Assume we have a locally convex topology $\tau$ induced by semi-norms $\mathcal P$, on some real vectorspace $E$. Let $\sigma$ be the locally convex topology induced by the semi-norms $$ \mathcal Q := \{|f| : f \text{ is $\tau$-continous and linear} \} $$ Let $A \subset E$ be convex. Is is true that $A$ is $\tau$ closed iff $A$ is $\sigma$ closed.

If $A$ is $\sigma$ closed, the claim is easy to prove, but I am wondering if the opposite is true.


Yes, it's true. That is a consequence of the Hahn-Banach theorem(s).

If $A$ is $\tau$-closed and convex, and $x \in E\setminus A$, there exists a $\tau$-continuous linear functional $f_A$ such that

$$S := \sup \{ f_A(a) : a \in A\} < f_A(x).$$


$$\left\{ y\in E : \lvert f_A(y-x)\rvert < f_A(x) - S\right\}$$

is a $\sigma$-neighbourhood of $x$ that doesn't intersect $A$.

  • $\begingroup$ How can we make $f_A$ with Hahn-Banach ? $\endgroup$ – user42761 Jun 18 '14 at 18:42
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    $\begingroup$ It's part (b) of theorem 3.4 in Rudin's Functional Analysis (note that Rudin assumes all TVSs Hausdorff, but that isn't necessary for the result). Generally, how to get $f_A$ from Hahn-Banach depends on what formulation(s) of the theorem(s) you have. $\endgroup$ – Daniel Fischer Jun 18 '14 at 18:47
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    $\begingroup$ Probably, you have a version that you can extend a linear functional dominated by a gauge functional. Then take an absolutely convex (convex and balanced) neighbourhood $U$ of $0$ such that $U \cap (x - A) = \varnothing$, and take the Minkowski functional of $U$ as the gauge functional. $\endgroup$ – Daniel Fischer Jun 18 '14 at 18:50

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