Let be $F_N$ the sequence of the arithmetic means of Dirichlet kernels $D_N (x)$ defined by $$ F_N := \frac{1}{N+1} (D_0 (x) +D_1 (x)+..+D_N(x)) $$
Where the Dirichlet kernel is defined by $$D_N (x)= \sum_{n=-N}^N e^{inx} $$
I have no ideas of ways to prove that
$$ F_N \geq 0 $$
and that even for $x \not\in 2\pi \mathbb{Z} $ it holds that: $$ \lim_{N \rightarrow \infty} F_N (x)=0 $$
I appreciate any help of you guys.
Available in many places, the idea is to sum the $D_N$ using a geometric series fomula \begin{align} D_N(x) &= e^{-iNx}\sum_{k=0}^{2N} e^{ikx} \\ & =e^{-iNx}\frac{e^{(2N+1)x}-1}{e^{ix}-1} \\ &= \frac{e^{(N+1)x}-e^{-iNx}}{e^{ix}-1}\\ & = \frac{e^{(N+1/2)x}-e^{-i(N+1/2)x}}{e^{ix/2}-e^{-ix/2}}\\ & = \frac{\sin((N+1/2)x)}{\sin(x/2)}\end{align} Alternatively, there's a nice telescoping sum argument here. Then $ \sum_{n=0}^N \sin((n+1/2)x) = \Im \sum_{k=0}^Ne^{i(n+1/2)x}$, and $$ \sum_{k=0}^Ne^{i(n+1/2)x} = e^{ix/2}\sum_{k=0}^Ne^{in x} = \frac{e^{i(N+1)x}-1}{e^{ix/2}-e^{-ix/2}} =\frac{e^{i(N+1)x}-1}{2i \sin(x/2)} $$ and therefore
$$F_N(x) = \frac1{(N+1)\sin(x/2)^2}\times (-1)\times \Re (\frac{e^{i(N+1)x}-1}2) =\frac{1-\cos((N+1)x)}{2(N+1)\sin(x/2)^2}$$ this already proves $F_N(x)\ge 0$, to put it in the "standard form" just use highschool trig $ \cos(2\theta) = 1 - 2\sin(\theta)^2$ to get $$ F_N(x) = \frac1{N+1} \left(\frac{\sin((N+1)x/2)}{\sin(x/2)}\right)^2\ge0.$$
For the convergence to zero, at every $x$ that is not a zero of $\sin(x/2)$, we have
$$ F_N(x) = \underbrace{\frac{1}{\sin(x/2)^2}}_{\text{constant in $N$}}\times \underbrace{\sin((N+1)x/2)^2}_{\le 1} \times \frac1{N+1}\to 0$$
