On the proof of Alaoglu's theorem

Question

Currently reading through a functional analysis text, namely over the weak topology and weak* topology. In the text, the author gives the following theorem and proof:

Theorem 1(Alaoglu): For a real Banach space $X$, the unit ball $\text{B}(X^*):=\{f\in X^*:\|f\|\leq1\}$ is a compact set in the $w^*$-topology.

Proof. Let $I_x:=[-\|x\|,\|x\|]$ and $K:=\prod I_x$ be the product of intervals indexed by elements $x\in X$ and equipped with the product topology. By Tikhonov theorem $K$ is a compact set in the product topology. We rename the elements of $K$ to be functions on $X$, not necessarily linear, such that $|f(x)|\leq\|x\|$ for each $x\in X$. Then $\text{B}(X^*)\subset K$ and the $w^*$-topology on $\text{B}(X^*)$ is exactly the restriction of the product topology defined on $K$. The set $\text{B}(X^*)$ is the intersection of all the sets $A(x,y)=\{f\in K:(x+y)=f(x)+f(y)\}$, $B(\lambda,x):=\{f\in K:f(\lambda x)=\lambda f(x)\}$, $x,y\in X$, $\lambda\in\mathbb{R}$. Each of these sets is closed in the $w^*$-topology. Hence $\text{B}(X^*)$ is a $w^*$-closed therefore $w^*$-compact.

There are two big things about this proof that I don't understand. Firstly, why can we "ename the elements of $K$ to be functions on $X$, not necessarily linear, such that $|f(x)|\leq\|x\|$ for each $x\in X$"? Secondly, I understand that a closed subset of a compact set is compact, and that is what we are doing in this proof. However, in my opinion, what we just showed is that $\text{B}(X^*)\subset K$ , $\text{B}(X^*)$ $w^*$-closed, but $K$ has the product topology. So don't we need this topologies to be the same in order to use the "a closed subset of a compact set is compact"? That is, we need that either $\text{B}(X^*)$ is closed in the product topology, or that $K$ is compact in the $w^*$-topology? Also, where do we need that $X$ is a Banach, i.e., why not relax the condition to $X$ being a normed vector space?

Answer 1

In set theory (and thus in analysis and topology too) a product $\prod_{i \in I} Y_i$ is just the set of all functions $f$ defined on $I$ that obey the requirement $f(i) \in Y_i$ for all $i \in I$. That just a definition and a given. I don’t see why your text says it “renames” a product, it’s just reminding the reader about what $\prod_{x \in X} [-\|x\|, \|x\|]$ actually means: namely all functions $f$, defined on $X$ (the index set) that take real values and so that $|f(x)| \le \|x\|$, which already implies that$\|f\|\le 1$ in the norm on $X^\ast$, but $f$ in the product need not be linear at all (so not a member of $X^\ast$ but the norm applies to all bounded functions on $X$) though it does satisfy $f(0)=0$ as$I_0 = \{0\}$, hence we have no choice there…

Now to topology: a product of spaces $Y=\prod_{i \in I} Y_i$ has a standard product topology: the minimal topology that makes all projections $p_i: Y \to Y_i$, defined by $p_i(f) = f(i)$ (well defined as $f$ is in the product) continuous from $Y$ to $Y_i$ (where the codomain space already has a topology, as in our $I_x$ case, which are subspaces of $\Bbb R$ after all). Recall that the weak$^\ast$ topology is defined on $X^\ast$ as the minimal topology that makes all point evaluations $f \to f(x)$, for all $x \in X$ continuous. So in fact (as the minimal topology that makes a set of functions continuous is unique (it’s called initial topology in topological theory) the weak$^\ast$ topology on $X^\ast$ is just the same as when we see $X^\ast$ as the a subspace of the product $\prod_{x \in X} \Bbb R_x$, where we have “$X$-many” copies of $\Bbb R$ as a space; this is often denotes as $\Bbb R^X$ as well (power as repeated multiplication, really). And as subspace topologies are also initial (wrt the inclusion map) a standard general fact from “categorical topology” tells us that the subspace topology of $K=\prod_{x \in X} I_x$ is just the same as that of $\Bbb R^X$, i.e. the weak$^\ast$ topology and the same holds for their respective subspaces $X^\ast \subseteq \Bbb R^X$ and $B(X^\ast) \subseteq K$ as well. For a proof of this categorical topology fact on initial topologies (long and boring and not at all deep) see my post here.

The previous is a justification of the remarks in your quoted proof (“the $w^\ast$ topology is exactly the restriction of the product topology defined on $K$” (the product topology on the intervals). It can also be seen by using net convergence (which also unique characterises a topology): $f_i \to f$ in the product $K$ iff for all $x \in X$ we have that $f_i(x) \to f(x)$ in $I_x$. And the exact same fact holds true (it’s often even defined that way) in the weak$^\ast$ topology.

Now the final part of the proof is to show that $B(X^\ast)$ is in fact a closed subset of $K$. As $K$ is a product of compact real intervals (for $\Bbb C$ we’d need closed balls around $0$ of radius $\|x\|$ instead but everything generalises ) $K$ is compact (Hausdorff) ad hence so will $B(X^\ast$ be after we’ve shown it to be closed. The sets $A(x,y)$ and $B(\lambda, x)$ are indeed closed in $K$ (or $\Bbb R^X$) and being in the intersection of all of them implies $f$ is linear and arbitrary intersections of closed sets are closed. So the linear functions in $K$ (a relatively small subset; many restrictions need to be met) is a closed subset and a member is in $B(X^\ast)$; the norm requirement is already met as we saw. The closedness is again easy if we use nets: if $f_i \in A(x,y)$ and $f_i \to f$ then $$f(x+y) = \lim_i f_i(x+y) = \lim_i( f_i(x) + f_i(y)) = \lim_i f_i(x) + \lim_i f_i(y)= f(x) + f(y) \text{ so } f \in A(x,y)$$ where we use continuity of $+$ as weak plus the fact that all $f_i \in A(x,y)$ of course, The argument for $B(\lambda,x)$is quite similar.

I hope this clarified the proof for you… And these notes show that we just need $X$ to be a normed space, but not necessarily complete, to address your final issue.