# Proving that the fourier coefficients for a pretty smooth function are pretty small

Let $f:[0,2\pi] \rightarrow \mathbb{R}$ be $C^k$ for some $k >0$. Prove that

$|\widehat{f}(n)|n|^k|$

is bounded above by some constant independent of $n$.

To do this, we've been given the Riemann-Lebesgue lemma and Bessel's inequality. What I tried was using integration by parts to express the Fourier coefficients of $f^{(k)}(n)$ in terms of $n^k \widehat{f}(n) n^k$:

$$\widehat{f^{(k)}}(n) = \frac{1}{2\pi} \sum_{j=1}^k (in)^{j-1} \left( f^{(k-j)}(2 \pi)-f^{(k-j)}(0) \right)+ \widehat{f}(n)(in)^k.$$

So, if this is right, all I need is to bound the sum. Bessel's inequality tells me the LHS is bounded, so the result will follow. Am I on the right track? How can I get Riemann-Lebesgue on it?

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You actually have to suppose that $f:\mathbb{R}\to\mathbb{R}$, satisfying $f(x+2\pi)=f(x)$, is $C^k$. The difference shows up only at the points $0$ and $2\pi$, but is is important - the statement you want to prove is not true (take e.g. $f(x)=x$ which satisfies your condition for every $n$). With this correct condition the sum in the equality you got disappears, you only get $\hat{f} (n) (in)^k$, and you're done. – user8268 Jun 5 '11 at 20:22

Let $f:\mathbb{R}\to\mathbb{C}$ be a $2\pi$-periodic $C^1$ function. We wish to compute the Fourier coefficients of $f'$. Note that $f'$ is a $2\pi$-periodic continuous function and therefore its Fourier coefficients exist. Let us now use integration by parts:
$\int f'(t)e^{-int}=e^{-int}\int f'(t)dt - \int (-in)e^{-int}(\int f'(t))dt = e^{-int}f(t)+(in)\int f(t)e^{-int}$
$\hat{f'}(n)=\frac{if(\pi)}{\pi}sin(-n\pi)+(in)\hat{f}(n) = in\hat{f}(n)$.
I leave it to you to compute the Fourier coefficients of higher derivatives of $f$ (assuming, of course, that $f$ is in the appropriate $C^k$).
Hint: If you use induction on $k$, you can compute that the $n$th Fourier coefficient of the $k$th derivative of a $C^k$ function is $(in)^k\hat{f}(n)$. – Amitesh Datta Jun 5 '11 at 23:32