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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

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    $\begingroup$ Fix $\epsilon$. Since $\phi$ is $\mathscr{C}^1$ it is Lipschitz on compact sets. Consider the compact set $K^\epsilon$ which equals the union of closed $\epsilon$-balls around points in $K$. This quickly gives you an upper bound on $N(\epsilon,\phi(K))$ in terms of the Lipschitz constant. In order to get an analogous lower bound, you would probably require bi-Lipschitzness of $\phi$. $\endgroup$ – Vidit Nanda Jan 17 '14 at 3:20
  • $\begingroup$ Okay, to be sure to understand, what you're saying is that if $C$ is the Lipschitz constant of $\phi$, we have always $N(\varepsilon,\varphi(K)) \leq N(\frac{\varepsilon}{C},K)$, just by transfering the covering, right ? $\endgroup$ – Ayman Moussa Jan 17 '14 at 9:06
  • $\begingroup$ Yes. And note also that $\phi$ must have a $\mathscr{C}^1$ inverse by assumption with a Lipschitz constant of its own, and so you can always map the covering along this inverse to get a similar inequality bounding $N(\epsilon,\phi(K))$ from the other side. $\endgroup$ – Vidit Nanda Jan 17 '14 at 11:06
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(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).

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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.

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  • $\begingroup$ I am not sure to understand : do you suggest that $N(\varepsilon,\varphi_{(\cdot)})$ is a continuous function ? Anyway, I think Vidit answered my question. $\endgroup$ – Ayman Moussa Jan 17 '14 at 9:11

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