Evaluation map is not continuous always.

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 $\{(f,e): a<f(e)<b\}$ 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 $$U'\times U\subset \{(f,e): a<f(e)<b\}$$
Now let $V:=\{e\in E: a<f(e)<b;\forall f\in U'\}$, 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.

• I find the presence of $a,b$ distracting. Since you are constructing a neighborhood of the origin, it's convenient to consider $\{(f,e) : |f(e)|<1\}$ and accordingly $V=\{e: |f(e)|<1 \ \forall f\in U'\}$. // More importantly, you have not really used the assumption that $E$ is locally convex: it carries a family of seminorms, which we can assume directed. // Take any open nbhd of zero $W\subset E$: it contains a nbhd of the form $\{p(x)<r\}$ where $p$ is a seminorm. Now you need a functional $f\in U'$ such that $|f(x)|<\epsilon \implies p(x)<r$. Think of Hahn-Banach. – user53153 Dec 31 '12 at 7:11
• What is $E'$? What topology do you take on $E'$? – Neutral Element Oct 26 '14 at 10:28
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.