Bounded linear map which is not continuous Definition: A subset $B$ of the TVS $E$ is said to be bounded if to every neighborhood of zero $U$ in $E$ there is a number $\lambda >0$ such that $B \subset \lambda U.$
Definition: Let $E,F$ be two TVS, and $u$ a linear map of $E$ into $F$. Let us say that $u$ is bounded if, for every bounded subset $B$ of $E$, $u(B)$ is a bounded subset of $F$.
We have the following result:
Theorem: Let $E$ be a metrizable space TVS. If a linear map of $E$ into a TVS $F$ is bounded, it is continuous.
My question: Is there a counterexample of a bounded linear map which is not continuous?
If this example exists, the space $E$ cannot be metrizable.
 A: For every Hausdorff locally convex space $(X,\mathcal T)$ the weak topology $\sigma(X,X')$ has the same bounded sets as $\mathcal T$, hence the identity $id: (X,\sigma(X,X')) \to (X,\mathcal T)$ is bounded but discontinuous whenever, e.g., $(X,\mathcal T)$ has a continuous norm. For example, $(X,\mathcal T)$ could be an infinite deimensional normed space.
Locally convex spaces $X$ with the property that every bounded linear map on $X$ is continuous are called bornological. Besides metrizable spaces locally convex inductive limits (aka colimits) of metrizable spaces belong to this class.
A: Edit: A large class of counterexamples is given in the accepted answer by Jochen. My answer does not actually answer the question, but provides a simple condition under which bounded does imply continuous. After googling bornology, it seems that I have described a specific case of a more general implciation. Namely, if every bornivorous subset of $E$ is a neighborhood of $0\in E$, then $E$ is bornological. But the existence of any bounded neighborhood of $0$ implies that every bornivorous subset of $E$ is a neighborhood of $0$. However, perhaps there is still some value in stating this coarser implication without any mention of bornology? I don't know.
Suppose $U\subset E$ is a bounded neighborhood of $0\in E$ and $T:E\rightarrow F$ is a bounded linear map. Let $V\subset F$ be a neighborhood of $0\in F$. Since $T(U)$ is bounded, there is some $\epsilon>0$ such that $T(\epsilon U)=\epsilon T(U) \subset V$. Then $$\epsilon U\subset T^{-1}(T(\epsilon U))\subset T^{-1}(V)$$ and $\epsilon U$ is a neighborhood of $0\in E$. Thus $T^{-1}(V)$ is a neighborhood of $0\in E$. This proves that $T$ is continuous at $0\in E$ and thus everywhere, under the assumption that $E$ contains a bounded neighborhood of the origin.
This assumption need not be true (as follows at once from Jochen's answer). To see an example that doesn't mention bornology, consider $\mathbb R^\mathbb{N}$ with the box topology. If $U$ is a neighborhood of $0$, then there exist $a_1,a_2,\ldots\in (0,\infty)$ such that $V\subset U$, where $V=\prod_i(-a_i,a_i).$
Now let $W=\prod_i(-a_i/i,a_i/i).$ Then there is no $\epsilon>0$ such that $\epsilon W\subset V$. Hence $V$ cannot be bounded and thus neither can $U$.
