I showed there exists a constant $c$ such that $\|D_N\|_1 \geq c \log N$ and $c$ is independent to $N$
using the fact that $$\| D_N\|_1= \frac 1 \pi \int_{[0,\pi]} \left|\frac{\sin(2N+1)y}{\sin y}\right|\,dt \geq \frac{1}{\pi}\int_{[0,\pi/2]} \left| \frac{\sin(2N+1)y}{\sin y} \right| \, dt$$
and $$\frac 1 {| y |} \leq \frac{1}{\sin(y)} \leq \frac{\pi}{2|y|} \text{ for } y \in [0, \frac{\pi}{2}].$$
Now I want to show that there is a upper bound (i.e there exists a constant $c'$ such that $\|D_N\|_1 \leq c' \log N$ for $N\geq2$)
but this time I can't deduce the interval and the function diverges to infinity near $\pi$, so i have no clue how to start. Am I sppose to divide $[0, 2\pi]$ into three subintervals such as $[0, \delta], [\delta, 2\pi-\delta],$ and $[2\pi-\delta, 2\pi]$ and show on each interval $L^1$ norm of Dirichlet kernel converges to $0$ or multiple of $\log N$ as $\delta \rightarrow 0$? I saw this trick a lot in the other examples.
Best Answer
$D_{n}(x)$ has the form $$D_{n}(x)=\dfrac{1}{2\pi}\dfrac{\sin(n+\frac{1}{2})x}{\sin\frac{1}{2}(x)},$$ so that $$D_{n}(2x)=\dfrac{1}{2\pi}\dfrac{\sin(2n+1)x}{\sin x}.$$
Since $\sin n\alpha\leq n\sin\alpha$, we know that $\sin(2n+1)x\leq (2n+1)\sin x$ and thus \begin{align*} (1)\ \ |D_{n}(2x)|=\dfrac{1}{2\pi}\dfrac{|\sin(2n+1)x|}{|\sin x|}&\leq \dfrac{1}{2\pi}\dfrac{(2n+1)|\sin x|}{|\sin x|}\\ &=\dfrac{2n+1}{2\pi}\leq 2n+1<4n\ \text{for all}\ n\geq 1. \end{align*}
On the other hand, note that $$|\sin\frac{x}{2}|>\dfrac{|x|}{\pi}\geq\dfrac{|x|}{2\pi}\ \text{for}\ 0<|x|<\pi,$$ and thus $$(2)\ \ |D_{n}(x)|=\dfrac{1}{2\pi}\dfrac{|\sin(n+\frac{1}{2})x|}{|\sin\frac{1}{2}x|}\leq\dfrac{1}{2\pi}\dfrac{1}{|\sin \frac{1}{2}x|}\leq\dfrac{1}{|x|}.$$
Now, let's compute: \begin{align*} \|D_{n}\|_{1}=\int_{-\pi}^{\pi}|D_{n}(x)|dx&=2\int_{0}^{\pi}|D_{n}(x)|dx\\ &=2\int_{0}^{\frac{\pi}{n}}|D_{n}(x)|dx+2\int_{\frac{\pi}{n}}^{\pi}|D_{n}(x)|dx,\\ &\text{replacing}\ y:=\dfrac{x}{2}\ \text{in the first integral, then}\\ &=2\int_{0}^{\frac{\pi}{2n}}|D_{n}(2y)|\cdot\dfrac{1}{2}dy+2\int_{\frac{\pi}{n}}^{\pi}|D_{n}(x)|dx. \end{align*}
Now we apply the inequality $(1)$ to the first term and inequality $(2)$ to the second term, and we have \begin{align*} \|D_{n}\|_{1}\leq \int_{0}^{\frac{\pi}{2n}}4ndy+2\int_{\frac{\pi}{n}}^{n}\dfrac{1}{|x|}dx&=\dfrac{4n\pi}{2n}+2\int_{\frac{\pi}{n}}^{\pi}\dfrac{1}{x}dx\\ &=2\pi+2(\log(\pi)-\log(\pi/n))\\ &=2(\log n+\pi)\\ &<12\log n\ \text{for all}\ n\geq 2, \end{align*} where the last inequality was obtained by $\pi<5\log(n)$ for all $n\geq 2$.