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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?

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Hint: Continuous linear functionals are continuous. –  Nate Eldredge Jan 28 '12 at 19:20
    
I think my problem boils down to this question: math.stackexchange.com/questions/103330/… –  Kyle Schlitt Jan 28 '12 at 19:27
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Ok, then I think my answer to that question may clear it up. –  Nate Eldredge Jan 28 '12 at 19:53
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1 Answer

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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 $\tau_{wo}\subset\tau_{so}$.

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If I may, I'm just going to take a stab at understanding your argument. Now given a basic open set $\theta\in B(\tau_{wo})$. We can write $\theta$ as an intersection of sets $U_{i} = \{T\in B(E,F) : p_{\phi_i,x_i}(x) < \epsilon_i\}$. –  Kyle Schlitt Jan 28 '12 at 21:02
    
Then we take an associated $V_i$ from your argument above correponding to each $U_i$. Then the set $V:=\cap_{i=1}^{n}V_i$ gives us a $V\in \tau_{so}$ such that $0\in V\subset U$. Therefore $\tau_{wo} \subset \tau_{so}$. –  Kyle Schlitt Jan 28 '12 at 21:06
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In fact we can check property I mentioned above only for sets of the form $U = \{T\in B(E,F) : p_{\phi,x}(T) < \varepsilon\}$. Collection of this sets is called subbase. –  Norbert Jan 28 '12 at 21:07
    
From now on I will only check them. :) Thanks for your help! –  Kyle Schlitt Jan 28 '12 at 21:27
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