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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?

  • 5
    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.2011-06-05

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