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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.

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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.

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Essentially the same argument as yours, but I find it a bit simpler than messing with $\limsup$: Since $f_j \to f$ weak$^\ast$ and the unit ball of $L^\ast$ is compact (hence closed) we have $\|f\| \leq 1$. Let $\varepsilon > 0$. Choose $n$ so large that $\|\hat{x}_n - \psi\| \lt \varepsilon$. Choose $j_0$ such that $|\hat{x}_n(f_j - f)| \lt \varepsilon$ for all $j \geq j_0$. Then $$\|\psi(f)\| = \|\psi(f) - \psi(f_j)\| \leq \|\psi - \hat{x}_n\|\|f\| + |\hat{x}_n(f - f_j)| + \|\hat{x}_n - \psi\| \|f_j\| \lt 3 \varepsilon.$$ As $\varepsilon$ was arbitrary we get $\psi(f) = 0$, as desired. – t.b. Mar 15 '12 at 5:17
Yes, that's simpler indeed. When I was writing I hesitated for a moment over whether my choices of $n$ and $j$ were independent, and so I switched to the limsups. What surprised me about my proof (if there's no mistake) is that it doesn't use that $\hat{x_n}\to\psi$ uniformly. – Martin Argerami Mar 15 '12 at 5:48
You use it implicitly when you take the lim sup over $n$ in the final part of the estimate (to argue that $|\psi(f) - \hat{x}_n(f)| \to 0$ I see no other way to justify this than using norm convergence $\hat{x}_n \to \psi$ -- of course you also have $\limsup_{m \to \infty} \|x_n - x_m\| = \|x_n - \psi\|$ but that this lim sup goes to zero can also be justified with the Cauchy property). Be that as it may, I see no flaw in the proof, very nice! – t.b. Mar 15 '12 at 6:04
Thanks. My point was that it looks like the estimates at the end work if $\hat{x_n}\to\psi$ weak-star (which is irrelevant regarding the result we want to prove). – Martin Argerami Mar 15 '12 at 6:22
I see what you're saying. But note that the second term needs that $\hat{x}_n$ is norm Cauchy, and $\hat{x}_n \to \psi$ weak$^\ast$ and $\hat{x}_n$ Cauchy in norm implies $\hat{x}_n \to \psi$ in norm. – t.b. Mar 15 '12 at 6:38

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