Consider a function $F\in L^2(0,1)$ whose Fourier coefficients are: \begin{align*} \widehat{F}(n)= \begin{cases}\frac{1}{n},\quad &n=1,2,3,\dots,\\ 0, \quad &n\le 0. \end{cases} \end{align*} $\textbf{Question}:$ Is $F$ a bounded function on $(0,1)$, with the possible exception of a null-set?
It resembles a little bit the Fourier series of the function $f(x)=x$, but in the case of $F$ the Fourier coefficients don't oscillate. Moreover, if we consider a formal series expansion $$ F(\theta)=\sum_{n=1}^\infty\frac{e^{in\theta}}{n}, $$ then it clearly diverges for $\theta=0$. It doesn't disqualify my question though, since $F$ doesn't have to be continuous.
I would appreciate any comments/hints.
Hint: The idea here is that $-\log(1 - z) = \sum_{n=1}^{\infty} {z^n \over n}$, so that $F(\theta) = -\log(1 - e^{i\theta})$ which is unbounded as $\theta$ goes to zero.