The question comes from Shakarchi and Stien's real analysis p.111, he said that the Poisson kernel for the disc approximates to the identity without proof. However, I don't see that fact immediately, could anybody give me some ideas? Here is the definitions.
A kernel $K_\delta $ (assumed to be Lebesgue integrable and $\delta >0$) can approximate to the identity if it satisfyies:
- $\int_{\mathbb R^d}K_\delta(x)dx=1$
- $\left |K_\delta(x)\right |\le A\delta^{-d}$ for all $\delta>0$
- $\left |K_\delta(x)\right |\le A\delta/|x|^{d+1}$ for all $\delta>0$ and $x\in \mathbb R^d$
The Poisson kernel is defined by
$$P_\delta(x)=\begin{cases} \frac1{2\pi}\frac{1-(1-\delta)^2}{1-2(1-\delta)\cos x+(1-\delta)^2} &\text{if }|x|\le\pi\\ 0 & \text{if } |x|>\pi \end{cases}$$
Here we need $0<\delta<1$.
\begin{align} & 1 - 2(1-\delta)\cos 2x + (1-\delta)^{2} \\ & = 1-2(1-\delta)\{1-2\sin^{2}x\}+(1-\delta)^{2} \\ & = (1-(1-\delta))^{2}+4(1-\delta)\sin^{2}x \\ & = \delta^{2}+4(1-\delta)\sin^{2}x \end{align} Therefore, $$ 0 \le P_{\delta}(x) = \frac{1}{2\pi}\frac{\delta(2-\delta)}{\delta^{2}+4(1-\delta)\sin^{2}x}. $$ So, you have $$ P_{\delta}(x) \le \frac{1}{2\pi}\frac{2\delta}{\delta^{2}}=\frac{1}{\pi}\delta^{-1} \\ P_{\delta}(x) \le \frac{1}{2\pi}\frac{2}{4(1-\delta)\sin^{2}x} \le \frac{1}{4\pi x^{2}},\;\;\delta\approx 0. $$