Xn l=1 sin lx

Undergraduate Analysis Handout 7 December 9, 2011 For each x ∈ R, N, n = 1, 2, . . . , define

Q(x, N, n) = 2 sin Nx Xn

l=1

sin lx l . Since 1

l & 0, and for each x 6= 2kπ, k ∈ Z, n = 1, 2, . . . ,

¯¯

¯¯

¯ Xn

l=1

sin lx

¯¯

¯¯

¯=

¯¯

¯¯cos 1

2x − cos(n + 1 2)x

¯¯

¯¯

¯¯

¯¯2 sin1 2x

¯¯

¯¯

≤ 1

¯¯

¯¯sin1 2x

¯¯

¯¯

, i.e. partial sums are bounded for each x,

we have X∞

l=1

sin lx

l converges for each x by the Dirichlet test. Hence, there exists a constant C such that

|Q(x, N, n)| ≤ C for all x, N, n.

Also, note that

Q(x, N, n) = 2 sin Nx Xn

l=1

sin lx l

= cos(N − n)x

n +cos(N − n + 1)x

n − 1 + · · · + cos(N − 1)x 1

− cos(N + 1)x

1 − · · · − cos(N + n − 1)x

n − 1 + cos(N + n)x n i.e. Q(x, N, n) contains terms ranging from cos(N − n)x to cos(N + n)x.

If we choose sequences N_k, n_k% ∞ such that

N_k+ n_k < N_k+1− n_k+1,

then the terms in Q(x, Nk, nk) and the terms in Q(x, Nk+1, nk+1) do not overlap since Q(x, Nk, nk) only contains those terms ranging from cos(N_k− n_k)x to cos(N_k+ n_k)x while Q(x, N_k+1, n_k+1) only contains those terms ranging from cos(N_k+1− n_k+1)x to cos(N_k+1+ n_k+1)x.

For example, if we choose

α_k = 1 k², N_k

2 = n_k= 2^k3, for each k = 1, 2, . . . , then N_k+ n_k= 3 2^k3 < 2^k3+3k2+3K+1 = N_k+1− n_k+1. Since

| X∞ k=1

α_kQ(x, N_k, n_k)| ≤ X∞ k=1

C k² < ∞ X∞

k=1

αkQ(x, Nk, nk) converges uniformly to a continuous function f. Let Sf(x) = X∞ ν=0

aνcos νx be the Fourier series of f, and let S_f^N(x) = a^N_ν cos νx denote the N^th partial sum of the Fourier series for f.

Then it is easy to see that S_f(x) = X∞ k=1

α_kQ(x, N_k, n_k) and since

|S_f^Nk+nk(0) − S_f^Nk(0)| =

nk

X

l=1

1

l ≥ log n_k → ∞

which implies that the Cauchy criterion cannot not hold for the series S_fⁿ(0), i.e. the Fourier series S_f diverges at x = 0.