In the book I am reading, the author gives an estimate for the Fourier coefficients and I don't get it.

In the following we consider integrable functions $f \in L^1(S^1)$ which are measurable functions $x \mapsto f(e^{2􏰃 \pi ix})$, periodic of period 1 and, abbreviating $f(e^{2 \pi 􏰃ix}) =􏰌 f(x)$, satisfy

$$ \int_0^1 \mid f(x) \mid dx < \infty. $$

The Fourier coefficients are defined as

$$ \widehat{f}(n) = \int_0^1 f(x)e^{-2 \pi inx} dx. $$

The statement is that for every integer $k \geq 0$, there exists a constant $c_k$ such that

$$ \mid \widehat{f}(n) \mid \leq \frac{c_k}{\mid n \mid^k}. $$

Since there's this $\mid n \mid^k$ in the denominator, I tried to use integration by parts repeatedly, but unfortunately I don't know how to conclude. Here's my attempt:

I derivate $f$ and integrate $e^{-2 \pi inx}$. The boundary terms cancels out, because of the periodicity, hence I get

$$\widehat{f}(n) = - \frac{i}{2 \pi n} \int_0^1 f'(x)e^{-2 \pi inx} dx$$

Is it the right path to follow ? After 3 integration by parts, I get

$$ \widehat{f}(n) = \frac{-i}{(2 \pi n)^3} \int_0^1 f'''(x)e^{-2 \pi inx} dx. $$

So I would get the following estimate

$$ \mid \widehat{f}(n) \mid \leq \frac{1}{(2 \pi n)^3} \int_0^1 \mid f'''(x) \mid dx. $$

Which seems to be near to what is stated but not quite. How can I conclude ?

Many thanks for your kind help.

  • $\begingroup$ You left out a crucial hypothesis - that estimate is not true for every $f\in L^1(S^1)$. $\endgroup$ – David C. Ullrich Mar 9 at 18:36
  • 1
    $\begingroup$ Take $C_k = \frac{1}{(2\pi)^k}\int_{0}^{1}f^k$. Also you need some sort of smoothness assumption. $\endgroup$ – rubikscube09 Mar 9 at 18:36
  • $\begingroup$ @DavidC.Ullrich Thanks you. You're right, the author take $f \in C^{\infty}(S^1,\mathbb{R}^n)$. $\endgroup$ – Alain Mar 9 at 18:40
  • $\begingroup$ @rubikscube09 but how can I be sure that $\int_0^1 f^{k}$ is fine ? Do I miss something ? $\endgroup$ – Alain Mar 9 at 18:41
  • $\begingroup$ It seems ok to me. It is a constant that depends only on $k,f$, and is well defined thanks to the smoothness of $f$. $\endgroup$ – rubikscube09 Mar 9 at 18:49

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