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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?

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  • $\begingroup$ Not true: look at $p_y(0)=\frac{1}{y}$. for $y$ near $0$. $\endgroup$ Commented Jun 8, 2019 at 2:09
  • $\begingroup$ 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 $h\to 0^+.$ $\endgroup$
    – User8128
    Commented Jun 8, 2019 at 2:29
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    $\begingroup$ @herbsteinberg But the question doesn't involve $y\to0$. $\endgroup$ Commented Jun 8, 2019 at 2:36
  • $\begingroup$ 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 $y$. $\endgroup$ Commented Jun 8, 2019 at 15:59

3 Answers 3

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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"...

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  • $\begingroup$ 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. $\endgroup$
    – MathGuy
    Commented Jun 8, 2019 at 4:23
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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.

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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 $h\to0^+$.

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