I'm trying to understand how the composition of a distribution with an infinitely differentiable function is defined and I was unable to find such a definition on the net.

I read in the wikipedia entry on distributions that the space of $C^{\infty}$ functions is dense in $D'(U)$, i.e

$\forall S \in D'(U) \exists \phi_n \in C^\infty(U)$

$\langle\phi_n,\psi\rangle\to \langle S,\psi\rangle$

using the sequence $\phi_n$, define $S\circ F$ as the limit of sequence $\phi_n\circ F$ i.e. $$\langle S\circ F, \psi\rangle := \lim_{n\to\infty} \langle\phi_n\circ F,\psi\rangle$$.

Am I correct in my reasoning

  • $\begingroup$ What should the composition be? Can you elaborate a bit on that? It's not entirely clear to me what you're after. $\endgroup$ Nov 5, 2013 at 15:11
  • $\begingroup$ Edited my question. Please go through it again. $\endgroup$
    – somitra
    Nov 5, 2013 at 15:19
  • $\begingroup$ So you have a smooth $F \colon V \to U$, where $U$ and $V$ are open subsets in some $\mathbb{R}^n$ (not necessarily the same $n$ for the two?), and want to know whether you can meaningfully define a map $F^\ast \colon \mathscr{D}'(U) \to \mathscr{D}'(V)$ that extends $F^\ast \colon C^\infty(U)\to C^\infty(V);\; F^\ast(\phi) = \phi\circ F$? $\endgroup$ Nov 5, 2013 at 15:26
  • $\begingroup$ Can you define the objects and operations you are dealing with more clearly. What is $D'(U)$, what is $F$, what is the composition... $\endgroup$
    – Wintermute
    Nov 5, 2013 at 15:26
  • $\begingroup$ @DanielFischer Fischer yeah...specificlly for the case when F is not one to one but an injection $\endgroup$
    – somitra
    Nov 5, 2013 at 15:28

1 Answer 1


A short and not very useful answer is that the limit doesn't exist in general. A longer and more useful answer is that it exists under some conditions, namely if the wave front of the distribution $S$ is transverse to the map $F$ (some explanations below).

This kind of problems belong to the microlocal analysis. A similar problem is whether/when we can multiply two distributions: again, we can multiply a distribution by a smooth function, and any distribution is the limit of a sequence of smooth functions. The condition is again formulated in terms of the wave front sets (namely that their sum should not contain any zero covector).

Now some details: every distribution $S$ on $M$ has its singular support $sing(S)\subset M$, away from which $S$ is a smooth function. The wave front of a distribution is a more detailed notion: $WF(S)$ is a (conical) subset of $T^*M$; one takes care not only of the singular points, but also of the (co)directions in which $S$ is singular at the point (see the wikipedia link for details). [The projection of $WF(S)$ to $M$ is, as you'd guess, $sing(S)$.]

So finally: If $F:N\to M$ is a smooth map then the composition $S\circ F$ (more often called "the pullback of $S$ by $F$", denoted $F^*S$) is defined if $F$ is transverse to $WF(S)$, which means that $F^*(\lambda)\neq 0$ for every $\lambda\in WF(S)$. Likewise, $S_1S_2$ is defined if $WF(S_1)+WF(S_2)\subset T^*M$ doesn't contain the zero covector at any point of $M$. [A little 'hint': if $F^*S$ exists then $WF(F^*S)\subset F^*WF(S)$, likewise $WF(S_1S_2)\subset WF(S_1)+WF(S_2)$; wave fronts contain (by their definition) no zero covectors.]

  • $\begingroup$ I'm new to this stuff...but still it gave me the direction to follow...in case the limit exists is my recipe valid $\endgroup$
    – somitra
    Nov 6, 2013 at 17:25
  • $\begingroup$ @somitra yes, your formula is correct $\endgroup$
    – user8268
    Nov 6, 2013 at 17:49

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