$\sum_{n=1}^{\infty}\frac{\cos(nx)}{\sqrt{n}}$ Fourier series

Question

Does there exist any continuous function $f$ such that its Fourier series is $$\sum_{n=1}^{\infty}\frac{\cos(nx)}{\sqrt{n}}?$$

If I suppose that there is, it should follow that $$a_0=\frac{1}{\pi}\int_0^{2\pi}{f(x)dx}=0,$$ $$a_n=\frac{1}{\pi}\int_0^{2\pi}{f(x)\cos(nx)dx}=\frac{1}{\sqrt{n}}\text{ and}$$ $$b_n=\frac{1}{\pi}\int_0^{2\pi}{f(x)\sin(nx)dx}=0,$$ By $a_0$ I have that $f$ is the constant $0$ function or is a function with same area positive and negative. By $a_n$ I know that $f$ isn't the $0$ constant function, but I don't know how to continue (I suspect that there exist) but how do I prove it?

Answer 1

No, not even of a function in $L^2$, since the Fourier coefficients are not square summable: $$\sum_{n\ge 1} (\frac{1}{\sqrt{n}} )^2 = \sum \frac{1}{n} = \infty$$ (see Parseval's theorem).

It is interesting if there exists a function in $L^1$ with these Fourier coefficients.

Added: There does not exists a continuous function on $\mathbb{R}$, with period $2\pi$ with this Fourier series, that is, a continuous function on $[0, 2\pi]$ with equal values at ends. However, there exists a continuous function on $(0, 2\pi)$. Note that the above Fourier series converges uniformly on compact subsets of $(0, 2\pi)$ and the sum is this continuous function. It seems that the function is in fact analytic (see Polylogarithm).