[proofplan] We use the conformality relation $\mathcal{L}_\phi^*\nu=\lambda\nu$ to express the measure of a cylinder through the exponential weight $e^{S_n\phi}$ along its inverse branch. Hölder continuity gives a uniform distortion bound for $S_n\phi$ on each cylinder of length $n$, so the weight may be evaluated at any chosen point of the cylinder up to a fixed multiplicative constant. The image of a cylinder under $\sigma^n$ belongs to a finite family of follower sets, all of which have positive and finite $\nu$-measure. Finally, since $h$ is continuous and strictly positive on the [compact space](/page/Compact%20Space) $\Sigma_A$, replacing $\nu$ by $\mu=h\nu$ changes the estimate only by another uniform multiplicative constant. [/proofplan]

[step:Extract uniform bounds from Hölder continuity and positivity of the eigenfunction]Because $\Sigma_A$ is compact and $h: \Sigma_A \to (0,\infty)$ is continuous, define \begin{align*} h_- := \min_{y \in \Sigma_A} h(y) \end{align*} and \begin{align*} h_+ := \max_{y \in \Sigma_A} h(y). \end{align*} Then $0<h_- \leq h_+ < \infty$, and for every Borel set $B \subset \Sigma_A$, \begin{align*} h_- \nu(B) \leq \mu(B) \leq h_+ \nu(B). \end{align*} Let $D_\phi \geq 1$ be a distortion constant for the Hölder potential $\phi$, meaning that for every $n \geq 1$, every admissible word $w$ of length $n$, and all $x,z \in [w]$, \begin{align*} D_\phi^{-1} \leq \exp(S_n\phi(z)-S_n\phi(x)) \leq D_\phi. \end{align*} Such a constant exists because $\phi$ is Hölder and points in the same length-$n$ cylinder remain exponentially close for the first $n$ iterates.[/step]

[guided]The first issue is that the desired Gibbs estimate is stated for $\mu$, while the conformality relation is stated for $\nu$. We therefore record a uniform comparison between the two measures. Since $\Sigma_A$ is compact and $h$ is continuous and strictly positive, the extreme value theorem gives finite constants \begin{align*} h_- := \min_{y \in \Sigma_A} h(y) \end{align*} and \begin{align*} h_+ := \max_{y \in \Sigma_A} h(y) \end{align*} with $0<h_- \leq h_+ < \infty$. For every Borel set $B \subset \Sigma_A$, the definition \begin{align*} \mu(B) = \int_B h \, d\nu(y) \end{align*} therefore gives \begin{align*} h_- \nu(B) \leq \int_B h \, d\nu(y) \leq h_+ \nu(B), \end{align*} so \begin{align*} h_- \nu(B) \leq \mu(B) \leq h_+ \nu(B). \end{align*} The second issue is distortion. We need to compare $S_n\phi(z)$ and $S_n\phi(x)$ for two points $x,z$ lying in the same cylinder $[w]$ of length $n$. Since $\phi$ is Hölder, there are constants $K_\phi>0$, $\theta \in (0,1)$, and a compatible symbolic metric $d$ such that \begin{align*} |\phi(u)-\phi(v)| \leq K_\phi d(u,v)^\theta \end{align*} for all $u,v \in \Sigma_A$. If $x,z \in [w]$, then for $0 \leq k \leq n-1$, the points $\sigma^k x$ and $\sigma^k z$ agree in their first $n-k$ symbols, so their distance is bounded exponentially in $n-k$. Summing the Hölder bounds over $k$ gives a finite constant $V_\phi>0$, independent of $n$, $w$, $x$, and $z$, such that \begin{align*} |S_n\phi(z)-S_n\phi(x)| \leq V_\phi. \end{align*} Define $D_\phi := e^{V_\phi}$. Exponentiating the preceding inequality gives \begin{align*} D_\phi^{-1} \leq \exp(S_n\phi(z)-S_n\phi(x)) \leq D_\phi. \end{align*} This is the bounded distortion estimate that lets us evaluate the cylinder weight at an arbitrary point of the cylinder.[/guided]

[step:Rewrite the conformality relation on a cylinder inverse branch] Let $w=w_0\cdots w_{n-1}$ be an admissible word of length $n$, and let $[w]\subset\Sigma_A$ be its cylinder. Define the follower set \begin{align*} F_w := \sigma^n([w]). \end{align*} The map $\sigma^n|_{[w]}: [w]\to F_w$ is a bijection. Let \begin{align*} \tau_w: F_w \to [w] \end{align*} be its inverse branch, so that $\sigma^n(\tau_w(y))=y$ for all $y\in F_w$. The relation $\mathcal{L}_\phi^*\nu=\lambda\nu$ implies, by iteration, that $\mathcal{L}_\phi^{*n}\nu=\lambda^n\nu$. Let $\mathbb{1}_{[w]}: \Sigma_A \to \{0,1\}$ denote the indicator function of the cylinder $[w]$. By the definition of the adjoint transfer operator applied to $\mathbb{1}_{[w]}$, \begin{align*} \lambda^n\nu([w]) = \int_{\Sigma_A} \mathcal{L}_\phi^n\mathbb{1}_{[w]}(y) \, d\nu(y). \end{align*} For each $y\in\Sigma_A$, the $n$-step transfer operator is \begin{align*} \mathcal{L}_\phi^n\mathbb{1}_{[w]}(y) = \sum_{\sigma^n z=y} e^{S_n\phi(z)}\mathbb{1}_{[w]}(z). \end{align*} Because $\sigma^n|_{[w]}: [w]\to F_w$ is a bijection, this sum has exactly one nonzero term when $y\in F_w$, namely $z=\tau_w(y)$, and has no nonzero terms when $y\notin F_w$. Hence \begin{align*} \lambda^n\nu([w]) = \int_{F_w} e^{S_n\phi(\tau_w(y))} \, d\nu(y). \end{align*} Dividing by $\lambda^n$ gives \begin{align*} \nu([w]) = \lambda^{-n}\int_{F_w} e^{S_n\phi(\tau_w(y))} \, d\nu(y). \end{align*} [/step]

[step:Use bounded distortion to estimate the conformal measure of the cylinder] Fix $x\in[w]$. For every $y\in F_w$, the point $\tau_w(y)$ lies in $[w]$, so the distortion bound gives \begin{align*} D_\phi^{-1} e^{S_n\phi(x)} \leq e^{S_n\phi(\tau_w(y))} \leq D_\phi e^{S_n\phi(x)}. \end{align*} Integrating this inequality over $F_w$ with respect to $\nu$ yields \begin{align*} D_\phi^{-1}\lambda^{-n}e^{S_n\phi(x)}\nu(F_w) \leq \nu([w]) \leq D_\phi\lambda^{-n}e^{S_n\phi(x)}\nu(F_w). \end{align*} Because $\Sigma_A$ is a finite subshift of finite type, the image $F_w=\sigma^n([w])$ depends only on the terminal symbol $w_{n-1}$: explicitly, it is the set of all one-sided sequences whose first symbol may legally follow $w_{n-1}$. Thus the collection of possible nonempty follower sets is a finite collection of finite unions of one-symbol cylinders. The [Ruelle-Perron-Frobenius theorem](/theorems/6817) supplies a positive eigenmeasure $\nu$ with full support on $\Sigma_A$; therefore every nonempty open cylinder has positive $\nu$-measure, and every nonempty follower set has positive $\nu$-measure because it contains at least one nonempty cylinder. Define \begin{align*} m_F := \min\{\nu(F): F \text{ is a nonempty follower set}\} \end{align*} and \begin{align*} M_F := \max\{\nu(F): F \text{ is a nonempty follower set}\}. \end{align*} Then $0<m_F\leq M_F\leq 1$, and therefore \begin{align*} D_\phi^{-1}m_F \lambda^{-n}e^{S_n\phi(x)} \leq \nu([w]) \leq D_\phi M_F \lambda^{-n}e^{S_n\phi(x)}. \end{align*} [/step]

[step:Transfer the estimate from $\nu$ to $\mu$ and identify the pressure] Using the comparison $h_-\nu(B)\leq \mu(B)\leq h_+\nu(B)$ with $B=[w]$, the preceding estimate gives \begin{align*} h_-D_\phi^{-1}m_F \lambda^{-n}e^{S_n\phi(x)} \leq \mu([w]) \leq h_+D_\phi M_F \lambda^{-n}e^{S_n\phi(x)}. \end{align*} Since $P=\log\lambda$, we have \begin{align*} \lambda^{-n}e^{S_n\phi(x)} = \exp(S_n\phi(x)-nP). \end{align*} Define \begin{align*} C := \max\left\{ \frac{D_\phi}{h_-m_F}, h_+D_\phi M_F, 1 \right\}. \end{align*} Then $C\geq 1$ and, for every admissible word $w$ of length $n\geq 1$ and every $x\in[w]$, \begin{align*} C^{-1}\exp(S_n\phi(x)-nP) \leq \mu([w]) \leq C\exp(S_n\phi(x)-nP). \end{align*} Equivalently, \begin{align*} C^{-1} \leq \frac{\mu([w])}{\exp(S_n\phi(x)-nP)} \leq C. \end{align*} This is exactly the Gibbs property for $\mu$ with potential $\phi$. [/step]