# Estimating the Dirichlet Kernel

• Mar 6th 2011, 03:01 PM
mgarson
Estimating the Dirichlet Kernel
Hi!

I need help with the following problem:

Let D_n (theta) = sum(k=-N to N) e^ik(theta)= sin ((N+1/2)theta)/sin(theta/2) and define L_n = 1/2Pi integral (from - Pi to Pi) |D_n (theta)| d(theta)
prove that L_N is greater than or equal to c log (N) for some constant c>0

Hint: show that |D_n(theta)| is greater than or equal to c sin ((n+1/2)theta)/|theta| change variables and prove that
L_n greater than or equal to c intergral (from Pi to N Pi) |sin(theta)|/|theta| d(theta) +O(1). Write the integral as sum (k=1 to N-1) integral (k pi to (k+1) Pi. To conclude use the fact that sum (k=1 to n) 1/k is greater than or equal to c log n.

(It can also be found on Fourier analysis: an introduction - Google Books)

The first part is quite easy (use |theta| is greater than or equal to |sin(theta)|)
• Mar 7th 2011, 01:13 AM
Opalg
This is a longish proof, but the hint takes you through it in fairly manageable steps. Which step (or steps) is giving you difficulties?

One thing to look out for is that the letter c is being used for a constant (that is, it must be independent of N and theta), but it's not necessarily the same constant at each step of the proof.
• Mar 7th 2011, 02:27 AM
mgarson
Hi!

I only have the "last" step left, i.e. I've managed to write the integral as the sum (k=1 to N-1) integral (k pi to (k+1) Pi) |sin(theta)|/|theta| d(theta). How do I continue? I guess I somehow need to show that my integrals are bigger than or equal to 1/k... But how?
• Mar 7th 2011, 02:55 AM
Opalg
Quote:

Originally Posted by mgarson
I only have the "last" step left, i.e. I've managed to write the integral as $\sum_{k=1}^{N-1} \int_{k \pi}^{(k+1)\pi} \frac{|\sin\theta|}{|\theta|}\, d\theta$. How do I continue? I guess I somehow need to show that my integrals are bigger than or equal to 1/k... But how?

Use the fact that $|\sin\theta|\geqslant \frac12$ in the interval from $(k+\frac16)\pi$ to $(k+\frac56)\pi$. Also, in that interval $\frac1\theta \geqslant \frac1{(k+1)\pi}.$ This tells you that

$\displaystyle\int_{k \pi}^{(k+1)\pi} \frac{|\sin\theta|}{|\theta|}\, d\theta \geqslant \int_{(k+\frac16) \pi}^{(k+\frac56)\pi} \frac1{2(k+1)\pi}\, d\theta = \frac{4/3}{k+1}.$