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 .

**Proof**

Let . Since is continuous, is finite. Now

Since when , . Finally,

.