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Let $E$ be a Hausdorff locally convex topological vectorspace. Consider the source $p_f\colon E\to\mathbb{R}$, where $p_f(x)=\lvert f(x)\rvert$ and $f\in E'$, the continuous dual of $E$. The maps $p_f$ are semi-norms, so they generate an initial topology on $E$, which we denote by $\sigma(E, E')$. This is the coarsest topology that makes all $p_f$ continuous. Check, I get this.

Now, I am supposed to show that $\sigma(E, E')$ is the coarsest topology that makes all linear maps in the dual $E'$ continuous. But aren't the maps in $E'$ continuous by definition? Or is there a typo in my notes and do they mean the algebraic dual $E^*$? In my notations $$E^*:=\{f\colon E\to\mathbb{R} \mid f \text{ linear}\},\\ E':=\{f\colon E\to\mathbb{R} \mid f \text{ linear and continuous}\}.$$

Probably this is really easy, but I'm completely confused for the moment.

Jeroen
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  • The answer is not immediate. Although it is essentially the same thing you are considering $E$ as locally convex space? – user288972 Dec 23 '15 at 22:08
  • Okay, you have the basics on locally convex spaces? and you know the product topology even if Infinity uncountable? is basically the same thing, Zorn's lemma helps us in this. – user288972 Dec 23 '15 at 23:05

1 Answers1

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I finally get it thanks to a bit of sleep and my teacher's explanation. First of all, the notation was indeed correct. For a Hausdorff locally convex space $(E, \tau)$, we define $$E':=\{f\colon (E, \tau)\to \mathbb{R} \mid f\text{ is linear and continuous}\}.$$

The subtility is that these maps can become discontinuous for another topology on $E$.

Now, note that $f$ is continuous if and only if $p_f$ is continuous. It is sufficient to show that this equivalence is true for $0$. In order to do this, simply write down the inverse images of an open neighbourhood of $0$: $$p_f^{-1}([0, a[)=\{x\in E \mid \lvert f(x)\rvert=p_f(x)<a\}=\{x\in E \mid -a<f(x)<a\}=f^{-1}(]-a, a[).$$

It follows that the initial topologies for the sources $(f\colon E\to \mathbb{R})_{f\in E'}$ and $(p_f\colon E\to \mathbb{R})_{f\in E'}$ are exactly the same, so $\sigma (E, E')$ is also initial for $(f\colon E\to \mathbb{R})_{f\in E'}$, which means by definition that it is the coarsest topology that makes all the $f\in E'$ continuous.

Jeroen
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  • Do you understand how the construction of your topology works for given seminorms? – C-star-W-star Dec 24 '15 at 15:40
  • To understand the weak topology in locally convex spaces is necessary to understand what is the topology of pointwise convergence, and then to know the product topology for infinite product spaces. You must know also how to define the vectorial topology in locally convex spaces. – user288972 Dec 24 '15 at 20:00