# Banach-Alaoglu theorem, Rudin's functional analysis.

Few questions about the theorem

If $$V$$ is a neighborhood of $$0$$ in a topological vector space $$X$$ and if $$K = \left\{\lambda \in X^* : |\Lambda x | \leq 1 \; \text{for every} \; x \in V \right\}$$ then $$K$$ is weak* compact.

I'll comment on the proof

Proof: Since neighborhoods of $$0$$ are abosrbing, there corresponds to each $$x \in X$$ a number $$\gamma(x) < \infty$$ such that $$x \in \gamma(x)V$$. Hence $$|\Lambda x|\leq \gamma(x) \; (x \in X, \Lambda \in K) \;\;\;\;(1)$$ Let $$D_x$$ be the set of all scalars $$\alpha$$ such that $$|\alpha| \leq \gamma(x)$$. Let $$\tau$$ be the product topology on $$P$$, the cartesian product of all $$D_x$$, one for each $$x \in X$$. Since each $$D_x$$ is compact, so is $$P$$, by Tychinoff's theorem. The elements of $$P$$ are the functions $$f$$ on $$X$$ (linear or not) that satisfy $$|f(x)| \leq \gamma(x) \;\;\; (x \in X) \;\;\;\;\; (2).$$

Why are such $$f$$ the elements of $$P = \prod_{x \in X} D_x$$?

Thus $$K \subset X^* \cap P$$. It follows that $$K$$ inherits two topologies, one from $$X^*$$ (it's weak* topology, to which the conclusion of the theorem refer) and the other, $$\tau$$, from $$P$$. We will see that

(a) these two topologies coincide on $$K$$, and

(b) $$K$$ is a closed subset of $$P$$

Since $$P$$ is compact, (b) implies that $$K$$ is $$\tau$$ compact, and then (a) implies that $$K$$ is weak*-compact. Fix $$\Lambda_0 \in K$$. Choose $$x_i \in X, 1 \leq i \leq n$$; choose $$\delta > 0$$. Put $$W_1 = \left\{\Lambda \in X^* : |\Lambda x_i - \Lambda_0 x_i | < \delta \; for \; 1 \leq i \leq n \right\} \;\;\;\; (3)$$ and $$W_2 = \left\{f \in P : |f(x_i) - \Lambda_0 x_i | < \delta \; for \; 1 \leq i \leq n \right\} \;\;\;\; (4)$$ Let $$n, x_i$$ and $$\delta$$ range over all admissible values. The resulting sets $$W_1$$ form a local base for the weak* topology of $$X^*$$ at $$\Lambda_0$$ and the sets $$W_2$$ form a local base for the product topology of $$P$$ at $$\Lambda_0$$. Since $$K \subset P \cap X^*$$, we have $$W_1 \cap K = W_2 \cap K.$$ This proves (a).

Why do we have both $$K \subset P \cap X^*$$ and $$W_1 \cap K = W_2 \cap K.$$ ?

The rest of the theorem is clear to me.

Let $$F=\{f:X\rightarrow \mathbb{R}: |f(x)|\leq \gamma(x)\}$$. $$P=\prod_{x\in X} D_x$$ this implies that an element of $$P$$ is a family $$(a_x)_{x\in X}$$ where $$a_x\in D_x$$, which is equivalent to saying that $$|a_x|\leq \gamma(x)$$. Write $$f(x)=a_x$$ then $$f(x)\in F$$.

Conversely, if you have $$f\in F$$, $$|f(x)|\leq \gamma(x)$$ implies that $$f(x)\in D_x$$, so the map $$F\rightarrow \prod_{x\in X}D_x$$ which assigns $$(f(x))_{x\in X}$$ is well defined and is a bijection.

$$K=\{\lambda\in X^*:|\lambda(x)|\leq 1, x\in V\}$$. Let $$y\in X$$, there exists $$\gamma(y)$$ and $$z\in V$$ with $$y=\gamma(y)z$$, we have $$|f(y)|=|\gamma(y)||f(z)|\leq |\gamma(y)|$$ since $$|f(z)|\leq 1$$, we deduce that $$(f(y))_{y\in X}\in P$$, this implies that $$f\in P$$.

$$W_1=\{\Lambda\in X^*:|\Lambda x_i-\Lambda x_0|\leq \delta, 1\leq i\leq n\}$$, this implies that $$W_1\cap K=\{\Lambda\in X^*\cap K:|\Lambda x_i-\Lambda x_0|\leq \delta, 1\leq i\leq n\}=\{\Lambda\in K:|\Lambda x_i-\Lambda x_0|\leq \delta, 1\leq i\leq n\}$$. since $$K\subset X^*\cap P\subset X^*$$.

$$W_2=\{\Lambda\in P^*:|\Lambda x_i-\Lambda x_0|\leq \delta, 1\leq i\leq n\}$$, this implies that $$W_2\cap K=\{\Lambda\in P\cap K:|\Lambda x_i-\Lambda x_0|\leq \delta, 1\leq i\leq n\}=\{\Lambda\in K:|\Lambda x_i-\Lambda x_0|\leq \delta, 1\leq i\leq n\}$$ since $$K\subset X^*\cap P\subset P$$.

We deduce that $$W_1\cap P=W_2\cap P$$.

• What about the second bit? – user8469759 Jan 10 at 17:39