Inequality for convolution $| f *g| \le \|f\|_1 \|g\|_1$

Question

Let $f \in L^1$ and $g \in L^1 \cap C^{\infty}_c, $ is it then true that a.e. we have

$$ |f *g| \le \|f\|_1 \|g\|_1?$$

Answer 1

I'm assuming you are talking about functions on $\mathbb{R}^n$.

So you are asking whether $$\|f*g\|_\infty\le \|f\|_1\|g\|_1$$

You can already see by scaling that this can't be true. Namely, replace $f$ by $f_\lambda(x)=f(\lambda x)$ and $g$ by $g_\lambda$ ($\lambda>0$). Also let's say that $f,g$ are such that $\frac{\|f\|_1 \|g\|_1}{\|f*g\|_\infty}=1$ (just a normalization). We have

$$\lambda^{-n} \|f*g\|_\infty=\|f_\lambda*g_\lambda\|_\infty \le \|f_\lambda\|_1 \|g_\lambda\|_1=\lambda^{-2n} \|f\|_1 \|g\|_1$$

Then, $$\lambda^n \le \frac{\|f\|_1 \|g\|_1}{\|f*g\|_\infty}=1$$

For $\lambda>1$ this is a contradiction!

But $\|f*g\|_\infty\le \|f\|_1 \|g\|_\infty$ is true (simply pull out the $L^\infty$ norm).

In general, there is Young's convolution inequality:

$$\|f*g\|_r\le \|f\|_p \|g\|_q$$

if $1+\frac1r=\frac1p+\frac1q$, $1\le p,q,r\le \infty$.