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Statement

Theorem 6 (p. 6), quoted: "Let $a>0$ and $\kappa>0$ be real numbers. Suppose that for all $t\geqslant0$ one has

$$J_0(t)+J_0(\kappa t)+J_0((1+\kappa)t)>-1\,.$$

Then for any measurable coloring of the plane $\Pi$ into two colors there is a monochromatic collinear triple $\{x,y,z\}$ such that $y\in[x,z]$ and $\|y-x\|=a$, $\|z-y\|=\kappa a$."

The displayed hypothesis is the paper's (11); $\Pi=\mathbf R^2$ with the Euclidean norm, and $J_0$ is the zeroth Bessel function of the first kind ((9), p. 6). The hypothesis does not involve $a$, so a $\kappa$ that meets it gives such a triple at every scale $a>0$.

Source. I. D. Shkredov, On some problems of Euclidean Ramsey theory, arXiv:1507.02727v2 (22 July 2015), Theorem 6, p. 6. The copy read is identified in the source digest.

Read depth. Claims checked: the statement was read clause by clause on the page image; the proof (pp. 6--7) was read for structure only, and none of its estimates was checked. Nothing here is independently reviewed.

Proof pointer

Pp. 6--7, following the finite-field argument of Theorem 3. Suppose neither color contains such a triple. The two colors are replaced by periodic sets with nearly the same upper densities that still avoid the triples. One counts the triples $x$, $x+s$, $x-\kappa s$ with $s$ on the circle of radius $a$ through a trilinear average $\sigma$, split into a main term, three two-function terms and a cubic term; the cubic terms of the two colors cancel. In Fourier space the circle's transform is a multiple of $J_0$, so the three middle terms are bounded below by $2\pi a$ times $\sum_t\alpha(t)(J_0(at)+J_0(\kappa at)+J_0((1+\kappa)at))$ with $\alpha(t)\ge0$, and Parseval gives $\sum_t\alpha(t)=\delta-\delta^2$. With the densities summing to $1$, this yields $(2\pi a)^{-1}(\sigma(A_*)+\sigma(B_*))\geqslant(J+1)/4>0$, where $J$ is the minimum of the Bessel sum, a contradiction. Not checked here.

Dependencies

The proof uses the formula $J_0(\|u\|)=\frac1{2\pi}\langle\mathcal S_1(x),e^{iux}\rangle$ for the unit circle ((10), p. 6), cited as well known, and the method of de Oliveira Filho and Vallentin (the paper's [9]).

Used by

• Theorem 1, second part.
• Corollary 7, the case $\kappa=1$.

Bears on

• Problem 173: gives, for measurable two-colorings only, monochromatic congruent copies of the degenerate triangle with collinear points at steps $a$ and $\kappa a$, for each $\kappa$ meeting the Bessel condition. It says nothing about non-measurable colorings.