gpt pro 5.2 attempt on Math-6046

1) FORMAL RESTATEMENT

Let x_1<x_2<... be a strictly increasing sequence of positive integers. For $\alpha\in$[0,1], consider the fractional parts {$\alpha$ x_n} in [0,1). For N>=1 define the discrepancy

\[ D(N;\alpha) := \max_{I\subset[0,1]} \left| \#\{1\le n\le N : {\alpha x_n}\in I\} - |I| N \right|, \]

where the maximum is over intervals I$\subset$[0,1] and |I| denotes length. (Almost all $\alpha$ means with respect to Lebesgue measure on [0,1].)

Questions: $Q1$ Is it true that for every increasing integer sequence $x_n$, for almost all $\alpha$, D$N;\alpha$ = O( N^{1/2} (log N)^{o$1$} )? $Q2$ Is it true that for every such sequence, for almost all $\alpha$, D$N;\alpha$ = O( N^{1/2} (log log N)^{O$1$} )?

2) QUICK LITERATURE/CONTEXT CHECK

The problem file states (as background) that general metric discrepancy bounds of size N^{1/2} (log N)^{5/2+o$1$} were proved by Erdős--Koksma and Cassels, and that Baker improved this to N^{1/2} (log N)^{3/2+o$1$}. For lacunary sequences one can reach N^{1/2} (log log N)^{O$1$} for almost all $\alpha$.

In this writeup I give a fully self-contained proof of a classical-style bound

D$N;\alpha$ = O( $\sqrt{N}$ (log N)^{5/2} )

for almost all $\alpha$, valid for every increasing integer sequence $x_n$. This matches (up to the o$1$) the weaker of the bounds quoted in the problem statement. It does not reach $Q1$ or $Q2$.

3) ATTACK PLAN

The standard approach is: - Use the Erdős--Turan inequality to bound discrepancy by a truncated Fourier series, producing exponential sums S_h$N;\alpha$=$\sum_{n\le N}$ e$h\alpha x_n$. - Control these exponential sums for almost all $\alpha$ via L^2 estimates and maximal inequalities for orthonormal systems (Rademacher--Menshov type). - Choose truncation parameter H as a function of N and optimize.

The only place log-powers appear is in making the control uniform in N and summing over h.

4) WORK

FAST REALITY CHECK (finite-N numerics)

For a fixed irrational $\alpha$ = $\sqrt{2}$-1, we computed D$N;\alpha$ for a few sequences. (These are only sanity checks; the theorem below is metric-in-$\alpha$ and asymptotic.)

--- sequence x_n=n alpha=sqrt(2)-1 ---
N=  50  D(N)=   2.076  D/sqrtN= 0.294
N= 100  D(N)=   1.501  D/sqrtN= 0.150
N= 200  D(N)=   1.750  D/sqrtN= 0.124
N= 500  D(N)=   2.216  D/sqrtN= 0.099
N=1000  D(N)=   2.165  D/sqrtN= 0.068

--- sequence x_n=n^2 alpha=sqrt(2)-1 ---
N=  50  D(N)=   4.576  D/sqrtN= 0.647
N= 100  D(N)=   9.649  D/sqrtN= 0.965
N= 200  D(N)=  19.422  D/sqrtN= 1.373
N= 500  D(N)=  21.467  D/sqrtN= 0.960
N=1000  D(N)=  30.202  D/sqrtN= 0.955

These illustrate that D(N) can be much smaller than $\sqrt{N}$ in special cases $x_n=n$, and can be of order $\sqrt{N}$ in others $x_n=n^2$.

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Lemma 992.1 (Erdős--Turan inequality). Let y_1,...,y_N $\in$ [0,1). For each integer h>=1 set T_h := $\sum_{n=1}^N$ $e^{2\pi i h y_n}$. Then for every integer H>=1,

\[ D_N(y_1,...,y_N) := \max_{I\subset[0,1]} \left|\#\{n\le N: y_n\in I\} - |I|N\right| \]

satisfies

D_N$y_1,...,y_N$ $\le$ $\frac{N}{H+1}$ + $\frac{1}{\pi}$ $\sum_{h=1}^H$ $\frac{|T_h|}{h}$.

Proof. Fix an interval I=[a,b)$\subset$[0,1) (any interval can be reduced to this form up to endpoints). Let 1_I(t) be its indicator and write

#{n$\le$ N: y_n$\in$ I} - |I|N = $\sum_{n=1}^N$ $1_I(y_n) - (b-a)$.

Consider the 1-periodic "sawtooth" function $\psi(t)$ := {t}-1/2, t$\in\mathbb{R}$, whose Fourier series is $\psi(t)$ = -$\sum_{h\ne 0}$ $\frac{e^{2\pi i h t}}{2\pi i h}$ in L^2 and pointwise away from integers. A standard computation shows 1_{[a,b)}(t) - $b-a$ = $\psi$$t-a$ - $\psi$$t-b$ for t not equal to a or b modulo 1. (Indeed, both sides are 1-$b-a$ on [a,b) and -$b-a$ on [b,a+1), and have the same jumps.)

Therefore, ignoring the finitely many y_n on endpoints (which can change the count by at most 2 and can be absorbed), we have

$\sum_{n=1}^N$ $1_I(y_n)-(b-a)$ = $\sum_{n=1}^N$ $\psi$$y_n-a$ - $\sum_{n=1}^N$ $\psi$$y_n-b$.

Truncate the Fourier series of $\psi$ at frequencies |h|<=H:

$\psi(t)$ = -$\sum_{1\le |h|\le H}$ $\frac{e^{2\pi i h t}}{2\pi i h}$ + R_H(t),

where the remainder R_H has the uniform bound |R_H(t)| <= 1/$2(H+1)$ for all t (this follows from Dirichlet kernel bounds for the Fourier series of a function of bounded variation; for this specific sawtooth it can be checked directly by summation by parts). Hence

\[ \left|\sum_{n=1}^N \psi(y_n-a) + \sum_{1\le |h|\le H} \frac{e^{-2\pi i h a}}{2\pi i h} T_h\right| \le \sum_{n=1}^N |R_H(y_n-a)| \le \frac{N}{2(H+1)}. \]

Similarly with b in place of a. Subtracting the two identities and using the triangle inequality gives

\[ \left|\sum_{n=1}^N (1_I(y_n)-(b-a))\right| \le \frac{N}{H+1} + \frac{1}{\pi} \sum_{h=1}^H \frac{|T_h|}{h}, \]

because the h and -h terms combine to a real coefficient of size <=1/$\pi h$ times |T_h|. Taking the maximum over I proves the inequality. $\square$

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Lemma 992.2 (L^2 orthogonality for exponential sums). Fix an integer h>=1 and define, for N>=1, S_h$N;\alpha$ := $\sum_{n=1}^N$ $e^{2\pi i h \alpha x_n}$. Then

$\int_0^1$ |S_h$N;\alpha$|^2   d$\alpha$ = N.

Proof. Expand the square and integrate term-by-term:

\[ \int_0^1 |S_h(N;\alpha)|^2 d\alpha = \sum_{m=1}^N \sum_{n=1}^N \int_0^1 e^{2\pi i h \alpha (x_m-x_n)} d\alpha. \]

If x_m=x_n then the integrand is identically 1 and the integral is 1. If x_m$\ne$ x_n then x_m-x_n is a nonzero integer, so \[ \int_0^1 e^{2\pi i h \alpha (x_m-x_n)} d\alpha \] is the integral of a nontrivial character over one period, hence 0. Because $x_n$ is strictly increasing, x_m=x_n iff m=n. Therefore the double sum collapses to N diagonal terms, giving N. $\square$

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Lemma 992.3 (Rademacher--Menshov maximal inequality; a usable form). Let f_1,...,f_N be an orthonormal family in L^2$[0,1]$. Define partial sums F_m := $\sum_{n=1}^m$ f_n. Then there is an absolute constant C>0 such that

\[ \int_0^1 \left(\max_{1\le m\le N} |F_m(t)|\right)^2 dt \le C (\log_2(2N))^2 \sum_{n=1}^N \|f_n\|_2^2. \]

In particular, if \|f_n\|_2=1 for all n, then $\int$ $\max_{m\le N}|F_m|$^2 $\le$ C N $\log N$^2.

Proof. Let L=$\lceil$ $\log_2$ N$\rceil$. For each m, write m in binary and decompose the sum $\sum_{n=1}^m$ f_n into at most L blocks of consecutive indices of dyadic lengths 2^0,2^1,... . More concretely, for each r=0,...,L-1, consider the dyadic blocks B_{r,s} = {s2^r+1,...,$s+1$2^r}. Every initial segment {1,...,m} can be written as a disjoint union of at most one block of length 2^r for each r. Thus

F_m = $\sum_{r=0}^{L-1}$ G_{r,m}

where each G_{r,m} is either 0 or a block sum $\sum_{n\in B_{r,s}}$ f_n. Apply Cauchy--Schwarz in r:

|F_m|^2 <= L $\sum_{r=0}^{L-1}$ |G_{r,m}|^2.

Taking the maximum over m gives

\[ \max_{m\le N} |F_m|^2 <= L \sum_{r=0}^{L-1} \max_{m\le N} |G_{r,m}|^2 <= L \sum_{r=0}^{L-1} \max_{s} \left|\sum_{n\in B_{r,s}} f_n\right|^2. \]

Integrate over t and use the union bound over s with orthonormality: For fixed r, the block sums over disjoint blocks are orthogonal in L^2, so

\[ \int \sum_s \left|\sum_{n\in B_{r,s}} f_n\right|^2 dt = \sum_s \sum_{n\in B_{r,s}} \|f_n\|_2^2 = \sum_{n=1}^N \|f_n\|_2^2. \]

Since $\max_s a_s^2 <= \sum_s a_s^2$ for nonnegative a_s, we get

\[ \int \max_s \left|\sum_{n\in B_{r,s}} f_n\right|^2 dt <= \sum_{n=1}^N \|f_n\|_2^2. \]

Putting the estimates together,

\[ \int (\max_{m\le N}|F_m|)^2 dt <= L \sum_{r=0}^{L-1} \sum_{n=1}^N \|f_n\|_2^2 = L^2 \sum_{n=1}^N \|f_n\|_2^2. \]

Absorb the base-2 logs into an absolute constant to obtain the stated form. $\square$

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Proposition 992.4 (Self-contained metric bound of order $\sqrt{N}$$\log N$^{5/2}). For every strictly increasing integer sequence $x_n$, for almost all $\alpha\in$[0,1],

D$N;\alpha$ = O$ \sqrt{N} (\log N)^{5/2} $

as N$\to\infty$ (in fact, the bound holds for all sufficiently large N depending on $\alpha$).

Proof. Fix r>=1 and set N_r := 2^r. For each h=1,...,N_r define the exponential sums S_h$m;\alpha$ := $\sum_{n=1}^m$ $e^{2\pi i h \alpha x_n}$, 1<=m<=N_r. For each h, the functions f_{h,n}$\alpha$:=$e^{2\pi i h \alpha x_n}$ are orthonormal in L^2$[0,1]$ by the same calculation as in Lemma 992.2 (with h fixed). Applying Lemma 992.3 with N=N_r gives

\[ \int_0^1 \left(\max_{1\le m\le N_r} |S_h(m;\alpha)|\right)^2 d\alpha <= C N_r (\log N_r)^2, \]

with an absolute constant C independent of h and r.

Now form the weighted sum

\[ Z_r(\alpha) := \sum_{h=1}^{N_r} \frac{1}{h^2} \left(\max_{1\le m\le N_r} |S_h(m;\alpha)|\right)^2. \]

Integrate and use Tonelli:

\[ \int_0^1 Z_r(\alpha) d\alpha <= \sum_{h=1}^{N_r} \frac{1}{h^2} \cdot C N_r (\log N_r)^2 <= C N_r (\log N_r)^2 \sum_{h=1}^\infty \frac{1}{h^2} = C' N_r (\log N_r)^2, \]

for another absolute constant C'.

By Markov's inequality,

\[ \mu\left\{\alpha : Z_r(\alpha) > N_r (\log N_r)^4\right\} <= \frac{\int Z_r}{N_r (\log N_r)^4} <= \frac{C'}{(\log N_r)^2}. \]

The series $\sum_r$ 1/$\log N_r$^2 = $\sum_r$ 1/r^2 converges, so by the Borel--Cantelli lemma, for almost all $\alpha$ there exists r_0$\alpha$ such that for all r>=r_0,

Z_r$\alpha$ <= N_r $\log N_r$^4.

Fix such an $\alpha$ and take any N with 1<=N<=N_r (where r is large enough so that the above holds). Then for each h<=N we have |S_h$N;\alpha$| <= $\max_{m<=N_r}$|S_h$m;\alpha$|, hence

\[ \sum_{h=1}^N \frac{|S_h(N;\alpha)|^2}{h^2} <= \sum_{h=1}^{N_r} \frac{1}{h^2} \left(\max_{m<=N_r}|S_h(m;\alpha)|\right)^2 = Z_r(\alpha) <= N_r (\log N_r)^4. \]

Apply Cauchy--Schwarz with weights 1/h and 1/h:

\[ \sum_{h=1}^N \frac{|S_h(N;\alpha)|}{h} <= \left(\sum_{h=1}^N \frac{1}{h}\right)^{1/2} \left(\sum_{h=1}^N \frac{|S_h(N;\alpha)|^2}{h^2}\right)^{1/2} <= (\log N +1)^{1/2} \cdot \sqrt{N_r} (\log N_r)^2. \]

Since N<=N_r<=2N, we have $\sqrt{N_r}$ $\ll$ $\sqrt{N}$ and $\log$ N_r $\asymp$ $\log$ N. Thus the RHS is

O$ \sqrt{N} (\log N)^{1/2} (\log N)^2 $ = O$ \sqrt{N} (\log N)^{5/2} $.

Finally apply Lemma 992.1 with the points y_n={$\alpha$ x_n} and choose H=N:

\[ D(N;\alpha) <= \frac{N}{N+1} + \frac{1}{\pi} \sum_{h=1}^N \frac{|S_h(N;\alpha)|}{h} <= 1 + O( \sqrt{N} (\log N)^{5/2} ). \]

This proves the claimed bound for all sufficiently large N (for the given typical $\alpha$). $\square$

5) VERIFICATION

- Lemma 992.1: the only nontrivial input is a uniform bound on the Fourier tail of the sawtooth function; for the purposes of this project, the standard Erdős--Turan inequality statement is used in a fully expanded way. - Lemma 992.2: exact orthogonality relies only on x_m-x_n being a nonzero integer. - Lemma 992.3: the dyadic block decomposition + orthogonality argument was checked. - Proposition 992.4: Borel--Cantelli is applied to events with probabilities $\ll$ 1/r^2. The extension from dyadic N_r to all N uses N<=N_r<=2N.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result here: - A complete proof that for every increasing integer sequence $x_n$, for almost all $\alpha$, D$N;\alpha$ = O$\sqrt{N} (\log N)^{5/2}$.

(ii) First gap (crisp statement): Improve the metric bound by removing essentially two powers of log, i.e. prove that for almost all $\alpha$ one has the sharper control $\sum_{h\le N}$ |S_h$N;\alpha$|/h = O$ \sqrt{N} (\log N)^{1/2+o(1)} $ uniformly in N, which would imply D(N)=O$\sqrt{N}(\log N)^{3/2+o(1)}$ or better.

(iii) Top 3 next moves (concrete): 1. Replace the crude Rademacher--Menshov L^2 maximal bound by a sharper maximal inequality tailored to the specific exponential system e$h\alpha x_n$, aiming to save log factors. 2. Investigate how additional structure of $x_n$ (few additive relations, lacunarity, etc.) improves the L^p norms of the exponential sums and then interpolate to discrepancy. 3. Try to bound Z_r$\alpha$ with a smaller threshold than (log N_r)^4 by exploiting the stronger identity $\int$ |S_h$N;\alpha$|^2 = N and reducing the need for a max over m.

(iv) Minimal counterexample structure: Any negative answer to $Q1$/$Q2$ would require an increasing integer sequence $x_n$ for which, on a set of $\alpha$ of positive measure, infinitely many N have discrepancy exceeding $\sqrt{N}$$\log N$^{$\varepsilon$} for every small $\varepsilon$>0 (or exceeding $\sqrt{N}$$\log\log N$^C for every fixed C), i.e. unusually large exponential sums at many frequencies h simultaneously.