Given two functions and , their convolution is
Let us choose and . Now we have that
Let us set , which implies that , , and so . Also, as , and vice versa. Finally, we can integrate:
Now, what happens if we compute
instead? Let us take a look:
This is not a coincidence, since convolution commutes. To see that, set , which implies . Note also that as , . Now,
which is .
However, there exist pairs of functions for which convolution does not converge. Let us demonstrate an example. We will reuse , yet redefine to be . Here we go:
Setting , we have that , , and so . Once again, as , , and vice versa. Finally
which does not converge. In order to prove divergence, first observe that . Next, let us choose . Now,
Since such and exist, diverges by Lemma 1.
Lemma 1 Let be a continuous function. If there exist and such that when and there exists and such that when , then .
Let . Since is continuous, is finite. Now
Since when , . Finally,