Let $f,g:\mathbb{R} \to \mathbb{C}$ be $L^1$ functions.
Theorem. If $|f(x)| \leq |x|^{-a}$ and $|g(x)| \leq |x|^{-b}$ for all large $x$, where $a,b > 0$, then $$ |(f \ast g)(x)| \leq \int |f(y)g(x-y)| dy \leq C|x|^{-a} + C|x|^{-b} \leq 2C\max\{|x|^{-a},|x|^{-b}\} $$ for all large $x$.
We give the proof here. But see also Decay of Convolution.
\begin{align} \int |f(y)g(x-y)| dy &= \int_{|y| \geq \frac{1}{2}|x|} |f(y)g(x-y)| dy + \int_{|y| \leq \frac{1}{2}|x|} |f(y)g(x-y)| dy \\ &= \int_{\substack{|y| \geq \frac{1}{2}|x| \\ }} |f(y)g(x-y)| dy + \int_{\substack{|y| \leq \frac{1}{2}|x| \\ |y-x| \geq \frac{1}{2}|x| }} |f(y)g(x-y)| dy \\ &\leq \int_{\substack{|y| \geq \frac{1}{2}|x| \\ }} |y|^{-a}g(x-y) dy + \int_{\substack{|y| \leq \frac{1}{2}|x| \\ |y-x| \geq \frac{1}{2}|x| }} f(y)|x-y|^{-b} dy \\ &\leq 2^a|x|^{-a} \| g \|_{1} + 2^b |y|^{-b} \| f \|_1 \\ \end{align}
The theorem says $(f \ast g)(x)$ decays at least at as fast as the slower of the functions $f$, $g$.
Question. Can $(f \ast g)(x)$ decay faster than the the slowest of $f$, $g$? For example, suppose we know that $|f(x)| \geq |x|^{-c}$ for infinitely many $x$ and $g$ is a Schwartz function. Is it possible to have $$ |(f \ast g)(x)| \leq C|x|^{-c} $$ for all large $x$?
Yes, the convolution can involve cancellation that makes it a lot smaller than the larger of $f,g$. For example, let $$ F(x) = (x^2+1)^{-2}\sin(x^3), \quad f=F', \quad g(x) = \exp(-x^2) $$ Note that $f(x)$ is often of size $\sim x^{-2}$ because it picks up the factor of $3x^2$ from the chain rule. On the other hand, $$f*g = F'*g = F*g' = O(x^{-4})$$ because both $F$ and $g'$ are $O(x^{-4})$ at infinity. This example can be amplified: more oscillation yields more cancellation in convolution.