Problem: Let $f, a$ be functions from $\mathbb R$ to $\mathbb R$ and $f$ is continuous. Assume that the convolution $g=f*a$ is well defined, i.e. $g(x)=\int_{t\in \mathbb R} f(t)a(x-t)dt<\infty$ for all $x\in \mathbb R$. I am interested in rewriting $g$ as a sum of $a$ functions, i.e. are there exits $t_1, ..., t_n$ such that $g(x)=\sum_{i=1,...,n} a(x-t_i)$ for all $x\in \mathbb R$.
My guess is that it is impossible. Because I observe that if
- If $a$ is discontinuous, $g$ is still continuous.
- If $a$ is continuous, $g$ is differentiable
- If $a$ is $k$-diffrentiable then $g$ is $(k+1)$-differentiable for $k\geq 1$
So I guess that the answer is impossible. But how to formalize my idea if it is correct?
You can certainly disprove the general assertion by inspecting a case like $f(x)=e^{-x^2}$ and $a(x)=\begin{cases}0&\text{if }\lvert x\rvert\ge1\\ e^{(x^2-1)^{-1}}&\text{if }\lvert x\rvert<1\end{cases}$. Then, $f*a(x)>0$ for all $x$, but all finite sums of translations of $a$ have compact support.