preprint

Inserted: 20 aug 2020

Year: 2019

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 \begin{align} \
u\
_{L^{d(d-\alpha),1}(\mathbb{R}^d)} \leq C \
D^\alpha u\
_{L¹(\mathbb{R}^{d;\mathbb{R}d)}} \end{align} for all $u \in L^q(\mathbb{R}^d)$ for some $1 \leq q<d/(1-\alpha)$ 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. $\
D^\alpha u\
_{L^1(\mathbb{R}^d;\mathbb{R}^d)}$ cannot control the integral of $u$ with respect to the Hausdorff content $\mathcal{H}^{d-\alpha}_\infty$. 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 $L^1(\mathcal{H}^{d-\beta}_\infty)$, $\beta \in [1,d)$. It is an open question whether this failure of a weak-type estimate for the Riesz transforms extends to $\beta \in (0,1)$.