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If $f$ is a $C^1$ function of period $2\pi$ and $f(0)=0$, and $g(x) = f(x)/(e^{ix} -1)$,

then let $C_n$ be the complex fourier coefficients of $f(x)$ and $D_n$ the coefficients of $g(x)$. How can we show that $D_n \to 0$?

ii) And how can we show that $C_n = D_{n-1} - D_n$, so that the series $\sum C_n$ is telescoping.

So, for the first one, I'm thinking of uniform continuity and how to show that $g(x)$ is continuous in it.

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$D_n \to 0$ looks like the Riemann-Lebesgue lemma... – Siminore Oct 28 '12 at 9:50

For ii), write $g(x)=f(x)/(\def\eix{\mathrm e^{\mathrm ix}}\eix-1)$ as $f(x)=g(x)(\eix-1)$. Now if you compute the Fourier coefficients on both sides, the factor $\eix$ shifts the coefficients of $g$ by $1$, yielding $C_n=D_{n-1}-D_n$.

For the first question, you can conclude from $C_n=D_{n-1}-D_n$ that $\sum_{k=j}^nC_k=D_j-D_n$ by induction, then let $j\to-\infty$ and $n\to\infty$ to get $\lim_{n\to\infty}(D_{-n}-D_n)=\sum_{k=-\infty}^\infty C_k=f(0)=0$. Contrary to what I wrote in the first version of this answer, that only establishes that the imaginary part of $D_n$, which is odd, goes to $0$. To conclude that all of $D_n$ goes to $0$, either note that $g$ is continuous at $0$ (and thus everywhere), and conclude that $\sum_{k=-\infty}^\infty D_k=g(0)$, so $D_n$ must go to zero for the series to converge; or invoke the Riemann-Lebesgue lemma, as Siminore suggested in a comment under the question.

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@mary: Didn't or doesn't? Does it make sense now? – joriki Oct 28 '12 at 19:57
@mary: Could you please a) use $\TeX$ in the comments, too, to make them readable, and b) explain how your comment relates to my answer? Neither of those, as far as I can decypher them, seems to be related to what I wrote. – joriki Oct 28 '12 at 21:44
@mary: I've updated the answer. Apologies for the error in the first version. – joriki Oct 29 '12 at 0:17
It looks good, can you make the answer more complete by helping out with computing the fourier coefficients and induction? I'm sure I will make a mistake on this part – mary Oct 29 '12 at 0:39
@mary: To be honest, it's beginning to feel like I'm doing your work for you. How about you do it and then if you're not sure about some part you can ask again? – joriki Oct 29 '12 at 0:51

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