5
$\begingroup$

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.

Improve this question
$\endgroup$
8
  • $\begingroup$ 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?) $\endgroup$
    – Harald Hanche-Olsen
    Commented Nov 12, 2009 at 13:10
  • $\begingroup$ 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. $\endgroup$
    – Harald Hanche-Olsen
    Commented Nov 12, 2009 at 13:36
  • $\begingroup$ 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. $\endgroup$
    – Harald Hanche-Olsen
    Commented Nov 12, 2009 at 15:02
  • $\begingroup$ 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). $\endgroup$
    – Andrew Stacey
    Commented Nov 13, 2009 at 10:32
  • $\begingroup$ 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? $\endgroup$
    – Theo Johnson-Freyd
    Commented Feb 5, 2010 at 3:59

1 Answer 1

Reset to default
1
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ I appreciate you taking the time to take a look at this very old question. It's one I'm still very interested in knowing the answer to. However, I don't see that what you wrote gets me very far. The space I'm interested in has the bornological lctvs so that issue isn't relevant; and I'm quite happy to pass to the bornological lctvs anyway in these matters - the point of my question is to distinguish the smooth topology from a lctvs topology but I don't care which lctvs so I may as well take the bornological one (ctd) $\endgroup$
    – Andrew Stacey
    Commented Feb 7, 2011 at 10:50
  • $\begingroup$ (ctd) Your example of a discontinuous smooth map is a well-known and very important one, but I fail to see how it applies to this case. I want to know if there is a discontinuous smooth map on a very specific space, so the presence of one on a related space might help but isn't enough to answer the specific question. I don't think that there's going to be an easy answer to this since, as I said, the answer is true for the countable case so it's going to involve the cardinality at some point if it is false (the proof that it is false for the product is quite involved). $\endgroup$
    – Andrew Stacey
    Commented Feb 7, 2011 at 10:53
  • $\begingroup$ 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 $\sum_{\mathbb R}\mathbb R$ ? $\endgroup$
    – TaQ
    Commented Feb 7, 2011 at 13:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .