Proof of $L_N\geq c\log N$

Question

Let $D_N$ denote the Dirichlet kernel
$$D_N(\theta)=\sum_{n=-N}^Ne^{ik\theta}=\frac{\sin((N+\frac 1 2) \theta)}{\sin (\frac {\theta} 2)}$$ And define $$L_N=\frac 1 {2\pi}\int_{-\pi}^{\pi}|D_N(\theta)|d\theta$$ Prove that $L_N\geq c\log N$

Could you help me understand these 4 steps that I remarked below, they literally do not ring any bell to me..

Since $\frac x {\sin x}\geq 1 $ for $x\in [-\frac {\pi}2,\frac {\pi}2]$, it follows that
$$|D_N(\theta)|\geq 2\frac {|{\sin((N+\frac 1 2) \theta})|}{\theta}$$ Then, $$\int_{-\pi}^{\pi}|D_N(\theta)|d\theta \\ \geq 4\int_0^{\pi}\frac {|{\sin((N+\frac 1 2) \theta})|}{\theta} d\theta $$ $$=4\int_0^{(N+\frac 1 2)\pi}\frac {|{\sin \theta}|}{\theta} d\theta \\ \geq 4\int_0^{N\pi}\frac {|{\sin \theta}|}{\theta} d\theta$$ $$=4\sum_{k=0}^{N-1}\int_{k\pi}^{(k+1)\pi}\frac {|{\sin \theta}|}{\theta} d\theta $$ $$\geq 4\sum_{k=0}^{N-1} \frac 1 {(k+1)\pi}\int_{k\pi}^{(k+1)\pi}{|{\sin \theta}|} d\theta \tag 1$$ $$=\frac 8 {\pi}\sum_{k=0}^{N-1} \frac 1 {(k+1)} \tag 2$$ $$=\frac 8 {\pi}\log(N+1) \tag 3\\ \geq \frac 8 {\pi}\log N \\ \therefore L_N\geq c\log N$$

If none of these 5 steps make sense for you, fourier analysis might be too complicated as of now — mathworker21, Mar 15 '18 at 23:25
One was a change of variable that I realised later.. But others still need to be clarified @mathworker21 — Leyla Alkan, Mar 15 '18 at 23:56
(1) is just saying $\theta \le (k+1)\pi$ in the range of integration $[k\pi, (k+1)\pi]$. — mathworker21, Mar 16 '18 at 00:04
Since answers are referring to your numberings, it might be more useful to keep them fixed. — Atbey, Mar 16 '18 at 00:09

score 0 · Accepted Answer · answered Mar 16 '18 at 00:05

0

(1) $\frac{|\sin\theta|}{\theta} \geq \frac{|\sin\theta|}{(k+1)\pi}$ for $\theta\in [k\pi,\ (k+1)\pi]$.

(2) Calculate the integral of sine, which is either always positive or negative depending on $k$. (which is 2).

(3) I am not sure about equality, but you can use $\sum_1^{N}\frac{1}{k} \geq \int_1^{N+1} 1/k = \log(N+1)$

answered Mar 16 '18 at 00:05

Atbey

1,497

You should be sure that equality does not hold (for all N). One is irrational, one is rational. – mathworker21 Mar 16 '18 at 00:40

Proof of $L_N\geq c\log N$

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