# When does the continuous dual of the weak operator topology consist only of finite linear combinations of evaluations?

Let $V$ and $W$ be topological vector spaces, say, over $\mathbb C$. Let $L(V, W)$ be the vector space of continuous linear maps equipped with the weak operator topology, which is the initial topology for the maps \begin{align*} \phi_{v, \mu} : L(V, W) &\to \mathbb C \\ T & \mapsto \mu(Tv) \end{align*} for $v \in V$ and $\mu \in W^*$ (the continuous dual of $W$). Equivalently, it is the topology induced by the seminorms $p_{v, \mu} = |\phi_{v, \mu}|$. In particular, $L(V, W)$ is locally convex, and Hausdorff iff $W^*$ separates points of $W$.

By construction, the $\phi_{v, \mu} : L(V, W) \to \mathbb C$ are continuous linear maps, and so are finite linear combinations of the $\phi_{v, \mu}$. Under which conditions do the $\phi_{v, \mu}$ span the continuous dual $L(V, W)^*$?

I know that this is the case when $V = W$ is a Hilbert space. (Takesaki, Theory of Operator Algebras I, Chapter II Theorem 2.6.)

In these notes, Paul Garrett suggests that this is more generally true for certain topological vector spaces: the proof of the second corollary on page 3 uses that this is true when $V$ is an LF-space and $W$ quasi-complete and locally convex.)

It does not give a precise statement, which makes we wonder if this is true for general topological vector spaces:

Is there any good reference which addresses the case of topological vector spaces (not just normed spaces)?

• It looks to me that you are reading Takesaki's II.2.6 the wrong way. What it says is that if $\omega$ is wot, sot, or sot*-continuous, then it is a sum of point functionals. It doesn't say that every functional in the dual is of that form. – Martin Argerami Aug 27 '18 at 2:38

Let $V,W$ be topological over a field $\mathbb K$ equal to $\mathbb R$ or $\mathbb C$ (or any field with an absolute value), and $\theta : L(V, W) \to \mathbb K$ continuous.
Then by definition, there exists $\delta>0$, $v_1, \ldots, v_n \in V$ and $\mu_1, \ldots, \mu_n \in W^*$ such that $|\phi_{v_i, \mu_i}(T)| < \delta$ for all $i$ implies $|\theta(T)| < 1$. By linearity, $|\phi_{v_i, \mu_i}(T)| < \delta \epsilon$ for all $i$ implies $|\theta(T)| < \epsilon$. In particular, $\theta(T)$ is determined by the $\phi_{v_i, \mu_i}(T)$. Define $\Phi : L(V, W) \to \mathbb K^n$ by $$\Phi(T) = (\phi_{v_1, \mu_1}(T), \ldots, \phi_{v_n, \mu_n}(T))$$ Then $\Phi$ is injective and continuous, and $\theta$ factors through $\Phi$. Write $\theta = f_0 \circ \Phi$ for some $f_0$ defined on the image of $\Phi$. We can extend it to a linear map $f : \mathbb K^n \to \mathbb K$, which is automatically continuous because $\mathbb K^n$ is finite-dimensional. It follows that $\theta$ is a linear combination of the $\phi_{v_i, \mu_i}$.
Note. There is nothing special about $L(V, W)$; a similar result holds for topological vector spaces whose topology is induced by a family of linear forms.