Aufgaben:Problem 4

1) Let \( f,g:\mathbb{R}^n \rightarrow \mathbb{C} \) be continuous, bounded. Further, \( f,g \in L^1(\mathbb{R}^n) \). Define the convolution \(h:\mathbb{R}^n \rightarrow \mathbb{C} \) as

$$ h(x) = (f * g)(x) = \int_{\mathbb{R}^n } f(x-y) g(y) dy $$

Part a)

Show that:

$$ (f*g)(x) = (g*f)(x) $$

Solution part a)

$$ (f*g)(x) = \int_{\mathbb{R}^n } f(x-y) g(y) dy $$

$$ = \int_{-\infty}^{\infty} ... \int_{-\infty}^{\infty} f(x-y) g(y) dy_1 ... dy_n $$

The claim follows simply by substitution:

$$ t = x-y $$

$$ y = x-t $$

$$ dy_i = -dt_i $$

$$ \int_{\infty}^{-\infty} ... \int_{\infty}^{-\infty} f(t) g(x-t) (-dt_1) ... (-dt_n) $$

We can now change the orientation of the integral if we multiply (-1) n-times. These will cancel out all the n minus-signs from the \( dt_i\)`s. Thus we get:

$$ \int_{-\infty}^{\infty} ... \int_{-\infty}^{\infty} f(t) g(x-t) dt_1 ... dt_n $$

$$ = \int_{\mathbb{R}^n } f(t) g(x-t) dt = (g*f)(x) $$

This proves the claim. \( \square \)