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I want to bound the $ L^{1} $ norm of the Dirichlet kernel from below: It is well known that $$\frac{1}{2\pi}\int _0^{2\pi} \left| \frac{\sin((N+1/2)x)}{\sin(x/2)}\right| dx \geq C \log N$$ for some $C>0$ . But some steps in the calculation are not clear:

1) $\displaystyle\frac{1}{2\pi}\int _0^{2\pi} \left|\frac{sin((N+1/2)x)}{\sin(x/2)}\right| dx = \frac{1}{\pi}\int _0^{\pi} \left| \frac{sin((N+1/2)x)}{sin(x/2)}\right| dx$

2) $\displaystyle\frac{1}{\pi}\int _0^{\pi} \frac{\vert \sin((N+1/2)x)\vert}{\vert \sin(x/2)\vert} dx \geq \frac{1}{\pi}\int _0^{\pi} \frac{\vert \sin((N+1/2)x)\vert}{x} dx$

are these steps right? and why are they true...? sorry, but I am not good in such stuff

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  • $\begingroup$ For 2) you mean $\displaystyle\frac{1}{\pi}\int _0^{\pi} \frac{\vert \sin((N+1/2)x)\vert}{\vert \sin(x/2)\vert} dx \geq \frac{1}{\pi}\int _0^{\pi} \frac{\vert \sin((N+1/2)x)\vert}{x/2} dx$ $\endgroup$ – reuns Dec 1 '16 at 14:53
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  1. This is due to the symmetry of the sine function. By periodicity the integral can be done on $[-\pi,\pi]$, and the integrand is an even function.
  2. This is because $\sin x\le x$, $0\le x\le\pi$.
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  • $\begingroup$ "by periodicty" is unclear : we are considering the $L^1([-\pi,\pi])$ norm here $\endgroup$ – reuns Dec 1 '16 at 14:51
  • $\begingroup$ If you like it better, because of the symmetries of the sine function, $\int_0^{2\pi}=2\int_0^\pi$. $\endgroup$ – Julián Aguirre Dec 1 '16 at 14:57
  • $\begingroup$ Yes I meant $f(x)$ is $2\pi$ periodic so we look at its $L^1([0,2\pi])$ norm, and by even-ness $\|f\|_{L^1([0,2\pi]) }= 2\|f\|_{L^1([0,\pi]) }$ $\endgroup$ – reuns Dec 1 '16 at 15:34

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