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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.

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    $\begingroup$ See Nate Eldredge's answere here, specifically the part about Fubini's Theorem. $\endgroup$ – mattos Jan 10 '16 at 12:34
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    $\begingroup$ Your text offers better justification than the physics prof who introduced me to the technique did. Upon being asked how one changes the order of summation and integration, he very patiently explained to us that "you write the symbols in the opposite order from the previous line." :) $\endgroup$ – David H Jan 10 '16 at 12:54
  • $\begingroup$ @Mattos I figured an justification on my own. But I still got some confusion. I posted my justification below because it's a bit too long to put in comments. $\endgroup$ – Hua Jan 10 '16 at 14:36
  • $\begingroup$ @DavidH Physicist have too much other important stuff to do to focus on these details, I think:-) $\endgroup$ – Hua Jan 10 '16 at 14:38
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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 $$ \left|\int_{0}^{2\pi}f(\theta')P_{N}(r,\theta-\theta')d\theta'-\int_{0}^{2\pi}f(\theta')P(r,\theta-\theta')d\theta'\right| \\ \le\int_{0}^{2\pi}|f(\theta')|d\theta'\sup_{\theta'\in\mathbb{R}}|P_{N}(r,\theta')-P(r,\theta)| \rightarrow 0 \mbox{ as } N\rightarrow\infty. $$

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  • $\begingroup$ Thanks for this detailed answer. I can see clearly from your derivation that the uniform convergence of the series $\sum r^{|n|}e^{in\theta}$ justifies the interchange. This clearifies my confusion in the text. By the way, do you think my proof in another post is OK? $\endgroup$ – Hua Jan 11 '16 at 7:52
  • $\begingroup$ I guess there should be a "n" in the exponent of "e" term in the third line. I cannot edit because an edit must be at least 6 characters. $\endgroup$ – Hua Jan 11 '16 at 7:58
  • $\begingroup$ @Hua : I added the missing "n" but forgot to respond. $\endgroup$ – Disintegrating By Parts Jan 11 '16 at 22:25
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@Mattos Thanks for comments. I tried to justify the interchange in my question by dominated convergence theorem. Let $g_{N}(\varphi)=\sum_{n=-N}^{N}r^{|n|}e^{in(\varphi-\theta)}f(\varphi)$. Then we need only show \begin{align} \lim_{N\to\infty}\int g_{N}(\varphi) d\varphi=\int \lim_{N\to\infty}g_{N}(\varphi) d\varphi \end{align}. By DCT, we need find an integrable function $h(\varphi)$ that dominates $g_{N}(\varphi)$. So we check the following,

\begin{align} |g_{N}(\varphi)|&\leq \sum_{n=-N}^{N}|r^{|n|}e^{in(\varphi-\theta)}f(\varphi)| \\ &\le \sum_{n=-N}^{N}|r^{|n|}f(\varphi)| \\ &\le |f(\varphi)|\sum_{n=-N}^{N}r^{|n|} \\ &\le |f(\varphi)|M = h(\varphi) \end{align} where M is a constant (Note $0\le r \lt 1$). Since $f(\varphi)$ is integrable as assumed, $h(\varphi)=f(\varphi)M$ is also integrable and dominates $g_{N}(\varphi)$. Thus by DCT, the interchange is justified.

Do you think this proof is OK? The problem is I still cannot understand 1) why the text says uniform convergence can justify the interchange. 2) In the post you point to, I am not sure how Fubini theorem applis here.

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