Proving that Poisson kernel is an approximation of identity Let $P_r (\theta) $ be the Poisson kernel; proving that it is an approximation of identity consists of proving that: $\int_0^{2\pi}P_r (\theta)d\theta=1$ and $||P_r||_{L^1}\le M $, for every $r\in [0,1)$, with $M $ real; $\lim _ {r\to 1}\int_\delta^{2\pi}P_r (\theta)d\theta=0$ for every $\delta\in (0,2\pi]$. I have two problems: first, how come $||P_r||_{L^1}\le M $? If $r \to 1$ the value of  $||P_r||_{L^1} $ approaches a sum of infinitely many $1$, since $|e^ {in\theta}|=1$. Moreover, I know that to prove the last condition I should use this inequality: $1-2r\cos\theta-r^2\le c_\delta$, that hold for every $r \in [\frac 1 2,1)$ and $\theta \in [\delta , \pi ]$; however I don't see how this helps me. In order to show that the integral from $\delta $ to $2\pi $ tends to zero, it would be better to know that the denominator of $P_r $ is greater than some constant (for $r $ and $\theta $ in the same intervals as above, for example), so that as $r $ approaches $1$ we would be sure that the integral goes to zero. Can you give me a clarify about this? Thanks in advance
 A: The usual definition of $P_r(\theta)$ is $P_r(\theta)=\sum_{n \in \mathbb Z} r^{|n|} e^{in\theta}$ but using the formula for the sum of a geometric series we can show that $P_r(\theta)=\frac {1-r^{2}} {1+r^{2}-2r\cos (\theta)}$. From this formula it follows that $P_r(\theta)$ is a non-negative function. From the definition of $P_r(\theta)$ we see that $\int_0^{2\pi} P_r(\theta)d\theta=2\pi$ since  $\int_0^{2\pi} e^{in\theta}d\theta=0$ for $n \neq 0$. Hence $\int_0^{2\pi} |P_r(\theta)| d\theta=\int_0^{2\pi} P_r(\theta) d\theta =2\pi$. This proves the first two properties (with $M=1$).
Let us prove the last property: $1+r^{2}-2r \cos (\theta) \geq  1+r^{2}-2r \cos (\delta)$ for $\theta \in [\delta, \pi]$ because $\cos (\theta)$ is decreasing on $[0, \pi]$. Hence $\int_{\delta}^{\pi} P_r(\theta) d\theta \leq \frac {(1-r^{2})} {1+r^{2}-2r \cos (\delta)} (\pi -\delta) \to \frac  0 {2-2\cos (\delta)}= 0$ as $r \to 1-$. Now conclude (from the fact that $P_r(\theta)$ is an even function) that $\int_{-\pi} ^{-\delta} P_r(\theta) d\theta \to 0$ and Hence $\int_{|\theta| >\delta} P_r(\theta) d\theta \to 0$ for  every $\delta \in (0, \pi)$. 
[Do not use integral from $\delta$ to $2\pi$; for a periodic function with period $2\pi$ the integral from $0$ to $2\pi$ is same as the integral from $-\pi$ to $+\pi$]. 
