Function with positive Fourier coefficients decaying as $\frac{1}{n}$ Is there an example (I'm not looking for a sufficient or necessary condition but just for an example) of a bounded Rieman-integrable function $f\colon [-\pi,\pi]\rightarrow\mathcal{R}$ with Fourier coefficients 
$c_n(f) = \int_{-\pi}^\pi{f(t)e^{-int}\,dt}$ that are positive and decay as $\frac{1}{n}$ to zero (so that the Fourier coefficients are not summable)?

From this discussion a sufficient condition would be to find a function which has Fourier coefficients decaying as $n^{-1/2}$ and I could then take the convolution of this function with itself to obtain what I want.
No, there is no such function.


  
    Suppose $f(t)\sim\sum_nc_ne^{int}$ where $c_n\ge0$ and $\sum c_n=\infty$. Then $f$ is not bounded (hence not Riemann integrable).
  


Not-quite proof: Let $t=0$: it follows that $f(0)=+\infty$.

Of course that's not quite a proof, since there's no reason a priori that the series should converge to $f(0)$ for $t=0$. It does make it clear that the answer must be no, and if you have any feeling for real analysis it seems clear that the condition $c_n\ge0$ means that it can't be hard to fix the argument.

Hint for an actual proof: Show that if $f$ is bounded then the Fejer means (or the Abel means) of the Fourier series must be uniformly bounded.

Edit: It appears that the hint was not sufficient. The argument is very simple. First, if $f$ is bounded then the Fejer means are uniformly bounded, as hinted: $\sigma_n=f*K_n$, so $$||\sigma_n||_\infty\le||f||_\infty||K_n||_1=||f||_\infty.$$

On the other hand, it's obvious that if $c_n\ge0$ and $\sum c_n=\infty$ then the Fejer means are not uniformly bounded:
$$\sigma_n(0)=\sum_j(1-|j|/n)^+c_j\ge\frac12\sum_{|j|&lt;n/2}c_j.$$
Now if I additionally assume $c_k = c_{-k}$ (which is the case for my problem, as $c_k$ is the autocovariance sequence of a real-valued time series), then $f$ will be symmetric. In that case Rieman-integrability would imply that the function admits right and left limits at zero, and by symmetry the left and right limits should be the same implying that $f$ can be made continuous at 0 without changing the Fourier coefficients, leading to a contradiction as the Fejer means diverges at zero. So I think that solves my problem in such a case.