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Let $(X,\tau)$ be a topological vector space (TVS for short). Denote by $X^*$ the topological dual of $(X,\tau)$. If there exists a locally convex topology $\mu$ on $X$ compatible with the duality $(X,X^*)$ (that is, $(X,\mu)^*=X^*$) and $\mu$ is finer than $\tau$ then is $(X,\tau)$ a locally convex space?

An (equivalent) reformulation of the above question would be: - Is there an (infinite dimensional) locally convex space $(X,\mu)$, such that in between the weak topology, $w$ and $\mu$ there exists a linear topology $\tau$ on $X$ which is NOT locally convex?

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This is an interesting question which I did not find answered in the literature.

Here is an example for a non-locally convex topology between the weak and the Mackey topology: The space is the Banach space $E:=L_1(0,1)$, and let $\sigma$ be the weak topology. Let $0<q<1$, define the metric $d_q(f,g):=\int|f-g|^q$ on $E$, and denote by $\tau_q$ the topology defined by this metric. (In other words, $\tau_q$ is the topology on $E$ induced by the non-locally convex linear topology of $L_q(0,1)$.) Define $\tau$ on $E$ as the initial topology on $E$ with respect to the mappings ${\rm id}\colon E\to(E,\sigma)$ and ${\rm id}\colon E\to(E,\tau_p)$. Then $\tau\supseteq\sigma$. But also $\mu=\tau_{\|\cdot\|_1}\supseteq\tau$, because the mappings ${\rm id}\colon(E,\|\cdot\|_1)\to(E,\sigma)$ and ${\rm id}\colon(E,\|\cdot\|_1)\to(E,\tau_q)$ are continuous.

It remains to verify that the topology is not locally convex. For this fact I only give a sketch. The first observation is that a neighbourhood basis of $0$ for $\tau$ is given by $\{U_{F,\;\epsilon};\ F\subseteq(E,\|\cdot\|_1)'=L_\infty(0,1)\text{ finite},\ \epsilon>0\}$, where $$ U_{F,\;\epsilon}:=\{f\in E;\ \sup_{\eta\in F}|\eta(f)|<\epsilon,\ d_q(f,0)<\epsilon\}. $$ And then one shows that the $\tau$-neighbourhood $U:=\{f\in E;\ d_q(f,0)<1\}$ of $0$ does not contain the convex hull of any of the $0$-neighbourhoods $U_{F,\;\epsilon}$. More precisely, one shows that the convex hull of $U_{F,\;\epsilon}$ contains elements $f\in\bigcap_{\eta\in F}\eta^{-1}(0)$ with $d_q(f,0)$ arbitrarily large. (This step is done in a way somewhat similar to the proof that the topology $\tau_q$ is not locally convex.) This proves that $U$ does not contain any convex neighbourhood of $0$, and therefore the topology $\tau$ is not locally convex.

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