I'm reading Fourier analysis an introduction by Stein, and I have a problem from section 5.4 about the Poisson kernel. For the following equations
\begin{align}
A_{r}(f)(\theta)&=\sum_{n=-\infty}^{\infty}r^{|n|}a_{n}e^{in\theta} \\
&=\sum_{n=-\infty}^{\infty}r^{|n|} \bigg(\frac{1}{2\pi}\int_{\pi}^{\pi}f(\varphi)e^{-in\varphi}d\varphi \bigg)e^{in\theta} \\
&=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(\varphi) \bigg(\sum_{n=-\infty}^{\infty}r^{|n|}e^{in(\varphi-\theta)} \bigg)d\varphi
\end{align}
I don't understand why we can interchange the integral and infinite sum in the last equation. The text says it is "justified by the uniform convergence of the series." I am not sure which series it means and why uniform convergence can justify this interchange.
Best Answer
You can apply the Weierstrass M-test to the following for fixed $0 \le r < 1$: $$ \sum_{n=-\infty}^{\infty}|r^{|n|}e^{in\theta}|=\sum_{n=-\infty}^{\infty}r^{|n|} = 2\sum_{n=0}^{\infty}r^{n}-1 = \frac{2}{1-r}-1 < \infty. $$ The conclusion is that, for fixed $0 \le r < 1$, the series $\sum_{n=-\infty}^{\infty}r^{|n|}e^{in\theta}$ converges absolutely and uniformly in $\theta$ for all $\theta\in\mathbb{R}$. The limit is the Poisson kernel $P(r,\theta)$.
Let $P_{N}(r,\theta)=\sum_{n=-N}^{N}r^{|n|}e^{in\theta}$. Then, $P_{N}$ converges uniformly to $P(r,\theta)$ as $N\rightarrow\infty$. That is, for fixed $0 \le r < 1$, $$ \sup_{\theta'\in\mathbb{R}}|P_{N}(r,\theta')-P(r,\theta')| \rightarrow 0 \mbox{ as } N\rightarrow \infty. $$ If $f$ is absolutely integrable and $0 \le r < 1$, then $$ \left|\int_{0}^{2\pi}f(\theta')P_{N}(r,\theta-\theta')d\theta'-\int_{0}^{2\pi}f(\theta')P(r,\theta-\theta')d\theta'\right| \\ \le\int_{0}^{2\pi}|f(\theta')|d\theta'\sup_{\theta'\in\mathbb{R}}|P_{N}(r,\theta')-P(r,\theta)| \rightarrow 0 \mbox{ as } N\rightarrow\infty. $$