# Uniform convergence of Poisson Kernel (of the upper half plane)

For $$y>0$$ and $$t\in\mathbb{R}$$, define $$p_y(t)=\frac{y}{t^2+y^2}.$$ This is essentially (up to a multiplicative constant $$\frac{1}{\pi}$$) the Poisson kernel of the upper half plane. Since $$p_y$$ is continuous in $$y$$, $$p_{y+h}(t)\to p_y(t)$$ as $$h\to0^+$$. I am interested in a uniform version of this convergence: Is the following true: for each fixed $$y>0$$, $$\lim_{h\to 0^+}\sup_{t\in\mathbb{R}}|p_{y+h}(t)-p_{y}(t)|=0$$

if true, how to prove it?

• Not true: look at $p_y(0)=\frac{1}{y}$. for $y$ near $0$. Commented Jun 8, 2019 at 2:09
• To expand on the comment by @herbsteinberg, the continuity is not uniform in $y$, since $p_0(0)= 0$ but $p_h(0) = \frac 1 h$ and so $$\sup_t \lvert p_h(t) - p_0(t)\rvert \ge \frac 1 h \not\to 0$$ as $h\to 0^+.$ Commented Jun 8, 2019 at 2:29
• @herbsteinberg But the question doesn't involve $y\to0$. Commented Jun 8, 2019 at 2:36
• I misinterpreted the statement. I was assuming that "uniform convergence" meant that the sup would hold for all $y$ at the same time, not just for each $y$. Commented Jun 8, 2019 at 15:59

It's been commented that $$p_y$$ does not converge uniformly as $$y\to0$$. Nonetheless, if $$y>0$$ then yes, $$p_{y+h}\to p_y$$ uniformly as $$h\to0$$. One can "just work it out"; I did that in an answer just now and made a mistake in the algebra. Or one can note that $$p_y$$ is (essentially) the Fourier transform of $$k_y$$, if $$k_y(t)=e^{-y|t|},$$so it's enough to show $$||k_{y+h}-k_y||_1\to0.$$The algebra there seems simpler; or one can just mumble "DCT"...
• thanks. we can work out the difference $|p_{y+h}(t)-p_{y}(t)|$ and try to bound this difference by something that doesn't depend on t but converges to 0 as $h\to0+$. This only involves some basic algebraic manipulations. Commented Jun 8, 2019 at 4:23
The derivative of $$p_y(t)$$ w.r.t. $$y$$ is $$\frac {t^{2}-y^{2}} {(t^{2}+y^{2})^{2}}$$ which is bounded by $$\frac 1 {t^{2}+y^{2}}$$ hence by $$\frac 1 {y^{2}}$$. MVT finishes the proof.
Let me answer my own question. $$y>0$$ is fixed.let $$h>0$$ be sufficiently small so that $$y-h>0$$. A little bit algebra shows that. for any $$t\in\mathbb{R}$$, \begin{align}|p_{y}(t)-p_{y-h}(t)|=&|\frac{yh(h-y)}{(t^2+y^2)(t^2+(y-h)^2)}+\frac{t^2h}{(t^2+y^2)(t^2+(y-h)^2)}|\\ \le &\frac{yh(y-h)}{(t^2+y^2)(t^2+(y-h)^2)}+\frac{t^2h}{(t^2+y^2)(t^2+(y-h)^2)}\\ \le & \frac{h}{y(y-h)}+\frac{h}{(y-h)^2}\end{align}
hence $$\sup_{t\in\mathbb{R}} |p_{y}(t)-p_{y-h}(t)|\le \frac{h}{y(y-h)}+\frac{h}{(y-h)^2}\to 0$$ as $$h\to0^+$$.