## Unique MMEs with specification – an alternate proof

The variational principle for topological entropy says that if ${X}$ is a compact metric space and ${f\colon X\rightarrow X}$ is a continuous map, then ${h_{\mathrm{top}}(f) = \sup_\mu h_\mu(f)}$, where ${h_{\mathrm{top}}}$ is the topological entropy, and the supremum is taken over all ${f}$-invariant Borel probability measures. A measure achieving this supremum is called a measure of maximal entropy (MME for short), and it is interesting to understand when a system has a unique MME.

Let’s look at this question in the setting of symbolic dynamics. Let ${A}$ be a finite set, which we call an alphabet, and let ${A^{\mathbb N}}$ be the set of all infinite sequences of symbols in ${A}$. This is a compact metric space in a standard way, and the shift map ${\sigma}$ defined by ${(\sigma(x))_n = x_{n+1}}$ is continuous. We consider a compact ${\sigma}$-invariant set ${X\subset A^{\mathbb N}}$ and ask whether or not ${(X,\sigma)}$ has a unique MME.

When ${X}$ is a mixing subshift of finite type (SFT), this was answered in the 1960’s by Parry; there is a unique MME, and it can be obtained by considering the transition matrix for ${X}$ and using some tools from linear algebra. A different proof was given by Bowen in 1974 using the specification property; this holds for all mixing SFTs, but also for a more general class of shift spaces.

The purpose of this note is to describe a variant of Bowen’s proof; roughly speaking, we follow Bowen’s proof for the first half of the argument, then give a shorter version of the second half of the argument, which follows comments made in conversation with Dan Thompson and Mike Hochman.

1. The strategy

The language of the shift space ${X}$ is the set of all finite words that appear in some element of ${X}$. That is, given ${n\in {\mathbb N}}$ we write ${\mathcal{L}_n = \{w\in A^n \mid [w]\neq\emptyset\}}$, where ${[w] = \{x\in X \mid x_1\cdots x_n = w\}}$. Then we write ${\mathcal{L} = \bigcup_n \mathcal{L}_n}$. Write ${|w|}$ for the length of ${w}$.

In the setting of symbolic dynamics, the topological transitivity property takes the following form: ${X}$ is transitive if and only if for every ${u,v\in \mathcal{L}}$ there is ${w\in \mathcal{L}}$ such that ${uwv\in \mathcal{L}}$. Transitivity by itself gives no control over the length of ${w}$. We say that shift space ${X}$ has specification if there is ${\tau\in {\mathbb N}}$ such that for every ${u,v\in \mathcal{L}}$ there is ${w\in \mathcal{L}}$ with ${|w|=\tau}$ such that ${uwv\in \mathcal{L}}$; thus specification can be thought of as a uniform transitivity property.

Let ${h = h_{\mathrm{top}}(f) = \lim_{n\rightarrow\infty} \frac 1n \log \#\mathcal{L}_n}$ be the topological entropy of ${(X,\sigma)}$. Bowen’s proof of uniqueness has the following structure:

1. Show that there is ${C>0}$ such that ${e^{nh} \leq \#\mathcal{L}_n \leq C e^{nh}}$ for every ${n}$.
2. Prove that for every ${\alpha>0}$ there is ${\beta>0}$ such that if ${\mu}$ is an MME and ${\mathcal{D}_n \subset \mathcal{L}_n}$ is a collection of words with ${\mu(\mathcal{D}_n) \geq \alpha}$, then we have ${\#\mathcal{D}_n \geq \beta e^{nh}}$. This can be thought of as a uniform version of the Katok entropy formula.
3. Follow the proof of the variational principle to explicitly construct an MME ${\mu}$; then use the specification property and the counting estimates on ${\#\mathcal{L}_n}$ to show that ${\mu}$ has the Gibbs property and is ergodic.
4. Show that if ${\nu}$ is another ergodic MME, then the uniform Katok entropy formula for ${\nu}$ and the Gibbs property for ${\mu}$ cannot hold simultaneously; this contradiction proves uniqueness.

The proof we give here follows the above structure exactly for steps 1 and 2. Instead of steps 3 and 4, though, we give the following argument; if ${\nu\neq \mu}$ are two distinct ergodic MMEs, then we can use the uniform Katok entropy formula for ${\nu,\mu}$ together with the specification property to create more entropy in the system, a contradiction. In the next section we make all of this precise.

The advantage of this proof is that it allows us to replace step 3 of Bowen’s proof with a different argument that may be easier and less technical (depending on one’s taste). The disadvantage is that it does not include a proof of the Gibbs property for ${\mu}$, which is useful to know about in various settings.

2. The proof

2.1. Counting estimates

Write ${\mathcal{L}_{m} \mathcal{L}_n}$ for the set of all words of the form ${vw}$, where ${v\in \mathcal{L}_m}$ and ${w\in \mathcal{L}_n}$. Then it is clear that ${\mathcal{L}_{m+n} \subset \mathcal{L}_m \mathcal{L}_n}$, so ${\#\mathcal{L}_{m+n} \leq (\#\mathcal{L}_m)(\#\mathcal{L}_n)}$. In particular, ${\log \#\mathcal{L}_n}$ is subadditive, so by Fekete’s lemma ${h = \lim \frac 1n \log \#\mathcal{L}_n = \inf_n \frac 1n \log \#\mathcal{L}_n}$, and we get ${\#\mathcal{L}_n \geq e^{nh}}$ for every ${n}$.

The upper bound requires the specification property. Define a map ${(\mathcal{L}_n)^k \rightarrow \mathcal{L}_{k(n+\tau)}}$ by sending ${w_1,\dots,w_k}$ to ${w_1 u_1 w_2 u_2 \cdots w_k u_k}$, where ${u_i \in \mathcal{L}_\tau}$ are provided by specification. This map is 1-1 so ${\#\mathcal{L}_{k(n+\tau)} \geq (\#\mathcal{L}_n)^k}$. Taking logs gives

$\displaystyle \frac 1{k(n+\tau)} \log\#\mathcal{L}_{k(n+\tau)} \geq \frac 1{n+\tau} \log \#\mathcal{L}_n,$

and sending ${k\rightarrow\infty}$ takes the left-hand side to ${h}$, so ${\#\mathcal{L}_n \leq e^{(n+\tau)h}}$; this gives the counting bounds we claimed earlier, with ${C= e^{\tau h}}$.

The gluing construction in the previous paragraph will be used later on when we need to derive a contradiction by producing extra entropy.

2.2. Uniform Katok formula

Now suppose ${\mu}$ is any MME, and given ${w\in \mathcal{L}}$ write ${\mu(w) = \mu([w])}$. Similarly write ${\mu(\mathcal{D}_n) = \mu(\bigcup_{w\in \mathcal{D}_n} [w])}$. Applying Fekete’s lemma to the subadditive sequence ${H_n(\mu) = \sum_{w\in \mathcal{L}_n} -\mu(w) \log \mu(w)}$, we get ${nh \leq H_n(\mu)}$.

Given ${\mathcal{D}_n}$ with ${\mu(\mathcal{D}_n)\geq \alpha}$, we have

\displaystyle \begin{aligned} nh &\leq \sum_{w\in \mathcal{D}_n} - \mu(w)\log \mu(w) + \sum_{w\in \mathcal{D}_n^c} -\mu(w)\log\mu(w) \\ &= \mu(\mathcal{D}_n)\sum_{w\in \mathcal{D}_n} - \frac{\mu(w)}{\mu(\mathcal{D}_n)} \log \frac{\mu(w)}{\mu(\mathcal{D}_n)} + \mu(\mathcal{D}_n^c)\sum_{w\in \mathcal{D}_n^c} - \frac{\mu(w)}{\mu(\mathcal{D}_n^c)} \log \frac{\mu(w)}{\mu(\mathcal{D}_n^c)} \\ &\qquad - \mu(\mathcal{D}_n)\log\mu(\mathcal{D}_n) - \mu(\mathcal{D}_n^c)\log\mu(\mathcal{D}_n^c) \\ &\leq \mu(\mathcal{D}_n) \log\#\mathcal{D}_n + \mu(\mathcal{D}_n^c) \log \#\mathcal{D}_n^c + \log 2 \\ &\leq \mu(\mathcal{D}_n) \log\#\mathcal{D}_n + (1-\mu(\mathcal{D}_n)) (nh + \log C) + \log 2 \\ &= \mu(\mathcal{D}_n) (\log \#\mathcal{D}_n - nh - \log C) + nh + \log (2C). \end{aligned}

Solving for ${\#\mathcal{D}_n}$ gives

$\displaystyle \log\#\mathcal{D}_n \geq nh + \log C - \frac{\log(2C)}{\mu(\mathcal{D}_n)} \geq nh + \log C - \frac{\log(2C)}{\alpha},$

so taking ${\beta}$ with ${\log\beta = \log C - \log(2C)/\alpha}$ gives ${\log \#\mathcal{D}_n \geq \beta e^{nh}}$, verifying the uniform Katok formula.

Note that the argument here does not rely directly on the specification property; it only uses the fact that ${\mu}$ is an MME and that we have the uniform upper bound ${\#\mathcal{L}_n \leq Ce^{nh}}$. In fact, if one is a little more careful with the computations it is possible to show that we can take ${\beta = \beta(\alpha) = C^{1-\frac 1\alpha} e^{-\frac{h(\alpha)}\alpha}}$, where ${h(\alpha) = -\alpha\log\alpha - (1-\alpha)\log(1-\alpha)}$. This has the pleasing property that ${\beta\rightarrow 1}$ as ${\alpha\rightarrow 1}$, so the lower bound for ${\#\mathcal{D}_n}$ converges to the lower bound for ${\#\mathcal{L}_n}$.

2.3. Producing more entropy

Now suppose ${\mu,\nu}$ are two distinct ergodic MMEs. Then ${\mu\perp \nu}$ so there are disjoint sets ${Y,Z\subset X}$ with ${\mu(Y) = 1}$ and ${\nu(Z)=1}$. Fix ${\delta>0}$ and let ${Y'\subset Y}$ and ${Z'\subset Z}$ be compact sets such that ${\mu(Y')> 1-\delta}$ and ${\nu(Z')>1-\delta}$. Then the distance between ${Y',Z'}$ is positive, so there is ${N\in {\mathbb N}}$ such that for every ${n\geq N-\tau}$, no ${n}$-cylinder intersects both ${Y',Z'}$. In particular, putting ${\mathcal{Y}_n = \{w\in \mathcal{L}_n \mid [w] \cap Y' \neq\emptyset\}}$, and similarly for ${\mathcal{Z}_n}$ with ${Z'}$, we have

• ${\mathcal{Y}_n \cap \mathcal{Z}_n = \emptyset}$ for every ${n\geq N-\tau}$, and
• ${\mu(\mathcal{Y}_n) \geq 1-\delta}$ and ${\nu(\mathcal{Z}_n) \geq 1-\delta}$, so by the uniform Katok formula we have ${\#\mathcal{Y}_n,\#\mathcal{Z}_n \geq \beta e^{nh}}$, where ${\beta = \beta(1-\delta) > 0}$.

Up to now all of the arguments that we have made appear in Bowen’s proof; the approximation argument just given appears in step 4 of his proof, where one uses the Gibbs property of the constructed MME to derive a contradiction with the uniform Katok formula. It is at this point that our argument diverges from Bowen’s, since we have not proved a Gibbs property. Instead, we apply the specification property to the collections of words ${\mathcal{Y},\mathcal{Z} \subset \mathcal{L}}$ to get a lower bound on ${\#\mathcal{L}_n}$ that grows more quickly than ${e^{nh}}$, a contradiction.

Before giving the correct argument, let me describe three incorrect arguments that illustrate the main ideas and also show why certain technical points arise in the final argument; in particular this will describe the process of arriving at the proof, which I think is worthwhile.

First wrong argument

Here is a natural way to proceed. With ${N,\mathcal{Y},\mathcal{Z}}$ as above, consider for each ${k\in {\mathbb N}}$ and each ${\omega\in \{0,1\}^k}$ the set of all words

$\displaystyle w^1 u^1 w^2 u^2 \cdots w^k,$

where ${w^i \in \mathcal{Y}_{N-\tau}}$ if ${\omega_i = 0}$ and ${\mathcal{Z}_{N-\tau}}$ if ${\omega_i=1}$, and ${u^i \in \mathcal{L}_\tau}$ are the gluing words provided by specification. Note that each choice of ${\omega}$ produces at least ${(\beta e^{(N-\tau)h})^k}$ words in ${\mathcal{L}_{kN}}$. Moreover, the collections of words produced by different choices of ${\omega}$ are disjoint, because ${\mathcal{Y}_{N-\tau}}$ and ${\mathcal{Z}_{N-\tau}}$ are disjoint. We conclude that

$\displaystyle \#\mathcal{L}_{kN} \geq 2^k \beta^k e^{k(N-\tau)h},$

so

$\displaystyle \tfrac 1{kN}\log \#\mathcal{L}_{kN} \geq h + \tfrac 1N(\log(2\beta) - \tau h).$

If ${\log(2\beta) > \tau h}$ then this would be enough to show that ${h_\mathrm{top} > h}$, a contradiction. Unfortunately since ${\beta< 1}$ this argument does not work if ${\tau h \geq 2}$, so we must try again…

Second wrong argument

Looking at the above attempted proof, one may describe the problem as follows: each of the words ${u^i}$ makes us lose a factor of ${e^{-\tau h}}$ from our estimate on ${\#\mathcal{L}_{kN}}$. If ${\omega_i = \omega_{i+1}}$, then instead of letting ${w^i}$ and ${w^{i+1}}$ range over ${\mathcal{Y}_{N-\tau}}$ and then gluing them, we could replace the words ${w^i u^i w^{i+1}}$ with the words ${v\in \mathcal{Y}_{2N-\tau}}$. In particular, this would replace the estimate ${\beta^2 e^{2(N-\tau)}}$ with the estimate ${\beta e^{2N - \tau}}$.

This suggests that given ${\omega\in \{0,1\}^k}$, we should only keep track of the set ${J \subset \{1,\dots, k\}}$ for which ${\omega_{j+1} \neq \omega_j}$, since if ${\omega_{i+1} = \omega_i}$ we can avoid losing the factor of ${e^{-\tau h}}$ by avoiding the gluing process.

So, let’s try it. Given ${J = \{j_1 < \cdots < j_\ell \} \subset \{1, \dots, k\}}$, let ${m_i = j_{i+1} - j_i}$ (with ${m_0=j_1}$ and ${m_\ell = k - j_\ell}$), and consider the map

$\displaystyle \pi_J \colon \mathcal{Y}_{m_0 N - \tau} \times \mathcal{Z}_{m_1 N - \tau} \times \mathcal{Y}_{m_2 N - \tau} \times \cdots \times \mathcal{Y}_{m_{\ell-1}N - \tau} \times \mathcal{Z}_{m_\ell N} \rightarrow \mathcal{L}_{kN}$

given by specification (whether the product ends with ${\mathcal{Y}}$ or ${\mathcal{Z}}$ depends on the parity of ${\ell}$).

Let ${\mathcal{D}^J_{kN}}$ be the image of ${\pi_J}$, and note that

$\displaystyle \#\mathcal{D}_{kN}^J \geq \prod_{i=0}^\ell \beta e^{(m_iN - \tau) h} \geq \beta^{\ell + 1} e^{(kN - \ell\tau)h}. \ \ \ \ \ (1)$

If the collections ${\mathcal{D}_{kN}^J}$, ${\mathcal{D}_{kN}^{J'}}$ were disjoint for different choices of ${J,J'}$, then we could fix ${\gamma>0}$ and sum (1) over all ${J}$ with ${\#J \leq \gamma k}$ to get

$\displaystyle \#\mathcal{L}_{kN} \geq \left(\sum_{\ell = 0}^{\gamma k} {k \choose \ell} \right) \beta^{\gamma k} e^{(kN - \gamma k \tau)h} \geq e^{(-\gamma \log \gamma)k} e^{k(\gamma\log\beta + Nh - \gamma\tau h)},$

where we use Sitrling’s approximation for the last inequality. Taking logs and dividing by ${kN}$ gives

$\displaystyle \tfrac 1{kN} \log \#\mathcal{L}_{kN} \geq h + \tfrac \gamma N ( -\log \gamma + \log\beta - \tau h).$

For ${\gamma>0}$ sufficiently small this is ${> h}$, which gives ${h_\mathrm{top} > h}$, a contradiction.

The problem with this argument is that the collections ${\mathcal{D}_{kN}^J}$ need not be disjoint for different choices of ${J}$; this is because ${w\in \mathcal{Y}_{mN}}$ may have ${w_{[jN, (j+1)N)} \in \mathcal{Z}}$ for some value of ${j}$, so that we cannot necessarily recover ${J}$ uniquely from knowing ${w\in \mathcal{D}_{kN}^J}$.

Third wrong argument

Let’s try to address the issue just raised, that we cannot recover ${J}$ uniquely from ${w\in \mathcal{D}_{kN}^J}$ because ${w\in \mathcal{Y}_{mN}}$ may have subwords in ${\mathcal{Z}}$. We address this by only using words ${w}$ where there are not many’ such subwords. More precisely, given ${w\in \mathcal{Y}_{n}}$ for ${n\geq N-\tau}$, let

$\displaystyle B(w) = \{ 0 \leq j < \tfrac{n+\tau}N \mid w_{[jN, (j+1)N-\tau)}\in \mathcal{Z} \}$

be the set of bad’ times, and similarly with the roles of ${\mathcal{Y}}$ and ${\mathcal{Z}}$ reversed. Let ${b(w) = \#B(w)}$, and observe that since ${\mu(Z') < \delta}$, invariance of ${\mu}$ gives

$\displaystyle \mu\{w\in \mathcal{Y}_{n} \mid j\in B(w)\} \leq \delta$

for every ${0\leq j < \frac{n+\tau}N}$. We conclude that

$\displaystyle \sum_{w\in \mathcal{Y}_{n}} b(w)\mu(w) \leq \sum_{j=0}^{\lfloor\frac{n+\tau}N\rfloor-1} \mu\{w\in \mathcal{Y}_{n} \mid j\in B(w)\} \leq \delta \tfrac{n+\tau}N.$

Let ${\mathcal{Y}'_{n} = \{w\in \mathcal{Y}_{n} \mid b(w) \leq 2\delta \frac{n+\tau}N\}}$, and note that

$\displaystyle \delta \tfrac{n+\tau}N \geq \sum_{w\in \mathcal{Y}_{n} \setminus \mathcal{Y}_{n}'} \mu(w) 2\delta \tfrac{n+\tau}N,$

so

$\displaystyle \tfrac 12 \geq \mu(\mathcal{Y}_n) - \mu(\mathcal{Y}_n') \geq 1-\delta - \mu(\mathcal{Y}_n').$

We conclude that

$\displaystyle \mu(\mathcal{Y}_{n}') \geq \tfrac 12 - \delta,$

so taking ${\beta = \beta(\frac 12 - \delta)}$, the uniform Katok estimate gives ${\#\mathcal{Y}_{n}' \geq \beta e^{nh}}$. A similar definition and argument gives ${\mathcal{Z}_{n}'}$.

Now we repeat the argument from the previous section (the second wrong argument) using ${\mathcal{Y}',\mathcal{Z}'}$ in place of ${\mathcal{Y},\mathcal{Z}}$. Given ${J\subset \{1,\dots, k\}}$, let ${\mathcal{D}_{kN}^J}$ be the image of the map ${\pi_J}$ with the corresponding restricted domain, and note that the estimate (1) still holds (with the new value of ${\beta}$).

The final piece of the puzzle is to take ${v \in \mathcal{D}_{kN}^J}$ and estimate how many other collections ${\mathcal{D}_{kN}^{J'}}$ can contain it; that is, how many possibilities there are for ${J}$ once we know ${v}$. We would like to do the following: write ${v = w^1 u^1 w^2 u^2 \cdots w^\ell}$ and then argue that ${J'}$ must be contained in the union of the sets of times corresponding to the ${B(w^i)}$. The problem is that this only works when there is a disagreement between which of the sets ${\mathcal{Y}',\mathcal{Z}'}$ the maps ${\pi_J,\pi_{J'}}$ are trying to use, and so I cannot quite make this give us the bounds we want.

The correct argument

To fix this final issue we change the construction slightly; instead of letting ${J}$ mark the times where we transition between ${\mathcal{Y}',\mathcal{Z}'}$, we do a construction where at each ${j\in J}$ we transition from ${\mathcal{Y}'}$ to ${\mathcal{Z}}$ and then immediately back to ${\mathcal{Y}'}$. Then ${v = \mathcal{D}_{kN}^J \cap \mathcal{D}_{kN}^{J'}}$ will impose strong conditions on the set ${J'}$.

Let’s make everything precise and give the complete argument. As in the very beginning of this section we fix ${\delta>0}$ and let ${Y',Z'}$ be disjoint compact sets with ${\mu(Y'),\nu(Z')>1-\delta}$, where ${\mu,\nu}$ are distinct ergodic MMEs. Let ${N}$ be such that for every ${n\geq N-\tau}$, no ${n}$-cylinder intersects both ${Y',Z'}$.

Let ${\mathcal{Y}_n = \{w\in \mathcal{L}_n \mid [w] \cap Y' \neq \emptyset\}}$, and similarly for ${\mathcal{Z}_n}$. We repeat verbatim part of the argument from the last section; given ${w\in \mathcal{Y}_{n}}$ for ${n\geq N-\tau}$, let

$\displaystyle B(w) = \{ 0 \leq j < \tfrac{n+\tau}N \mid w_{[jN, (j+1)N-\tau)}\in \mathcal{Z} \}$

be the set of `bad’ times. Let ${b(w) = \#B(w)}$, and observe that since ${\mu(Z') < \delta}$, invariance of ${\mu}$ gives

$\displaystyle \mu\{w\in \mathcal{Y}_{n} \mid j\in B(w)\} \leq \delta$

for every ${0\leq j < \frac{n+\tau}N}$. We conclude that

$\displaystyle \sum_{w\in \mathcal{Y}_{n}} b(w)\mu(w) \leq \sum_{j=0}^{\lfloor\frac{n+\tau}N\rfloor-1} \mu\{w\in \mathcal{Y}_{n} \mid j\in B(w)\} \leq \delta \tfrac{n+\tau}N.$

Let ${\mathcal{Y}'_{n} = \{w\in \mathcal{Y}_{n} \mid b(w) \leq 2\delta \frac{n+\tau}N\}}$, and note that

$\displaystyle \delta \tfrac{n+\tau}N \geq \sum_{w\in \mathcal{Y}_{n} \setminus \mathcal{Y}_{n}'} \mu(w) 2\delta \tfrac{n+\tau}N,$

so

$\displaystyle \tfrac 12 \geq \mu(\mathcal{Y}_n) - \mu(\mathcal{Y}_n') \geq 1-\delta - \mu(\mathcal{Y}_n').$

We conclude that

$\displaystyle \mu(\mathcal{Y}_{n}') \geq \tfrac 12 - \delta,$

so taking ${\beta = \beta(\frac 12 - \delta)}$, the uniform Katok estimate gives ${\#\mathcal{Y}_{n}' \geq \beta e^{nh}}$.

Now given ${k\in {\mathbb N}}$ and ${J = \{j_1 < j_2< \cdots < j_\ell\} \subset \{1,\dots,k\}}$, let ${m_i = j_{i+1} - j_i}$ for ${0\leq i\leq \ell}$ (putting ${j_0=0}$ and ${j_{\ell+1} = k}$) and define a map

$\displaystyle \pi_J \colon \prod_{i=0}^{\ell+1} (\mathcal{Y}_{(m_i-1)N - \tau}' \times \mathcal{Z}_{N-\tau}) \rightarrow \mathcal{L}_{kN}$

by the specification property, so that for ${v^i \in \mathcal{Y}'_{(m_i-1)N-\tau}}$ and ${w^i\in \mathcal{Z}_{N-\tau}}$ we have

$\displaystyle \pi_J(v^0,w^0,\dots,v^\ell,w^\ell) = v^0 u^0 w^0 \hat u^0 v^1 \cdots v^\ell u^\ell w^\ell,$

where ${u^i,\hat u^i}$ have length ${\tau}$ and are provided by specification. Let ${\mathcal{D}_{kN}^J}$ be the image of ${\pi_J}$ and note that

$\displaystyle \#\mathcal{D}_{kN}^J \geq \beta^{2\ell+2} e^{(kN - (2\ell +1)\tau)h}. \ \ \ \ \ (2)$

Given ${J}$ and ${x\in \mathcal{D}_{kN}^J}$, let ${\mathcal{J}(x,J)}$ denote the set of ${J' \subset \{1,\dots, k\}}$ such that ${x\in \mathcal{D}_{kN}^{J'}}$. Writing ${x = v^0 u^0 w^0 \hat u^0 \cdots v^\ell u^\ell w^\ell}$, note that ${x\in \mathcal{D}_{kN}^{J'}}$ implies that for each ${j\in J'}$, either ${j\in J}$ or there are consecutive elements ${j_i < j_{i+1}}$ of ${J}$ such that ${j_i < j < j_{i+1}}$, and in this latter case we have that ${v^i_{[(j - j_i)N, (j-j_i + 1)N-\tau)} \in \mathcal{Z}_{N-\tau}}$, so ${j-j_i \in B(v^i)}$. We conclude that

$\displaystyle J' \subset J \cup \left(\bigcup_{i=0}^\ell j_i + B(v^i)\right).$

By our choice of ${\mathcal{Y}'}$, for each ${x}$ the set on the right has at most ${\ell + 2\delta k}$ elements. In particular, we have

$\displaystyle \#\mathcal{J}(x,J) \leq 2^{\ell + 2\delta k}.$

This bounds the number of ${\mathcal{D}_{kN}^J}$ that can contain a given ${x\in \mathcal{L}_{kN}}$, and since there are ${\geq e^{(-\gamma\log\gamma)k}}$ distinct choices of ${J}$ with ${\#J \leq \gamma k}$, the bound in (2) gives

$\displaystyle \#\mathcal{L}_{kN} \geq 2^{-(\gamma + 2\delta)k} e^{(-\gamma\log\gamma)k}\beta^{2\gamma k+2} e^{(kN - (2\gamma k+1)\tau)h}.$

Taking logs gives

$\displaystyle \log \#\mathcal{L}_{kN} \geq kNh -(\gamma + 2\delta)k\log 2 - (\gamma\log\gamma)k + (2\gamma k + 2)\log\beta - (2\gamma k + 1)\tau h.$

Dividing by ${kN}$ and sending ${k\rightarrow\infty}$ gives

$\displaystyle h_\mathrm{top} \geq h + \tfrac 1N(-\gamma\log\gamma - (\gamma + 2\delta)\log 2 + 2\gamma\log\beta - 2\gamma\tau h).$

Putting ${\gamma = \delta}$ this gives

$\displaystyle h_\mathrm{top} \geq h + \tfrac \delta N(-\log\delta - \log 8 + 2\log\beta - 2\tau h).$

Thus it suffices to make the appropriate choice of ${\delta}$ at the beginning of the proof. More precisely, let ${\beta' = \beta(\frac 14)}$ be as in the uniform Katok lemma, and let ${\delta\in (0,\frac 14)}$ be small enough that ${-\log\delta > \log 8 - 2\log\beta' + 2\tau h}$. Then ${\beta(\frac 12-\delta) \geq \beta'}$ and so the estimate above gives

$\displaystyle h_\mathrm{top} \geq h + \tfrac\delta N(-\log\delta - \log 8 + 2\log\beta' - 2\tau h) > h,$

which contradicts our original assumption that ${h}$ was the topological entropy. This contradiction shows that there is a unique measure of maximal entropy.