$\varepsilon$-covering number after diffeomorphism

Question

Consider a compact set $K \subset\mathbb{R}^d$ and a $\mathscr{C}^1$-diffeomorphism $\varphi:\mathbb{R}^d\rightarrow\mathbb{R}^d$.

Fix $\varepsilon>0$. Since $K$ is compact, we may define \begin{align*} N(\varepsilon,K) := \min\left\{n\in\mathbb{N}\,:\, \exists x_1,\dots,x_n \in K\text{ s.t. } K\subseteq\bigcup_i B(x_i,\varepsilon)\right\}. \end{align*}

Can something be said about $N(\varepsilon,\varphi(K))$ generally ? Or we have to impose some conditions on $K$ ?

More precisely if $(\varphi_t)_{t\in[0,T]}$ is a family of such diffeomorphisms, is there a simple condition to insure the boundedness of $[0,T] \ni t\mapsto N(\varepsilon,\varphi_t(K))$ ? Denoting $\Theta:(t,x) \mapsto \varphi_t(x)$, I expect $\Theta\in\mathscr{C}^0([0,T];\mathscr{C}^1(\mathbb{R}^d))$ to be sufficient but I am not sure and not really familiarized with that kind of issues !

Thanks in advance for any piece of information.

Ayman

Answer 1

(Since the OP appears satisfied with my comment, I am reproducing it here as an answer.)

Fix $\epsilon > 0$. Since $\phi$ is $\mathscr{C}^1$, it must also be Lipschitz-continuous on compact subsets of its domain. Let $C$ be the Lipschitz constant of $\phi$ on the set $K^\epsilon$ which equals the union $\bigcup_{x \in K}\overline{B}_\epsilon(x)$ of closed $\epsilon$-balls around points in $K$. Now, any cover of $\phi(K)$ by $\epsilon$-balls furnishes a cover of $K$ by $\frac{\epsilon}{C}$-balls and hence we get an inequality $$N(\epsilon,\phi(K)) \leq N\left(\frac{\epsilon}{C},K\right)$$.

A similar inequality going in the other direction follows from using the Lipschitz constant (let's call it $D$) of the inverse $\phi^{-1}$. So, we have a sandwich of covering numbers

$$N\left(D\epsilon,K\right) \leq N(\epsilon,\phi(K)) \leq N\left(\frac{\epsilon}{C},K\right)$$

Asking for something stronger would almost certainly require stricter assumptions on $\phi$ or $K$ (the inequalities become equalities when $\phi$ is the identity map).

Answer 2

As long as $\varphi_{(\cdot)}$ is a continuous function from $[0,T]\times\mathbb{R}^d$ to $\mathbb{R}^d$, the map $t\mapsto N(\varepsilon,\varphi_t(K))$ is necessarily bounded, because the continous image of the compact set $[0,T]\times K$ is compact.