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.


  • 1
    $\begingroup$ 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

(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).


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.