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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?

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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 $x_1,\dots,x_n$.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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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, $\phi(z)=0$.

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