Commutative Convolution. Problem 26 Royden 2 ed.

Question

Let $f$ and $g$ be functions in $L^1(—\infty,\infty)$, and define $f\ast g$ to be the function $h$ defined by $h(y) = \int f(y — x)g(x) dx$.

Why $f\ast g=g\ast f$?

I have this: If $y-x=z$ then $\int f(y-x)g(x)dx=\int f(z)g(y-z)(-dz)=-g\ast f$

Answer 1

The indefinite integral is not a value by a function (the anti-derivative of the integrand). Convolution is defined with a definite integral as follows: $$ (f\ast g)(y) = \int_{-\infty}^{\infty} f(y-x) g(x) \, dx. $$

Now let's do the substitution where $z=y-x$. Then $$ (f\ast g)(y) = \int_{-\infty}^{\infty} f(z) \, g(y-z) \, (-dz). $$ Bringing out the negative sign and reversing the limits of integration yields $$ (f\ast g)(y) = \int_{-\infty}^{\infty} g(y-z) \, f(z) \, dz = (g\ast f)(y). $$