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In this paper we give a streamlined proof of an inequality recently obtained by the author: For every there exists a constant such that
for all for some such that . We also give a counterexample which shows that in contrast to the case , the fractional gradient does not admit an trace inequality, i.e. cannot control the integral of with respect to the Hausdorff content . 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 , . It is an open question whether this failure of a weak-type estimate for the Riesz transforms extends to .
Cite this article
Daniel Spector, A noninequality for the fractional gradient. Port. Math. 76 (2019), no. 2, pp. 153–168DOI 10.4171/PM/2031