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Source. Theorem 3.1, p. 5, 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

Restricting the operator $\mathcal T$ of the Lemma 1.1 page to the unit circle, parametrised by $z=e^{2\pi i\phi}$, $\phi\in[0,1)$, the paper writes it, for $1$-periodic $g\in L^2_{loc}(\mathbb R,\mathbb C)$, as

$$\mathcal Tg(\phi)=\tfrac12\Bigl(g\bigl(\tfrac\phi2\bigr)+g\bigl(\tfrac{\phi+1}2\bigr)\Bigr) +\tfrac{e^{\pi i\phi}}2\Bigl(g\bigl(\tfrac{3\phi}2\bigr)-g\bigl(\tfrac{3\phi+1}2\bigr)\Bigr)$$

(p. 5), that is $\mathcal T=L(I+B)$ with $L$ the doubling-map transfer operator $Lg(\phi)=\frac12\bigl(g(\frac\phi2)+g(\frac{\phi+1}2)\bigr)$ and $Bg(\phi)=e^{2\pi i\phi}g(3\phi)$.

Theorem 3.1 (p. 5). Lebesgue measure $dx$ on the circle, identified with $[0,1)$, is invariant for $\mathcal T:L^2(\mathbb S^1,\mathbb C)\to L^2(\mathbb S^1,\mathbb C)$, that is, $\int_{\mathbb S^1}\mathcal Tf\,dx=\int_{\mathbb S^1}f\,dx$ for every $f\in L^2(\mathbb S^1,\mathbb C)$.

Read depth. Claims checked: the statement and its short proof were read on p. 5.

Proof pointer

Page 5. Lebesgue measure is invariant for $L$, so it suffices that $\int Bf\,dx=0$; after the substitution $\psi=3\phi$ and periodicity, that integral carries the factor $1+\xi+\xi^2=0$, $\xi=e^{2\pi i/3}$.

Dependencies

None.

Bears on

No Erdős problem page cites it.