2
$\begingroup$

I am working on a homework problem to which we were given a hint: "use the fact that the Strong operator topology is stronger than the Weak operator topology".

The setting is this: Let $E,F$ be normed spaces, $\phi:B(E,F)\to \mathbb{F}$.

Define $P_s = \{p_x:x\in E\}$, where $p_x:B(E,F)\to \mathbb{F}$ is given by $p_x(T) = ||Tx||$ and define $P_w = \{p_{x,\phi}: x\in E, \phi\in F^{*}\}$ where $p_{x,\phi}:B(E,F)\to \mathbb{F}$ is given by $p_{x,\phi}(T) = |\phi(Tx)|$.

$P_s$ turns $B(E,F)$ into a locally convex space with the strong operator topology, and $P_w$ turns $B(E,F)$ into a locally convex space with the weak operator topology.

Forgetting about the problem itself, I am not sure how to verify this hint. Several references either state the fact as a remark without proof, or as their own exercise. Wikipedia says it follows from continuity of the inner product, but I believe they defined these topologies in the setting of $B(H,H)$ for a Hilbert space $H$, so I can't use that fact in my setting.

Any advice on where to find or how to come up with a proof for this fact?

  • 1
    Ok, then I think my answer to that question may clear it up.2012-01-28

2 Answers 2

1

General remark. Let $X$ be some set with topologies $\tau_1$, $\tau_2$. Obviously $\tau_2$ is stronger than$\tau_1$ iff for each $\tau_1$-open set $U$ for each $x\in U$ there exist $\tau_2$ open set $V$ such that $x\in V\subset U$. Denote by $B(\tau)$, any subbase of topology $\tau$, then to check whether $\tau_1\subset\tau_2$ it is enough to prove for each $U\in B(\tau_1)$ and each $x\in U$ there exist $V\in \tau_2$ such that $x\in V\subset U$. If $X$ is a topological vector spaces it is sufficient to prove that each neighborhood $U\in B(\tau_1)$ of zero contains zeros neighborhood $V\in\tau_2$.

Take $U=\{T\in\mathcal{B}(E,F):p_{\phi,x}(T)<\varepsilon\}\in B(\tau_{wo})$. Consider $V=\{T\in\mathcal{B}(E,F):p_x(T)<\varepsilon/(1+\Vert\phi\Vert)\}\in \tau_{so}$. Then for each $T\in V$ we have $ p_{\phi,x}(T)=|\phi(T(x))|\leq\Vert\phi\Vert \Vert T(x)\Vert<\frac{\varepsilon\Vert\phi\Vert}{1+\Vert\phi\Vert}<\varepsilon $ so $T\in U$. Since $T\in V$ is arbitrary $V\subset U$. From remark given above we conclude that $\tau_{wo}\subset\tau_{so}$.

  • 0
    From now on I will only check them. :) Thanks for your help!2012-01-28
1

More easily: Sot generated by seminorms $p_e(x):= \|xe\|$ for $e\in H$ and wot generated by seminorms $p_\xi,\eta(x) : = \langle x\xi,\eta\rangle$ for every $\xi,\eta \in H$. To show sot is stronger than wot, I show that inclusion map $i:(B(H) , sot) \to (B(H) , wot) $ is continuous. Suppose $x_i\to 0$ (sot) we show that $x_i\to 0$(wot). for $\xi,\eta \in H$ $|\langle x\xi,\eta\rangle|\leq \|x\xi\|\|\eta\|\to 0$