Hardy–Littlewood–Sobolev inequality

Remarks

A short proof is Hedberg's. Fix $x$ and $R > 0$, and split the integral defining $I_\alpha \abs{f}(x)$ into $\abs{x - y} < R$ and $\abs{x - y} \ge R$. The near part is at most $C R^\alpha Mf(x)$, where $M$ is the Hardy–Littlewood maximal function, by summing over dyadic shells. The far part is at most $C R^{-n/q} \norm{f}_p$, by Hölder's inequality. Choosing $R$ to balance the two terms gives $\abs{I_\alpha f(x)} \le C \, Mf(x)^{p/q} \norm{f}_p^{1 - p/q}$, and the maximal inequality for $M$ gives both parts.

The relation between $p$, $q$ and $\alpha$ is forced by scaling. If $f_\delta(x) = f(\delta x)$, then $I_\alpha f_\delta(x) = \delta^{-\alpha} I_\alpha f(\delta x)$, and an inequality $\norm{I_\alpha f}_q \le C \norm{f}_p$ for all $f$ and all $\delta > 0$ needs $\frac{1}{q} = \frac{1}{p} - \frac{\alpha}{n}$. At $p = 1$ the strong bound fails: if $f \ge 0$ is integrable with $\int f > 0$, then $I_\alpha f(x)$ behaves like $c \abs{x}^{\alpha - n}$ at infinity, which is in weak $L^{q}$ but not in $L^{q}$ for $q = n/(n - \alpha)$.

Going deeper

For a student

The inequality implies the Sobolev inequality for $1 < p < n$. A smooth function $u$ with compact support satisfies $\abs{u(x)} \le C \, I_1(\abs{\nabla u})(x)$, and the case $\alpha = 1$ then gives $\norm{u}_{np/(n - p)} \le C \norm{\nabla u}_p$. There is also a bilinear form: $\abs{\iint f(x) g(y) \abs{x - y}^{-\lambda} \, dx \, dy} \le C \norm{f}_p \norm{g}_r$ when $0 < \lambda < n$, $1 < p, r < \infty$ and $\frac{1}{p} + \frac{1}{r} + \frac{\lambda}{n} = 2$. Lieb found the sharp constant of this bilinear form in the case $p = r$.