weak* continuous linear functional is in the predual 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?
 A: 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 $x_1,\dots,x_n$. 
A: As $f$ is waek$^{\star}$ continuous then it is  weak$^{\star}$ continuous in $0$, therefore, there exists $x_{1},\ldots,x_{n}\in X$ such that
\begin{equation}\begin{aligned}V(x_{1},\ldots,x_{n})&:=\left\{z\in X^{*} \::\: |z(x_{i})|\leq1 \: i=1,\ldots,n \right\}\\ &\subseteq \left\{z\in X^{*} \: :\: |\phi(z)|<1\right\} \end{aligned}\tag{$\bigstar$}\end{equation}
We show that $$\bigcap_{i=1}^{n}\mathrm{ker}x_{i}\subseteq\ker f.$$
In fact, let $z \in \ker x_{i}$, 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, $\phi(z)=0$.
