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Source. Theorem 4.5, p. 8, 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 and $\mathcal F$ the Berg--Meinardus operator of Definition 4.1 (p. 6), as on the Theorem 4.4 page; $A(\Omega)$ is the space of analytic functions on $\Omega$ and $D$ the open unit disc.

Theorem 4.5 (p. 8). Let $l\in\mathbb N\cup\{0\}$ and let $\hat\phi=\sum_{n\ge0}\phi(n)z^n\in A(D)$ satisfy $|\phi(n)|\le C(n+1)^l$ for $n\ge0$ (4.4). Put

$$F^{\hat\phi}(z,w)=\sum_{m,n\ge0}\phi(T^m(n))z^nw^m,\qquad z,w\in\mathbb C.$$

Then $F^{\hat\phi}\in A\bigl(D\times\frac{2^l}{3^l}D\bigr)$ and $F^{\hat\phi}(z,w)=(I-w\mathcal F)^{-1}\hat\phi$ (4.5).

Read depth. Claims checked: the statement was read clause by clause on p. 8, and the two-step proof was read through, not checked step by step.

Proof pointer

Page 8. Convergence on compact subsets of $D\times\frac{2^l}{3^l}D$ follows from (4.4) and the bound $T^m(n)\le(3/2)^m(n+1)$. Applying $\mathcal F$ term by term with (4.2) shifts $m$ by one and gives $\mathcal FF^{\hat\phi}=\frac1w(F^{\hat\phi}-\hat\phi)$.

Dependencies

Formula (4.2) of the same paper (p. 6).

Bears on

No Erdős problem page cites it. Example 4.10 (p. 9) takes $\phi(n)=n$ and recovers the Berg--Meinardus identity $F^{\hat\phi}-w\mathcal FF^{\hat\phi}=z/(1-z)^2$; Example 4.8 (p. 9) takes $\phi(n)=\delta_{n,1}$ and leads to Theorem 4.9.