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Question: For $f \in C^{1},$ show that the Fourier coefficients, $c_n(f),~n \in \mathbb{Z}$ satisfies $$\lim_{n \rightarrow \infty} n \cdot c_n(f)=0.$$

My approach: WLOG assume that the period of $f$ to be $1.$ The Fourier series of is $$f(x)=\sum_{ n \in \mathbb{Z}}~c_n \cdot e^{2 \pi i n t}~dt$$ and the corresponding Fourier coefficients are given by $$c_n(f)=\int_{0}^{1}~e^{-2 \pi i n t} \cdot f(t)~dt.$$ Using integrating by parts with $u=f(t)$ and $dv=e^{-2 \pi i n t}~dt.$ Then $du=f'(t)~dt$ and $v=e^{-2 \pi i n t}/{-2 \pi i n}.$ Thus we have $$c_n(f)=\frac{1}{-2 \pi i n} \cdot \big\{ f(1) - f(0) \big\}+\frac{1}{2 \pi i n}\int_{0}^{1} e^{-2 \pi i n t} \cdot f'(t)~dt.$$

Now when I multiplied by the $n$ I don't get the desired result. Any help in proving this is much appreciated. Thank you.

  • You should look at a proof of the Riemann-Lebesgue lemma, it is very interesting. – reuns Sep 23 '16 at 22:14
  • I fail to see how this is true if $f$ if $f(0)\neq f(1)$. Take $f(x)=x$, so that its fourier coefficients are something like $1/n$, which doesn't satisfy the required limit. – Alex R. Sep 23 '16 at 22:16
  • @AlexR. Yes, you are right there is an ambiguity. It is true only if $f$ is $C^1$ AND $1$-periodic, i.e. $f(0) = f(1)$. Otherwise, in the context of Fourier series, if $f(0) \ne f(1)$ then it is not considered continuous. – reuns Sep 23 '16 at 22:23

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If $f$ is periodic $f(1)=f(0)$. Then use Riemann-Lebesgue Lemma to show your second term vanishes as well since $f'\in L^1$.

If $f$ is not periodic. The conclusion does not stand. For example, $c_n[x]\sim \frac{2(-1)^n}{n\pi}$.

Hans
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  • Periodicity was not mentioned in the problem. May be it is an inherent being the Fourier series. I'll check that out. Thank you. –  Sep 23 '16 at 21:46
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    @ManMath: Did you not say "WLOG assume that the period of $f$ to be $1$" in your question? – Hans Sep 23 '16 at 21:50
  • Yeah, I made that assumption. –  Sep 23 '16 at 21:52
  • How is $c_n(f)$ even defined if $f$ isn't periodic? – arkeet Sep 23 '16 at 22:09
  • @ManMath the problem mentions the Fourier coefficients of $f$, so implicitly $f$ is $T$-periodic and we are given its values on one period. Of course WLOG we can assume $T=1$. – reuns Sep 23 '16 at 22:09
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    I fail to see how this is true if $f$ is not periodic, specifically if $f(-1)\neq f(1)$. Take $f(x)=x$, so that its fourier coefficients are something like $1/n$, which doesn't satisfy the required limit. – Alex R. Sep 23 '16 at 22:16
  • Thanks everyone for all the helpful suggestions. I didn't use my own assumption and ignorance of Riemann Lebesgue lemma was why I couldn't get the answer. So periodicity is the key part here. –  Sep 23 '16 at 22:31
  • I did accept the answer @Hans. This didn't allow me to post the comment then. –  Sep 23 '16 at 23:42