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Like in the title written I need to show that $$|\hat{f}(n)|\leq\int_{0}^{2\pi}\frac{|\partial_{k}f(x)|dx}{2\pi|n|^k}$$ for any $n\neq0$ where $f$ is a $2\pi$ periodic function on $C^{k}$.

Now I've arrived with the Chauchy-Schwarz inequality at $$|\hat{f}(n)|\leq\int_{0}^{2\pi}\frac{|f(x)|dx}{2\pi}$$ but I don't see how to proceed and show that $$|f(x)|\leq\frac{|\partial_{k}f(x)|}{|n|^k}$$ Can someone please help?

Edit: we have adopted the following notation for the Fourier coefficients: $$\hat{f}(n)=\int_{0}^{2\pi}\frac{e^{-inx}f(x)dx}{2\pi}$$

Matthew Cassell
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N. Maks
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    Integrate by parts on your Fourier coefficients integral $k$ times with $u = f, dv = e^{-inx}$, assuming you have 'enough' smoothness. – Matthew Cassell Mar 05 '17 at 09:52

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Like @Mattos suggested I have found the proof by using integration by parts with $u=f(x)$ and $v'=e^{-inx}$ for a total of k times (which is allowed, since we assume that $f$ is in $C^{k}$).

My first try on using the Cauchy-Schwarz inequality was to hasty and I could not obtain the desired result. However I've ended up using the C-S inequality it to get the $\frac{1}{|n|}$ into the denominator out of the $\frac{1}{in}$ which one gets when applying the integration by parts.

N. Maks
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