Young's convolution inequality

Remarks

The case $p = 1$ and $q = r$ follows from Minkowski's inequality for integrals: $f * g$ is an average of translates of $g$, each of norm $\norm{g}_q$. The case $r = \infty$, with $p$ and $q$ conjugate, is Hölder's inequality applied for each fixed $x$; then $f * g$ is defined everywhere. The general case follows from these two by the Riesz–Thorin theorem, or directly by a three-factor Hölder inequality.

Example: with $p = q = 1$, one gets $r = 1$ and $\norm{f * g}_1 \le \norm{f}_1 \norm{g}_1$. So $L^1(\R^n)$ is a Banach algebra under convolution.

The exponent relation is forced by scaling. If $f_\lambda(x) = f(\lambda x)$, then $\norm{f_\lambda}_p = \lambda^{-n/p} \norm{f}_p$ and $f_\lambda * g_\lambda = \lambda^{-n} (f * g)_\lambda$. An inequality $\norm{f * g}_r \le C \norm{f}_p \norm{g}_q$ for all $\lambda > 0$ therefore requires $1 + 1/r = 1/p + 1/q$.

The same inequality holds with the same proof on any unimodular locally compact group with Haar measure. The constant $1$ is not sharp on $\R^n$ when $1 < p, q, r < \infty$; the sharp constant is known.

Going deeper

For a student

Young's inequality is the basic estimate for convolution operators $f \mapsto f * g$ with a fixed kernel $g$. With $g \in L^1$ it shows that mollification $f * \varphi_\varepsilon$ is bounded on every $L^p$, which is the first step of approximation by smooth functions. With $g$ a power such as $\abs{x}^{-\alpha}$, which is not in any $L^q(\R^n)$, Young's inequality does not apply, and one needs the Hardy–Littlewood–Sobolev inequality, which uses weak $L^q$ instead.