This is a statement which I found without proof and maybe it's obvious, but I can't understand why it should be true:

Let $X$ be a locally convex vector space, $\mathcal{W}$ a neighbourhood basis of zero, consisting of balanced, convex, closed sets. Then $\bigcup_{W \in \mathcal{W}} W^\circ = X' $.

$W^\circ$ does mean the polar of $W$ and $X'$ the topological dual space, so all linear, continuous functionals on $X$. So actually my problem is: Why is a function $f$ from the algebraic dual space $X^*$ continuous if and only if there is a $W \in \mathcal{W}$ such that $|<x,x^*>| \le 1$ with $x \in W$...?


The set $U=\{a\in\mathbb K:|a|\le 1\}$ (either an interval or a disc depending on the field $\mathbb K$) is certainly a neighbourhood of $0$ in $\mathbb K$. Since every continuous linear map $f$ maps $0\in X$ to $0\in\mathbb K$, there is a $0$-neighbourhood $W$ in $X$ with $f(W)\subseteq U$, hence $f\in W^\circ$.

You might want to try the other implication yourself.

  • $\begingroup$ Thank you! For the other implication: $| \langle x,f \rangle | \leq 1$ with $x \in W, f \in X^* $. Because $W$ is a balanced zero neighbourhood, for any $n \in \mathbb{C}$ with $ |n| \ge 1$: $\frac{1}{n}x$ is also $\in W$. Then, together with the linearity of $f$, the equation becomes $| \langle \frac{1}{n} x,f \rangle | \leq \frac{1}{n}$. With $n \rightarrow \infty $: f is continuous in $0$ and because it's a linear functional, it's continuous everywhere. Does that make sence? $\endgroup$ – FeWa May 8 '18 at 19:09
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    $\begingroup$ I would write the continuity of $f\in W^\circ$ at $0$ in the following way: Given $\varepsilon >0$ the set $V=\varepsilon W$ is a $0$-neighborhood in $X$ such that $|f(x)|\le \varepsilon$ for all $x\in U$. $\endgroup$ – Jochen May 9 '18 at 7:03
  • $\begingroup$ Thank you again! $\endgroup$ – FeWa May 9 '18 at 17:20

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