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I'd love your help proving that if $f$ is an infinitely differentiable function, then $\lim_{n \to \infty} n^k \hat f = 0$, where $\hat f (n)$ is the Fourier coefficient for $n$. I wanted to use the Riemann-Lebesgue theorem that $$\hat f (n)_{n \to \infty} \to 0$$ and the fact that $\hat f\,' (n)= in\hat f(n)$, and to use L’Hôpital's Theorem, but it didn't work.

Any help?

Thanks a lot!

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1 Answer 1

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Assume $f$ is periodic over $[0,2\pi]$ and infinitely differentiable.

Then: $$ \eqalign{ \hat {f' }(r) &={1\over 2\pi}\int_0^{2\pi} f' (t)\exp(-irt)\,dt\cr &= {1\over 2\pi} f (t){\exp(-irt) }\Bigl|_0^{2\pi} -{1\over 2\pi} \int_0^{2\pi}(-ir) f (t) {\exp(-irt) }\,dt\cr &=0+ { ir\over 2\pi} \int_0^{2\pi} f (t) {\exp(-irt) }\,dt \cr &= { ir} \hat{f }(r). } $$

So

$$ \tag{1}ir\hat{f }(r) =\hat {f'}(r). $$

Applying (1) with $\hat {f'}(r)$ on the left hand side: $$ ir\hat{f'}(r) =\hat {f''}(r); $$ whence $$ -r^2\hat{f }(r)=\hat {f''}(r).$$ Successive iterations yield: $$ (ir)^n \hat {f }(r) =\widehat {f^{(n)}}(r) . $$

Since the Riemann-Lebesgue Theorem implies $\widehat {f^{(n)}}(r)$ tends to 0 as $r$ tends to infinity, we have that $r^n \hat f(r) $ tends to 0 as $r$ tends to infinity.

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I think that there's a problem with the integration above, I think that it is suppose to be $\hat f\,' (n)= in\hat f(n)$. The function in the integral in the second row suppose to be $f''(x)$,no? –  Jozef Dec 27 '11 at 8:31
    
@Jozef Thanks. This was off... It should be fine now. –  David Mitra Dec 27 '11 at 10:40

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