= Solution
Use the geometric properties of the <Feigenbaum geometric potential>: for some constants $0<b\leq B<\infty$,
$$
-B\leq U\leq-b<0,\qquad \operatorname{var}_rU\leq C\vartheta^r,\quad0<\vartheta<1.
$$
Here $\operatorname{var}_r$ means variation between sequences agreeing in their first $r$ symbols. In particular $D=\sum_{r\geq1}\operatorname{var}_rU<\infty$. These are the standard bounds for the <Feigenbaum geometric potential>; the <symbolic potential> is not the singular expression $-\log|g'|$ at a quadratic critical point.
First allow all $n$ spins and use a fixed remote tail $\eta$, writing $Z_n(\gamma;\eta)$ for the weighted <partition function>. In a concatenated block of lengths $n$ and $m$, replacing the history before the second block by $\eta$ changes its energy by at most $D$: the $j$th summand has the same first $j$ symbols, so its difference is at most $\operatorname{var}_jU$. Therefore
$$
e^{-|\gamma|D}Z_n(\gamma;\eta)Z_m(\gamma;\eta)
\leq Z_{n+m}(\gamma;\eta)
\leq e^{|\gamma|D}Z_n(\gamma;\eta)Z_m(\gamma;\eta).
$$
For $a_n=\log Z_n$, the <almost-additive partition-function limit> follows explicitly: $a_n+|\gamma|D$ is subadditive and $a_n-|\gamma|D$ superadditive, so <Fekete's lemma> gives the same finite limit for both after division by $n$. The finiteness follows from bounded $U$ and the $2^n$ words.
Changing the fixed remote tail also changes each full energy by at most $D$. The question fixes the first spin to one and sums the remaining $n-1$ spins; its <partition function> is $e^{\gamma U(1,0,0,\ldots)}Z_{n-1}(\gamma;(1,0,0,\ldots))$. The tail comparison and bounded first contribution show that its normalized logarithm has the same limit. Thus the requested $f(\gamma)$ exists and equals the <topological pressure> of $\gamma U$.
For any $\delta>0$, every length-$n$ energy lies between $-Bn$ and $-bn$, hence
$$
e^{-Bn\delta}Z_n(\gamma)\leq Z_n(\gamma+\delta)\leq e^{-bn\delta}Z_n(\gamma).
$$
Taking normalized logarithms and limits gives
$$
\boxed{-B\delta\leq f(\gamma+\delta)-f(\gamma)\leq-b\delta.}
$$
So $f$ is <Lipschitz continuous> and strictly decreasing. For each finite $n$, differentiation of the log-sum gives $f_n''(\gamma)=n^{-1}\operatorname{Var}_{n,\gamma}(H_n)\geq0$. Thus $f_n$ is <convex>, and passing to its finite pointwise limit proves that $f$ is <convex> too.
At zero weight the prescribed sum has $2^{n-1}$ terms, so $f(0)=\log2$. For $\gamma>0$ the preceding bounds give $\log2-B\gamma\leq f(\gamma)\leq\log2-b\gamma$. The function therefore becomes negative at sufficiently large positive $\gamma$. Continuity and strict decrease prove
$$
\boxed{\text{There is exactly one }\gamma_*>0\text{ with }f(\gamma_*)=0,\qquad
\frac{\log2}{B}\leq\gamma_*\leq\frac{\log2}{b}.}
$$
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