The Poisson kernel on the circle is a positive approximate identity

Let $z=e^{2\pi it}$. By [L1], $P_r(t)=1+{\sum }_{k=1}^{\mathrm{\infty }}{r}^k{z}^k+{\sum }_{k=1}^{\mathrm{\infty }}{r}^k{z}^{-k}\mathrm{.}$ Both geometric series converge absolutely, so $P_r(t)=1+\frac{rz}{1-rz}+\frac{r{z}^{-1}}{1-r{z}^{-1}}=\frac{1-r^2}{(1-rz)(1-r{z}^{-1})}\mathrm{.}$ Since $z+z^{-1}=2\cos (2\pi t)$, this is exactly $P_r(t)=\frac{1-r^2}{1-2r\cos (2\pi t)+r^2}\mathrm{.}$

Step 1.1 shows $P_r(t)\ge 0$ because $1-2r\cos (2\pi t)+r^2=(1-r{)}^2+2r(1-\cos (2\pi t))\ge 0,$ and the numerator is positive for $r<1$. Uniform convergence permits termwise ordinary Riemann integration of the defining series. Every nonconstant character has Riemann integral zero on a period, so ${\int }_0^1{P}_r(t)dt=1.$

The limit only concerns $r\uparrow 1$, so it is enough to consider $r\in [1\mathrm{/}2,1)$. If $t\in [\delta ,1-\delta ]$, then $1-2r\cos (2\pi t)+r^2=(1-r{)}^2+4r{\sin }^2(\pi t)\ge 2{\sin }^2(\pi \delta )\mathrm{.}$ Therefore step 1.1 gives $P_r(t)\le \frac{1-r^2}{2{\sin }^2(\pi \delta )}\mathrm{.}$ As $r\uparrow 1$, the right-hand side tends to $0$, so the displayed supremum tends to $0$. Multiplying that supremum bound by the interval length at most $1$ gives the same limit for the tail integral.