I'm reading a proof of Banach Alaoglu Theorem from Functional Analysis by S Keshavan on page 142:
Theorem: Let $V$ be a Banach space. Then $B^{*}$, the closed unit ball in $V^*$ is weak$^*$-compact.
Proof: For each $x$ in $V$, we let $U_{x}$ be the closed ball of radius $||x||_{b}$ in $\mathbb{C}$. Then, we can take $U = \Pi_{x\in V} U_{x}$,which is clearly compact by Tychonoff theorem.
The map $$ \tau : f \mapsto (f(x))_{x\in V} $$ defines a continuous injection from the unit ball of $B^*$ to $U$. Let $A$ denote the image of $\tau$, so it suffices to prove that $A$ is closed in $U$.
Let $(f_x)_{x\in V} \in cl(\tau(B^*))$. Define for $x\in V$, $f(x)=f_x$. The proof will be complete if we can show that $f$ is linear.
let $\epsilon >0$ be given. Then given $x$ and $y$ in $V$,
we can find $g \in B^{*}$ such that $ \vert g(x)-f(x) \vert < \epsilon/3$,$ \vert g(y)-f(y) \vert < \epsilon/3$,$ \vert g(x+y)-f(x+y) \vert < \epsilon/3$
Can someone explain me how do we get $g$ mentioned in last paragraph?
Note that, by definition $\tau(B^*)$ is dense in $\def\c{\mathop{\rm cl}}\c(\tau B^*)$. Given now $x,y \in V$ and $\epsilon > 0$, the set $$ W := \{h \in \c(\tau B^*) : |h(x) - f(x)|, |h(y) - f(y)|, |h(x+y) - f(x+y)| < \epsilon/3 \} $$ is an open - by the definition of the product topology - set containing $f$. As $f \in \c(\tau B^*)$, there is an $g \in B^*$ such that $\tau(g) \in W$.