# An example of Fefferman-Stein estimate

By | January 2, 2018

We deal only with the elliptic case assuming that .

Lemma. Let . Then on , Proof. Set . Then we have where , are certain constants, and Take an and a such that in and outside and set where is obtained in Note that and are well-deifned since have compact supports and . Recalling some facts on Newtonian potential, we have Note also that are infinitely differentiable and Furthermore So for any , we have We estimate the integral oscillation of . Since in , we see that if , then and Note that and for , , we have .
Therefore, Note that where is the surface measure on .

We write in polar coordinate. Then by using previous identity By using mean value theorem, we obtain Hence So Note that this estmiate allows shifting the origin. For this reason for any and any ball such that , we have By taking the supremum of the left-hand side over all balls containing , we obtain the desired result.

The following result can be obtained by Calderon-Zygmund theory of singular integral by using the -boundedness of Riesz transform. However, by using Fefferman-Stein theorem, we can obtain it elegantly without using kernel method.

Theorem. Let , and . Then We first prove the case . For the case , it can be obtained by using Fourier transform. For , by the Fefferman-Stein theorem and previous lemma and the -boundedness of Hardy-Littlewood maximal function, we obtain For , we use duality argument. Assume that . Then integration by parts and Holder’s inequality give Since is dense in ,
by density, we conclude that for any ,
we have So by taking supremum to with , we obtain 