Convolution maps $L^p\times L^q$ to $L^r$ when $1/r=1/p+1/q-1$, with $\norm{f\ast g}_r\le\norm{f}_p\norm{g}_q$.Suggest a slogan
Let \(1\le p,q,r\le\infty\) satisfy \[ \frac1r = \frac1p + \frac1q - 1 . \] If \(f\in L^p(\RR^n)\) and \(g\in L^q(\RR^n)\), then the convolution \(f\ast g\) is defined almost everywhere, \(f\ast g\in L^r(\RR^n)\), and \[ \norm{f \ast g}_{r} \leq \norm{f}_{p} \norm{g}_{q} \] [@Fol13].
The following are the cases \(p=q=r=1\); \(p=1\) and \(q=r\); \(r=\infty\) with \(\frac1p+\frac1q=1\); and \(p=q=2\), \(r=\infty\): \[ \begin{aligned} \norm{f\ast g}_1 & \leq \norm{f}_1 \norm{g}_1, \\ \norm{f\ast g}_q & \leq \norm{f}_1 \norm{g}_q, \\ \norm{f\ast g}_\infty & \leq \norm{f}_p \norm{g}_q, \\ \norm{f\ast g}_\infty & \leq \norm{f}_2 \norm{g}_2. \end{aligned} \]
What is Young's inequality?
State Young's inequality for convolutions.