# 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.

1. 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.

2. 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).

3. 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.

• I'd be surprised if smoothness was a factor here. I think you might as well ask: If $f\circ c$ is continuos for every continuous curve $c$, is $f$ continuous? But maybe my intuition that continuous curves can be approximated by smooth ones is wrong in this setting. I am not at all familiar with these spaces. (Minor snark: It's written Frölicher, isn't it?) – Harald Hanche-Olsen Nov 12 '09 at 13:10
• One question, though – what do you mean by smoothness here? $C^1$ or $C^\infty$? If $C^1$, I think my comment above is right. If $C^\infty$, I would question if the answer is yes even for two dimensions. I think the counterexample goes something like this: Let $f\colon\mathbb{R}^2\to\mathbb{R}$ be given by $f(x,y)=y^2x^3/(x^6+y^4)$, $f(0,0)=0$. Then it seems (but I am not sure) that $f\circ c$ is $C^\infty$ for every $C^\infty$ curve $c$, but $f$ is discontinuous. – Harald Hanche-Olsen Nov 12 '09 at 13:36
• Er, I wasn't thinking carefully through that example. It's probably ok for curves $c$ with $c'(t)\ne0$ whenever $c(t)=0$, but put $c(t)=(0,0)$ for $t\le0$ and $c(t)=(e^{-2/t},e^{-3/t})$ for $t>0$, and then $c\in C^\infty$ but $f\circ c$ is discontinuous. I'll stop rambling now. – Harald Hanche-Olsen Nov 12 '09 at 15:02
• In some parts of the web, umlauts are boojums and I didn't want to risk the danger of someone softly and silently vanishing away. On the serious points, for $C^\infty$ then you can check on smooth curves (a paper in Math Scand, no less!) - that's the starting point of the whole Frolicher machine. But you're right for $C^1$. Turns out that if you impose the condition that the derivatives be Lipschitz, then you can check on the corresponding class of curves (details in the early chapters of Kriegl and Michor's book). – Andrew Stacey Nov 13 '09 at 10:32
• I'm confused about the statement in 2. Currently, it reads that a curve $c: \mathbb R \to X$ is smooth iff $\psi \circ c : \mathbb R \to \mathbb R$ is smooth for all $\psi \in X$. I assume you mean for all linear maps $\psi: X \to \mathbb R$? (Of course, $X$ is your big space.) Or is there something I'm missing? – Theo Johnson-Freyd Feb 5 '10 at 3:59

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 $\varepsilon B\subseteq U-x$ .
• I only came to answer the question since I first thought that my idea would give the example for the specific space $\sum_{\mathbb R}\mathbb R$ you are interested in, but I did not think it through carefully enough before sending the answer. Only after sending the answer I noticed that it wasn't really an answer to your question. Unfortunately, I could not totally remove the answer, so the best I could do was then to give the explanation which now stands there. May I ask why it is so important to get the example for $\sum_{\mathbb R}\mathbb R$ ? – TaQ Feb 7 '11 at 13:15