1

Does one need the condition $\int f(x-y)g(y)dy<\infty$ to could use commutativity and write $\int f(y)g(x-y)dy$?

In my notes I found that this condition is essential to use this property but it is not very considered in practice.

sky90
  • 1,558
  • So you're asking: if one is infinite, must the other also be infinite, so they're both undefined at the same time? You can try hitting f,g with the characteristic functions of an appropriate sequence of sets to obtain $f_n \to f, g_n \to g$ (monotonic increasing), and for each of these $f_n,g_n$ you should be able to prove commutativity of the convolution, and then you simply take limits. – Elchanan Solomon Jan 22 '14 at 06:16

1 Answers1

3

One defines convolution product on $L^1(\mathbb{R})$ by the formula you described. The same formula is valid for any other locally compact group $G$. However, since $\mathbb{R}$ is an abelian group, $L^1(\mathbb{R})$ with convolution product as a Banach algebra is a commutative Banach algebra.

The boundedness of the integral is necessary for the definition of the convolution product. To get the second integral from the first one you can simply use the change of variable $y\to x-y$. Since $\mathbb{R}$ is an abelian locally compact group, this change of variable is measure preserving, and so does not change the value of the integral.