Okay, so our Fourier analysis book gives us this example of a continuous function with a Fourier series that diverges at a point, by breaking the natural symmetry of the Fourier series:

Define f_N(x) = (sum 1<=|n|<=N) e^(inx)/n and
f~_N(x) = (sum -N<=n<=-1) e^(inx)/n
So that the former is symmetric in n but the latter isn't. These are just the partial sums of the sawtooth function, defined to be 0 (discontinuously) at 0.

Now define P_N(x) = e^(2Nix) f_N(x) and P~_N(x) = e^(2Nix) f~_N(x). Then the former will be symmetric with respect to 2N, but the latter is not. A simple calculation gives us that the partial sums S_M(P_N) are
P_N if M>=3N
P~_N if M=2n
0 if MSo that S_2N breaks the symmetry of P_N, but not for (many) other M.

Now find a convergent series of positive terms a_k and a sequence of integers N_k such that
1) N_(k+1) > 3N_k
2) a_k log(N_k) --> infty as k --> infty.

Now let f(x) = (sum k>=1) a_k P_(N_k)(x). The idea is that f(x) is a continuous function, but that
|S_(2N_m)(f)(0)| >= ca_k log(N_m) + O(1) --> infty as m --> infty

I understand everything but the statement that f(x) is continuous at 0. We constructed f(x) from the (slightly modified) partial sums of the Fourier series of a function which is *discontinuous* at the origin. Maybe I'm just stupid, but I haven't been able to figure out a good proof of why f(x) is continuous at 0. Help?