I have a follow-up question to [On the Riesz representation theorem](https://mathoverflow.net/q/215523/56920) .
Let $V$ be a subspace of a Hilbert space, and let $V^\times$ be the space of all antilinear functionals on $V$, equipped with the weak-\* topology. 


Let $\Psi$ be a weak-\* continuous antilinear functional on $V^\times$. Is there always a vector $\psi\in V$ such that
$$\Psi(\phi)=\phi(\psi)~~~\forall~ \phi\in V^\times?$$.



On the Riesz representation theorem II
Yes, a more general version of this is a well-known theorem in functional analysis. Specifically, let $W$ be any vector space and let $V$ be any vector space of functionals on $W$ which separates points of $W$. Equip $W$ with the weak topology with respect to $V$. Then every continuous functional on $W$ is in $V$.


To prove this, let $\Psi:W\to\mathbb{C}$ be continuous. This means that there are finitely many vectors $\psi\_1,\dots,\psi\_n\in V$ such that whenever $|\psi\_i(\phi)|<1$ for $i=1,\dots,n$, $|\Psi(\phi)|<1$ as well. It follows that $\Psi$ vanishes on $\bigcap \ker(\psi\_i)$ and thus factors through the map $(\psi\_1\dots,\psi\_n):W\to\mathbb{C}^n$. Every functional on $\mathbb{C}^n$ is a linear combination of the coordinate functionals, so it follows that $\Psi$ is a linear combination of the $\psi\_i$.


