# A Noninequality for the Fractional Gradient

created by spector on 20 Aug 2020

[BibTeX]

preprint

Inserted: 20 aug 2020

Year: 2019

ArXiv: 1906.05541 PDF

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 $\beta \in (0,1)$.

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