$$ \sum_{k=-N}^N |a_k| \le \frac1{\sqrt{2\pi}} \int_0^{2\pi} |P_n(x)| \mathrm dx+\frac{\pi}{\sqrt 3} \int_0^{2\pi} |P'_N(x)|^2 \mathrm dx $$
I need to prove that inequality above:
My approach was to write $|a_k| = |a_k| \cdot k \cdot \frac{1}{k}$ for all $k \neq 0$. Then, using the QM-GM inequality, we obtain:
$$|a_k| \leq \frac{1}{2}\left( |a_k|^2 k^2 + \frac{1}{k^2} \right).$$
Now, we know that the absolute values of the Fourier coefficients of $P'_N(x)$ are $|a_k|^2 k^2$, so we have:
$$ \sum_{k=-N}^{N} |a_k|^2 k^2 = \frac{1}{2\pi} \int_0^{2\pi} |P'_N(x)|^2 \, dx.$$
We can write this since we have Parseval's identity for the finite sum. also note that the $0$'s fourier coefficient of the derivative is $0$ by $\hat{f'(n)}=in\hat f(n)$
and $P_N(x)$ is certainly continuous. So I did manage to work out the second integral (the derivative one) but how can I attack the first one? and get rid of the sum of $\frac{1}{k^2}$
since $a_0 = \frac{1}{2\pi} \int_0^{2\pi} P_N(x) \, dx$we must have $|a_0| \leq \frac{1}{2\pi} \int_0^{2\pi} |P_N(x)| \, dx$
thus$|a_0| \leq \frac{1}{\sqrt{2\pi}} \int_0^{2\pi} |P_N(x)| \, dx$
now for the derivative integral:
We know from Cauchy-Schwarz and Parseval's theorem that:
$\sum_{k \neq 0} |a_k| \leq \sqrt{\sum_{k \neq 0} \frac{1}{k^2}} \sqrt{\sum_{k \neq 0} k^2 |a_k|^2}.$
and$\sum_{k \neq 0} \frac{1}{k^2} = 2 \sum_{k=1}^{\infty} \frac{1}{k^2} = 2 \cdot \frac{\pi^2}{6} = \frac{\pi^2}{3}.$
plugging in everything we obtain:
$\sum_{k \neq 0} |a_k| \leq \frac{\pi}{\sqrt{3}} \sqrt{\frac{1}{2\pi} \int_{0}^{2\pi} |P'_N(x)|^2 dx}.$which can be simplified to
$\sum_{k \neq 0} |a_k| \leq \frac{\pi}{\sqrt{3}} \sqrt{\int_{0}^{2\pi} |P'_N(x)|^2 dx}.$
thus in the end we get the following:
$\sum_{k=-N}^{N} |a_k| = |a_0| + \sum_{k \neq 0} |a_k| \leq \frac{1}{\sqrt{2\pi}} \int_{0}^{2\pi} |P_N(x)| dx + \frac{\pi}{\sqrt{3}} \sqrt{\int_{0}^{2\pi} |P'_N(x)|^2 dx}.$
So close yet still not there yet!
any ideas would be very much appreciated!