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. A more careful estimate gives L_n =(4/Pi^2) log N + O(1).
This is problem 2a on this page.
Complete the proof.