$\frac{4}{\pi^{2}}\sum_{k~=1}^{N}\frac{1}{k}\le\|D_{N}\|_{L^{1}({\bf T})}\le3~-\frac{2}{\pi}+\frac{4}{\pi^{2}}\sum_{k~=1}^{N}\frac{1}{k}$

Question

Let $D_{N}(t)=\frac{\sin((2N+1)\pi t)}{\sin(\pi t)}$ be defined on ${\bf T}={\bf R}/{\bf Z}~=[0,1]$ , which is the so-called Dirichlet kernel . Prove that for each integer $N>0~$ we have $$\frac{4}{\pi^{2}}\sum_{k~=1}^{N}\frac{1}{k}\le\|D_{N}\|_{L^{1}({\bf T})}\le3~-\frac{2}{\pi}+\frac{4}{\pi^{2}}\sum_{k~=1}^{N}\frac{1}{k}$$

Here's my attempt for the first inequality but I stuck with the latter inequality :

\begin{align} \|D_{N}\|_{L^{1}(\bf T)}&=\int_{0}^{1}\bigg|\frac{\sin((2N+1)\pi x)}{\sin(\pi x)}\bigg|~dx\\ &=\int_{-\frac{1}{2}}^{\frac{1}{2}}\bigg|\frac{\sin((2N+1)\pi x)}{\sin(\pi x)}\bigg|~dx\\ &=2\int_{0}^{\frac{1}{2}}\bigg|\frac{\sin((2N+1)\pi x)}{\sin(\pi x)}\bigg|~dx\\ &=2\bigg(\sum_{k=0}^{N-1}\int_{\frac{k}{2N+1}}^{\frac{k+1}{2N+1}}~\bigg|\frac{\sin((2N+1)\pi x)}{\sin(\pi x)}\bigg|~dx+\int_{\frac{N}{2N+1}}^{\frac{1}{2}}~\bigg|\frac{\sin((2N+1)\pi x)}{\sin(\pi x)}\bigg|~dx\bigg)\\ &\geq2\sum_{k=0}^{N-1}\int_{\frac{k}{2N+1}}^{\frac{k+1}{2N+1}}~\bigg|\frac{\sin((2N+1)\pi x)}{\sin(\pi x)}\bigg|~dx\\ &\geq2\sum_{k=0}^{N-1}\int_{\frac{k}{2N+1}}^{\frac{k+1}{2N+1}}~\frac{|\sin((2N+1)\pi x)|}{\pi x}~dx\\ &\geq2\sum_{k=0}^{N-1}\int_{\frac{k}{2N+1}}^{\frac{k+1}{2N+1}}~\frac{2N+1}{(k+1)\pi}|\sin((2N+1)\pi x)|~dx\\ &=2\sum_{k=0}^{N-1}\int_{k\pi}^{(k+1)\pi}\frac{1}{(k+1)\pi^{2}}|\sin x|~dx\\ &=2\sum_{k=0}^{N-1}\int_{0}^{\pi}\frac{1}{(k+1)\pi^{2}}|\sin x|~dx\\ &=2\sum_{k=1}^{n}~\frac{1}{k\pi^{2}}\int_{0}^{\pi}~\sin x~dx\\ &=\frac{4}{\pi^{2}}\sum_{k=1}^{N}\frac{1}{k} \end{align}

For the second inequality I tried but fail . Any comments or valuable suggestions I will be grateful .

Answer 1

For the second part , we have this following : \begin{align} \|D\|_{L^{1}({\bf T})}&=2\int_{0}^{\frac{1}{2}}\bigg|\frac{\sin((2N+1)\pi x)}{\sin \pi x}\bigg|~dx\\ &=\int_{0}^{\frac{\pi}{2}}\frac{2}{\pi}\bigg|\frac{\sin((2N+1) x)}{\sin x}\bigg|~dx\\ &=\underbrace{\int_{0}^{\frac{\pi}{2}}\frac{2}{\pi}|\sin((2N+1) x)|\bigg(\frac{1}{\sin x}-\frac{1}{x}\bigg)~dx}_{\color{blue}{(1)}}~+\underbrace{\int_{0}^{\frac{\pi}{2}}\frac{2}{\pi}\frac{|\sin((2N+1)x)|}{x}~dx}_{\color{blue}{(2)}} \end{align}

We now estimate the blue one $\color{blue}{(1)}$ :

Firstly , note that $0\le\big(\frac{1}{\sin x}-\frac{1}{x}\big)\le1-\frac{2}{\pi}~$for $x\in(0,\frac{\pi}{2}]~.$ \begin{align} \int_{0}^{\frac{\pi}{2}}\frac{2}{\pi}|\sin((2N+1)x)|\bigg(\frac{1}{\sin x}-\frac{1}{x}\bigg)~dx&\le\int_{0}^{\frac{\pi}{2}}\frac{2}{\pi}\bigg(\frac{1}{\sin x}-\frac{1}{x}\bigg)~dx\\ &\le\int_{0}^{\frac{\pi}{2}}\frac{2}{\pi}\bigg(1-\frac{2}{\pi}\bigg)~dx\\ &=1-\frac{2}{\pi} \end{align}

For the latter one $\color{blue}{(2)}~:$ \begin{align} \int_{0}^{\frac{\pi}{2}}\frac{2}{\pi}\frac{|\sin ((2N+1)x)|}{x}~dx &=\bigg(\underbrace{\int_{0}^{\frac{\pi}{2N+1}}}_{\color{blue}{(3)}}+\underbrace{\sum_{k=1}^{n-1}\int_{\frac{k\pi}{2N+1}}^{\frac{(k+1)\pi}{2N+1}}}_{\color{blue}{(4)}}+\underbrace{\int_{\frac{N\pi}{2N+1}}^{\frac{\pi}{2}}}_{\color{blue}{(5)}}~\bigg)\frac{2|\sin ((2N+1)x)|}{\pi x}dx \end{align} For $\color{blue}{(3)}~:$

$$\frac{2}{\pi}\int_{0}^{\frac{\pi}{2N+1}}\frac{|\sin ((2N+1)x)|}{x}dx\le\frac{2}{\pi}\int_{0}^{\frac{\pi}{2N+1}}\frac{(2N+1)x}{x}~dx=\frac{2}{\pi}\cdot\pi=2$$ For $\color{blue}{(4)}~:$ \begin{align} \frac{2}{\pi}\sum_{k=1}^{n-1}\int_{\frac{k\pi}{2N+1}}^{\frac{(k+1)\pi}{2N+1}}\frac{|\sin ((2N+1)x)|}{x}~dx&\le\frac{2}{\pi}\sum_{k=1}^{n-1}\int_{\frac{k\pi}{2N+1}}^{\frac{(k+1)\pi}{2N+1}}~\frac{2N+1}{k\pi}~|\sin((2N+1)x)|~dx\\ &=\frac{2}{\pi}\sum_{k=1}^{N-1}\int_{k\pi}^{(k+1)\pi}\frac{1}{k\pi}|\sin x|~dx\\ &=\frac{2}{\pi}\sum_{k=1}^{N-1}\int_{0}^{\pi}\frac{1}{k\pi}|\sin x|~dx\\ &=\frac{2}{\pi}\sum_{k=1}^{N-1}\frac{2}{k\pi}\\ &=\frac{4}{\pi^{2}}\sum_{k=1}^{N-1}\frac{1}{k}\\ \end{align} For $\color{blue}{(5)}~:$ \begin{align} \int_{\frac{N\pi}{2N+1}}^{\frac{\pi}{2}}\frac{2|\sin ((2N+1)x)|}{\pi x}dx&\le\int_{\frac{N\pi}{2N+1}}^{\frac{\pi}{2}}\frac{2}{\pi}\frac{2N+1}{N\pi}|\sin((2N+1)x)|~dx\\ &=\frac{2}{N\pi^{2}}\int_{\frac{N\pi}{2N+1}}^{\frac{\pi}{2}}(2N+1)|\sin((2N+1)x)|~dx\\ &=\frac{2}{N\pi^{2}}\int_{N\pi}^{(N+\frac{1}{2})\pi}~|\sin x|~dx\\ &\le\frac{2}{N\pi^{2}}\int_{N\pi}^{(N+1)\pi}~|\sin x|~dx\\ &=\frac{2}{N\pi^{2}}\int_{0}^{\pi}~|\sin x|~dx\\ &=\frac{4}{N\pi^{2}} \end{align}

Combine $\color{blue}{(3)}$ , $\color{blue}{(4)}$ and $\color{blue}{(5)}$ to get \begin{align} \int_{0}^{\frac{\pi}{2}}\frac{2}{\pi}\frac{|\sin ((2N+1)x)|}{x}~dx&\le2+\frac{4}{\pi^{2}}\sum_{k=1}^{N-1}\frac{1}{k}+\frac{4}{N\pi^{2}}\\ &=2+\frac{4}{\pi^{2}}\sum_{k=1}^{N}\frac{1}{k}\color{blue}{~~-(6)} \end{align} Finally, we associate the results $\color{blue}{(1)}$ and $\color{blue}{(6)}$ to obtain our desired consequence , \begin{align} \|D\|_{L^{1}({\bf T})}&=2\int_{0}^{\frac{1}{2}}\bigg|\frac{\sin((2N+1)\pi x)}{\sin \pi x}\bigg|~dx\\ &=\int_{0}^{\frac{1}{2}}\frac{2}{\pi}\bigg|\frac{\sin((2N+1) x)}{\sin x}\bigg|~dx\\ &\le\int_{0}^{\frac{\pi}{2}}\frac{2}{\pi}|\sin((2N+1)x)|\bigg(\frac{1}{\sin x}-\frac{1}{x}\bigg)~dx+\int_{0}^{\frac{\pi}{2}}\frac{2}{\pi}\frac{|\sin ((2N+1)x)|}{x}~dx\\ &\le(1-\frac{2}{\pi})+\bigg(2+\frac{4}{\pi^{2}}\sum_{k=1}^{N}\frac{1}{k}\bigg)\\ &=3-\frac{2}{\pi}+\frac{4}{\pi^{2}}\sum_{k=1}^{N}\frac{1}{k} \end{align}