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Source. Theorem 4.4, p. 7, of Mikhail Neklyudov, Functional analysis approach to the Collatz conjecture, arXiv:2106.11859v9 (2022), published in Results Math. 79 (2024), no. 4, Paper No. 140, in the edition identified on the source card.

Statement

$T$ is the reduced Collatz map, $T(n)=(3n+1)/2$ for odd $n$ and $n/2$ for even $n$, and $\mathcal T$ the operator with $\mathcal T(z^n)=z^{T(n)}$, here taken in $\mathcal L(H^2(D))$, the bounded operators on the Hardy space of the open unit disc. Its adjoint there is the Berg--Meinardus operator

$$\mathcal F(g)(z)=g(z^2)+\frac{z^{-1/3}}3\Bigl(g(z^{2/3}) +e^{2\pi i/3}g(z^{2/3}e^{2\pi i/3})+e^{4\pi i/3}g(z^{2/3}e^{4\pi i/3})\Bigr)$$

(Definition 4.1, p. 6; Lemma 4.3, p. 7), which sends $z^n$ to $z^{2n}$ when $n\equiv0,1\pmod 3$ and to $z^{2n}+z^{(2n-1)/3}$ when $n\equiv2\pmod3$ (4.2).

Theorem 4.4 (p. 7). Quoted: "Number of cycles of Collatz map $T:\mathbb N\to\mathbb N$ (including trivial one) is bounded above by index of operator $Id-\mathcal T\in\mathcal L(H^2(D))$."

The proof uses the index in the form $\operatorname{ind}(Id-\mathcal T)=\dim\operatorname{Ker}(Id-\mathcal T) -\dim\operatorname{Ker}(Id-\mathcal F)$. The paper does not discuss whether this index is finite; if $\operatorname{Ker}(Id-\mathcal T)$ is infinite-dimensional, the bound says nothing.

Read depth. Claims checked: the statement was read clause by clause on p. 7, and the proof and Proposition 4.2 (p. 6) were read through, not checked step by step.

Proof pointer

Page 7. By Proposition 4.2 (p. 6), $\mathcal F$ is expansive on $H^2(D)$, $\|f\|\le\|\mathcal Ff\|$, with $\|\mathcal F\|\le\sqrt2$; an $H^2$ fixed point $h$ of $\mathcal F$ therefore satisfies $\mathcal Fh(z)=h(z^2)$ by (4.3) and is constant, so $\dim\operatorname{Ker}(Id-\mathcal F)=1$. The polynomials of Lemma 1.1 attached to distinct cycles are linearly independent fixed points of $\mathcal T$, since cycles are disjoint. The proof leaves implicit that the constant $1$, from the fixed point $0$ of $T$, is a further fixed point of $\mathcal T$ not counted among the cycles in $\mathbb N$; that is what makes the count at most the index rather than the index plus one.

Dependencies

Lemma 1.1, Proposition 4.2 and Lemma 4.3 of the same paper.

Bears on

• #1135: the problem's map $f$ is the paper's $T$ on $\mathbb N$, and a positive answer requires that $\{1,2\}$ be its only cycle. The theorem bounds the number of cycles by an operator index that the paper does not compute; it proves no case of the problem.