TITLE: Evaluation map is not continuous always.
QUESTION [6 upvotes]: 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.
Thanks for your time.

REPLY [3 votes]: 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.