Proof of young's inequality for convolution

Question

Suppose $f \in L^p(\mathbb{R}^n), g \in L^q(\mathbb{R}^n)$.

I would like to show that $|| f*g||_{L^{\infty}} \leq ||f||_{L^p}||g||_{L^q}$ for $\frac{1}{p}+\frac{1}{q}=1.$

my main idea was to use Holder's inequality. This means that I have

$||fg||_{L^1} \leq ||f||_{L^p} ||g||_{L^q}$ Now I need to show $||f*g||_{L^\infty} \leq || fg||_{L^1}$

Is this the case? Why is this true?

No, it is not true: It's trivial to come up with examples of functions $f$ and $g$ such that $|fg|_1 = 0$ but the convolution is non-zero. — , Mar 02 '18 at 16:41
Daniel Fischer's answer to this previous Question gives details for the case $p=q=2$ and then sketches the necessary steps to prove Young's inequality in general. However your Question concerns validity of the inequality $||f*g||{L^\infty} \leq || fg||{L^1}$. — hardmath, Mar 02 '18 at 16:51

score 1 · Accepted Answer · answered Mar 02 '18 at 16:50

1

answered Mar 02 '18 at 16:50

Umberto P.

but wouldnt that imply that the $L^\infty$ norm of $f*g$ is smaller than the $L^1$ norm of fg? – Epsilon Mar 02 '18 at 16:59
That would contradict what @user296602 said – Epsilon Mar 02 '18 at 17:00
$G$ depends on $x$. There is no reason to expect that $|fg|_1 = |fG|_1$. – Umberto P. Mar 02 '18 at 18:53
Why is $|G|_q = |g|_q$? When I try to make a change of variables I end up with an extra negative ... – Hrafn Magnus Sep 03 '21 at 04:35

1 Answers1