Skip to main content
1 of 3
Chris Wendl
  • 486
  • 2
  • 11

strong topologies on $C_c^\infty$

My question is similar to this question about relating different topologies on $C_c^\infty(M)$, but I'm wondering about a few specific issues that weren't cleared up by that question. The short version of my question is:

Which of the commonly used "strong" topologies on the space of smooth compactly supported functions are equivalent to each other?

I have developed a definite opinion on what the answer should be, but it conflicts in part with things I've read elsewhere, including in Hirsch's Differential Topology book -- so if Hirsch is right and I am wrong, I need someone to tell me why.

Concretely, fix an open set $\Omega \subset {\mathbb R}^n$ and consider the space $C_c^\infty(\Omega)$ of smooth real-valued functions with compact support. (We could also talk about functions on a smooth manifold, or maps from one manifold to another, but let's not make this more complicated than it needs to be.) Borrowing some terminology from Daniel Bruegmann's excellent answer to the other question, I would like to compare the following topologies on $C_c^\infty(\Omega)$:

  1. The Whitney $C^\infty$-topology: ${\mathcal U} \subset C_c^\infty(\Omega)$ open means that for every $\varphi \in {\mathcal U}$, there exists an integer $k \ge 0$ and a collection of continuous functions $f_\alpha : \Omega \to (0,\infty)$ such that every $\psi \in C_c^\infty(\Omega)$ with $|\partial^\alpha(\psi - \varphi)| < f_\alpha$ for all multi-indices $\alpha$ of order at most $k$ belongs to ${\mathcal U}$. Equivalently, this is the coarsest topology such that for every $k \ge 0$, the $k$-jet map $$ j^k : C_c^\infty(\Omega) \to C^0(\Omega,J^k(\Omega,{\mathbb R})) $$ is continuous, with the space $C^0(\Omega,J^k(\Omega,{\mathbb R}))$ of continuous sections of the $k$-jet bundle endowed with the strong Whitney $C^0$-topology.
  2. The strong $C^\infty$-topology: a neighborhood base of $\varphi \in C_c^\infty(\Omega)$ is given by all sets of the form $$ \left\{ \psi \in C_c^\infty(\Omega)\ \Big|\ \|\psi - \varphi\|_{C^{k_i}(\Omega_i)} < \epsilon_i \text{ for all $i \in I$} \right\} $$ where $\{\Omega_i\}_{i \in I}$ is an arbitrary locally finite open covering of $\Omega$, and $\{k_i\}_{i \in I}$ and $\{\epsilon_i\}_{i \in I}$ are arbitrary collections of nonnegative integers and positive real numbers respectively. Unless I'm mistaken, this is the same as what has sometimes been called the very strong $C^\infty$-topology, e.g. in this paper, in order to contrast it with 1 (which is also sometimes called the "strong $C^\infty$-topology").
  3. This one is equivalent to either 1 or 2, depending on whom you ask: the coarsest topology such that the infinity-jet map $$ j^\infty : C_c^\infty(\Omega) \to C^0(\Omega,J^\infty(\Omega,{\mathbb R})) $$ is continuous, where $C^0(\Omega,J^\infty(\Omega,{\mathbb R}))$ is endowed with the strong $C^0$-topology and $J^\infty(\Omega,{\mathbb R})$ with the inverse limit topology with respect to the sequence of natural projections $\Omega \times {\mathbb R} = J^0(\Omega,{\mathbb R}) \leftarrow J^1(\Omega,{\mathbb R}) \leftarrow J^2(\Omega,{\mathbb R}) \leftarrow \ldots \leftarrow J^\infty(\Omega,{\mathbb R})$.
  4. The inductive limit topology: endowing $C_K^\infty(\Omega) := \{ \varphi \in C_c^\infty(\Omega)\ |\ \text{supp}(\varphi) \subset K \}$ for each compact subset $K \subset \Omega$ with its natural Fréchet $C^\infty$-topology, one defines ${\mathcal U} \subset C_c^\infty(\Omega)$ to be open if and only if ${\mathcal U} \cap C_K^\infty(\Omega)$ is an open subset of $C_K^\infty(\Omega)$ for every $K \subset \Omega$ compact. This topology does not seem to be mentioned often by differential topologists, but is well known to analysts as the topology of the space of test functions in the theory of distributions.

Question 1: Are 1 and 3 equivalent? This impression emerges from Hirsch's book, which defines the strong topology for $C^\infty$ maps between two manifolds as 1 on page 36, and then on page 62 casually states without proof that it is equivalent to 3. Perhaps I am misunderstanding what Hirsch intended to say, but I believe this is wrong.

Here are some things that I believe to be true, and I shall be very grateful if anyone can confirm or refute them:

  • If the definition of 2 were modified to require that the families $\{k_i\}_{i \in I}$ are always bounded, then it would become equivalent to 1. As written, however, 2 is a strictly finer topology than 1, even though they have the same notion of convergent sequences (which must always have support in a fixed compact subset).
  • Topologies 2 and 3 are equivalent. Appendix C of this paper appears to give a proof of this. The tricky part is to understand what exactly the inverse limit topology on $J^\infty(\Omega,{\mathbb R})$ is; I think the key point is that an open subset of $J^\infty(\Omega,{\mathbb R})$ constrains only finitely many derivatives over sufficiently small neighborhoods of any point in $\Omega$, and the same is therefore true over any compact subset $K \subset \Omega$, but it may still constrain derivatives of unboundedly high order as one moves out toward infinity.
  • Topology 4 is also equivalent to 2 (and therefore 3).

Actually, let's reformulate that as Question 2: is 4 equivalent to 2 and/or 3?

It seems almost obvious if one stares long enough at the family of seminorms for 4 given in this answer to another question. Nonetheless, I have been struggling to find a clear and unambiguous statement of this equivalence, and my suspicion is that the only reason it's so hard to find is that topologists and analysts do not always talk to each other as much as they should. The rest of my confusion is probably due to the fact that the words "strong $C^\infty$-topology" are often used in the literature without clearly specifying whether 1 or 2 is meant -- indeed, it had not occurred to me on first glance that they are different. But I have in the mean time convinced myself that there exist distributions $C_c^\infty({\mathbb R}^n) \to {\mathbb R}$ that are not continuous with respect to 1, so if I'm correct about 4 and 2 being equivalent, this proves 2 to be strictly finer than 1.

Chris Wendl
  • 486
  • 2
  • 11