JournalspmVol. 76, No. 2pp. 153–168

A noninequality for the fractional gradient

  • Daniel Spector

    National Chiao Tung University, Hsinchu, Taiwan and Okinawa Institute of Science and Technology Graduate University, Jap
A noninequality for the fractional gradient cover
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Abstract

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

uLd/(dα),1(Rd)CDαuL1(Rd;Rd)\|u\|_{L^{d/(d-\alpha),1}(\mathbb{R}^d)} \leq C \| D^\alpha u\|_{L^1(\mathbb{R}^d;\mathbb{R}^d)}

for all uLq(Rd)u \in L^q(\mathbb{R}^d) for some 1q<d/(1α)1 \leq q < d/(1-\alpha) such that Dαu:=I1αuL1(Rd;Rd)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 α=1\alpha =1, the fractional gradient does not admit an L1L^1 trace inequality, i.e. DαuL1(Rd;Rd)\| D^\alpha u\|_{L^1(\mathbb{R}^d;\mathbb{R}^d)} cannot control the integral of uu with respect to the Hausdorff content Hdα\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 L1(Hdβ)L^1(\mathcal{H}^{d-\beta}_\infty), β[1,d)\beta \in [1,d). It is an open question whether this failure of a weak-type estimate for the Riesz transforms extends to β(0,1)\beta \in (0,1).

Cite this article

Daniel Spector, A noninequality for the fractional gradient. Port. Math. 76 (2019), no. 2, pp. 153–168

DOI 10.4171/PM/2031