Are smooth functions on an uncountable sum continuous? Consider the linear space $\sum_{\mathbb{R}} \mathbb{R}$.  As in the Frolicher-Kriegl-Michor view, we make this into a Frolicher space as follows.


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*Equip it with the locally convex topology of the colimit.  Specifically, it is given the finest locally convex topology so that all of the inclusions of finite summands are continuous.  What is important is that a linear functions $\phi \colon \sum_{\mathbb{R}} \mathbb{R} \to \mathbb{R}$ is continuous if and only if its restriction to any finite summand is continuous.  Thus all linear functionals $\sum_{\mathbb{R}} \mathbb{R} \to \mathbb{R}$ are continuous.  Technical note: with this topology, it is complete.

*Now define the family of smooth curves to be those curves $c \colon \mathbb{R} \to \sum_{\mathbb{R}} \mathbb{R}$ with the property that $\psi \circ c \colon \mathbb{R} \to \mathbb{R}$ is smooth for all $\psi \in \prod_{\mathbb{R}} \mathbb{R}$ (that is, for all continuous, linear maps $\psi : \sum_{\mathbb{R}} \mathbb{R} \to \mathbb{R}$).  Note that curves are not assumed to be continuous (though they will be).

*Now define the family of smooth functions to be those $f \colon \sum_{\mathbb{R}} \mathbb{R} \to \mathbb{R}$ for which $f \circ c \colon \mathbb{R} \to \mathbb{R}$ is smooth for all smooth curves $c$.  Again, smooth functions are not assumed to be continuous.
Here's the question: are all smooth functions continuous?
If we took a countable sum then this would be true.  For the uncountable product, Kriegl and Michor show (Example 4.8, pp37-38 of A Convenient Setting of Global Analysis) that this is not true.  So I suspect the answer to be false, but don't know if this is known or not.
 A: Not all smooth functions are continuous. It is a fact of the Frölicher−Kriegl−Michor theory that bounded multilinear maps are smooth. For example the canonical bilinear evaluation $E\times E'\to\mathbb R$ given by $(x,u)\mapsto u(x)$ is bounded, hence smooth, but discontinuous when $E=\sum_{\mathbb R}\mathbb R$. I too quickly thought that this would give the required smooth discontinuous map as a composite $E\to E\times E\to E\times E'\to\mathbb R$.
Using Jarchow's notation, and istead considering the space $F=\mathbb R^{\ \mathbb N}\times\mathbb R^{\ (\mathbb N)}=\prod_{\mathbb N}\mathbb R\times\sum_{\mathbb N}\mathbb R$ , then one has the Frölicher−Kriegl smooth discontinuous map $F\to\mathbb R$ given by $(x,y)\mapsto\sum_{i\in\mathbb N}(x_i\cdot y_i)$ .
It should be noted that this discontinuity is with respect to the locally convex topology. Frölicher−Kriegl smooth maps are always continuous with respect to the Mackey−closure topology whose open sets are precisely the $U\subseteq F$ such that for every $x\in U$ and every bounded set $B$ in $F$ there is $\varepsilon>0$ with $\varepsilon B\subseteq U-x$ . 
The Frölicher−Kriegl theory is essentially a bornological theory. One may observe that in Frölicher's and Kriegl's book one uses a canonical topology corresponding to the bornology, namely the strongest, bornological one, whereas in Kriegl's and Michor's book one allows any locally convex topology with the same bounded sets. In this sense, the KM−approach to smoothness is a bit floppy since the spaces are topological but bornology is the only one that matters.
