# Convolution is uniformly continuous and bounded

Suppose $f\in L^\infty(\mathbb{R})$ and $K\in L^1(\mathbb{R})$ with $\int_\mathbb{R}K(x)dx=1$. Show that the convolution $f\ast K$ is a uniformly continuous and bounded function.

The definition of the convolution is $(f\ast K)(x)=\int_\mathbb{R}f(x-y)K(y)dy$.

There is the inequality $\|f\ast K\|_\infty\leq\|f\|_\infty\|K\|_1$, which yields that $f\ast K$ is bounded. But what about uniform continuity?

## 1 Answer

First, assume that $K$ is continuous with compact support. Then using $$|(f\star K)(x)-(f\star K)(x')|\leqslant\int_\mathbb R|f(y)|\cdot |K(x-y)-K(x'-y)|\mathrm dy,$$ boundedness of $f$, and uniform continuity of $K$, we get what we want.

Then we conclude by a density argument: if $K_n\to K$ in $L^1$, then $(f\star K_n)_n$ converges uniformly on the real line to $f\star K$. Indeed, we have $$|f\star K_n(x)-f\star K(x)|\leqslant \left|\int_\mathbb Rf(x-t)K_n(t)\mathrm dt-\int_\mathbb Rf(x-t)K(t)\mathrm dt\right|\leqslant\lVert f\rVert_\infty\lVert K_n-K\rVert_1.$$