Suppose $f,x_1^*,\ldots,x_n^*\in E^*$ such that $\bigcap\ker x_j^*\sbe\ker f$. Prove that $f$ is a linear combination of $x_1^*,\ldots,x_n^*$.
1. If $\dim E < \infty$ we may argue as follows:
If $f\notin\lhull{x_1^*,\ldots,x_n^*}$, then there is some $x^{**}\in E^{**}$ such that $x^{**}(f)=1$ and $x^{**}(x_j^*)=0$. Since $E$ is finite dimensional, there is some $x\in E$ such that $x^{**}=J(x)$; it follows that $f(x)=1$ and $x_j^*(x)=0$.
2. The following argument also applies to the infinite dimensional case:
Define a linear map $u:E\rar\R^n$ by $u(x)=\sum x_j^*(x)e_j$. On the space $\im u$ we define a linear functional $y^*$ by $y^*(u(x))=f(x)$. This is indeed well defined, since $u(x)=0$ implies by assumption $f(x)=0$. Since any linear functional on $\im u$ is the restriction of a linear functional on $\R^n$, we conclude that $y^*=\sum\l_ke_k^*$ and thus
$$
f(x)
=y^*(u(x))
=\sum\l_ke_k^*(u(x))
=\sum\l_ku^*(e_k^*)(x)
$$
and by definition of $u^*$: $u^*(e_k^*)(x)=e_k^*(u(x))=x_k^*(x)$, i.e. $f=\sum\l_k x_k^*$.