Let $V$ be a topological $k$-vector space.

Let $V^{\star}$ denote the space of all linear functionals $V \rightarrow k$ and $V' \subset V^{\star}$ the subspace of all continuous linear functionals.

The weak (Mackey) topology is the weakest (strongest) topology on the vector space $V$, such that all continuous functionals $v' \in V'$ remain continuous, and all discontinuous functionals $v' \in V^{\star} \setminus V'$ remain discontinuous w.r.t. the new weak (Mackey) topology.

Let $V_{\sigma} (V_{\tau})$ denote the underlying vector space $V$ equipped with its weak (Mackey) topology. In my opinion its clear that the weak topology on $V$ is an initial topology w.r.t. the family V', i.e. a function $f: U \rightarrow V_{\sigma}$ is continuous iff $v' \circ f$ is continuous for all $v' \in V'$.

By construction, if $f: U \rightarrow V_{\tau}$ is continuous, also $v' \circ f$ is continuous for all $v' \in V'$.

Does the inverse also hold? I.e. is the Mackey topology also initial w.r.t. V'? (No its not final)

Thank you.

  • Every topology $T$ on a set $X$ is initial, e.g. w.r.t. the identity $X \to (X,T)$. It is thus not clear what you are asking for. – Jochen Wengenroth Nov 16 at 8:05
  • Thank you for your answer. I will edit my question: is the Mackey topology initial w.r.t. V' (in the above sense)? – user120487 Nov 16 at 11:55
  • If the Mackey topology were the initial topology w.r.t. $V'$ it would be equal to the weak topology which (in many case, e.g., for infinite dimensional Banach spaces) is not the case. – Jochen Wengenroth Nov 16 at 12:27
  • This seems to be correct by the usual (and trivial, sorry) argument of taking $f:= id: V_{\sigma} \rightarrow V_{\tau}$ and the symmetric variant of it. – user120487 Nov 16 at 13:10

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