A noninequality for the fractional gradient

Abstract

In this paper we give a streamlined proof of an inequality recently obtained by the author: For every $\alpha \in (0,1)$ there exists a constant $C=C(\alpha ,d)>0$ such that

for all $u\in L^q({\mathbb{R}}^d)$ for some $1\le q<d/(1-\alpha )$ such that $D^{\alpha }u:=\nabla I_{1-\alpha }u\in L^1({\mathbb{R}}^d;{\mathbb{R}}^d)$. We also give a counterexample which shows that in contrast to the case $\alpha =1$, the fractional gradient does not admit an $L^1$ trace inequality, i.e. $\Vert D^{\alpha }u{\Vert }_{L^1({\mathbb{R}}^d;{\mathbb{R}}^d)}$ cannot control the integral of $u$ with respect to the Hausdorff content ${\mathcal{H}}_{\infty }^{d-\alpha }$. The main substance of this counterexample is a result of interest in its own right, that even a weak-type estimate for the Riesz transforms fails on the space $L^1({\mathcal{H}}_{\infty }^{d-\beta })$, $\beta \in [1,d)$. It is an open question whether this failure of a weak-type estimate for the Riesz transforms extends to $\beta \in (0,1)$.