I want to show that the series
$\sum_1^\infty \frac{\sin(nx)}{\sqrt{n}}$ is not the Fourier series of a Riemann integrable function on $[-\pi,\pi]$.
I was going to do this by showing that the partial Fourier sums of $f$ do not converge to $f$ in $L^2 [-\pi,\pi]$ as $n\to \infty$.
So I was going about it like this:
$\int_{-\pi}^\pi|\sum_1^\infty \frac{\sin(nx)}{\sqrt{n}}-\sum_1^m \frac{\sin(nx)}{\sqrt{n}}|^2dx=\int_{-\pi}^\pi|\sum_{m+1}^\infty \frac{\sin(nx)}{\sqrt{n}}|^2 dx
$
I want to show that this goes to infinity but I am struggling with how to go further. I was also wondering if I was correct in just allowing the function to be the Fourier series.
Any help would be much appreciated. Cheers!
Best Answer
First of all the series converges for all $x$ in $[-\pi,\pi]$. Therefore it is the Fourier series of its sum function. It is continuous except at $x = 0$. Moreover since ${a_n} = \frac{1}{{\sqrt n }}$ is not of O(1/n), (i.e. $n{a_n}$ is not bounded) the Fourier series does not converge boundedly, consequently the sum function is unbounded and hence not Riemann integrable. It is nevertheless Lebesgue integrable.
See Theorem 14 in Fourier Cosine and Sine Series.