An example of Fefferman-Stein estimate

By | January 2, 2018

We deal only with the elliptic case assuming that d\ge3.

Lemma. Let u\in C_{0}^{\infty}. Then on \mathbb{R}^{d},

    \[ \left(u_{xx}\right)^{\sharp}\le C\left(d\right)M\left(\left|\triangle u\right|^{2}\right)^{\frac{1}{2}}. \]

Proof. Set f=-\triangle u. Then we have

    \[ u_{x^{i}x^{j}}\left(x\right)=C_{1}\left(d\right)K_{ij}f\left(x\right)+f\left(x\right)N_{ij}, \]

where C_{1}\left(d\right), K_{ij} are certain constants, and

    \begin{align*} K_{ij}f\left(x\right) & =p.v.\int_{\mathbb{R}^{d}}K_{ij}\left(y\right)f\left(x+y\right)\myd{y}\\ K_{ij}\left(x\right) & =\frac{y^{i}y^{j}d-\delta^{ij}\left|y\right|^{2}}{\left|y\right|^{d+2}}. \end{align*}

Take an r>0 and a \zeta\in C_{0}^{\infty} such that \zeta=1 in B_{3r} and \zeta=0 outside B_{4r} and set

    \begin{align*} g & =\zeta f,\quad h=\left(1-\zeta\right)f,\\ v\left(x\right) & =C\left(d\right)\int_{\mathbb{R}^{d}}\frac{1}{\left|y\right|^{d-2}}g\left(x-y\right)\myd{y},\\ w\left(x\right) & =C\left(d\right)\int_{\mathbb{R}^{d}}\frac{1}{\left|y\right|^{d-2}}h\left(x-y\right)\myd{y}, \end{align*}

where C\left(d\right) is obtained in

    \[ \frac{1}{\left(4\pi\right)^{\frac{d}{2}}}\int_{0}^{\infty}t^{-\frac{d}{2}}e^{-\frac{\left|y\right|^{2}}{4t}}\myd{t}=C\left(d\right)\frac{1}{\left|y\right|^{d-2}}. \]

Note that v and w are well-deifned since f,g,h have compact supports and d\ge3. Recalling some facts on Newtonian potential, we have

    \begin{align*} u\left(x\right) & =C\left(d\right)\int_{\mathbb{R}^{d}}\frac{1}{\left|y\right|^{d-2}}f\left(x-y\right)\myd{y}\\  & =C\left(d\right)\int_{\mathbb{R}^{d}}\frac{1}{\left|y\right|^{d-2}}\left[g\left(x-y\right)+h\left(x-y\right)\right]\myd{y}\\  & =v\left(x\right)+w\left(x\right). \end{align*}

Note also that v,w are infinitely differentiable and

    \[ \triangle v=-g\quad\triangle w=-h. \]

Furthermore

    \[ \int_{\mathbb{R}^{d}}\left|v_{xx}\right|^{2}\myd{x}\le C\int_{\mathbb{R}^{d}}\left|g\right|^{2}\myd{x}\le C\int_{B_{4r}}\left|f\right|^{2}\myd{x}. \]

So for any x_{0}\in B_{r}, we have

    \begin{align*}  & \fint_{B_{r}}\left|v_{xx}-\left(v_{xx}\right)_{B_{r}}\right|\myd{x}\\  & \le\left(\fint_{B_{r}}\left|v_{xx}-\left(v_{xx}\right)_{B_{r}}\right|^{2}\myd{x}\right)^{\frac{1}{2}}\\  & \le\left(\fint_{B_{r}}\left|v_{xx}\right|^{2}\myd{x}\right)^{\frac{1}{2}}\\  & \le C\left(\fint_{B_{4r}}\left|f\right|^{2}\myd{x}\right)^{\frac{1}{2}}\\  & \le CM\left(\left|f\right|^{2}\right)\left(x_{0}\right)^{\frac{1}{2}}. \end{align*}

We estimate the integral oscillation of w_{xx}. Since h=0 in B_{3r}, we see that if x\in B_{r}, then

    \[ w_{x^{i}x^{j}}\left(x\right)=\int_{\left|y\right|\ge2r}K_{ij}\left(y\right)h\left(x+y\right)\myd{y} \]

and

    \begin{align*} w_{x^{i}x^{j}x^{k}}\left(x\right) & =\int_{\left|y\right|\ge2r}K_{ij}\left(y\right)h_{x^{k}}\left(x+y\right)\myd{y}\\  & =-\int_{\left|y\right|\ge2r}K_{ijx^{k}}\left(y\right)h\left(x+y\right)\myd{y}. \end{align*}

Note that

    \[ \left|K_{ijx^{k}}\left(y\right)\right|\le\frac{C}{\left|y\right|^{d+1}} \]

and for \left|x\right|<r, \left|y\right|\ge2r, we have r\le\left|x+y\right|\le\left|x\right|+\left|y\right|\le2\left|y\right|.
Therefore,

    \begin{align*}  & \left|w_{x^{i}x^{j}x^{k}}\left(x\right)\right|\\  & \le C\int_{\left|y\right|\ge2r}\frac{\left|h\left(x+y\right)\right|}{\left|x+y\right|^{d+1}}\myd{y}\\  & \le C\int_{\left|x+y\right|\ge2r}\frac{\left|h\left(x+y\right)\right|}{\left|x+y\right|^{d+1}}\myd{y}\\  & =C\int_{\left|y\right|\ge r}\frac{\left|h\left(y\right)\right|}{\left|y\right|^{d+1}}\myd{y}:=CI. \end{align*}

Note that

    \begin{align*} \frac{\partial}{\partial\rho}\int_{B_{\rho}}\left|h\left(y\right)\right|\myd{y} & =\frac{\partial}{\partial\rho}\int_{0}^{\rho}\int_{\mathbb{S}^{d-1}}\left|h\left(r\gamma\right)\right|r^{d-1}\myd{S_{1}}\myd{r}\\  & =\int_{\mathbb{S}^{d-1}}\left|h\left(\rho\gamma\right)\right|\rho^{d-1}\myd{S_{1}}\\  & =\int_{\partial B_{\rho}}\left|h\left(\gamma\right)\right|\myd{S_{\rho}}, \end{align*}

where \myd{S_{\rho}} is the surface measure on \partial B_{\rho}.

We write I in polar coordinate. Then by using previous identity

    \begin{align*} I & =C\int_{r}^{\infty}\rho^{-d-1}\int_{\partial B_{\rho}}\left|h\left(y\right)\right|\myd{S_{\rho}}\myd{\rho}\\  & =C\int_{r}^{\infty}\rho^{-d-2}\int_{B_{\rho}}\left|h\left(y\right)\right|\myd{y}\myd{\rho}-Cr^{-d-1}\int_{B_{r}}\left|h\left(y\right)\right|\myd{y}\\  & \le C\int_{r}^{\infty}\rho^{-2}\fint_{B_{\rho}}\left|h\left(y\right)\right|\myd{y}\myd{\rho}\\  & \le CMh\left(x_{0}\right)\int_{r}^{\infty}\rho^{-2}\myd{\rho}\\  & \le Cr^{-1}Mf\left(x_{0}\right)\\  & \le Cr^{-1}M\left(\left|f\right|^{2}\right)\left(x_{0}\right)^{\frac{1}{2}}. \end{align*}

By using mean value theorem, we obtain

    \begin{align*}  & \fint_{B_{r}}\left|w_{xx}\left(x\right)-\left(w_{xx}\right)_{B_{r}}\right|\myd{x}\\  & =\fint_{B_{r}}\left|\frac{1}{\left|B_{r}\right|}\int_{B_{r}}w_{xx}\left(x\right)\myd{y}-\frac{1}{\left|B_{r}\right|}\int_{B_{r}}w_{xx}\left(y\right)\myd{y}\right|\myd{x}\\  & \le\fint_{B_{r}}\frac{1}{\left|B_{r}\right|}\int_{B_{r}}\left|w_{xx}\left(x\right)-w_{xx}\left(y\right)\right|\myd{y}\myd{x}\\  & \le\fint_{B_{r}}\fint_{B_{r}}\left|w_{xxx}\left(x^{*}\right)\right|\left|x-y\right|\myd{y}\myd{x}. \end{align*}

Hence

    \begin{align*} \fint_{B_{r}}\left|w_{xx}-\left(w_{xx}\right)_{B_{r}}\right|\myd{x} & \le Cr\sup_{B_{r}}\left|w_{xxx}\right|\\  & \le CM\left(\left|f\right|^{2}\right)\left(x_{0}\right)^{\frac{1}{2}}. \end{align*}

So

    \[ \fint_{B_{r}}\left|u_{xx}-\left(u_{xx}\right)_{B_{r}}\right|\myd{x}\le CM\left(\left|f\right|^{2}\right)\left(x_{0}\right)^{\frac{1}{2}} \]

Note that this estmiate allows shifting the origin. For this reason for any x_{0} and any ball B such that x_{0}\in B, we have

    \[ \fint_{B}\left|u_{xx}-\left(u_{xx}\right)_{B}\right|\myd{x}\le CM\left(\left|f\right|^{2}\right)\left(x_{0}\right)^{\frac{1}{2}}. \]

By taking the supremum of the left-hand side over all balls B containing x_{0}, we obtain the desired result.

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

Theorem. Let d\ge3, u\in C_{0}^{\infty}, and p\in\left(1,\infty\right). Then

    \[ \Norm{u_{xx}}_{\Leb p}\le C\left(p,d\right)\Norm{\triangle u}_{\Leb p}. \]

We first prove the case 2\le p<\infty. For the case p=2, it can be obtained by using Fourier transform. For 2<p<\infty, by the Fefferman-Stein theorem and previous lemma and the \Leb p-boundedness of Hardy-Littlewood maximal function, we obtain

    \begin{align*} \Norm{u_{xx}}_{\Leb p} & \le C\Norm{\left(u_{xx}\right)^{\sharp}}_{\Leb p}\\  & \le C\Norm{M\left(\left|\triangle u\right|^{2}\right)^{\frac{1}{2}}}_{\Leb p}\\  & \le C\Norm{\left|\triangle u\right|^{2}}_{\Leb{\frac{p}{2}}}^{\frac{1}{2}}\\  & =C\Norm{\triangle u}_{\Leb p}. \end{align*}

For 1<p<2, we use duality argument. Assume that u,v\in C_{0}^{\infty}\left(\mathbb{R}^{d}\right). Then integration by parts and Holder’s inequality give

    \begin{align*} \int_{\mathbb{R}^{d}}u_{xx}\triangle vdx & =\int_{\mathbb{R}^{d}}v_{xx}\triangle udx\\  & \le\Norm{v_{xx}}_{\Leb{p^{\prime}}}\Norm{\triangle u}_{\Leb p}\\  & \le C\Norm{\triangle v}_{\Leb{p^{\prime}}}\Norm{\triangle u}_{\Leb p}. \end{align*}

Since \triangle C_{0}^{\infty} is dense in \Leb{p^{\prime}}\left(\mathbb{R}^{d+1}\right),
by density, we conclude that for any g\in\Leb{p^{\prime}}\left(\mathbb{R}^{d}\right),
we have

    \[ \int_{\mathbb{R}^{d}}u_{xx}gdx\le\Norm g_{\Leb{p^{\prime}}}\Norm{\triangle u}_{\Leb p}. \]

So by taking supremum to g with \Norm g_{\Leb{p^{\prime}}}\le1, we obtain \Norm{u_{xx}}_{\Leb p}\le C\Norm{\triangle u}_{\Leb p}

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