Can we expect $g(f\ast h)= gf \ast gh$ for some $g\in C_{c}^{\infty}(\mathbb R)$? Let $f,g:\mathbb R \to \mathbb C$ be nice functions so that their convolution make sense.

My question: Is it possible to choose $0\neq g\in C_{c}^{\infty}(\mathbb R)$ (= the space of smooth functions with compact support) so that
  $g(f\ast h)= gf \ast gh$? Or atleast: $|g(f\ast h)| \leq C |gf \ast gh|$ for some constant $C$?  Or $g(f\ast h)= gf \ast h$?

Edit: Let $f,h \in L^{p}(\mathbb R)\cap C_{0}(\mathbb R), (2\leq p \leq \infty),$ where $C_{0}(\mathbb R)$ is the space of continuous functions vanishing at infinity.
Or even if we can choose $g=g_{1}g_{2}$ so that $g(f\ast h)= g_{1}f \ast g_{2}h$ for $g_{1}, g_{2}\in C_{c}^{\infty}(\mathbb R)$ is also o.k.($g_{1}, g_{2}$ can be depends on $f$ and $h$)
 A: First of all, I don't really understand why you assume $f,h\in L^{p}\left(\mathbb{R}\right)\cap C_{0}\left(\mathbb{R}\right)$
with $p\in\left[2,\infty\right]$, since in general the convolution
$f\ast h$ will not be well-defined with these assumptions, consider
e.g.
$$
f\left(x\right)=h\left(-x\right)=\frac{1}{\ln x}\cdot\chi_{\left(e,\infty\right)}\left(x\right),
$$
suitably modified on $\left(-10,10\right)$ to make it continuous.
For $x<0$, this yields
$$
\int\left|f\left(y\right)h\left(x-y\right)\right|\, dy\geq\int_{\left(1000,\infty\right)}\frac{1}{\ln y}\frac{1}{\ln\left(y-x\right)}\, dy=\infty.
$$
Anyway, if $f,h$ have compact support, let $K_{1}:={\rm supp}\, f$
and $K_{2}:={\rm supp}\, h$ as well as $K_{3}:=K_{1}+K_{2}$. Since
$f,h$ have compact support, we have $f,h\in C_{c}\left(\mathbb{R}\right)$.
Any $x\in\mathbb{R}$ with $0\neq f\ast h\left(x\right)$ satisfies
$$
0\neq f\ast h\left(x\right)=\int f\left(y\right)\cdot h\left(x-y\right)\, dy,
$$
so that there is some $y\in\mathbb{R}$ with $y\in K_{1}$ and $x-y\in K_{2}$.
Hence, $x=\left(x-y\right)+y\in K_{2}+K_{1}$, i.e. ${\rm supp}\left(f\ast h\right)\subset K_{1}+K_{2}$.
Now, choose any $g\in C_{c}^{\infty}\left(\mathbb{R}\right)$ with
$g\equiv1$ on $K_{1}\cup K_{2}\cup\left(K_{1}+K_{2}\right)$. This
implies
$$
g\cdot\left(f\ast h\right)=f\ast h=\left(gf\right)\ast\left(gh\right),
$$
since $g\equiv1$ on the supports of all involved functions.
Finally, I think the claim is false in general. To see this, let
$$
f\left(x\right)=h\left(x\right)=e^{-x^{2}}
$$
for $x\in\mathbb{R}$. Then $f,h\in L^{p}\left(\mathbb{R}\right)\cap C_{0}\left(\mathbb{R}\right)$
for all $p\in\left(0,\infty\right]$. Now assume that there is some
$g\in C_{c}^{\infty}\left(\mathbb{R}\right)$ with
$$
g\cdot\left(f\ast h\right)=\left(gf\right)\ast\left(gh\right).
$$
Note that $f\ast h\left(x\right)=\int f\left(y\right)\cdot h\left(x-y\right)\, dy>0$
for all $x\in\mathbb{R}$, since the integrand is everywhere positive.
Thus, the support of the left-hand side is ${\rm supp}\, g$.
By the Titchmarsh convolution theorem (see http://en.wikipedia.org/wiki/Titchmarsh_convolution_theorem ;
note that if you require $g\geq0$, the following is clear and we do not have
to invoke that theorem), we have
\begin{eqnarray*}
\inf\left[{\rm supp}\, g\right] & = & \inf\left[{\rm supp}\left(gf\right)\ast\left(gh\right)\right]\\
 & = & \inf\left[{\rm supp}\left(gf\right)\right]+\inf\left[{\rm supp}\left(gh\right)\right]\\
 & = & \inf\left({\rm supp}\, g\right)+\inf\left({\rm supp}\, g\right)\\
 & = & 2\cdot\inf\left({\rm supp}\, g\right),
\end{eqnarray*}
which implies
$$
\inf\left[{\rm supp}\, g\right]=0.
$$
The same statement holds with "$\sup$" instead of "$\inf$",
i.e. $\sup\left[{\rm supp}\, g\right]=0$. But this means ${\rm supp}\, g\subset\left\{ 0\right\} $,
in contradiction to $g\neq0$ and $g\in C_{c}^{\infty}\left(\mathbb{R}\right)$.
Of course, it could still be possible that one can choose $g$ with
$$
\left|g\cdot\left(f\ast h\right)\right|\lesssim\left|\left(gf\right)\ast\left(gh\right)\right|.
$$
I will have to think about that.
A: Suppose that such a $g$ exists :
$$g(f*g)=gf*gh\text{ for all } f,h\in C^{\infty}_c(\mathbb{R})$$
Then we have for each $x\in\mathbb{R}$ :
$$g(x)\int f(t)h(x-t)dt=\int g(t)f(t)g(x-t)h(x-t)  $$
i.e. :
$$\int(g(x)-g(t)g(x-t))f(t)h(x-t)dt=0 $$
Set for all $x$, $\psi_x(t):=g(x)-g(t)g(x-t)$. Then $\psi_x$ is a function in $C^{\infty}(\mathbb{R})$ verifying :
$$\int\psi_x(t)f(t)h(x-t)dt=0 $$
Taking $h$ a function wich is $1$ on $x-Supp(f)=\{x-y|y\in Supp(f)\}$ and $0$ at infinity (you can always construct such a function) you get that for each $f\in C^{\infty}_c(\mathbb{R})$ :
$$\int\psi_x(t)f(t)dt=0 $$
Suppose now that $\psi_x\neq 0$, for instance : $\psi_x(x_0)>0$ (the case $<0$ works the same way), then you can find a little open set $]x_0-\delta,x_0+\delta[$ such that $\psi_x(t)>0$ for $t$ in it. Take $f>0$ a non-zero continuous function such that $f(t)=0$ for $t\leq x_0-\delta$ and $t\geq x_0+\delta$. Then :
$$\int\psi_x(t)f(t)dt>0$$
Because it is the integral of a positive continuous function over  $]x_0-\delta,x_0+\delta[$, this contradicts the fact that :
$$\int\psi_x(t)f(t)dt=0 $$
This being true for each $x\in \mathbb{R}$ we have shown that :
$$g(x)=g(t)g(x-t)\text{ for all } x,t\in\mathbb{R} $$
In particular for all $x$ :
$$g(x)=g(-1)g(x+1)=...=g(-1)^ng(x+n) $$
But because $g$ itself is assume to be of compact support you get that $g(x+n)=0$ for $n$ big enough and hence $g(x)=0$. It follows that $g$ must be null.
