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?