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Let $\mathbb E$ and $\mathbb F$ be two Banach spaces and $T:\mathbb E\to \mathbb F$ a linear operator. Use the definition of the weak topology to show that $T$ is continuous with respect $\mathbb E$ being equipped with the weak topology $\sigma(\mathbb E,\mathbb E^*)$ and $\mathbb F$ being equipped with the weak topology $\sigma(\mathbb F,\mathbb F^*)$ if and only if it is continuous at $0\in \mathbb E$ with respect to these weak topologies.

I think the idea behind is that continuity (weak) implies continuity (strong) and we know function is continuous iff it is continuous at one point. so this is also continuous(strong) at $0$, then again continuity(strong) implies continuity(weak) and we are done.

I am not sure whether my rough sketch is right or wrong? Also couldn't quite get the way of perfectly writing it. Any help would be highly appreciated. Thanks in advance.

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    $\begingroup$ No, you don't need anything about the relationship between weak and strong continuity. It's simpler than that. This fact will be true for any two topological vector spaces. You probably know how to prove it for the norm topology; if you rephrase the proof in terms of open sets or nets (so as not to assume your space is sequential) then it should go through in general. $\endgroup$ Apr 4, 2017 at 3:24
  • $\begingroup$ @Nate Eldredge, Still struggling with the composing. A little bit more help would greatly appreciated. Thanks a lot. $\endgroup$
    – mint
    Apr 4, 2017 at 14:28

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You don't need anything about the relationship between weak and strong continuity. It's simpler than that. This fact will be true for any linear operator $T$ between any two topological vector spaces $E,F$.

To get you started, let $x \in E$ and let $V$ be an open set containing $Tx$. (For purposes of your problem, we are taking $E$ equipped with the weak topology, so "open" here can be read as "weakly open".) We have to show there is an open set $U$ containing $x$ such that $T(U) \subset V$. Now since $F$ is a topological vector space, the set $V - x = \{v - x : v \in V\}$ is open and contains $0$. We are assuming $T$ is continuous at $0$, so... (I'll let you take it from here.)

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