Relation between the integral of the imaginary part and the real part of trigonometric polynomial I am trying to solve the following problem: 
Given a trigonometric polynomial $F(t) = \sum_{k=1}^{n} c_k e^{2\pi i k t}$ with $F(0)=0$.
i) Prove that $\int_0^1 (Im(F(t))^2 dt \leq \int_0^1 (Re(F(t))^2 dt$
My idea is the following:
$$\int_0^1\sum_{k=1}^n \sum_{j=1}^n c_k c_j(cos(2\pi k t) cos(2 \pi j t) - sin(2 \pi k t) sin(2 \pi j t)) dt = \sum_{k=1}^n \sum_{j=1}^n c_k c_j \int_0^1 cos(2 \pi (k+j)t) dt = 0$$
I am not sure if this is correct because then everything would be equal and we don't use the assumption $F(0)=0$. I especially have problems with the second part, since I can't use any 
ii) Prove that for every integer $m \geq 1$ there exists $K_m>0$ independent of $F$ such that 
$$\int_0^1 (Im(F(t))^{2m} dt \leq K_m \int_0^1 (Re(F(t))^{2m} dt$$
I would be grateful for anybody who can give me a hint, since I couldn't find anything in my textbooks.
 A: The first point is correct (as noted in the OP we actually have equality if the $c_0$ term is missing and that is $\hat F(0)=0$ which is the condition we actually need not $F(0)=0$ which is irrelevant)
let $G(z)=\sum_{k=1}^n c_kz^k=u(z)+iv(z), z=re^{i\theta}$, while $u(e^{2\pi it})=\Re F(t), v(e^{2\pi it})=\Im f(e^{it})$ so we need to prove $\int_0^1 v^{2m}(e^{2\pi it})dt \le C_m \int_0^1 u^{2m}(e^{2\pi it})dt$
We assume $u,v$ not identically zero (equivalently not all $c_k$ are zero) and we use the fact that $G(0)=0$ hence $G^{2m}(0)=0$ hence $\Re G^{2m}(0)=0$ hence $\int_0^{2\pi}\Re G^{2m}(e^{i\theta})d\theta=0$ by the harmonic mean property
Changing variables to $\theta=2\pi t$ we get $\int_0^1 \Re (u+iv)^{2m}(e^{2\pi it})dt=0$ and expanding the binomial we get:
$\int_0^1 v^{2m}(e^{2\pi it})dt=\sum_{k=0}^{m-1}B_{k,m}\int_0^1 u^{2m-2k}(e^{2\pi it})v^{2k}(e^{2\pi it})dt$ where $B_{k,m}$ are $\pm$ binomial coefficients so are independent on $u,v$ and depend only on $m$
By the Holder inequality $\int_0^1 u^{2m-2k}(e^{2\pi it})v^{2k}(e^{2\pi it})dt \le (\int_0^1 u^{2m}(e^{2\pi it})dt)^{\frac{2m-2k}{2m}}(\int_0^1 v^{2m}(e^{2\pi it})dt)^{\frac{2k}{2m}}$ 
($\int|ab| \le (\int|a|^p)^{1/p}(\int|b|^q)^{1/q}, 1/p+1/q=1, a=u^{2m-2k}, b=v^{2k}, p=\frac{2m}{2m-2k}, q=\frac{2m}{2k}$)
So letting $(\int_0^1 v^{2m}(e^{2\pi it})dt)^{1/2m}=Y, (\int_0^1 u^{2m}(e^{2\pi it})dt)^{1/2m}=X$ we get:
$Y^{2m} \le \sum_{k=0}^{m-1}|B_{k,m}|X^{2m-2k}Y^{2k}$ and since $XY \ne 0$ we get that $Y/X$ is less or equal than $R_m$, the positive root of the equation $R^{2m}-\sum_{k=0}^{m-1}|B_{k,m}|R^{2k}=0, |B_{k,m}|>0$ depending only on $m$,  so we finally get:
$\int_0^1 v^{2m}(e^{2\pi it})dt=Y^{2m} \le R_m^{2m}X^{2m}=R_m^{2m}\int_0^1 u^{2m}(e^{2\pi it})dt$ and we are done!
Note that this inequality (M Riesz theorem) holds in full generality for any power $1<p<\infty$ and any $F$ Fourier series that is in $H^p$ (in other words it is a Taylor series in $e^{it}$ in the usual $L^p$) and reflects the fact that the harmonic conjugation operator is well defined and continuous in $h^p, 1<p<\infty$ and actually one proof goes as above (showing it first for even integers pretty much as above after reduction to $\hat F(0)=0$ as we haven't used that $F$ is a trigonometric polynomial and not a Taylor series anywhere and then interpolating by some standard functional analysis stuff to show it for any $2 \le p < \infty$ with duality getting the $1<p<2$ case) though the standard textbook proof goes the other way (starts with $1<p<2$ and dualizes that being direct (no interpolation) but more complicated)
