[proofplan] We compare the separated-set growth for the flow with the separated-set growth for the sampled map $\varphi_a$. Separation under the first $n$ iterates of $\varphi_a$ is immediately separation along the flow segment of length $an$, giving one inequality. For the reverse inequality, [uniform continuity](/page/Uniform%20Continuity) of the action on the compact set $[0,a]\times X$ says that closeness at sampling times $ja$ forces closeness during the entire interval $[ja,(j+1)a]$. Thus a flow-separated set over time $T$ must be separated for $\varphi_a$ over about $T/a$ iterates, and the ceiling error disappears after dividing by $T$. [/proofplan]

[step:Fix the separated-set notation for the flow and for the sampled map] For $T>0$, define the continuous-time Bowen metric $d_T^\varphi:X\times X\to[0,\infty)$ by \begin{align*} d_T^\varphi(x,y)=\sup_{0\le t\le T} d(\varphi_t(x),\varphi_t(y)). \end{align*} For $\varepsilon>0$, let $s_T(\varphi,\varepsilon)$ denote the maximal cardinality of a subset $E\subset X$ such that $d_T^\varphi(x,y)>\varepsilon$ for all distinct $x,y\in E$. For a continuous map $f:X\to X$ and $n\in\mathbb{N}$, define the discrete-time Bowen metric $d_n^f:X\times X\to[0,\infty)$ by \begin{align*} d_n^f(x,y)=\max_{0\le k\le n-1} d(f^k(x),f^k(y)). \end{align*} For $\varepsilon>0$, let $s_n(f,\varepsilon)$ denote the maximal cardinality of a subset $E\subset X$ such that $d_n^f(x,y)>\varepsilon$ for all distinct $x,y\in E$. With these conventions, \begin{align*} h_{\mathrm{top}}(\varphi)=\lim_{\varepsilon\downarrow 0}\limsup_{T\to\infty}\frac{1}{T}\log s_T(\varphi,\varepsilon) \end{align*} and \begin{align*} h_{\mathrm{top}}(f)=\lim_{\varepsilon\downarrow 0}\limsup_{n\to\infty}\frac{1}{n}\log s_n(f,\varepsilon). \end{align*} We shall apply the second definition to the map $f=\varphi_a:X\to X$. [/step]

[step:Compare discrete-time separation with flow separation along the same sampled orbit segment]Fix $\varepsilon>0$ and $n\in\mathbb{N}$. If $E\subset X$ is $\varepsilon$-separated for $\varphi_a$ over $n$ iterates, then for every distinct $x,y\in E$ there exists an integer $k$ with $0\le k\le n-1$ such that \begin{align*} d(\varphi_a^k(x),\varphi_a^k(y))>\varepsilon. \end{align*} Since $\varphi_a^k=\varphi_{ak}$ by the flow law and $ak\in[0,an]$, we get \begin{align*} d_{an}^\varphi(x,y)\ge d(\varphi_{ak}(x),\varphi_{ak}(y))>\varepsilon. \end{align*} Thus $E$ is $\varepsilon$-separated for the flow over time $an$, and hence \begin{align*} s_n(\varphi_a,\varepsilon)\le s_{an}(\varphi,\varepsilon). \end{align*} Taking logarithms, dividing by $n$, and passing to $\limsup_{n\to\infty}$ gives \begin{align*} \limsup_{n\to\infty}\frac{1}{n}\log s_n(\varphi_a,\varepsilon) \le a\limsup_{n\to\infty}\frac{1}{an}\log s_{an}(\varphi,\varepsilon). \end{align*} The limsup over the subsequence $T=an$ is bounded above by the limsup over all $T\to\infty$, so \begin{align*} \limsup_{n\to\infty}\frac{1}{n}\log s_n(\varphi_a,\varepsilon) \le a\limsup_{T\to\infty}\frac{1}{T}\log s_T(\varphi,\varepsilon). \end{align*} Letting $\varepsilon\downarrow0$ yields \begin{align*} h_{\mathrm{top}}(\varphi_a)\le a\,h_{\mathrm{top}}(\varphi). \end{align*}[/step]

[guided]Fix a scale $\varepsilon>0$ and an integer $n\in\mathbb{N}$. The discrete map $\varphi_a$ samples the flow at the times $0,a,2a,\dots,(n-1)a$. Thus any pair of points separated by the first $n$ iterates of $\varphi_a$ is also separated somewhere along the continuous flow segment from time $0$ to time $an$. Formally, let $E\subset X$ be $\varepsilon$-separated for $\varphi_a$ over $n$ iterates. For distinct $x,y\in E$, the definition of $d_n^{\varphi_a}$ gives an integer $k$ with $0\le k\le n-1$ such that \begin{align*} d(\varphi_a^k(x),\varphi_a^k(y))>\varepsilon. \end{align*} The flow identity implies $\varphi_a^k=\varphi_{ak}$. Since $ak\in[0,an]$, this sampled time is one of the times included in the supremum defining $d_{an}^\varphi$. Therefore \begin{align*} d_{an}^\varphi(x,y) = \sup_{0\le t\le an} d(\varphi_t(x),\varphi_t(y)) \ge d(\varphi_{ak}(x),\varphi_{ak}(y)) > \varepsilon. \end{align*} So the same set $E$ is $\varepsilon$-separated for the flow over time $an$. Taking maximal cardinalities gives \begin{align*} s_n(\varphi_a,\varepsilon)\le s_{an}(\varphi,\varepsilon). \end{align*} Now pass from separated-set counts to entropy. Taking logarithms and dividing by $n$ gives \begin{align*} \frac{1}{n}\log s_n(\varphi_a,\varepsilon) \le \frac{1}{n}\log s_{an}(\varphi,\varepsilon) = a\frac{1}{an}\log s_{an}(\varphi,\varepsilon). \end{align*} Passing to $\limsup_{n\to\infty}$ and observing that the sequence of times $an$ is only a subsequence of all times $T\to\infty$, we obtain \begin{align*} \limsup_{n\to\infty}\frac{1}{n}\log s_n(\varphi_a,\varepsilon) \le a\limsup_{T\to\infty}\frac{1}{T}\log s_T(\varphi,\varepsilon). \end{align*} Finally, the entropy definitions take the limit as $\varepsilon\downarrow0$, so \begin{align*} h_{\mathrm{top}}(\varphi_a)\le a\,h_{\mathrm{top}}(\varphi). \end{align*}[/guided]

[step:Use uniform continuity to control the flow between consecutive sampled times] Fix $\varepsilon>0$. Define the continuous map $G:[0,a]\times X\times X\to[0,\infty)$ by \begin{align*} G(u,x,y)=d(\varphi_u(x),\varphi_u(y)). \end{align*} The domain $[0,a]\times X\times X$ is compact, so $G$ is uniformly continuous. Since $G(u,x,x)=0$ for every $(u,x)\in[0,a]\times X$, there exists $\delta=\delta(\varepsilon,a)>0$ such that whenever $x,y\in X$ satisfy $d(x,y)\le\delta$, one has \begin{align*} d(\varphi_u(x),\varphi_u(y))\le\varepsilon \end{align*} for every $u\in[0,a]$. [/step]

[step:Convert flow separation over time $T$ into sampled separation for $\varphi_a$] Let $T>0$, and define the integer $N_T\in\mathbb{N}$ by \begin{align*} N_T=\left\lfloor\frac{T}{a}\right\rfloor+1. \end{align*} We claim that \begin{align*} s_T(\varphi,\varepsilon)\le s_{N_T}(\varphi_a,\delta). \end{align*} Indeed, suppose $E\subset X$ is $\varepsilon$-separated for the flow over time $T$. Let $x,y\in E$ be distinct. If $d_{N_T}^{\varphi_a}(x,y)\le\delta$, then \begin{align*} d(\varphi_{ja}(x),\varphi_{ja}(y))\le\delta \end{align*} for every integer $j$ with $0\le j\le N_T-1$. Now fix $t\in[0,T]$. Let $j=\lfloor t/a\rfloor$ and let $u=t-ja$. Then $0\le u<a\le a$, and the definition of $N_T$ gives $0\le j\le N_T-1$. Using the flow law, \begin{align*} \varphi_t=\varphi_u\circ\varphi_{ja}. \end{align*} Applying the choice of $\delta$ to the points $\varphi_{ja}(x)$ and $\varphi_{ja}(y)$ gives \begin{align*} d(\varphi_t(x),\varphi_t(y)) = d(\varphi_u(\varphi_{ja}(x)),\varphi_u(\varphi_{ja}(y))) \le \varepsilon. \end{align*} Taking the supremum over $t\in[0,T]$ yields $d_T^\varphi(x,y)\le\varepsilon$, contradicting the $\varepsilon$-separation of $E$. Hence $d_{N_T}^{\varphi_a}(x,y)>\delta$ for every distinct $x,y\in E$, so $E$ is $\delta$-separated for $\varphi_a$ over $N_T$ iterates. Taking maximal cardinalities proves the claim. [/step]

[step:Pass to entropy in the reverse comparison] From the previous step, for every $T>0$, \begin{align*} \frac{1}{T}\log s_T(\varphi,\varepsilon) \le \frac{1}{T}\log s_{N_T}(\varphi_a,\delta) = \frac{N_T}{T}\cdot \frac{1}{N_T}\log s_{N_T}(\varphi_a,\delta). \end{align*} Since \begin{align*} \frac{N_T}{T} = \frac{\lfloor T/a\rfloor+1}{T} \to \frac{1}{a} \end{align*} as $T\to\infty$, and since $N_T\to\infty$ as $T\to\infty$, taking $\limsup_{T\to\infty}$ gives \begin{align*} \limsup_{T\to\infty}\frac{1}{T}\log s_T(\varphi,\varepsilon) \le \frac{1}{a}\limsup_{n\to\infty}\frac{1}{n}\log s_n(\varphi_a,\delta). \end{align*} Because $\delta=\delta(\varepsilon,a)>0$, the right-hand side is bounded above by $\frac{1}{a}h_{\mathrm{top}}(\varphi_a)$. Therefore \begin{align*} \limsup_{T\to\infty}\frac{1}{T}\log s_T(\varphi,\varepsilon) \le \frac{1}{a}h_{\mathrm{top}}(\varphi_a). \end{align*} Letting $\varepsilon\downarrow0$ gives \begin{align*} h_{\mathrm{top}}(\varphi)\le \frac{1}{a}h_{\mathrm{top}}(\varphi_a). \end{align*} Equivalently, \begin{align*} a\,h_{\mathrm{top}}(\varphi)\le h_{\mathrm{top}}(\varphi_a). \end{align*} [/step]

[step:Combine the two entropy inequalities] The first comparison gave \begin{align*} h_{\mathrm{top}}(\varphi_a)\le a\,h_{\mathrm{top}}(\varphi), \end{align*} and the reverse comparison gave \begin{align*} a\,h_{\mathrm{top}}(\varphi)\le h_{\mathrm{top}}(\varphi_a). \end{align*} Hence equality holds: \begin{align*} h_{\mathrm{top}}(\varphi_a)=a\,h_{\mathrm{top}}(\varphi). \end{align*} This proves the theorem for every $a>0$. [/step]