If $L$ is normed and for every hyperplane $M \cap \operatorname{ball} L^*$ weak*-closed implies $M$ weak*-closed then $L$ is a Banach space If $L$ is a normed space with the property that if $M$ is a hyperplane in $L^*$ and $M \cap \operatorname{ball} L^*$ is weak-star closed $\implies$ $M$ itself is weak star closed, then how do I show $L$ is a Banach space?
I think I should should identify $L$ with $L^{**}$, but I'm not sure how to prove this.
 A: Let $\{x_n\}$ be a Cauchy sequence in $L$. Letting $i:L\to L^{**}$ be the canonical inclusion, we write $i(x_n)=\hat{x_n}$. Then $\{\hat{x_n}\}$ is a Cauchy sequence in $L^{**}$ and, as a dual is complete, there exists $\psi\in L^{**}$ with $\|\hat{x_n}-\psi\|\to0$. 
Now let $M=\ker\psi\subset L^*$, and let $M_1=M\cap\text{ball}L^*$. We want to show that $M_1$ is weak-star closed. Let $\{f_j\}$ be a net in $M_1$ such that $f_j\to f$ weak-star in $L^*$ (i.e. $f_j(y)\to f(y)$ for all $y\in L$). 
We have, for any $n,m$ with $m\geq n$ and any $j$,
\begin{eqnarray}
|\psi(f)|&\leq&|\psi(f)-\hat{x_n}(f)|+|f(x_n)-f_j(x_n)|+|f_j(x_n)-f_j(x_m)|+|f_j(x_m)|\\
&\leq&|\psi(f)-\hat{x_n}(f)|+|f(x_n)-f_j(x_n)|+\|x_n-x_m\|+|f_j(x_m)|.
\end{eqnarray}
Taking first $\limsup$ when $m\to\infty$, we get
$$
|\psi(f)|\leq|\psi(f)-\hat{x_n}(f)|+|f(x_n)-f_j(x_n)|+\limsup_m\|x_n-x_m\|
$$
(note that $\lim_mf_j(x_m)=\lim_m\hat{x_n}(f_j)=\psi(f_j)=0$). Now taking $\limsup$ over $j$,
$$
|\psi(f)|\leq|\psi(f)-\hat{x_n}(f)|+\limsup_m\|x_n-x_m\|.
$$
Finally, taking $\limsup$ over $n$, we get
$$
|\psi(f)|\leq0,
$$
so $\psi(f)=0$. This shows that $f\in M_1$, so $M_1$ is weak-star closed. The hypothesis implies then that $M$ is weak-star closed. So $\psi$ is weak-star continuous. 
It is a well-known fact that the dual of $L^*$ considered with the weak-star topology is $L$ itself. This means that the weak-star continuous elements in $L^{**}$ are precisely those in $i(L)$. So there exists $x\in L$ such that $\psi=\hat{x}$. As the inclusion map $i$ is isometric, we have
$$
\lim_n\|x_n-x\|=\lim_n\|\hat{x_n}-\hat{x}\|=0.
$$
We have shown that every Cauchy sequence in $L$ is convergent, so $L$ is complete.
