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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 $C_c^\infty(G)=\text{Span }C_c^\infty(G)*C_c^\infty(G)$?

2. I also came across this post by Marc Palm where he gives an answer to my question below (in the comments).

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 $C_c^\infty(G)=\text{Span }C_c^\infty(G)*C_c^\infty(G)$?

2. I also came across this post by Marc Palm where he gives an answer to my question below (in the comments).

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 $C_c^\infty(G)=\text{Span }C_c^\infty(G)*C_c^\infty(G)$?

2. I also came across this post by Marc Palm where apparently he gives an answer to my question below (in the comments) about writing a compactly supported function as a linear combination of "symmetric" convolutions. Marc writes:

A convolution product is the linear combination of four positive functions.

however his "polarization" identity doesn't seem valid to me because in this context $f^*(x):=\overline{f(x^{-1})}$ (when $G$ is unimodular) and hence $(if)^*=-if^*$ etc. Am I missing something trivial here or there is a mistake in the answer referenced above?

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 $C_c^\infty(G)=\text{Span }C_c^\infty(G)*C_c^\infty(G)$?

2. I also came across this post by Marc Palm where he gives an answer to my question below (in the comments).

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.

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 $C_c^\infty(G)=\text{Span }C_c^\infty(G)*C_c^\infty(G)$?

2. I also came across this post by Marc Palm where apparently he gives an answer to my question below (in the comments) about writing a compactly supported function as a linear combination of "symmetric" convolutions. Marc writes:

A convolution product is the linear combination of four positive functions.

however his "polarization" identity doesn't seem valid to me because in this context $f^*(x):=\overline{f(x^{-1})}$ (when $G$ is unimodular) and hence $(if)^*=-if^*$ etc. Am I missing something trivial here or there is a mistake in the answer referenced above?