Alaoglu's Theorem goes as follows: If $X$ is a normed space, then $B^* := \{x^* \in X^* \mid \|x^*\| \leq 1\}$ is weak-star compact.
The proof starts something like this:
For $x \in B =\{x \in X \mid \|x\| \leq 1\}$, let $D_x = \{ \alpha \in \mathbb{F} \mid |\alpha|\leq1\}$ and put $D:= \prod \{D_x \mid x \in B\}$. By Tychonoff, $D$ is compact. Now define $\tau \colon B^* \to D$ by $x^* \mapsto \tau(x^*)$, where $\tau(x^*) \colon B \to \mathbb{F}$ is given by $\tau(x^*)(x) = x^*(x)$. It is not hard to see that $\tau$ is a continuous injection and we can consider $\tau^{-1} \colon \tau(B^*) \to B^*$. Somehow I am stumped at proving that $\tau^{-1}$ is also continuous. I know that it would suffice to prove that $ev_x \circ \tau^{-1}$ is continuous for all $x \in B$ (since the weak*-topology is the initial topology on $X$ wrt. the evalution maps $\{ev_x \mid x \in X\} \subset X^{**}$, where $ev_x(x^*) = x^*(x)$ for all $x^* \in X^*$, and since therefore we know that the relative topology on $B^*$ is just the initial topology induced by the evaluations $\{ev_x \mid x \in B\} \subset B^{**}$). Now if we take $y \in \tau(B^*)$, then for $x \in X$ we have $ev_x \circ \tau^{-1}(y) = ev_x(y^*) = y^*(x)$, where $\tau(y^*) = y$ for some $y^* \in B^*$. Some books now claim that it is evident, that $\tau^{-1}$ is continuous.
Can someone clarify on why this is justified? Or can someone give an alternative argument on why $\tau^{-1}$ is continuous?
To show that $\tau^{-1}: \tau(B^*) \rightarrow B^*$ is continuous, consider a net $(\tau(y_i))_{i \in I}$ in $\tau(B^*)$, (where $y \in B^*, \forall i \in I: y_i \in B^*$) which converges to $\tau(y) \in \tau(B^*)$. Then by definition of convergence in the product, we have that the image under every evaluation map converges i.e. we have $\forall x \in B: y_i(x) \rightarrow y(x)$.
We want to show that $\tau^{-1}(\tau(y_i)) = y_i \rightharpoonup^* y = \tau^{-1}(\tau(y))$ i.e. by definition of the weak topology, that $\forall x \in X: y_i(x) \rightarrow y(x)$.
Now it is just left to prove that convergence on the unit sphere implies convergence on the entire normed space.
Actually, if we defined $D_x := \{ \alpha \in \mathbb{F} : \vert \alpha \vert \leq \Vert x \Vert\}$, then we could define $\prod \{D_x : x \in X\}$ (note the different indexing set) and thus the convergence on $D$ were $\forall x \in X: y_i(x) \rightarrow y(x)$ which then immediately gives the result.