# Weak Bernstein inequality for trigonometric polynomials

Could somebody help me to complete or fix my solution to the problem 6.7 from the book Number Theory, Fourier Analysis and Geometric Discrepancy by Giancarlo Travglini.

Problem: Let $$P_{N}=\sum_{k=0}^{N} a_{k} e^{2 \pi i k\cdot x}$$ be a trigonometric polynomial on $$\mathbb{T}$$ or [0,1] if you prefer. Prove that $$\left\|P_{N}^{\prime}\right\|_{L^{P}(\mathbb{T})} \leq 2 \pi N\left\|P_{N}\right\|_{L^{p}(\mathbb{T})}$$ where $$1 \leq p \leq \infty$$. You can use the hint proposed by the author, that is, the inequality $$\|f * g\|_{L^{p}\left(\mathbb{T}^{d}\right)} \leq\|f\|_{L^{p}\left(\mathbb{T}^{d}\right)}\|g\|_{L^{1}\left(\mathbb{T}^{d}\right)} \quad if \quad f\in L^{p}\left(\mathbb{T}^{d}\right), \quad 1 \leq p \leq \infty$$ My "almost" solution:

Firstly we observe that $$P'_{N}(x)=2\pi i \sum_{k=0}^{N} k a_{k}e^{2\pi i k\cdot x}$$

Choosing the function $$Q(x):=2\pi i \sum_{k=0}^{N} k e^{2\pi i k x}$$ we have \begin{aligned} P_{N}*Q(t)&=\int_{\mathbb{T}}P_{N}(t-y)Q(y)dy\\ &=\int_{\mathbb{T}}\sum_{k_{1}=0}^{N}a_{k_{1}}e^{2\pi i k_{1}\cdot (t-y)}2\pi i \sum_{k_{2}=0}^{N} k_{2} e^{2\pi i k_{2} \cdot y}dy\\ &=2\pi i\sum_{k_{1}=0}^{N}a_{k_{1}}e^{2\pi i k_{1}\cdot t}\sum_{k_{2}=0}^{N}k_{2}\int_{\mathbb{T}}e^{2\pi i (k_{2}-k_{1}) \cdot y}dy\\ \end{aligned} But , if $$k_{1}\neq k_{2}$$ we have $$\int_{\mathbb{T}}e^{2\pi i (k_{2}-k_{1}) \cdot y}dy=\frac{e^{2\pi i (k_{2}-k_{1}) \cdot y}}{2\pi i (k_{2}-k_{1})}\Biggr|_{0}^{1}=0$$ Then, \begin{aligned} P_{N}*Q(t)&=2\pi i\sum_{k_{1}=0}^{N}a_{k_{1}}e^{2\pi i k_{1}\cdot t}\sum_{k_{2}=k{1}}\int_{\mathbb{T}}1dy\\ &=2\pi i\sum_{k_{1}=0}^{N}a_{k_{1}}e^{2\pi i k_{1}\cdot t}k_{1}\\ &=P'_{N}(t) \end{aligned} Now, using proposition 6.10 \begin{aligned} \|P'_{N}\|_{L^{p}(\mathbb{T})}&=\|P_{N}*Q\|_{L^{p}(\mathbb{T})}\\ &\leq\|P_{N}\|_{L^{p}(\mathbb{T})}\|Q\|_{L^{1}(\mathbb{T})}\\ &=\|P_{N}\|_{L^{p}(\mathbb{T})}\int_{\mathbb{T}}\left|2\pi i \sum_{k=0}^{N} k e^{2\pi i k x}\right|dx\\ &=2\pi \|P_{N}\|_{L^{p}(\mathbb{T})}\int_{\mathbb{T}}\left|\sum_{k=0}^{N} k e^{2\pi i k x}\right|dx\\ \end{aligned} If we show that $$\int_{\mathbb{T}}\left|\sum_{k=0}^{N} k e^{2\pi i k x}\right|dx\leq N$$ we did. We know that for $$a\neq1$$, $$\sum_{k=0}^{N} k a^{k}=\frac{Na^{N+2}-(N+1)a^{N+1}+a}{(a-1)^{2}}$$ Taking $$a=e^{2\pi i x}$$ for $$a\neq1$$, we have $$\left|\sum_{k=0}^{N} k a^{k}\right|=\frac{|Na^{N+1}-(N+1)a^{N}+1|}{4\cdot(\text{sen}(\pi x))^{2}}=\frac{|Ne^{2\pi i (N+1)x}-(N+1)e^{2\pi i N x}+1|}{4\cdot(\text{sen}(\pi x))^{2}}$$ By the triangular inequality, we obtain $$\int_{\mathbb{T}}\left|\sum_{k=0}^{N} k e^{2\pi i k x}\right|dx\leq \frac{N(N+1)}{2}$$ hence $$\|P'_{N}\|_{L^{p}(\mathbb{T})}\leq \pi N(N+1) \|P_{N}\|_{L^{p}(\mathbb{T})}$$ It is necessary to complete this step or redefine the function $$Q$$ to sharp the bound $$\frac{N(N+1)}{2}$$ for $$N$$, but apparently this bound can not be improved more than $$O\left(N\log(N)\right)$$ using this polynomial $$Q$$. One inequality more sharp than this one, avoiding the factor $$2\pi$$, is the Bernstein inequality. A elegant proof of latter using Riez' interpolation polynomilal is found in Approximation of Functions of Several Variables and Imbedding Theorems by S. M. Nikol’skii.

I used the new function $$Q(x)=[K_{N}(x)(1+e^{-2\pi i (N+1) x}+e^{2\pi i (N+1) x})]'$$ and got the wished bound. This function $$K_{N}(x)$$ is defined in the same book and is called Fejer Kernel, which seem very useful.