Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.