fourier analysis - Does Weyl's Inequality prove equidistribution?

This response is in answer to David's further question about whether it is possible to bound the rate at which S_N/N tends to zero, as he was wanting to use Weyl's inequality to do.
This is not possible, even in the case d=2 and f(n)=θn². (for d=1 it is not hard to show that S_N is bounded so $S_N/N=O(N^{-1})$).
Set

$$S_N(theta)=sum_{n=1}^Ne^{2pi itheta n^2}$$
in the following. Given any function h: ℕ → ℝ₊ with liminf_nh(n) = 0, I show that there are irrational θ with
$$begin{array}{}displaystylesup_Nvert S_N(theta)/(h(N)N)vert=infty.&&(*)end{array}$$

[Note: The following is a much simpler argument than the original version]. I'll use the Baire category theorem to find counterexamples

For any countable collection A_n of open dense subsets of ℝ, the intersection A = ∩_nA_n is dense in ℝ.

In particular, any such A is nonempty. We can say more than this; if S is a countable subset of the reals then $Asetminus S=left(bigcap_nA_nright)capleft(bigcap_{sin S}mathbb{R}setminus{s}right)$ is an intersection of dense open sets, so is dense. In particular, A will contain a dense set of irrational values.

To construct counterexamples then, it is only necessary to show that the set of all θ at which the sequence diverges to infinity is an intersection of countably many open sets, and show that it contains a dense set of rational numbers. The Baire category theorem implies that it will also diverge at a dense set of irrationals.

In fact, for any sequence x_n(θ) depending continuously on a real parameter θ, the set of values of θ for which it diverges to infinity is an intersection of countably many open sets

$${thetacolonsup_nvert x_n(theta)vert=infty}=bigcap_nbigcup_m{thetacolonvert x_m(theta)vert>n\}.$$

So, we only need to find a dense set of rational numbers at which (*) holds.

Let θ = a/b for integers a,b with b > 0. Setting $x=S_b(theta)/b$ then $S_N(theta)/Nto x$ as $Ntoinfty$.

Proof:
If m ≡ n (mod b) then θm² - θn² is an integer, and $e^{2pi itheta m^2}=e^{2pi i theta n^2}$. So $nmapsto e^{2pi itheta n^2}$ has period b, giving

$$S_{bN}(theta)=sum_{j=0}^{N-1}sum_{k=1}^{b}e^{2pi itheta(jb+k)^2}=Nsum_{k=1}^be^{2pi itheta k^2}.$$
So, S_bN(θ) = NS_b(θ). Now, any N can be written as N = bM + R for some R < b. Then, $vert S_N-MS_bvertle R$ and, dividing by N gives $vert S_N/N-S_b/bvertto0$ as N goes to infinity.

As |S_N(θ)/(h(N)N)| ∼ |x|/h(N) → ∞ whenever x is nonzero, the following shows that (*) holds whenever θ is of the form a/p for an odd prime p not dividing a. Such rationals are dense, so the existence of irrational θ for which (*) holds follows from the Baire category theorem.

Let θ = a/p for integers a,p with p an odd prime not dividing a. Then $x=S_p(theta)/p$ is nonzero.

Proof:
Note that $u=e^{2pi i a/p}$ is a primitive p'th root of unity with minimal polynomial $X^{p-1}+X^{p-2}+cdots+X+1$ over the rationals. Then, all proper subsets of ${1,u,u^2,ldots,u^{p-1}}$ are linearly independent over the rationals and

$$S_p(theta)=sum_{k=1}^{p}u^{k^2}=1+2sum_{k=1}^{(p-1)/2}u^{k^2}$$
is nonzero.

In fact as pointed out by David below, S_p is a Gauss sum and has size √p.

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