Young's convolution inequality

Statement

Let $G$ be a unimodular locally compact group with a fixed Haar measure. Suppose $1\leq p,q,r\leq\infty$ satisfy

\frac1p+\frac1q=1+\frac1r.

For $f\in L^p(G)$ and $g\in L^q(G)$ , their convolution, initially defined where the integral exists, has an $L^r(G)$ representative and obeys Young's convolution inequality

\lVert f*g\rVert_r\leq \lVert f\rVert_p\lVert g\rVert_q.

Consequently the integral defines a bounded bilinear map $L^p(G)\times L^q(G)\to L^r(G)$ .

Endpoint estimates

The case $p=1$ and $q=r$ says that averaging left translates of $g$ against $f$ costs at most $\lVert f\rVert_1$ . The case $p=q=2$ , $r=\infty$ , follows directly from the integral Hölder inequality. Taking $p=q=r=1$ proves that $L^1(G)$ is closed under convolution and supplies its Banach-algebra norm estimate.

Proof mechanism and extension

For compactly supported continuous functions, translation invariance of Haar measure, Hölder's inequality, and Tonelli's theorem give the endpoint bounds. Hölder and Tonelli give the full range by splitting the integrand into factors with the prescribed powers; interpolation is another route. Density justifies extensions from compactly supported functions when the input exponents are finite. At an infinite input exponent, use the integral and its direct bound: compactly supported continuous functions are not norm-dense in general in $L^\infty$ , so density alone does not give uniqueness of an extension there.

Scope and nonunimodular groups

Euclidean Young inequality is the case $G=\mathbb R^n$ . The same constant-one estimate holds on discrete groups with counting measure and on compact groups with normalized Haar measure.

References

G. B. Folland, A Course in Abstract Harmonic Analysis, 2nd ed., CRC Press, 2016. DOI record. Relevant: Chapter 2, convolution estimates on locally compact groups.
E. Hewitt and K. A. Ross, Abstract Harmonic Analysis, Volume I, Springer, 1963. DOI record. Relevant: convolution and $L^p$ -space inequalities.