Let's start with an interesting story. In his celebrated Partial Differential Relations (p. 146), the great Misha Gromov gives a nice exercise of which the following is a (strict) part.

Exercise. Consider the action of $\mathrm{Diff}(\mathbb R^n)$ on the space $C^\infty(\mathbb R^n)$ of smooth functions $\mathbb R^n \to \mathbb R$. Then

  • if $n\geq 2$, then there are exactly six (non-trivial) invariant subspaces;

  • if $n = 1$, then there are ten of them.

It's quite easy to find the invariant subspaces Gromov is thinking about (I will give no spoiler...). The problem is that, according to these two messages of C. McMullen, the exercise is actually false.

McMullen's counterexample is the following: take a sequence space $S \subset \mathbb R^\mathbb N$ and define the function subspace $$\mathscr E[S] = \left\{ f \in C^\infty(\mathbb R) \,\middle|\, \forall (x_n)_n \in \mathbb R^\mathbb N, |x_n| \xrightarrow[n\to\infty]{} \infty \Longrightarrow (f(x_n))_n \in S\right\}.$$ This is quite clearly a $\mathrm{Diff}(\mathbb R)$-invariant subspace of $C^\infty(\mathbb R)$. McMullen claims that the sequence space $$\mathrm{bv} = \left\{(a_n)_{n\in\mathbb N}\,\middle |\, \sum_{k} |a_{k+1} - a_k| < +\infty \right\}$$ already gives an example of an "exotic" invariant subspace $\mathscr E[\mathrm{bv}]$ (i.e. one different from Gromov's ten examples) and that it is easy to modify it to give uncountably many such invariant subspaces.

I have two questions:

  1. Can you prove McMullen's claims?
  2. If $n\geq 2$, is there an invariant subspace which isn't of the form $\mathscr E[S]$? (if $n = 1$, the space $\mathscr E[S]$ doesn't make any difference between the two ends $\pm \infty$ of $\mathbb R$ so only six of Gromov's ten subspaces are of this form).

1 Answer 1


EDIT: I left the original answer below, but I think the sequence space $\mathcal{E}[bv]$ as defined in the question and its variation defined below is just the space of functions which are constant in some neighborhood of $\infty$, and I assume that this is one of the "standard" spaces Gromov mentions. Just to clarify, I used $m$ as an index since $n$ is used for the dimension of the space.

If a function $f$ is not constant in some neighborhood of $\infty$, there exists a sequence of points $y_k \to \infty$ such that $f(y_k) \ne f(y_l)$ for $k \ne l$. Now pick integers $N_k > 1/|f(y_k) - f(y_{k+1})|$, and define the sequence $(x_m)$ by concatenating $N_1$ pairs $y_1, y_2$, then $N_2$ pairs $y_2, y_3$, in general $N_k$ pairs $y_k, y_{k+1}$. By the choice of integers, the variation in the sequence $f(x_m)$ over the $N_k$ pairs $f(y_k), f(y_{k+1})$ is greater than 1 for every $k$, so the total variation is infinite.

In the case $n=1$ one could modify the definition and restrict it to monotone sequences $(x_m)$, in which case the spaces of finite $p$-variation are all distinct, which can be shown using functions of the form $f(x) = \frac{\sin x}{x^\alpha}$ for $x \ge 1$. However, I see no similar modification which would work in dimension greater than 1.

ORIGINAL ANSWER: I don't know the answer to the second question, but for the first part it seems that replacing finite variation by finite $p$-variation for $p>1$ leads to an uncountable (nested) family of invariant subspaces. I.e., instead of bv consider the sequence space $$ \mathrm{bv}^p = \left\{ (a_n) : \sum_n |a_{n+1}-a_n|^p < \infty \right\} $$

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    $\begingroup$ Seems like a good idea. How do you show that the $\mathscr E[\mathrm{bv}^p]$ are not equal for two different choices of the parameter $p$? $\endgroup$
    – PseudoNeo
    May 31, 2015 at 15:19
  • $\begingroup$ @PseudoNeo: Thinking about it a little more, I found a major problem even with the original construction, edited the post to explain it. $\endgroup$ May 31, 2015 at 19:41
  • $\begingroup$ Your answer is great. (I think that your "modification" of the definition must be what McMullen had in mind, even if his message is quite unclear). $\endgroup$
    – PseudoNeo
    Jun 1, 2015 at 10:07
  • $\begingroup$ Seems unlikely that I get a better answer between now and in two days. I give you the bounty with pleasure. $\endgroup$
    – PseudoNeo
    Jun 3, 2015 at 12:21

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