Let $f$ be a real-valued trigonometric polynomial on $\mathbb T$ with $\hat f (0) = 0.$ Let $\widetilde f$ be the Hilbert transform or the conjugate transform of $f.$ Then $\widetilde f$ is also real-valued and hence $(f + i \widetilde f)$ only contains all the positive frequencies. Therefore,
$$\int_{-\pi}^{\pi} (f(t) + i \widetilde f(t))^{2k}\ dt = 0$$ for any $k \in \mathbb N.$
This has been claimed in the book Classical Fourier Anslysis by Grafakos. Could anyone please shed some light on how does the integral being zero follow from the fact that the integrand has only positive Fourier coefficients? Any help would be greatly appreciated.
Thanks for your valuable time.
EDIT $:$ The integrand can be easily seen to be $2^{2k} \left (\sum\limits_{n=1}^{N} \hat f (n) e^{int} \right )^{2k}.$ I think it is zero because $\int_{-\pi}^{\pi} e^{im t}\ dt = 0$ for any $m \in \mathbb N.$ But then it should be true for any odd powers as well.
Consider a trig polynomial $\phi(t) = \sum_{m} a_m e^{2\pi imt}$ with only positive frequencies. Note this does not mean that the coefficients $a_m$ are positive, but rather that $a_m = 0$ if $m\le 0$ (in other words, "positive" refers to the index $m$). Since $\phi(t)$ is a trig polynomial with only positive frequencies, $a_m = 0$ if $m \le 0$ or if $m > M$ for some large enough positive $M$.
As you noted, $\phi(t)^n$ can be expanded out for any $n\in\mathbb Z^+$, and we get $$ \int_{\mathbb T}\phi(t)^n\,dt = \sum_{m_1,\dots,m_n=1}^Ma_{m_1}\dots a_{m_n}\int_{\mathbb T} e^{i(m_1+\dots+m_n)t}\,dt = 0, $$ the last equality holding because $m_1+\dots+m_n\ge n >0$ is a nonzero integer.
I think there may be a separate reason for choosing $2k$ (rather than arbitrary $n$) in the proof, possibly owing to the organization of the proof. The fact itself does not rely on $2k$ being even.