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A famous factorization theorem of Cohen states that for any locally compact group $G$, $$L^1(G)=L^1(G)*L^1(G).$$ I want to know if analogous results exist for the class of smooth functions when $G$ is a Lie group. For instance, is it true that $$C^\infty(G)=C^\infty(G)*C^\infty(G)?$$

(Feel free to impose any simplifying assumption such as compactness of $G$, compactly supported functions, or $$C^\infty(G)=\text{Span } C^\infty(G)*C^\infty(G)$$ etc.)

It would be great if someone could point out some of the known positive or negative results or open problems that might exist. References are also highly appreciated.

Added in Edit:

  1. While browing the web, I came across the thesis of Marc Palm, where he mentions on page 65 that

Every smooth function is the convolution product of smooth functions, briefly denoted by $$C_c^\infty(G)=C_c^\infty(G)*C_c^\infty(G).$$

This is clearly stronger than the D-M result, and in contrast, in my opinion, to what Paul writes below in the comments:

A technical point is that finite sums are necessary in that context, so not every test function is exactly a single convolution of two. Does Marc really mean $C_c^\infty(G)=\text{Span }C_c^\infty(G)*C_c^\infty(G)$?

  1. I also came across this post by Marc Palm where he gives an answer to my question below (in the comments).
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    $\begingroup$ Isn't this exactly the Dixmier-Malliavin result? $\endgroup$
    – paul garrett
    Commented Jun 19, 2015 at 17:59
  • $\begingroup$ @paulgarrett The version of Dixmier-Malliavin that I have in mind is (roughly) the following: "Every smooth vector in a Frechet representation $(\pi, V)$ belongs to the Garding space." $\endgroup$
    – EPS
    Commented Jun 19, 2015 at 18:02
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    $\begingroup$ So the D-M result is not addressing some aspect of things relevant to your purposes? Yes, the D-M more directly addresses test functions, i.e., compactly-supported, not all smooth, but that's not a fatal problem. A technical point is that finite sums are necessary in that context, so not every test function is exactly a single convolution of two. Can you clarify your issue? $\endgroup$
    – paul garrett
    Commented Jun 19, 2015 at 18:26
  • $\begingroup$ @paulgarrett You are absolutely right, D-M result is completely relevant to my question. My interpretation of D-M in this context is that any function in $C_c^\infty(G)$ can be approximated by a linear combination of the functions in $C_c^\infty(G)*C_c^\infty(G)$. Now I have two questions. (1) is this the right conclusion from D-M? (2) if the answer to the previous question is yes, then isn't the same conclusion clear from the fact that any function in $f\in C_c^\infty(G)$ can be approximated by $g_t*f$, where $g_t$ is an approximate unit? Sorry if these are too naive questions. $\endgroup$
    – EPS
    Commented Jun 19, 2015 at 21:02
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    $\begingroup$ It's not only approximated, but equal. (The weaker and much easier "approximation" assertion is essentially Garding's from 1947 or so, which would also cope directly with approximation of smooth by convolutions). But then, in any case, you can certainly approximate arbitrary smooth by test functions, if that's enough for you. To get exact equality with $C^\infty*C^\infty_c$ seems tricky... Is approximation good enough for you? $\endgroup$
    – paul garrett
    Commented Jun 19, 2015 at 21:08

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