(Folland's page 169) Let $X$ be a normed vector space. Let $B^*:= \{ f \in X^* \,: \, ||f|| \le 1 \}$ be the unit ball in $V^*$ under the operator norm. $B^*$ is compact in $X^*$ in the weak$^*$ topology.
In his proof, he stated that
We may identify $B^*$ with $D:= \prod D_x$, where $D_x:= \{ z \in \mathbb{C} \, : \, |z| \le ||x|| \}$. We would like to show that $B^*$ is closed in $D$.
Here is my attemt of proof :
Define $$ \phi_{x,y,\lambda}: D \rightarrow \mathbb{F}, f \mapsto f(x+\lambda y)-f(x) -\lambda f(y) $$
$\phi_{x,y,\lambda}$ is continuous as it is the composition of the maps (supposing, $x,y,x+\lambda y$ are distinct) $$ D \rightarrow \mathbb{F}_x \times \mathbb{F}_y \times \mathbb{F}_{x+\lambda y } \rightarrow \mathbb{F}$$ The first map $f \mapsto (f(x), f(y),f(x+\lambda y))$ is continuous by definition universal property/definition of product topology. Continuity of last map follows from property of topological vector space.
$B^* = \bigcap_{x,y \in X, \lambda \in \mathbb{F}} \ker \phi_{x,y,\lambda} $ is closed in inherited product topology.
Is this proof correct?
For any $x\in X$, let \begin{equation} D_{x}=\lbrace z\in ℂ:\vert z\vert\leqslant\Vert x\Vert \rbrace\\ \end{equation}
and $D=\prod_ {x\in X} D_{x}$. Since $D_{x}$ is a compact subset of ℂ, D is compact in product topology by Tychonoff theorem.
We prove the theorem by finding a homeomorphism that maps the closed unit ball $ B_{{X^\ast}} $ of $X^{*}$ onto a closed subset of D. Define \begin{equation} \varphi_{x}:B_{X^{\ast}} \longrightarrow D_{x} , \varphi x(f)=f(x) , \varphi:B_{X^{\ast}}\longrightarrow D\\ \end{equation} by \begin{equation} \varphi=\prod_{x\in X} \varphi{x}, \end{equation} so that \begin{equation} \varphi\left(f \right)=\left(f(x)\right) x\in X.\\ \end{equation} Obviously, $\varphi $ is one to one, and a net $\left(f_{α}\right)\in B_{X^{*}}$ converges to f in weak$^{*}$ topology of $X^{*}$ if $\varphi \left(f_{α}\right)$ converges to $\varphi(f)$ in product topology, therefore $ \varphi $ is continuous and so is its inverse \begin{equation} \varphi^{-1}:\varphi\left(B_{X^{*}}\right) \longrightarrow B_{X^{*}}\\ \end{equation} It remains to show that $\varphi \left(B_{X*}\right)$ is closed. If \begin{equation} \varphi \left(f_{α}\right) \end{equation} is a net in $\varphi \left(B_{X*}\right)$, converging to a point $ d=(d_{x}) $, $ x\in X\in D$, we can define a function \begin{equation} f: X \longrightarrow C , f(x)=d_{x}.\\ \end{equation} As $\lim_{α}\varphi\left(f{α(x)}\right)=d_{x}$ for all $ x\in X$ by definition of weak$^{*}$ convergence, one can easily see that f is a linear functional in $ B_{X*}$ and that $\varphi(f)=d$. This shows that d is actually in $\varphi(B_{X^{*}})$ and finishes the proof.