[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*}
[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]
[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]
