# Mackey topology characterising property

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