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We prove uniqueness in the sense of at most one Gibbs measure satisfying the printed conditional distributions. No shift-invariance assumption is needed in advance. Write a history as $\eta =({ϵ}_{s-1},{ϵ}_{s-2},\dots )$. The variation hypothesis makes $U$ continuous and bounded on the finite-alphabet sequence space; let $u_{-}\le U\le u_{+}$ and $V=u_{+}-u_{-}$. The conditional distribution of the first spin, obtained by marginalizing the prescribed two-spin law, must beIn particular $q(a\mid \eta )\ge \delta :=e^{-2V}/k>0$ for every symbol and history. Any two candidate measures have this same time-homogeneous one-step kernel, by conditional marginalization and the tower property.

If $\eta ,{\eta }^{\prime }$ agree in their most recent $r$ symbols, each two-spin energy in this formula changes by at mostBoth numerator and denominator therefore change by factors between $e^{-D{\lambda }^{-r}}$ and $e^{D{\lambda }^{-r}}$. Consequently the total variation distance of the two kernels is at most $e^{2D{\lambda }^{-r}}-1\le B{\lambda }^{-r}$, where $B=e^{2D}-1$. Uniform positivity also bounds it by $1-k\delta$. PutThese bounds are decreasing, strictly below one, and summable. We now give the summable-memory coupling argument rather than assuming a small one-step contraction coefficient.

Starting from any two histories at a remote time, use a maximal coupling of the next-spin laws. If the actual histories have agreed for $r$ consecutive symbols, their probability of disagreement is at most $b_r$. Use shared independent uniform random numbers to construct a dominating match-length process $R$: it increases from $r$ to $r+1$ with probability $1-b_r$ and resets to zero with probability $b_r$. Since $b_r$ decreases, the actual consecutive-match length can be kept at least $R$ at every step. When the dominating process does not reset, the actual spins can be coupled to agree.

A new run from zero has probabilityof never resetting. The product is positive because $b_r<1$ and $\sum b_r<\infty$: its tail logarithm converges absolutely, and its finite initial factors are positive. After each reset a fresh run has the same success probability, so the probability of $m$ failed runs is $(1-p_{\ast }{)}^m$. There are almost surely only finitely many resets. Every failed run has finite duration, and thus the time of the final reset is finite almost surely. Its distribution depends only on $(b_r)$, not on the two starting histories. It follows that the probability of any disagreement in a fixed future block tends uniformly to zero as the block's distance from the initial time tends to infinity.

Now take two candidate Gibbs measures, sample their respective histories before time $-N$, and couple their forward kernels as above. Their finite-block laws really are obtained by these kernels: successive conditional expectations use the common conditional distribution $q$. For any fixed coordinate block $[s,t]$, the coupling inequality for total variation bounds the distance between the two candidate block laws by the dominating probability of a reset at or after step $N+s$. That bound tends to zero as $N\to \infty$. Hence all probabilities of finite cylinder sets agree; cylinder sets generate the product Borel sigma-algebra, so the measures agree.

Thus there is at most one measure satisfying these conditional specifications. There is a separate compatibility issue: for an arbitrary unnormalized symbolic potential the finite-horizon conditional distributions printed in the question need not be consistent as the right endpoint changes. The argument proves the requested uniqueness whenever such a law exists; it does not silently infer existence from the variation estimate. A normalized one-sided kernel or the usual compatible Gibbs specification supplies the standard existence formulation. For a concrete incompatibility, take two symbols and $U(a,\eta )=\log W_{{\eta }_1,a}$ with $W=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)$. This has zero variation once two symbols agree. From a history ending in symbol one, the first-symbol probability obtained from a two-spin block is $2/5$, but from a three-spin block it is $5/13$. Thus the printed finite-horizon specification can indeed have no measure, even though its uniqueness implication remains valid.

A qualification is necessary: an arbitrary symbolic Gibbs measure does not satisfy the claim. Consider the doubling map with its two binary branches and a Bernoulli symbolic Gibbs measure with probabilities $2/3$ and $1/3$. This is the Gibbs law for a locally constant symbolic potential, while every level-$n$ interval has length $2^{-n}$. Along the nonboundary alternating itinerary, for even $n$,No finite $g$ can approximate these $g_n$ with a decaying uniform error. This proves the reusable fact that an arbitrary Gibbs measure need not be absolutely continuous.

For the intended geometric version, take the Gibbs measure for the geometric potential of an expanding map $-\log |T^{\prime }|$, with the standard sufficiently smooth finite full-branch hypotheses. The allowed transfer-operator result supplies a strictly positive Lipschitz continuous invariant probability density function $h$, normalized by $\int h=1$, and its symbolic weights of cylinder sets are $\nu (C_w)={\int }_{{\Delta }_w}h(x)dx$. State the needed regularity explicitly: $h\ge h_{\ast }>0$ and $|h(x)-h(y)|\le L_h|x-y|$. Expansion gives $\mathrm{diam}{\Delta }_w\le {\lambda }^{-n}$.

Let $\pi (ϵ)$ denote the point coded by the infinite sequence, and define $g(ϵ)=\log h(\pi (ϵ))$. The expression defining $g_n$ is the logarithm of the average of $h$ over the cylinder. For $x=\pi (ϵ)\in {\Delta }_w$,Since the logarithm is Lipschitz continuous with constant $1/h_{\ast }$ on this positive range, the cylinder averages of an invariant density giveThis proves the intended estimate. Expansion alone and an unspecified Gibbs symbolic potential are insufficient; the geometric choice and the regularity producing a positive Lipschitz continuous probability density function are substantive hypotheses.

Use zero-based indices $0\le i<2^n$; the two index conventions in the printed question enumerate the same partition. Let $O_n(\gamma )$ be the sum of length powers over odd intervals and $H_n(\gamma )$ the sum over all intervals. For odd-address words, the given symbolic potential estimate has $c\le |{\Delta }_i^{(n)}|e^{-H_n^{\mathrm{pot}}(i)}\le C$, with constants independent of $n,i$. Raising these bounds to a fixed real power $\gamma$ and reversing the order when necessary gives finite positive constants $c_{\gamma },C_{\gamma }$ such thatTherefore $n^{-1}\log O_n(\gamma )\to f(\gamma )$. It remains to justify the even intervals, since the supplied estimate covers only half of the partition.

In the standard period-doubling partition, ${\Delta }_0^{(n)}={\beta }^n[-1,1]$ and ${\Delta }_i^{(n)}=g^i({\Delta }_0^{(n)})$. All odd-address intervals lie in the first image restrictive interval, a compact set away from the unique critical point zero. Hence there are uniform constants $0<d\le |g^{\prime }|\le D_1$ there. For an odd $i<2^n-1$, its image is the next even interval, and the mean value theorem gives $d|{\Delta }_i^{(n)}|\le |{\Delta }_{i+1}^{(n)}|\le D_1|{\Delta }_i^{(n)}|$.

For the wraparound pair the image of the last interval is a subinterval of ${\Delta }_0^{(n)}$, not necessarily the whole of it. Iterating the renormalization identity givesThus $|g({\Delta }_{2^n-1}^{(n)})|=|\beta {|}^n(1-\beta )$, while $|{\Delta }_0^{(n)}|=2|\beta {|}^n$. The image is the fixed fraction $(1-\beta )/2$ of the central interval. Combining this with the derivative bounds proves uniform comparability of the last odd interval and the central even interval as well. This establishes the odd-even comparison for Feigenbaum partition lengths for every pair, including wraparound.

Consequently, for each fixed real $\gamma$, the even sum is bounded above and below by constant multiples of $O_n(\gamma )$. The same is true of $H_n(\gamma )$, soFinally take $\gamma >{\gamma }_{\ast }$. Strict decrease gives $f(\gamma )<0$, so this length-power sum tends to zero exponentially. The cover mesh also tends to zero: negative bounded $U$ and the odd-interval estimate give $|{\Delta }_i^{(n)}|\le C{e}^{-bn}$ on odd intervals, and the odd-even comparison transfers this bound to all intervals. Since the intervals cover the Feigenbaum attractor, they are admissible arbitrarily fine covers in the definition of Hausdorff measure, and ${\mathcal{H}}^{\gamma }(F)=0$. Therefore the pressure upper bound for Hausdorff dimension yieldsThe covering argument supplies the requested upper bound; it does not assume a dimension equality or a lower bound not proved here.

Use the geometric properties of the Feigenbaum geometric potential: for some constants $0<b\le B<\infty$,Here ${\mathrm{var}}_r$ means variation between sequences agreeing in their first $r$ symbols. In particular $D={\sum }_{r\ge 1}{\mathrm{var}}_{r}U<\infty$. These are the standard bounds for the Feigenbaum geometric potential; the symbolic potential is not the singular expression $-\log |g^{\prime }|$ 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 ${\mathrm{var}}_{j}U$. ThereforeFor $a_n=\log Z_n$, 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 $e^{\gamma U(1,0,0,\dots )}{Z}_{n-1}(\gamma ;(1,0,0,\dots ))$. 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 $-bn$, henceTaking normalized logarithms and limits givesSo $f$ is Lipschitz continuous and strictly decreasing. For each finite $n$, differentiation of the log-sum gives $f_n^{\prime \prime }(\gamma )=n^{-1}{\mathrm{Var}}_{n,\gamma }(H_n)\ge 0$. 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 $f(0)=\log 2$. For $\gamma >0$ the preceding bounds give $\log 2-B\gamma \le f(\gamma )\le \log 2-b\gamma$. The function therefore becomes negative at sufficiently large positive $\gamma$. Continuity and strict decrease prove