0

I'm reformulating the proof of Banach–Alaoglu theorem in Brezis's book of Functional Analysis. My goal is make his argument clearer. I'm very happy to use net convergence to characterize compactness and continuity.

Let $(E, | \cdot |)$ be a normed linear space and $E^\star$ its topological dual. We denote by $E^\star_\mathrm{w}$ the set $E$ with its weak$^\star$ topology. Let $\mathbb B_{E^\star} := \{f \in E^\star \mid \| f\| \le 1 \}$. Then $\mathbb B_{E^\star}$ is compact in the weak$^\star$ topology.

I posted my proof as an answer below. Could you have a check on my attempt?

PS: I posted my proof separately so that I can accept my own answer to remove my question from unanswered list. Surely, if other people post answers, I will hapily accept theirs.

Analyst
  • 5,637

1 Answers1

0

Let $F := \mathbb R^E$ be the collection of all maps from $E$ to $\mathbb R$. Let $\mathbb R_x := \mathbb R$ for all $x\in E$. Then we can write $$F = \prod_{x\in E} \mathbb R_x.$$ In this way, we endow $F$ with the product topology. Of course, $E^\star \subseteq F$. Let $i:E^\star_\mathrm{w} \to F, f \mapsto f$ be the canonical injection. Let's prove that $i$ is continuous. For $x\in E$, let $\pi_x: F \to \mathbb R_x, f \mapsto f(x)$ be the canonical projection. It suffices to show that $\pi_x \circ i:E^\star_\mathrm{w} \to \mathbb R_x, f \mapsto \langle f, x \rangle$ is continuous for all $x\in E$. This is clearly true due to the construction of the weak$^\star$ topology.

Lemma: Let $(X, \tau)$ be a topological space, $A \subseteq X$, and $\tau_A$ the subspace topology of $A$. Let $a\in A$ and $(x_d)_{d \in D}$ is a net in $A$. Then $x_d \to a$ in $\tau$ if and only if $x_d \to a$ in $\tau_A$.

Clearly, $\operatorname{im} i = E^\star$. We denote by $E^\star_\tau$ the set $E^\star$ together with the subspace topology $\tau$ induced from $F$. Then $i:E^\star_\mathrm{w} \to E^\star_\tau$ is bijective. Let $f\in E^\star_\mathrm{w}$ and $(f_d)_{d\in D}$ be a net in $E^\star_\mathrm{w}$ such that $f_d \to f$. Because $i:E^\star_\mathrm{w} \to F$ is continuous, $f_d \to f$ in the topology of $F$. By our lemma, $f_d \to f$ in $E^\star_\tau$. Hence $i:E^\star_\mathrm{w} \to E^\star_\tau$ is indeed continuous.

Let $i^{-1}:E^\star_\tau \to E^\star_\mathrm{w}$ be the inverse of $i:E^\star_\mathrm{w} \to E^\star_\tau$. Let's prove that $i^{-1}$ is continuous. It suffices to show that $\varphi_x: E^\star_\tau \to \mathbb R, f \mapsto \langle f, x\rangle$ is continuous for all $x\in E$. This is indeed true because $\varphi_x = \pi_x \restriction E^\star$. Notice that continuous map sends compact set to compact set. Hence it suffices to prove that $\mathbb B_{E^\star}$ is compact in $\tau$. By our lemma, it suffices to prove that $\mathbb B_{E^\star}$ is compact in the topology of $F$.

Let $B_1 := \{f\in F \mid f \text{ is linear}\}$ and $B_2 := \prod_{x\in E}[-|x|, |x|]$. Then $\mathbb B_{E^\star} = B_1 \cap B_2$. The closed interval $[-|x|, |x|]$ is clearly compact. By Tychonoff's theorem, $B_2$ is compact.

Let $f\in F$ and $(f_d)_{d\in D}$ be a net in $B_1$ such that $f_d \to f$. Because convergence in product topology is equivalent to pointwise convergence, we get $f_d(x) \to f(x)$ for all $x\in E$. Then $f_d(x) + f_d(y) =f_d(x+y) \to f(x+y)$. On the other hand, $f_d(x) \to f(x)$ and $f_d(y) \to f(y)$. This implies $f(x+y)=f(x)+f(y)$. Similarly, $f(\lambda x) =\lambda f(x)$ for all $\lambda \in \mathbb R$. Hence $B_1$ is closed. The intersection of a closed set and a compact set is again compact. This completes the proof.

Analyst
  • 5,637