I have a follow-up question to On the Riesz representation theorem . 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 $$\Psi(\phi)=\phi(\psi)~~~\forall~ \phi\in V^\times?$$.


1 Answer 1


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 $\psi_i$.

  • $\begingroup$ Thanks. Do you have a reference that I could cite? $\endgroup$ Aug 25, 2015 at 12:23
  • 1
    $\begingroup$ This is Theorem 3.10 in Rudin's Functional Analysis. $\endgroup$ Aug 25, 2015 at 12:28

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