9
$\begingroup$

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?

$\endgroup$
9
$\begingroup$

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 $$|f\star K_n(x)-f\star K(x)|\leqslant \left|\int_\mathbb Rf(x-t)K_n(t)\mathrm dt-\int_\mathbb Rf(x-t)K(t)\mathrm dt\right|\leqslant\lVert f\rVert_\infty\lVert K_n-K\rVert_1.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.