Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'd like to check that I understand the proof in full detail. Can you tell me if the following is correct? Thanks for your help.

Claim: The closed unit ball $B_{\|\cdot\|_{op}}(0,1)$ in $X^\ast$ is compact in the weak$^\ast$ topology.

Proof: Define $Y = \prod_{x \in X} B(0, \|x\|)$, the product of closed balls of radius $\|x\|$ in $\mathbb R$ (or $\mathbb C$). Then by Tychonoff's theorem, $Y$ is compact.

Now define a map $\phi : B_{\|\cdot\|_{op}}(0,1) \to Y$ by $\ell \mapsto (\ell(x))_x$. Then the image $\mathrm{im} \phi$ are sequences with $x$-th entry $\|\ell(x)\| \leq \|x\|$. Hence if the things in the image are linear, then they are in the image of $\phi$. So everything outside the image, that is, lying in $Y \setminus \mathrm{im} \phi$ is non-linear. (That all of $Y$ are functionals, that is, maps $X \to \mathbb C$, is clear by definition).

So we have established that the image $\phi (B_{\|\cdot\|_{op}}(0,1))$ is $B_{\|\cdot\|_{op}}(0,1)$ itself. Hence $\varphi$ is surjective. That it is injective is clear. We proceed by showing that it's a homeomorphism and finish the proof by showing that its image is closed:

First, $\phi$ is a homeomorphism: We already know that $\phi$ is continuous, by definition (we put the weak$^\ast$ topology on its domain and $\phi$ is an evaluation map). Its inverse is also continuous: Let $y \in B_{\|\cdot\|_{op}}(0,1)$ be a point and $\prod_{x \in X} B(y(x),\varepsilon_x) = \prod_{x \in X} B(\ell(x),\varepsilon_x)$ an open set containing it. Then since we have the product topology, all but finitely many components are the entire space, that is, for all but finitely many $x$ we have $B(\ell(x),\varepsilon_x) = B(0, \|x\|)$. So that $\phi^{-1}$ of an open set like this is a finite intersection: $$ \phi^{-1}(\prod_{x \in X} B(\ell(x),\varepsilon_x)) = \bigcap_{i=1}^n \{ \ell^\prime \in B_{\|\cdot\|_{op}}(0,1) \mid |\ell (x) - \ell^\prime (x)| < \varepsilon_x \} \supset \bigcap_{i=1}^n \{ \ell^\prime \in B_{\|\cdot\|_{op}}(0,1) \mid |\ell (x) - \ell^\prime (x)| < \varepsilon \}$$ where $\varepsilon = \min(\varepsilon_x)$. So that $\phi^{-1}$ of an open set maps to a basic open set in $B_{\|\cdot\|_{op}}(0,1)$.

To finish we need to show that $\phi (B_{\|\cdot\|_{op}}(0,1))$ is closed: We show instead that its complement is open. To this end, let $y \in Y \setminus \phi (B_{\|\cdot\|_{op}}(0,1))$. Then $y$ is a non-linear functional so that there exist scalars $a,b$ and $x_1, x_2 \in X$ such that $y(ax_1 + bx_2) \neq a y(x_1) + b y(x_2)$. Choose $\varepsilon$ such that $B(y(ax_1 + bx_2), \varepsilon) \cap (a B(y(x_1), \varepsilon) + b B(y(x_2), \varepsilon)) = \varnothing$. If $\pi_x: Y \to B(0, \|x\|)$ is the projection then $\pi_x$ is continuous and hence its inverse maps open sets to open sets so that it's enough to show that the intersection of the inverse images of the open balls lies entirely in the complement of the image of $\varphi$: Let $z \in \pi_{ax_1 + bx_2}^{-1}(B(y(ax_1 + bx_2), \varepsilon)) \cap (\pi_{ax_1}^{-1} aB(y(x_1), \varepsilon) \cap \pi_{ax_1}^{-1} bB(y(x_2), \varepsilon)$. Then $z(ax_1 + bx_2) \neq a z(x_1) + b z(x_2)$.

share|cite|improve this question
For a different proof (but similar), see my answer there:… – M Turgeon Aug 18 '12 at 20:32
Your proof seems fine, although I find the end of the second paragraph rather confusing (e.g. "That all of $Y$ are functionals, that is, maps $X\to\mathbb C$, is clear by definition".) And you seem to be confusing image and codomain*. – M Turgeon Aug 18 '12 at 20:37
"Hence if the things in the image are linear, then they are in the image of Ï•." @Matt – M Turgeon Aug 18 '12 at 22:01
I always thought the name of this is Banach-Alaoglu. But I guess you lecturer has some reason to call it like this. (There's no doubt that Tychonoffs theorem about product of compact spaces is important in the proof.) – Martin Sleziak Aug 19 '12 at 6:21
I think that a possible way to see that the two topologies are the same is noticing that both are equal to the initial topology w.r.t. the same system of maps. (This is about the part of your proof about homeomorphism of the two spaces.) Weak${}^*$-topology is initial topology w.r.t. the maps $f\mapsto f(x)$ by definition. The other topology is subspace of a product - both subspace topology and product topology are initial topologies. (cont...) – Martin Sleziak Aug 19 '12 at 7:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.