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