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This is only part of a larger question.

I've already shown that if $f$ is $2\pi$ periodic, then $$c_n = \frac{1}{4\pi}\int_{-\pi}^\pi\left(f(x)-f(x+\pi/n)\right)e^{-inx}dx.$$

I'm asked to then show that if $f$ is also lipschitz continuous then there is some constant $M$ such that $|c_n|\leq \frac{M}{|n|}$.

Using the lipschitz property, we then have $$\frac{1}{4\pi}\int_{-\pi}^\pi |f(x)-f(x+\pi/n)|e^{-inx}dx\leq \frac{1}{4\pi}\int_{-\pi}^\pi M\left\vert \frac{\pi}{n}\right\vert e^{-inx}dx = \frac{M}{4|n|}\int_{-\pi}^\pi e^{-inx}dx.$$

However, $\int_{-\pi}^\pi e^{-inx}dx = 0$, so unless the fourier coefficients are all $0$, I'm pretty sure that I'm missing something.

Any thoughts would be really appreciated.

Thanks in advance!

Bears
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1 Answers1

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Your inequalities don't even make sense since there is no ordering for complex numbers. To correct your argumnet all you have to do is to insert absolute value sign for $e^{-inx}$.

Kavi Rama Murthy
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