[proofplan] We view pressure through the leading eigenvalue of the Ruelle transfer operator. The real-analytic dependence of the potentials implies real-analytic dependence of the transfer operators on the Hölder [Banach space](/page/Banach%20Space). Since the leading eigenvalue is simple and isolated throughout the interval, analytic perturbation theory for isolated simple eigenvalues gives a real-analytic eigenvalue branch $t \mapsto \lambda(t)$. The variational/Ruelle pressure formula identifies $P(\sigma,\beta_t)=\log \lambda(t)$, and the logarithm is real analytic on $(0,\infty)$, so the pressure is real analytic and therefore cannot have a first-order transition. [/proofplan] [step:Realize the pressure as the logarithm of the leading transfer eigenvalue] Fix $t_0 \in I$. Let $A$ be the finite transition matrix from the theorem statement, let $\Sigma_A$ be the associated one-sided subshift, let $\sigma:\Sigma_A\to\Sigma_A$ be the left shift, and let $\alpha\in(0,1]$ be the Hölder exponent from the statement. Let $X := C^\alpha(\Sigma_A;\mathbb{R})$ denote the real Banach space of real-valued $\alpha$-Hölder functions on $\Sigma_A$, equipped with its usual Hölder norm. For each $t \in I$, define the transfer operator \begin{align*} \mathcal{L}_t:X \to X, \qquad (\mathcal{L}_t f)(x)=\sum_{\sigma y=x} e^{\beta_t(y)} f(y). \end{align*} Because $(\Sigma_A,\sigma)$ is a subshift of finite type over a finite alphabet, the fiber $\sigma^{-1}\{x\}$ is finite for every $x \in \Sigma_A$. Since $\beta_t \in C^\alpha(\Sigma_A;\mathbb{R})$, multiplication by $e^{\beta_t}$ preserves Hölder regularity, and hence $\mathcal{L}_t$ is a [bounded linear operator](/page/Bounded%20Linear%20Operator) on $X$. By the spectral-gap hypothesis, understood here in the standard Ruelle-operator sense that the spectral radius is a positive simple eigenvalue separated from the rest of the spectrum, the spectral radius $\lambda(t)>0$ of $\mathcal{L}_t$ is a simple isolated eigenvalue. The Ruelle-Perron-Frobenius pressure formula for Hölder potentials on a topologically mixing subshift of finite type gives \begin{align*} P(\sigma,\beta_t)=\log \lambda(t). \end{align*} The uniqueness of the equilibrium state is compatible with this spectral description: it identifies the equilibrium state with the Gibbs state determined by the leading eigenfunction and eigenmeasure. Thus the uniqueness hypothesis rules out coexistence of distinct equilibrium states, while the analyticity of the pressure follows from the spectral-gap hypothesis together with the pressure formula. [guided] The purpose of this step is to translate a thermodynamic statement into a spectral statement. The pressure is initially defined dynamically, through the variational principle \begin{align*} P(\sigma,\beta_t)=\sup_{\mu \in \mathcal{M}_\sigma(\Sigma_A)}\left(h_\mu(\sigma)+\int_{\Sigma_A}\beta_t\,d\mu\right), \end{align*} where $\mathcal{M}_\sigma(\Sigma_A)$ denotes the set of $\sigma$-invariant Borel probability measures and $h_\mu(\sigma)$ denotes the measure-theoretic entropy of $\sigma$ with respect to $\mu$. For Hölder potentials on a topologically mixing subshift of finite type, the Ruelle-Perron-Frobenius pressure formula identifies this variational quantity with the logarithm of the maximal eigenvalue of the Ruelle transfer operator. For each $t \in I$, the transfer operator is \begin{align*} \mathcal{L}_t:X \to X, \qquad (\mathcal{L}_t f)(x)=\sum_{\sigma y=x} e^{\beta_t(y)} f(y), \end{align*} where $X=C^\alpha(\Sigma_A;\mathbb{R})$. The sum is finite because $\Sigma_A$ is a subshift of finite type over a finite alphabet. Since $\beta_t$ is Hölder, the function $e^{\beta_t}$ is Hölder, and the finite preimage sum preserves Hölder regularity. Thus $\mathcal{L}_t$ is a bounded linear operator on $X$. The spectral-gap hypothesis is being used in the standard Ruelle-operator sense: the spectral radius $\lambda(t)>0$ is not merely an element of the spectrum, but a positive simple isolated eigenvalue separated from the remaining spectrum. This is the exact spectral configuration needed for perturbation theory. The pressure formula then gives \begin{align*} P(\sigma,\beta_t)=\log \lambda(t). \end{align*} So it remains to prove that $t \mapsto \lambda(t)$ is real analytic. [/guided] [/step] [step:Show that the transfer operators depend real analytically on the parameter] Let $\mathcal{L}(X,X)$ denote the Banach space of bounded linear operators from $X$ to $X$, equipped with the operator norm. We claim that \begin{align*} I \to \mathcal{L}(X,X), \qquad t \mapsto \mathcal{L}_t \end{align*} is real analytic. Since $t \mapsto \beta_t$ is real analytic as an $X$-valued map, for every $t_0 \in I$ there exists $\varepsilon>0$ and elements $\gamma_k \in X$ such that, for $|t-t_0|<\varepsilon$, \begin{align*} \beta_t=\sum_{k=0}^{\infty}(t-t_0)^k\gamma_k \end{align*} with convergence in $X$. The exponential map $\operatorname{Exp}:X \to X$ defined by $\operatorname{Exp}(u)=e^u$ is real analytic because $X=C^\alpha(\Sigma_A;\mathbb{R})$ is a Banach algebra under pointwise multiplication and the [power series](/page/Power%20Series) for $e^u$ converges absolutely in the Banach algebra norm. Hence $t \mapsto e^{\beta_t}$ is real analytic as an $X$-valued map. For each $g \in X$, define the bounded-operator candidate $T_g:X \to X$ by \begin{align*} (T_g f)(x)=\sum_{\sigma y=x} g(y)f(y). \end{align*} The map $X \to \mathcal{L}(X,X)$ given by $g \mapsto T_g$ is linear. To see boundedness, let $N$ be the number of inverse branches of $\sigma$ on the finite-type shift, and let $C_\sigma \geq 1$ be a Hölder distortion constant depending only on the symbolic metric and the finite transition matrix. For the usual shift metric, each admissible inverse branch is a contraction; the only variation coming from changes in admissible predecessor sets is controlled by the supremum norm, and the finitely many predecessor symbols are absorbed into $N C_\sigma$. The Hölder product estimate gives \begin{align*} \|gf\|_{C^\alpha} \leq \|g\|_{C^\alpha}\|f\|_{C^\alpha}. \end{align*} Summing over the at most $N$ inverse branches and applying the contraction estimate for each branch yields \begin{align*} \|T_g f\|_{C^\alpha} \leq N C_\sigma \|g\|_{C^\alpha}\|f\|_{C^\alpha}. \end{align*} Hence \begin{align*} \|T_g\|_{\mathcal{L}(X,X)} \leq N C_\sigma \|g\|_{C^\alpha}, \end{align*} so $g \mapsto T_g$ is bounded. Therefore the composition \begin{align*} t \mapsto e^{\beta_t} \mapsto T_{e^{\beta_t}}=\mathcal{L}_t \end{align*} is real analytic from $I$ into $\mathcal{L}(X,X)$. [/step] [step:Apply analytic perturbation theory to the isolated leading eigenvalue] Fix $t_0 \in I$. Let $X_{\mathbb{C}}:=C^\alpha(\Sigma_A;\mathbb{C})$ be the complexification of $X$, and let $\mathcal{L}_{t,\mathbb{C}}:X_{\mathbb{C}}\to X_{\mathbb{C}}$ denote the complex-linear extension of $\mathcal{L}_t$. By hypothesis, $\lambda(t_0)>0$ is a simple isolated eigenvalue of $\mathcal{L}_{t_0,\mathbb{C}}$. Choose $r>0$ so small that the circle \begin{align*} \Gamma:=\{z \in \mathbb{C}: |z-\lambda(t_0)|=r\} \end{align*} contains no point of the spectrum of $\mathcal{L}_{t_0,\mathbb{C}}$ and encloses no spectral point of $\mathcal{L}_{t_0,\mathbb{C}}$ other than $\lambda(t_0)$. By analytic perturbation theory for isolated simple eigenvalues applied to the analytic family $t\mapsto \mathcal{L}_{t,\mathbb{C}}$, after shrinking to an open interval $J \subset I$ with $t_0 \in J$, the Riesz projection \begin{align*} \Pi_t:=\frac{1}{2\pi i}\oint_\Gamma (zI-\mathcal{L}_{t,\mathbb{C}})^{-1}\,dz \end{align*} depends real analytically on $t \in J$, has rank one, and its range is the eigenspace associated to a unique simple eigenvalue $\lambda_J(t)$ of $\mathcal{L}_{t,\mathbb{C}}$ enclosed by $\Gamma$. We now justify that this local branch is still the leading spectral-radius branch. Since $\lambda(t_0)>0$ is isolated and the spectral gap holds at $t_0$, define the positive modulus gap \begin{align*} \delta:=\lambda(t_0)-\sup\{|z|:z\in\sigma(\mathcal{L}_{t_0,\mathbb{C}})\setminus\{\lambda(t_0)\}\}>0. \end{align*} Choose $r>0$ small enough that $r<\delta/4$ and $\Gamma$ encloses only $\lambda(t_0)$. Norm-continuity of $t\mapsto\mathcal{L}_{t,\mathbb{C}}$ and upper semicontinuity of the spectrum imply that, after shrinking $J$, all spectral points of $\mathcal{L}_{t,\mathbb{C}}$ outside the disk bounded by $\Gamma$ have modulus at most $\lambda(t_0)-\delta/2$, while the enclosed eigenvalue satisfies $|\lambda_J(t)-\lambda(t_0)|<\delta/4$. Hence $|\lambda_J(t)|>\lambda(t_0)-\delta/4>\lambda(t_0)-\delta/2$, so the enclosed eigenvalue is the spectral-radius eigenvalue of $\mathcal{L}_{t,\mathbb{C}}$ on $J$. Moreover $t \mapsto \lambda_J(t)$ is real analytic on $J$ and $\lambda_J(t_0)=\lambda(t_0)$. Because the spectral radius eigenvalue is simple and isolated for every $t \in I$, the local analytic branches obtained in this way agree on overlaps: two such branches coincide at any common parameter where both represent the unique leading eigenvalue. Therefore the local branches glue to a globally defined real-analytic function \begin{align*} I \to (0,\infty), \qquad t \mapsto \lambda(t), \end{align*} where $\lambda(t)$ is the leading eigenvalue of $\mathcal{L}_t$. [guided] This is the spectral core of the proof. The operator family $t \mapsto \mathcal{L}_t$ is real analytic in the Banach-space operator norm, and at the fixed parameter $t_0$ the eigenvalue $\lambda(t_0)$ is isolated from the rest of the spectrum. We pass to the complexification $X_{\mathbb{C}}:=C^\alpha(\Sigma_A;\mathbb{C})$ because the Riesz projection is defined by a complex contour integral. Isolation lets us draw a small contour $\Gamma$ around $\lambda(t_0)$ and around no other spectral point of $\mathcal{L}_{t_0,\mathbb{C}}$: \begin{align*} \Gamma=\{z \in \mathbb{C}: |z-\lambda(t_0)|=r\}. \end{align*} The analytic perturbation theorem for isolated simple eigenvalues says that if an operator family depends analytically on a parameter, then the spectral data inside such a contour also depend analytically on the parameter, as long as the contour remains in the resolvent set. Thus, after restricting to a smaller interval $J$ around $t_0$, the Riesz projection \begin{align*} \Pi_t=\frac{1}{2\pi i}\oint_\Gamma (zI-\mathcal{L}_{t,\mathbb{C}})^{-1}\,dz \end{align*} is well-defined and real analytic in $t$. We shrink $J$ so that $\Gamma$ remains in the resolvent set and the spectral point enclosed by $\Gamma$ remains separated from the rest of the spectrum. To see that the enclosed point is still the spectral-radius eigenvalue, use the modulus gap at $t_0$. Define \begin{align*} \delta:=\lambda(t_0)-\sup\{|z|:z\in\sigma(\mathcal{L}_{t_0,\mathbb{C}})\setminus\{\lambda(t_0)\}\}>0. \end{align*} Choose the contour radius $r$ with $0<r<\delta/4$. By norm-continuity of $\mathcal{L}_{t,\mathbb{C}}$ and upper semicontinuity of spectra, after shrinking $J$ every spectral point outside the disk bounded by $\Gamma$ has modulus at most $\lambda(t_0)-\delta/2$, while the enclosed eigenvalue remains within $\delta/4$ of $\lambda(t_0)$. Therefore the enclosed eigenvalue has strictly larger modulus than every other spectral point and is the spectral-radius eigenvalue for every $t\in J$. The rank of a Riesz projection is locally constant under norm-continuous perturbation, so $\Pi_t$ has rank one on $J$ because $\Pi_{t_0}$ has rank one. Its range is therefore the eigenspace for a single simple eigenvalue, which we denote by $\lambda_J(t)$. The same perturbation theorem gives real analyticity of this eigenvalue branch: \begin{align*} J \to \mathbb{R}, \qquad t \mapsto \lambda_J(t). \end{align*} The branch is real-valued by conjugation invariance. Complex conjugation $C:X_{\mathbb{C}}\to X_{\mathbb{C}}$, defined by $C f=\overline{f}$, satisfies $C\mathcal{L}_{t,\mathbb{C}}=\mathcal{L}_{t,\mathbb{C}}C$ because $\mathcal{L}_t$ has real coefficients on the real Banach space. Hence the spectrum and the Riesz projection inside the conjugation-invariant circle $\Gamma$ are invariant under complex conjugation. Since the contour encloses exactly one simple eigenvalue, its conjugate must be the same eigenvalue. Therefore $\lambda_J(t)=\overline{\lambda_J(t)}$, so $\lambda_J(t)\in\mathbb{R}$. The positivity $\lambda_J(t)>0$ follows after shrinking $J$ if necessary, since $\lambda_J(t_0)>0$ and $\lambda_J$ is continuous. Finally, these local branches are compatible. If two neighborhoods overlap, both branches represent the simple isolated leading eigenvalue of the same operator $\mathcal{L}_t$ at each common parameter. Since that leading eigenvalue is unique by hypothesis, the branches agree on the overlap. Hence the local analytic functions glue to one real-analytic function \begin{align*} I \to (0,\infty), \qquad t \mapsto \lambda(t). \end{align*} [/guided] [/step] [step:Compose with the logarithm to obtain analytic pressure] The logarithm function \begin{align*} \log:(0,\infty)\to\mathbb{R} \end{align*} is real analytic. Since $t \mapsto \lambda(t)$ is real analytic and takes values in $(0,\infty)$, the composition $t \mapsto \log \lambda(t)$ is real analytic on $I$. Using the pressure identity from the first step, \begin{align*} P(\sigma,\beta_t)=\log \lambda(t), \end{align*} we conclude that \begin{align*} I \to \mathbb{R}, \qquad t \mapsto P(\sigma,\beta_t) \end{align*} is real analytic. [/step] [step:Exclude first-order pressure transitions] A first-order pressure transition at a parameter $t_0 \in I$ would require failure of differentiability of the pressure function at $t_0$, equivalently a jump discontinuity or nonexistence of the one-sided derivatives of $t \mapsto P(\sigma,\beta_t)$ at $t_0$. Since $I$ is an open interval by the theorem statement, real-analyticity on $I$ means real-analyticity in a neighbourhood of every parameter $t_0\in I$. Real-analytic functions on open intervals are differentiable of all orders. Since $t \mapsto P(\sigma,\beta_t)$ is real analytic on all of $I$, no such parameter $t_0$ exists. Therefore the family $(\beta_t)_{t \in I}$ has no first-order pressure transition on $I$. [/step]