# Is the following application of the Banach-Alaoglu theorem sound?

Let $X$ be a Banach space, $B_{X^*}$ denote the norm-closed unit ball of its dual. Let $E$ be a weak*-closed subset of $X^*$ such that $E \cap B_{X^*}$ is non-empty. Let $M$ be a subspace of $X$, not necessarily closed. Suppose for any finite-dimensional subspace $M'$ of $M$, there exists an element $x^* \in E \cap B_{X^*}$ such that $x^* \in (M')^\perp$ ($(x,x^*) = 0$ for all $x \in M'$),

then from the Banach-Alaoglu theorem (which states that $B_{X^*}$ is weak* compact), we can conclude that there exists $x^* \in E$ such that $x^* \in \overline{M}^\perp$.

Thanks for any help.

## 1 Answer

For any Finite subspace $M'$, there is a $x_{M'}^*\in E\cap B_{X^*}$ such that $x^*\in M'^\perp$. $\{x^*_{M'}\}$ form a net in $E\cap B_{X^*}$ by inclusion, so there is a convergent subnet $\{x^*_{M''}\}$ in $E\cap B_{X^*}$ since $E\cap B_{X^*}$ is compact. Pick $x^*$ as the convergence point, then $x^*\in E\cap B_{X^*}$ and $x^*\in({\cup M''})^\perp= M^\perp=\bar{M}^\perp.$