Proof*. Let $C = \text{conv}\bracs{x_n: n \in \nat}$ be the convex hull for the sequence, then the strong closure of $C$ is also convex. Since $x_n \to x$ weakly, $x$ is in the weak closure of $C$. As the weak closure with $C$ coincides with the strong closure of $C$, $x \in \ol{C}$, and there exists a sequence $\seq{y_n} \subset C$ such that $y_n \to x$. > > Since $C$ consists of the convex combinations of $\bracs{x_n: n \in \nat}$, $\seq{y_n}$ is a sequence of the desired convex combinations. # Topology > [!definition] > > Let $X$ be a set, $\bracs{(X_i, \topo_i)}_{i \in I}$ be a family of [[Topological Space|topological spaces]], and $\bracs{f_i: X \to X_i}_{i \in I}$ be a family of mappings. The topology $\topo = \topo(\ce)$ generated by > $ > \ce = \bracs{f_{i}^{-1}(U_i): U_i \in \topo_i, i \in I} > $ > being the smallest topology that makes all $f_i$s [[Continuity|continuous]] is the **weak topology** generated by $\bracs{f_i}_{i \in I}$. > [!theorem] > > Let $X$ be a set, $Y$ be a [[Hausdorff Space|Hausdorff space]]. $\cf$ be a collection of mappings from $X$ to $Y$, and $\topo$ be the weak topology generated by $\cf$. Then $\topo$ is Hausdorff if and only if for each $x, y \in X: x \ne y$, there exists $f \in \cf$ such that $f(x) \ne f(y)$. > > *Proof*. Suppose that $\topo$ is Hausdorff. Let $x, y \in \cx: x \ne y$, then there exists $U, V \in \topo$ such that $U \cap V = \emptyset$ and $x \in U, y \in V$. Since $\ce$ generates the topology, there exists $\seqf{f_i^{-1}(U_i)}$ (where $f_i \in \cf$ and $U_i$ is open in $Y$) such that $\bigcap_{i = 1}^{n}f_i^{-1}(U_i) \subset U$. Since $y \not\in \bigcap_{i = 1}^{n}f_i^{-1}(U_i)$, there exists $i$ such that $y \not\in f_i^{-1}(U_i)$. Therefore $f(y) \not\in U_i$ and $f(x) \ne f(y)$. > > Now suppose that for each $x, y \in X: x \ne y$, there exists $f \in \cf$ such that $f(x) \ne f(y)$. Since $Y$ is Hausdorff, let $U \in \cn(f(x))^o$ and $V \in \cn(f(y))^o$ such that $U \cap V = \emptyset$, then $f^{-1}(U) \in \cn(x)^o$, $f^{-1}(V) \in \cn(y)^o$ with $f^{-1}(U) \cap f^{-1}(V) = \emptyset$. > [!theorem] > > Let $X$ be a topological space and $\cf$ be a family of functions mapping $X$ to different topological spaces. If $\topo$ is the weak topology generated by $\cf$, then $\net{x}$ converges to $x$ if and only if $f(x_\alpha) \to f(x)$ for all $f \in \cf$. > > *Proof*. Suppose that $x_\alpha \to x$. Let $f \in \cf$, then for all $U \in \cn(f(x))^o$, there exists $\beta \in A$ such that $x_\alpha \in f^{-1}(U)$ for all $\alpha \gtrsim \beta$. Therefore $f(x_\alpha) \to f(x)$. > > Now suppose that $f(x_\alpha) \to f(x)$ for all $f \in \cf$. Let $U \in \cn(x)$, then there exists $\seqf{f_i^{-1}(U_i)}$ where $f_i \in \cf$ and each $U_i$ is open in the target of $f_i$, such that $\bigcap_{i = 1}^{n}f_i^{-1}(U_i) \subset U$. For each $i$, there exists $\beta_i$ such that $x_\alpha \in f_i^{-1}(U_i)$ for all $\alpha \gtrsim \beta_i$. Let $\beta \in A: \beta \gtrsim \beta_i$ for all $i$, then $x_\alpha \in \bigcap_{i = 1}^{n}f_i^{-1}(U_i) \subset U$ for all $\alpha \gtrsim \beta$. Therefore $x_\alpha \to x$.