Constructive proof of the Banach-Alaouglu theorem

Question

Is there a constructive (i.e., not using Axiom of choice, and at most Axiom of dependent choice) proof of the Banach-Alaoglu theorem in the case of separable Banach spaces? Even if it is needed assume that the dual is separable. Under even more assumptions - is there such a proof for infinite-dimensional Banach spaces.

We know such proof exists for Hahn-Banach in the separable case.

Do you know a reference for Hahn-Banach in separable case, as you stated in the last sentence of the question? — Hayk, Jan 24 '19 at 10:37
I don't know too much about set-theoretic foundations, but I think the proof in Dieudonnè's Foundations of Modern Analysis II (Section 12.15 in my copy) works without axiom of choice (maybe you need countable choice?). — MaoWao, Jan 24 '19 at 10:43
As I recall the first proof published by Banach himself only dealt with the separable case, and was constructive (he proved sequential compactness directly). Later Alaouglu published the general theorem which relies on Tychonoff's theorem. — pitariver, Jan 24 '19 at 10:54
After the answer, I want to point out something. Actually, countable product of metrizable sets is metrizable (with constructive metric), countable product of compacts is compact (constructively). Now if one follows the standard proof of Banach-Alaouglu and applies those two instead of Tychonoff, one arrives at constructive proof of it. — Stoyan Apostolov, Apr 29 '19 at 15:10

score 7 · Accepted Answer · edited Mar 30 '22 at 15:23

If $X$ is separable then the closed unit ball $B$ of $X^{*}$ is metrizable in the $weak^{*}$ topology. [ $d(f,g)=\sum\limits_{k=1}^{\infty} \frac 1 {2^{i}} \frac {|f(x_i)-g(x_i)|} {1+|f(x_i)-g(x_i)|}$ metrizes it if $\{x_i\}$ is dense in $X$]. Any sequence $\{f_n\}$ in $B$ has a subsequence $\{f_n'\}$ converging at each of the $x_j$'s (by a diagonal procedure). It is now fairly easy to see that if $\{x_{i_l}\}$ converges to some point $x$ then $f_n'(x_{i_l})$ is Cauchy. Call the limit $f(x)$. Obviously, $\|f\|\leq 1$. It follows that $f_n'(x_{i_l})$ converges in the weak* topology. Thus, $B$ is sequentially compact and metrizable, hence compact.

Constructive proof of the Banach-Alaouglu theorem

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