Let $H$ Lebesgue integrable. Let $f$ be measurable and bounded on $\mathbb{R}$ with $\lim_{\left|x\right|\rightarrow \pm \infty}f(x)=0$.
Let $F(x)=K\ast f=\int_{\mathbb{R}} K(x-s)f(s) \, ds$ be the convolution.
To show that $\lim_{\left|x\right|\rightarrow \pm \infty}F(x)=0$.
I have that $K(x-s)f(s)$ tends to $0$ when you let $s$ go to $\pm \infty$ since it obviously holds for all simple functions.
After this result I tried to deduce from the definition of the integral that it must then also hold for $F(x)$, but this limit is taken over $x$ which shifts $f(s)$ in relation to $K(x-s)$.
My spider sense is telling me Fubini should be applied. Any hints (w.r.t. Fubini)?
Best Answer
A simple substitution shows that $$ \int_{\mathbb R} K(x-s)f(s)\,ds = \int_{\mathbb R} K(s) f(x-s)\,ds. $$ Now use one of Lebesgue's convergence theorems.
(This is another example of the fact that if you convolve an integrable function with a well-behaved function, you get a well-behaved function. For example, if you convolve an integrable function with a continuous function, you get a continuous function, and similarly for "differentiable" or "polynomial" in place of "continuous".)