Further smoothness conditions on $f$ imply better decay of Fourier Coefficients.

Question

Suppose $f$ is a periodic function of period $2\pi$ which belongs to the class $C^{k}.$

Show that $$\hat{f}(n) = O(\frac{1}{|n|^{k}})$$ as $|n|\rightarrow \infty.$ How can I iterate to get that the above is true for $|n|^{k}$, Could anyone help me ?

Answer 1

First of all, $a_n=O(\frac{1}{|n|^k})$ means that $|a_n|\le C/|n|^k$ for some constant $C$. Now $$ \hat f(n)=\int_0^{2\pi}f(x)e^{inx}\,dx. $$ Taking absolute values $$ |\hat f(n)|\le\int_0^{2\pi}|f(x)|\,dx=O(1). $$ If $f$ is $C^1$, integrating by parts and using the periodicity we get $$ \hat f(n)=\int_0^{2\pi}f(x)e^{inx}\,dx=\frac{1}{i\,n}\int_0^{2\pi}f'(x)e^{inx}\,dx $$ and $$ |\hat f(n)|\le\frac{1}{|n|}\int_0^{2\pi}|f'(x)|\,dx=O\Bigl(\frac{1}{|n|}\Bigr). $$ Iterate to get the result for $f$ of class $C^k$.