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.