Suppose \(1\leq p,q,r \leq \infty\) with \[ {1\over p } + {1 \over q} = 1 + {1 \over r} .\]
Prove that \[ f \in L^p, g\in L^q \implies f \convolve g \in L^r \text{ and } \norm{f \convolve g}_r \leq \norm{f}_p \norm{g}_q .\]
Solution
Work on \(\RR^n\), fix \(f \in L^p\), let \(p'\) be the conjugate exponent of \(p\), and let \(Tg = f \ast g\).
1. \(\|f \ast g\|_p \le \|f\|_p\|g\|_1\) for \(g \in L^1\).
Write \(f \ast g(x) = \int f(x-y)g(y)\,dy\). By Minkowski’s inequality for integrals and translation invariance of the \(L^p\) norm, \(\|f\ast g\|_p \le \int \|f(\cdot - y)\|_p\,|g(y)|\,dy = \|f\|_p\|g\|_1\).
2. \(\|f \ast g\|_\infty \le \|f\|_p\|g\|_{p'}\) for \(g \in L^{p'}\).
By Hölder’s inequality, \(|(f\ast g)(x)| \le \int |f(x-y)||g(y)|\,dy \le \|f\|_p\|g\|_{p'}\) for every \(x\).
3. With \(\theta = 1 - p/r \in [0,1]\), \(\frac1q = \frac{1-\theta}{1} + \frac{\theta}{p'}\) and \(\frac1r = \frac{1-\theta}{p} + \frac{\theta}{\infty}\).
\(\frac{1-\theta}{p} = \frac{p/r}{p} = \frac1r\). Also \(1 - \theta + \theta\left(1 - \frac1p\right) = 1 - \frac\theta p = 1 - \frac1p + \frac1r = \frac1q\) by the hypothesis. Since \(\frac1q \le 1\), \(\frac1p \ge \frac1r\), so \(p \le r\) and \(\theta \in [0,1]\).
4. Q.E.D.
By steps 1 and 2, \(T\) is bounded \(L^1 \to L^p\) and \(L^{p'} \to L^\infty\), both with norm at most \(\|f\|_p\). By step 3 and the Riesz–Thorin interpolation theorem, \(T\) is bounded \(L^q \to L^r\) with norm at most \(\|f\|_p^{1-\theta}\|f\|_p^{\theta} = \|f\|_p\).