A Sum of Two Geometric Decays Can Misbehave Only Once

Consider the points

$$P_n=(r^n,r^{2n})$$

on the parabola $y=x^2$, where $0<r<1$. Their signed projections in the direction $(A,B)$ are

$$z_n=Ar^n+Br^{2n}.$$

The two coordinates of $P_n$ shrink rapidly. Their projection usually does as well, except when the two terms have opposite signs and nearly cancel.

The problem in this post is taken from a geometric lemma in Hannah Cairo’s paper A Counterexample to the Mizohata—Takeuchi Conjecture (arXiv:2502.06137). The paper uses points of this type to construct subset sums with few incidences with hyperplanes. We will only study the elementary sequence above.

There is a simple reason that rapid decay is useful. Suppose two subset sums of the $P_n$ have the same projection. Subtracting them gives

$$\sum_{n=1}^N \varepsilon_n z_n=0, \qquad \varepsilon_n\in\{-1,0,1\}.$$

If the first nonzero term is larger than the whole tail, this equality is impossible.

Statement

Fix $0<\theta\leq1$. For small enough $r$, all but at most one index $n$ have the property that every later term $z_m$ is smaller than $\theta\lvert z_n\rvert$. The result is sharp. We can find $A$ and $B$ such that one of the terms $z_n$ vanishes.

We use the non-sharp condition $r\leq\theta/4$ because it gives a short proof.

Theorem. Let $0<\theta\leq1$ and $0<r\leq\theta/4$. For

$$z_n=Ar^n+Br^{2n},\qquad n=1,\dots,N,$$

where $A,B\in\mathbb{R}$ are not both zero, there is a set $S\subseteq\{1,\dots,N\}$ with $\lvert S\rvert\leq1$ such that

$$\lvert z_m\rvert<\theta\lvert z_n\rvert$$

whenever $1\leq n<m\leq N$ and $n\notin S$.

Proof

If $A=0$, then

$$\frac{\lvert z_m\rvert}{\lvert z_n\rvert}=r^{2(m-n)}<\theta.$$

If $A\neq0$ and $B/A\geq0$, then

$$\frac{\lvert z_m\rvert}{\lvert z_n\rvert} =r^{m-n}\frac{1+(B/A)r^m}{1+(B/A)r^n} \leq r<\theta.$$

It remains to consider $AB<0$.

Write $B/A=-D$ with $D>0$, and set

$$x_n=Dr^n, \qquad f(x)=x\lvert 1-x\rvert.$$

Then

$$x_{n+1}=rx_n, \qquad \lvert z_n\rvert=\frac{\lvert A\rvert}{D}f(x_n).$$

The value $x=1$ corresponds to exact cancellation.

For $0<t<\theta$, define

$$a(t)=\frac{\theta-t}{\theta-t^2}, \qquad b(t)=\frac{\theta+t}{\theta+t^2}.$$

We claim that

$$f(tx)\geq\theta f(x) \quad\Longleftrightarrow\quad a(t)\leq x\leq b(t).$$

The verification is a three-case check:

$0<x<1$: this is $t(1-tx)\geq\theta(1-x)$, equivalent to $x\geq a(t)$.

$1<x<1/t$: this is $t(1-tx)\geq\theta(x-1)$, equivalent to $x\leq b(t)$.

$x\geq1/t$: this gives $t(tx-1)\geq\theta(x-1)$, hence $x\leq a(t)<1/t$, impossible.

This proves the equivalence.

On the interval $0<t\leq r\leq\theta/4$,

$$a'(t)=-\frac{\theta-2\theta t+t^2}{(\theta-t^2)^2}<0, \qquad b'(t)=\frac{\theta-2\theta t-t^2}{(\theta+t^2)^2}>0.$$

Therefore, for every $j\geq1$,

$$[a(r^j),b(r^j)]\subseteq[a(r),b(r)].$$

Set

$$I=[a(r),b(r)].$$

If $x_n\notin I$, then the equivalence above, applied with $t=r^{m-n}$, gives

$$f(x_m)<\theta f(x_n) \qquad(m>n).$$

It remains to show that $(x_n)$ has at most one term in $I$. The inequality $a(r)>rb(r)$ is equivalent to

$$r^3+\theta r^2+\theta r-\theta^2<0.$$

Since $r\leq\theta/4$ and $\theta\leq1$,

$$r^3+\theta r^2+\theta r \leq\theta^2\left(\frac1{64}+\frac1{16}+\frac14\right) <\theta^2.$$

Hence

$$\frac{b(r)}{a(r)}<\frac1r.$$

Suppose $n<m$ and $x_n,x_m\in I$. On the one hand,

$$\frac{x_n}{x_m}=r^{-(m-n)}\geq\frac1r.$$

On the other hand,

$$\frac{x_n}{x_m}\leq\frac{b(r)}{a(r)}<\frac1r,$$

a contradiction. Therefore, $\lvert S\rvert\leq1$, where $S$ is the set of indices for which $x_n\in I$.

The sharp threshold

The bound $r\leq\theta/4$ was used only to simplify the proof. The exact boundary $r_\star(\theta)$ is the unique root in $(0,\theta)$ of

$$r^3+\theta r^2+\theta r-\theta^2=0.$$

The conclusion holds for every $0<r<r_\star(\theta)$ and fails when $r\geq r_\star(\theta)$. As $\theta\to0^+$,

$$r_\star(\theta) =\theta-2\theta^2+10\theta^3-66\theta^4+O(\theta^5)\sim\theta.$$

A proof of the stronger statement requires tracking $I$ without replacing the root by the bound $\theta/4$.