Let $X$ be a Banach space and $X^*$ its dual. We know that the weak* topology is the least topology that makes every $x \in X$ continuous as an evaluation functional. However, this does not imply that every weak* continuous linear functional is something in $X$, even though this happens to be true.

The question is: how can we prove this?

What have I though is:

It is enough to show that $\cap_{i=1}^{k} Ker{x_i} \subset Ker{\phi}$ for some $x_i \in X, i=1,2,...,k$

I have shown this for infinitely many $x_i$s (easy, using the weak* continuity and that 0 is always in the ker) and in order to pass to finitely many I would need some kind of compactness result (probably by using Banach-Alaoglu somehow), but I do not know how to do this.

Can anyone help?


Since $\phi$ is weak*-continuous, the set $\{z\in X^*:|\phi(z)|<1\}$ is weak*-open. Therefore, it contains a weak*-neighborhood of $0$, which by the definition of weak* topology means there exist vectors $x_1,\dots,x_n$ such that $$ \{z\in X^*: |z(x_k)| <1,\quad k=1,\dots,n\}\subseteq \{z\in X^*:|\phi(z)|<1\} $$ By homogeneity, this implies $$ |\phi(z)| \le \max_{k=1,\dots,n} |z(x_k)| $$ and therefore $\bigcap_{k=1}^n \ker x_k \subset \ker \phi$.

It follows that $\phi$ is a linear combination of $x_1,\dots,x_n$.

  • $\begingroup$ Can explain why by homogeneity we obtain $|\phi(z)| \leq \max_{1 \leq k \leq n}{|z(x_k)|}$? $\endgroup$ – Idonknow Nov 23 '15 at 8:34
  • 2
    $\begingroup$ Let $M$ be greater than the right hand side. Then $w = z/M$ is such that $|w(x_k)|<1$ for all $k$. So $|\phi(w)|<1$. Hence $|\phi(z)|<M$. $\endgroup$ – user147263 Nov 23 '15 at 8:36
  • $\begingroup$ Do you mean 'Let $M$ be an integer smaller than the RHS'? $\endgroup$ – Idonknow Nov 23 '15 at 9:00
  • 1
    $\begingroup$ No, I mean greater. The point is, if you prove $|\phi(z)|<M$ for any $M$ that is greater than RHS, then it follows $|\phi(z) |\le RHS$ $\endgroup$ – user147263 Nov 23 '15 at 9:01
  • 1
    $\begingroup$ The argument in this answer has a problem: $M$ can be zero, which implies that it can not be divided by $M$. $\endgroup$ – Diego Fonseca Nov 3 '17 at 17:47

As $f$ is waek$^{\star}$ continuous then it is waek$^{\star}$ continuous in $0$, therefore, there exists $x_{1},\ldots,x_{n}\in X$ such that $$V(x_{1},\ldots,x_{n}):=\left\{z\in X^{*} \::\: |z(x_{i})|<=1 \: i=1,\ldots,n \right\}\subseteq \left\{z\in X^{*} \: :\: |\phi(z)|<1\right\} \tag{$\bigstar$}$$

We show that $$\bigcap_{i=1}^{n}\mathrm{ker}x_{i}\subseteq\mathrm{ker}f.$$ In fact, let $z \in \mathrm{ker}x_{i}$, then $\left| z(x_{i}) \right|=0$ for each $i=1,\ldots ,n$. Let $\varepsilon >0$, we consider $w=\frac{z}{\varepsilon}$, then $|w(x_{i})|=\frac{1}{\varepsilon}|z(x_{i})|=0$ for each $i=1,\ldots,n$. In particular, $w\in V(x_{1},\ldots,x_{n})$, then, by ($\bigstar$) we have $$|\phi(w)|<1 \quad \Longrightarrow \quad |\phi(z)|<\varepsilon.$$ But $\varepsilon$ is arbitrary, therefore, $\phi(z)=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.