I have a question about convolutions and compactness. Is it true that if we have a function $g$ which is just integrable (does not have to be continuous nor differentiable), and f which is smooth on all orders and compact (i.e. $f\in C^\infty_c$), that the convolution of the two:
$$(f*g)(x)=\int_{-\infty}^\infty f(y-x) g(x) dx$$ Is compact? I mean the way I look at it is: $f(y-x)$ is precisely $0$ outside of say $-a<x-y<a$, where $a\in\mathbb R$. Thus the whole integral would yield $0$ outside of that integral (maybe dependent on $y$), thus the convolution must be compact as well. Is my logic sound?
No, the resulting function may not have compact support. For example, suppose $g > 0$ on the whole line (say, $g = \exp(-x^2)$) and $$f = \begin{cases}\exp\left(\frac{-1}{1-|x|^2}\right) & |x|<1 \\ 0 & |x| \ge 1\end{cases}$$ - the standard approximate identity. Then for every $y \in \mathbb{R}$ we have $$(f * g)(y) = \int_{-\infty}^{\infty} g(x)f(y-x)\, dx = \int_{y-1}^{y+1} \exp(-x^2)\exp\left(\frac{-1}{1-|y-x|^2}\right)\, dx > 0.$$ In particular, $\mathrm{supp}(f*g)=\mathbb{R}$.
However, $f*g$ will, in fact, be smooth and integrable.