Trigonometric polynomial derivative upper bound

Question

Let $P$ be a trig poly of degree $N$ on the torus $\mathbb T$. Prove that $$\Vert P' \Vert_\infty \lesssim N\Vert P \Vert_\infty.$$ I'm not sure how to approach this problem, though I feel like some Fourier approach is probably in the right direction.

Answer 1

In fact, we can show that $ \|P_N'\|\leq 4\pi N\|P_N\|_{\infty} $. To prove this, we firstly claim that $ P_N'(x) $ can be represented by $$ \frac{P_N'(x)}{2\pi iN}=((e^{-2\pi iN(\cdot)}P_N)*F_{N-1})(x)e^{2\pi iNx}-((e^{2\pi iN(\cdot)}P_N)*F_{N-1})(x)e^{-2\pi iNx}. $$ If this claim is true, the improvement for the results can be easily got that $$ \left\|\frac{P'}{2\pi iN}\right\|_{\infty}\leq \|e^{-2\pi iN(\cdot)}P_N\|_{\infty}\|F_{N-1}\|_{1}+\|e^{2\pi iN(\cdot)}P_N\|_{\infty}\|F_{N-1}\|_{1}\leq 2\|P_N\|_{\infty}, $$ which completes the proof. Now we will show the claim. By the linearity, we only need to verify the result for $ P_N(x)=ae^{2\pi ikx} $. Indeed, $$ \begin{aligned} \int_0^1e^{-2\pi iN(x-y)}\cdot ae^{2\pi ik(x-y)}F_{N-1}(y)e^{2\pi iNx}dy&=\left\{\begin{aligned} &\frac{ka }{N}e^{2\pi ikx}&\text{ if }&k>0,\\ &0&\text{ if }&k\leq 0, \end{aligned}\right.\\ \int_0^1e^{2\pi iN(x-y)}\cdot ae^{2\pi ik(x-y)}F_{N-1}(y)e^{-2\pi iNx}dy&=\left\{\begin{aligned} &\frac{ka}{N}e^{2\pi ikx}&\text{ if }&k<0,\\ &0&\text{ if }&k\geq 0. \end{aligned}\right. \end{aligned} $$