No function $f$ with $\hat{f}=(-1)^k/\sqrt{k}$ exists for all $k\in\mathbb{Z}, k \neq0$

Question

I just started learning about the Fourier series, is this statement true or false?

Looking at $\mathcal {R}(-\pi,\pi)$. No function $f$ with $\hat{f}=(-1)^k/\sqrt{k}$ exists for all $k\in\mathbb{Z}, k \neq0$.

Answer 1

If $$ \sum_{k \in \mathbb Z} a_k e^{i k t} $$ is the Fourier series of a square-integrable Lebesgue measurable function $f$ on $[-\pi,\pi)$, then Parseval's theorem states $$ \frac{1}{2\pi}\int_{-\pi}^{\pi} |f(t)|^2\;dt = \sum_{k \in \mathbb Z} |a_k|^2 $$ Now a Riemann integrable function on $[\pi,\pi]$ is bounded and Lebesgue measurable, so $$ \frac{1}{2\pi}\int_{-\pi}^{\pi} |f(t)|^2\;dt < +\infty $$ But on the other hand, if $a_k = (-1)^k/\sqrt{k}$ for $k=1,2,3\dots$, then $$ \sum_{k \in \mathbb Z} |a_k|^2 \ge \sum_{k=1}^\infty \frac{1}{k} = +\infty . $$

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On the other hand, what if $f$ is not square integrable? Maple suggests $$ \sum_{k=1}^\infty \frac{(-1)^k}{\sqrt{k}} e^{ikt} = \operatorname{Li}_{1/2}(-e^{it}) $$ where $\operatorname{Li}_{1/2}$ is a polylogarithm . Of course the convergence is only conditional. Here is the graph
Real part in blue, imaginary part in red.