6
$\begingroup$

Let $E$ be a not normable locally convex space, define $F: E'\times E\to \mathbb R$ $(f,e)\to f(e)$ I have to show that $F$ is not continuous when $E'\times E$ is given product topology.

I was reading an article and i came across with this fact.. Please give me atleast a hint to start..

My try: I know that $E$ is normable if and only if origin has a convex bounded neighborhood. So i was trying to produce any such neighborhood to contradict to assumption. Assume $F$ is continuous, then we have $\{(f,e): a is open in product topology of $E'\times E$, for any $a,b\in \mathbb R$. This means there is some open set $U'$ in $E'$ and $U$ in $E$ such that $U'\times U\subset \{(f,e): a
Now let $V:=\{e\in E: a, this is open convex neighborhood of origin, but how to prove this is bounded. Or we have any other way to produce such a neighborhood.

Thanks for your time.

  • 0
    What is $E'$? What topology do you take on $E'$?2014-10-26

1 Answers 1

2

In fact it turns out that the evaluation map is discontinuous for any locally convex topology on $E'$ (if $E$ is not normable).

You can find a concise sketch of the argument on the top of p.2 of Kriegl and Michors book: The convenient setting of global analysis (available online 1): To prove your statement one argues by contradiction and assumes that $ev$ was continuous, i.e. $ev (V \times U) \subseteq [-1,1]$. Then $U$ is a bounded zero-neighbourhood $U$ in $E$ since it is contained in the polar of $V$. Thus $E$ must be normable.