(Obvious?) Half-Space Poisson Kernel Estimate

Question

In Stein's Singular Integrals and Differentiability Properties of Functions, Theorem 1 (a) of Chapter 8 (pg. 197) he makes the claim without proof that the Poisson Kernel for the half-space $\mathbb{R}^{n+1}_{+}$ $$P_{y}(x)=\frac{c_{n}y}{(|x|^{2}+y^{2})^{(n+1)/2}}$$ majorizes its $t$ translation $P_{y}(x-t)$ with absolute constant $A_{\alpha}$ (depending only on $\alpha$, which we are free to fix) so long as $|t|<\alpha y.$ That is, with $t$ so restricted, $$P_{y}(x-t)\leq A_{\alpha}P_{y}(x).$$

The setting here is $x\in\mathbb{R}^{n}$ and $y\in\mathbb{R}$. $c_{n}$ is just a constant depending on the dimension $n$ (the volume of the unit ball I believe). $x$ is regarded as being in $\partial\mathbb{R}^{n+1}_{+}$ and $y$ the perpendicular component of a point $(x,y)\in\mathbb{R}^{n+1}.$

This does not seem at all obvious to me as he says. Furthermore, this assertion is markedly similar to my previous (still unanswered) question here Some Scaling Estimate for Heat Kernel (notationally, just change $y$ to $\sqrt{y}$, add an aditional constant $\beta$ for $\beta y$ and then interchange the roles of $y$ and $t$). I guess I am just having a hard time understanding and proving these estimates and it's causing me some problems with what I'm trying to do. Any help is appreciated.

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So let us cancel $y$ and $c_{n}$ and raise both sides of the inequality to the $(n+1)/2$ power and put $C_{\alpha}=A_{\alpha}^{(n+1)/2}$. Then we have to show

$$\frac{1}{|x|^{2}+y^{2}-2x\cdot t+|t|^{2}}\leq\frac{C_{\alpha}}{|x|^{2}+y^{2}}$$ when $|t|\leq\alpha y.$ We obviously have to somehow control the $|t|^{2}-2x\cdot t$ term and adjust $C_{\alpha}$ accordingly, but I'm not sure how to do this because of $x$, which is unrestricted.

Answer 1

For real $x$, $y$, and $t$, $$ x^{2}+y^{2} = ((x-t)+t)^{2}+y^{2} = (x-t)^{2}+2(x-t)t+t^{2}+y^{2}. $$ For real $a$ and $b$, one has $ab \le |ab| \le \frac{1}{2}(a^{2}+b^{2})$. For $\rho > 0$, one also has $2ab \le \rho a^{2}+\frac{1}{\rho}b^{2}$. Hence, if $\alpha > 0$, $y \ge 0$, and $|t| \le \alpha y$, $$ x^{2}+y^{2} \le (x-t)^{2}+\alpha(x-t)^{2}+\frac{1}{\alpha}t^{2}+y^{2} \le (1+\alpha)\{(x-t)^{2}+y^{2})\}. $$ Finally, in terms of the Poisson kernel $P_{y}(x)=\frac{y}{x^{2}+y^{2}}$ for the upper half-plane, $$ P_{y}(x-t)=\frac{y}{(x-t)^{2}+y^{2}} \le \frac{A_{\alpha}y}{x^{2}+y^{2}}=A_{\alpha}P_{y}(x),\;\;\; y > 0,\;|t| \le \alpha y,\; x \in\mathbb{R}. $$ The argument can be modified to work for Poisson kernels on higher-dimensional half spaces.

BTW: This is also related to the heat kernel issue. It is possible to study non-tangential limits as $t \rightarrow 0$ for solutions of the heat equation, or to study non-tangential limits of harmonic functions on a half-space as one approaches the boundary of the half space.

One thing that can be shown is this: if $\phi(x,y)$ is a bounded harmonic function on the upper half-plane where $y > 0$, then $\phi(x,0+)=\lim_{y\downarrow 0}\phi(x,y)$ exists for almost every $x \in \mathbb{R}$, and $$ \phi(x,y)=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\phi(x,0+)y}{(x-t)^{2}+y^{2}}\,dt =\frac{1}{2\pi i}\int_{-\infty}^{\infty}\phi(x,0+)\left[\frac{1}{t-(x+iy)}-\frac{1}{t-(x-iy)}\right]dt. $$ The non-tangential limits of $\phi$ also exist, and equal the orthogonal limits.

Answer 2

I think the statement should be understood from the perspective of Possion kernel. The classical Possion kernel is given by

$$P_{y}(x)=\frac{y}{x^{2}+y^{2}},u(x+iy)=\frac{1}{\pi}\int^{\infty}_{-\infty}P_{y}(x-t)f(t)dt$$(see the wikipedia article http://en.wikipedia.org/wiki/Poisson_kernel)

In other words at here both $x,y$ are fixed and only $t$ is allowed to changed. Here we have to show $$\frac{1}{(|x-t|^{2}+y^{2})^{n+1}}\le C\frac{1}{(x^{2}+y^{2})^{n+1}},\forall t, |t|<\alpha y$$ Notice Stein's book has a typo where he wrote $f(z)dx$, which should be $f(z)dz$ instead. Now for fixed $x$, the maximum and minimum for $|x-t|^{2}$ is fixed as well. So the above statement must be true since the left hand side is bounded.