Proof. (⇒) Suppose $\( {f}_{α} \to f \)$ in $\( \mathop{\mathrm{w}^*} \)$. For $\( x\in X \)$, $\( ε\gt 0 \)$, define $\( V\mathopen{}\left( f, x, ε\right)\mathclose{}= \mathopen{}\left\{\, g\in X^{*}\,\middle\vert\, , \mathopen{}\left\lvert{}f\mathopen{}\left( x\right)\mathclose{}-g\mathopen{}\left( x\right)\mathclose{}\right\rvert\mathclose{}\lt ε, \,\right\}\mathclose{} \)$. Then $\( {f}_{α} \mathbin{\overset{\mathop{\mathrm{w}^*}}{\to}}f \)$ implies $\( {f}_{α}\in V\mathopen{}\left( f, x, ε\right)\mathclose{} \)$ eventually for all $\( x\in X \)$ and $\( ε\gt 0 \)$, which implies $\( \mathopen{}\left\lvert{}{f}_{α}\mathopen{}\left( x\right)\mathclose{}-f\mathopen{}\left( x\right)\mathclose{}\right\rvert\mathclose{}\lt ε \)$ eventually for all $\( x\in X \)$ and $\( ε\gt 0 \)$, which implies $\( {f}_{α}\mathopen{}\left( x\right)\mathclose{} \to f\mathopen{}\left( x\right)\mathclose{} \)$ for all $\( x\in X \)$.

(⇐) Suppose $\( {f}_{α}\mathopen{}\left( x\right)\mathclose{} \to f\mathopen{}\left( x\right)\mathclose{} \)$ for all $\( x\in X \)$. Let $\( {x}_{i}\in X \)$ for $\( i\in \mathopen{}\left\{\, 1, 2, \dotsc, n\,\right\}\mathclose{} \)$ and let $\( ε\gt 0 \)$ be given. $\( {f}_{α}\mathopen{}\left( x\right)\mathclose{} \to f\mathopen{}\left( x\right)\mathclose{} \)$ for all $\( x \)$ implies $\( {f}_{α}\mathopen{}\left( {x}_{i}\right)\mathclose{} \to f\mathopen{}\left( {x}_{i}\right)\mathclose{} \)$ for all $\( i \)$. This implies $\( {f}_{α}\in \mathop{\mathrm{N}}\mathopen{}\left( f, {x}_{i}, ε\right)\mathclose{} \)$ for all $\( i \)$ eventually. Then for all $\( i \)$, there exists $\( {α}_{i} \)$ such that for all $\( α\gt {α}_{i} \)$, $\( {f}_{α}\in \mathop{\mathrm{N}}\mathopen{}\left( f, {x}_{i}, ε\right)\mathclose{} \)$. Pick $\( β\gt {α}_{i} \)$ for all $\( i \)$. Then $\( {f}_{β}\in \bigcap_{i=1}^{n}{} \mathop{\mathrm{N}}\mathopen{}\left( f, {x}_{i}, ε\right)\mathclose{} = \mathop{\mathrm{N}}\mathopen{}\left( f, {x}_{1}, \dotsc, {x}_{n}, ε\right)\mathclose{} \)$ which gives $\( {f}_{α} \mathbin{\overset{\mathop{\mathrm{w}^*}}{\to}}f \)$.

Proof. For $\( x\in X \)$, define $\( {I}_{x}= \mathopen{}\left\{\, z\in \mathbb{C}\,\middle\vert\, , \mathopen{}\left\lvert{}z\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}x\right\rVert\mathclose{}, \,\right\}\mathclose{} \)$. Then $\( {I}_{x} \)$ is compact in $\( \mathbb{C} \)$. Let $\( I= \prod_{x\in X}{}{I}_{x} \)$. $\( I \)$ is compact by the Tychonoff product theorem (with the product topology). Let $\( B= \mathopen{}\left\{\, f\in X^{*}\,\middle\vert\, , \mathopen{}\left\lVert{}f\right\rVert\mathclose{}\leq 1, \,\right\}\mathclose{} \)$. Define $\( J : B \hookrightarrow I \)$, $\( f\mapsto \mathopen{}\left\{\, f\mathopen{}\left( x\right)\mathclose{}\,\middle\vert\, , x\in X, \,\right\}\mathclose{} \)$.

Claim 1: $\( B\simeq J\mathopen{}\left( B\right)\mathclose{} \)$. $\( J \)$ is injective, as $\( J\mathopen{}\left( f\right)\mathclose{}= J\mathopen{}\left( g\right)\mathclose{} \)$ implies $\( \mathopen{}\left\{\, f\mathopen{}\left( x\right)\mathclose{}\,\middle\vert\, x\in X\,\right\}\mathclose{}= \mathopen{}\left\{\, g\mathopen{}\left( x\right)\mathclose{}\,\middle\vert\, x\in X\,\right\}\mathclose{} \)$ implies $\( f= g \)$. To show $\( J \)$ is bicontinuous (both it and its inverse are continuous) onto $\( J\mathopen{}\left( B\right)\mathclose{} \)$, let $\( \mathopen{}\left({f}_{α}\right)\mathclose{} \)$ be a net in $\( B \)$ converging in $\( \mathop{\mathrm{w}^*} \)$ to $\( f\in B \)$. By Lemma III.32, this is if and only if $\( {f}_{α}\mathopen{}\left( x\right)\mathclose{} \to f\mathopen{}\left( x\right)\mathclose{} \)$ for all $\( x\in X \)$ if and only if $\( \mathopen{}\left\{\, {f}_{α}\mathopen{}\left( x\right)\mathclose{}\,\middle\vert\, x\in X\,\right\}\mathclose{} \to \mathopen{}\left\{\, f\mathopen{}\left( x\right)\mathclose{}\,\middle\vert\, x\in X\,\right\}\mathclose{} \)$ if and only if $\( J\mathopen{}\left( {f}_{α}\right)\mathclose{} \to J\mathopen{}\left( f\right)\mathclose{} \)$. This shows $\( J \)$ and $\( {J}^{-1} \)$ are continuous, and the claim is proven.

Thus $\( B \)$ is compact if and only if $\( J\mathopen{}\left( B\right)\mathclose{} \)$ is compact if and only if $\( J\mathopen{}\left( B\right)\mathclose{} \)$ is closed.

Claim 2: $\( J\mathopen{}\left( B\right)\mathclose{} \)$ is closed. Let $\( \mathopen{}\left\{\, J\mathopen{}\left( {f}_{α}\right)\mathclose{}\,\right\}\mathclose{} \)$ be a net in $\( J\mathopen{}\left( B\right)\mathclose{} \)$ and suppose $\( \mathopen{}\left\{\, J\mathopen{}\left( {f}_{α}\right)\mathclose{}\,\right\}\mathclose{} \to \mathopen{}\left\{\, f\mathopen{}\left( x\right)\mathclose{}\,\middle\vert\, x\in X\,\right\}\mathclose{} \)$. Then $\( \mathopen{}\left\{\, {f}_{α}\mathopen{}\left( x\right)\mathclose{}\,\middle\vert\, x\in X\,\right\}\mathclose{} \to \mathopen{}\left\{\, f\mathopen{}\left( x\right)\mathclose{}\,\middle\vert\, x\in X\,\right\}\mathclose{} \)$. We want to show $\( f\in B \)$.

First, we have $\( f\mathopen{}\left( x+y\right)\mathclose{}= f\mathopen{}\left( x\right)\mathclose{}+f\mathopen{}\left( y\right)\mathclose{} \)$ and $\( f\mathopen{}\left( λx\right)\mathclose{}= λf\mathopen{}\left( x\right)\mathclose{} \)$ because these are limits of nets and thus we can use properties of nets in topological vector spaces. That is $$\[ f\mathopen{}\left( x+y\right)\mathclose{}= \lim_{α}{} {f}_{α}\mathopen{}\left( x+y\right)\mathclose{} \text{,} \]$$ $$\[ f\mathopen{}\left( x\right)\mathclose{}+f\mathopen{}\left( y\right)\mathclose{}= \lim_{α}{} {f}_{α}\mathopen{}\left( x\right)\mathclose{} +\lim_{α}{} {f}_{α}\mathopen{}\left( y\right)\mathclose{} \text{,} \]$$ etc.

For $\( x\in X \)$, $\( f\mathopen{}\left( x\right)\mathclose{}\in {I}_{x} \)$ implies $\( \mathopen{}\left\lvert{}f\mathopen{}\left( x\right)\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}x\right\rVert\mathclose{} \)$ for all $\( x \)$. This gives $\( \mathopen{}\left\lVert{}f\right\rVert\mathclose{}= \sup_{ \mathopen{}\left\lVert{}x\right\rVert\mathclose{}\leq 1 }{}\mathopen{}\left\lvert{}f\mathopen{}\left( x\right)\mathclose{}\right\rvert\mathclose{}\leq 1 \)$. Thus we have $\( f\in B \)$. From this, we get $\( \mathopen{}\left\{\, f\mathopen{}\left( x\right)\mathclose{}\,\middle\vert\, , x\in X, \,\right\}\mathclose{}= J\mathopen{}\left( f\right)\mathclose{}\in J\mathopen{}\left( B\right)\mathclose{} \)$. Finally, then, $\( J\mathopen{}\left( B\right)\mathclose{} \)$ is closed.