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I've been trying to figure this out for days now, but I have no idea how to show this. It's from Partial Differential Equations: An Introduction by Walter A. Strauss.

Suppose $\int_{-\pi}^{\pi} [ |f(x)|^2 + |g(x)|^2 ] dx $ is finite where $g(x) = \frac{f(x)}{e^{ix} - 1}$. Let $c_n$ be the coefficients of the complex Fourier series of $f(x)$. Show that $\sum_{-N}^N c_n \rightarrow 0$ as $N \rightarrow 0$.

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Let $f(x) = \sum c_n e^{inx}$ and $g(x) = \sum d_n e^{inx}$. Since $\int_{-\pi}^\pi |f(x)|^2 \, dx < \infty$, and $\int_{-\pi}^\pi |g(x)|^2 \, dx < \infty$, by Parseval's identity, we know that $\sum |c_n|^2 < \infty$ and $\sum |d_n|^2 < \infty$.

Also, since $f(x) = g(x)(e^{ix}-1)$, we have $c_n = d_{n-1} - d_{n}$. Hence $$ d_{-n} = d_0 - c_0 - c_{-1} - \dots - c_{-n+1} \text{ for $n \ge 0$.}$$ Since $d_{-n}\to 0$ as $n \to \infty$, it follows that $$ d_0 = \sum_{k=0}^\infty c_{-k} .$$ Also $$ d_n = c_n + c_{n-1} + \dots + c_{1} + d_0 .$$ Since $d_{n}\to 0$ as $n \to \infty$, it follows that $$ d_0 + \sum_{k=1}^{\infty} c_k = 0 .$$

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  • $\begingroup$ What does this have to do with $\int_{-\pi}^\pi [|f(x)|^2 + |g(x)|^2] dx$? Also $f(x) = g(x) (e^{ix} - 1)$ not $g(x)(e^{inx} - 1)$. $\endgroup$ – jlc1112 Nov 28 '13 at 2:54
  • $\begingroup$ You use the integral with Parseval's Theorem to get the sum of squares formulas. As for the sign error, I'll fix it, but it won't substantially change the proof. $\endgroup$ – Stephen Montgomery-Smith Nov 28 '13 at 2:56
  • $\begingroup$ Can you show me how you used parsevals for this? $\endgroup$ – jlc1112 Nov 28 '13 at 2:58
  • $\begingroup$ $$ \frac1{2\pi} \int_{-\pi}^\pi |f(x)|^2 \, dx = \sum_{n=-\infty}^\infty |\hat f_n|^2 $$ en.wikipedia.org/wiki/Parseval%27s_identity $\endgroup$ – Stephen Montgomery-Smith Nov 28 '13 at 3:01
  • $\begingroup$ Oops, I had an extra $n$ there. That was a typo. $\endgroup$ – Stephen Montgomery-Smith Nov 28 '13 at 3:04

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