Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 1 down vote accepted

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

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


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

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.