Convolution of two exponentially decaying functions also decay exponentially Suppose $f, g:\mathbb{R}^n \to \mathbb R$ are defined by $f(x) = e^{-a|x|}$ and $g(x) = e^{-b|x|}$ for some $a,b>0$. Is the convolution $f*g(x) = \int_{\mathbb R^n} f(y) g(x-y) dy$ also of exponential decay?
I think $f * g$ will decay like $e ^ {- \min( a, b) |x|}$. This guess is based on the Fourier transform $$ \hat f (k) = c_n \frac{a}{(a^2 + |k|^2)^{(n+1)/2}}$$ and similar for $g$. The Fourier transform of $f*g$ will have the nearest singularity at $-|k|^2 = \min( a, b)^2$. But I don't know how to transfer this information back to $x$-space.
It would also be nice to have an argument that do not uses the Fourier transform.
Edit: Jose27 gave a very clear answer that $f*g(x)$ decays like $e^{-c|x|}$ for any $c<\min(a,b)$. Is there some other way to improve the decay to $c= \min(a,b)$?
 A: If all you want is exponential decay you can do this directly: Let $c>0$ to be determined
$$
|e^{c|x|}(f*g)(x)|\leq \int_{\mathbb{R}} e^{c|y|}|f(y)| e^{c(|x|-|y|)} |g(x-y)|\, dy\leq \int_{\mathbb{R}} e^{c|y|} |f(y)|e^{c|x-y|}|g(x-y)|\, dy.
$$
The quantity on the right is bounded whenever $c<\min(a,b)$, so you get exponential decay of order $c$.
If you want $c=\min(a,b)$, a different argument is required (one that I don't see at the moment).
A: WLOG we assume that $a=1$ and $b\geq 1$.
Firstly, we prove that if $b>1$ then there exists $C>0$ such that
$$|(f*g)(x)|\leq Ce^{-|x|},\qquad x\in\mathbb R^n. \tag{1}$$
By definition, using $|y|+|x-y|\geq |x|$ we have
\begin{align*}
(f*g)(x)&=\int_{\mathbb R^n}e^{-|y|}e^{-b|x-y|}\,dy\\
&=\int_{\mathbb R^n}e^{-|y|}e^{-|x-y|}e^{-(b-1)|x-y|}\,dy\\
&\leq e^{-|x|}\int_{\mathbb R^n}e^{-(b-1)|x-y|}\,dy\leq Ce^{-|x|}.
\end{align*}
If $b=1$, we only have
$$|(f*g)(x)|\leq C_\alpha e^{-\alpha|x|},\qquad x\in\mathbb R^n, \tag{2}$$
for $\alpha<1$. The proof is similar with above: for $\alpha\in(0,1)$ we have
\begin{align*}
(f*g)(x)&=\int_{\mathbb R^n}e^{-|y|}e^{-|x-y|}\,dy\\
&=\int_{\mathbb R^n}e^{-\alpha|y|}e^{-\alpha|x-y|}e^{-(1-\alpha)|y|}e^{-(1-\alpha)|x-y|}\,dy\\
&\leq e^{-\alpha|x|}\int_{\mathbb R^n}e^{-(1-\alpha)|y|}\,dy\leq C_\alpha e^{-\alpha|x|}.
\end{align*}
The bound $(2)$ fails for $\alpha=1$. As an example, considering $n=1$, we have, for $x>0$,
\begin{align*}
(f*g)(x)&=\int_{\mathbb R}e^{-|y|}e^{-|x-y|}\,dy\\
&=\int_{-\infty}^0e^{y}e^{-(x-y)}\,dy+\int_{0}^x e^{-y}e^{-(x-y)}\,dy+\int_{x}^\infty e^{-y}e^{-(y-x)}\,dy\\
&= e^{-x}\int_{-\infty}^0e^{2y}\,dy+\int_{0}^x e^{-x}\,dy+e^x\int_{x}^\infty e^{-2y}\,dy\\
&=(1+x)e^{-x}.
\end{align*}
