Lemma: Open-domain cocycle rigidity

Let $\phi :[0,1]\to [0,1]$ be continuous, with

Suppose

for all

Then $\phi (x)=x$ on $[0,1]$.

Proof. We first prove positivity. Choose $a=x$ and $m=2xu$. This is allowed when

and gives

If $\phi (x_0)=0$ for $x_0\in (0,1)$, then $\phi$ vanishes on

If it vanishes on $(0,\lambda )$ with $\lambda \le 1/2$, then for $x\in (\lambda /2,\lambda )$ we have $x<1/2$, hence $U(x)=1$; (eq:doubling) makes it vanish on $(0,2x)$. Taking the union gives vanishing on $(0,2\lambda )$. Finite repetition produces an interval containing $1/2$, contradicting the anchor. Thus $\phi >0$ on $(0,1)$.

Put $H=\log \phi$. For $x,y\in (0,1)$, set $p=xy$. The admissible intervals for the factorizations $(x,y)$ and $(\sqrt{p},\sqrt{p})$ are

Their intersection is nonempty. Choosing $m$ in that intersection and applying (eq:cocycle) twice gives

Hence, with $f(t)=2H(\sqrt{t})$,

Write $G(s)=H(e^{-s})$ and $F(t)=f(e^{-t})$. Then

For every $\tau >0$,

The common increment $\Delta (\tau )$ is continuous and additive on positive arguments. Additivity gives $\Delta (q)=q\Delta (1)$ for positive rational $q$; continuity gives the same formula for every positive real argument. Thus

Continuity at one and $\phi (1)=1$ give $d=0$. The half anchor gives $2^{-c}=1/2$, so $c=1$. The endpoint anchors complete the result.