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Source. Lemma 1.1, p. 2, 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

Throughout, $T:\mathbb Z\to\mathbb Z$ is the reduced Collatz map, $T(n)=(3n+1)/2$ for odd $n$ and $T(n)=n/2$ for even $n$, and $\mathcal T$ is the linear operator with $\mathcal T(z^n)=z^{T(n)}$, given on the Bergman space $H^2_{ber}(D)$ of the open unit disc $D$ by $\mathcal Tf(z)=(Sf)(\sqrt z)+\sqrt z\,(Af)(z^{3/2})$, where $Sf$ and $Af$ are the even and odd parts of $f$ (p. 2).

Lemma 1.1 (p. 2). If $(n_1,\dots,n_k)$ is a cycle of $T$, then $\sum_{i=1}^k z^{n_i}$ is a fixed point of $\mathcal T$. Further, a polynomial fixed point of $\mathcal T$ has this form up to an element of the kernel of $\mathcal T$, which the paper computes in (2.1) (p. 3) as the span of the binomials $z^{2k+1}-z^{6k+4}$, $k\ge0$, there for $\mathcal T$ acting on $H^2_{ber}(D)/X$ with $X=\operatorname{span}\{1,z,z^2\}$.

For the trivial cycle $\{1,2\}$ the fixed point is $z+z^2$.

Read depth. Claims checked: the statement was read clause by clause on p. 2.

Proof pointer

The paper calls the proof trivial and omits it (p. 2). The first part is the identity $\mathcal T(z^{n_i})=z^{T(n_i)}=z^{n_{i+1}}$ around the cycle.

Dependencies

None.

Bears on

• #1135: the lemma turns cycles of the problem's map $f$ (the paper's $T$ on the positive integers) into fixed points of a linear operator; it says nothing about which cycles exist.