# Proof of the Banach-Alaoglu theorem

I have some doubts about the point of this proof of the Banach-Alaouglu theorem:

"The cloesed ball $B_{E'}:=\lbrace u \in E' : \left \| u \right \|_{E'} \leq 1 \rbrace$ of dual $E'$ of a normed space $E$ is $\sigma(E',E)-$compatto (weak* compact). Moreover, if $E$ è separable, the weak* topology $\sigma(E',E)$ restricted to $B_{E'}$ is metrizable."

Proof.

(1) We remember that the weak* topology $\sigma(E',E)$ is the restriction to $E'$ of the product topology (or topology of pointwise convergence) on $\mathbb{K}^E$, and we have \begin{align*} \displaystyle B_{E'} \subseteq K:= \prod_{x \in E} \overline{D}(0,\left \| x \right \|)=\prod_{x \in E} \lbrace z \in \mathbb{K} : |z| \leq \left \| x \right \|_{E} \rbrace \end{align*} since if $\left \| u \right \|_{E'} \leq 1$ then $u=\lbrace u(x) \rbrace_{x \in E}$ satisfy $|u(x)| \leq \left \| u \right \|_{E'} \left \| x \right \|_E \leq \left \| x \right \|_E$ $\forall x \in E$. By Tychonoff theorem $K$ is a compact subset of $\mathbb{K}^E$, and \begin{align*} (2) \displaystyle B_{E'} = \bigcap_{x,y} \lbrace f \in K : f(x+y)=f(x)+f(y) \rbrace \cap \bigcap_{\lambda,x} \lbrace f \in K: f(\lambda x) = \lambda f(x) \rbrace \end{align*} (can you have a clarification on identity in (2)?) is the intersection of closed subsets defined by equalities with continuous functions $\pi_{x+y}(f)=\pi_x(f)+\pi_y(f)$ e $\pi_{\lambda x}= \lambda \pi_x(f)$ Then $B_{E'}$ is a closed in compact set $K$ and it is also $\sigma(E',E)-$compact as subspace and by point (1).

If $E$ is separable then exists $\lbrace x_n \rbrace_{n=1}^{\infty} \subset E$ that is dense in $E$, and we have that the seminorm family $p_{x_n}(u):=|u(x_n)|$ is separable on $E'$ since by continuity: $u(x_n)=0$ $\forall \lbrace x_n \rbrace \subset E$ $\Longrightarrow$ $\displaystyle u(x)=0$ $\forall x \in E$. This sequence of seminorm defines on $E'$ a locally convex metrizable (why metrizable?) topology $\mathcal{T}$ wich is clearly weaker than $\sigma(E',E)$ (so $\mathcal{T} \subset \sigma(E',E)$? why ?) Now, on $B_{E'}$, these topology concide since $I: (B_{E'},\sigma(E',E)) \longrightarrow (B_{E'}, \mathcal{T})$ is continuous and the image of a compact (or closed) set is $\mathcal{T}$-closed since it is $\mathcal{T}-$compact (so identity is a homeomorphism?).

Thanks for any help

Identity (2) just says that $B_{E'}$ is the subset of $K$ conssting of the linear elements of $K$.
The topology defined by any countable family $(\rho_n)$ of seminorms (that separates points) is metrizable; it's equivalent to the topology defined by the metric $$d(x,y)=\sum_{n=1}^\infty 2^{-n}\frac{\rho_n(x-y)}{1+\rho_n(x-y)}.$$You can check that if $x_\alpha$ is any net then $\rho_n(x_\alpha)\to0$ for every $n$ if and only if $d(x_\alpha,0)\to0$.
• Thanks, now I understand these points. So it is also in this sense we have that $\mathcal{T} \subset \sigma(E',E)$? and for last question is it correct that this check implies that identity is a homeomorphism? – user288972 Aug 16 '16 at 16:06
• I'm not sure I understand the question. $\Tau$ is contained in the weak* topology because it is; this is immediate from the definitions of the open sets in the two topologies. And it's a fact from general topology that any continuous bijection from a compact space onto a Hausdorff space is a homoemorphism. – David C. Ullrich Aug 16 '16 at 16:21