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.
