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I have a function $H_n: [0,1]^n\rightarrow [0,1]^n$ where $n$ is a positive integer $>1$.

I have to show that $H_n$ is a contraction mapping "uniformly in $n$".

Question 1: what does it mean "uniformly in $n$"? Is it correct that it means I have to prove that $\forall \theta_n, \tilde{\theta}_n$ $$ d(H_n(\theta_n),H_n(\tilde{\theta_n})) \leq k d(\theta_n, \tilde{\theta}_n) $$ with $0\leq k <1$, where $k$ is invariant over $n$?

Question 2: an hint of the exercise is showing that $$ (\star)\hspace{1cm}||\frac{\partial H_n(\theta_n)}{\partial \theta'_n}||_{\infty}\leq 1 $$ uniformly in $n$. Firstly, what does it mean that the inequality $(\star)$ should be satisfied "uniformly in $n$"? Secondly, why does it imply that $H_n$ is a contraction mapping "uniformly in $n$"? Which theorem are we using? Does using $\leq$ versus $<$ in $(\star)$ change something?

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    $\begingroup$ Do you have more conditions on $H_n$? Because if you take the identity, it will certainly not be a contraction... (Else, you can consider the product of open intervals, than it will work.) Other than that, the answer to question 1 should be yes. $\endgroup$ – Daniel Robert-Nicoud Sep 21 '17 at 13:04
  • $\begingroup$ Yes, of course, I have a specific $H_n$ to consider. My questions posted here are just about the general interpretation of the exercise. $\endgroup$ – STF Sep 21 '17 at 13:05
  • $\begingroup$ @DanielRobert-Nicoud Could you give me some help with question 2? $\endgroup$ – STF Sep 21 '17 at 13:06
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For question $1$, you've said nothing about a sequence of functions, which suggests that the standard interpretation of "uniform contraction applies.

In other words, $\|x-y\| \geq c \|f(x)-f(y)\|$ for all $x,y \in [0,1]^n$.

For the hint: use the bounded derivative to show that your function is locally lipschitz, with constant less than $1$ by the hint. Furthermore, since the set is compact, deduce that the function is globally lipschitz, with constant less than $1$, since you can choose a finite subcover, and hence choose a maximal constant $K < 1$ to bound your function.

More formally, show that in some $\delta$-neighborhood of $x$, we have $\|x-y\| \geq K_{\delta}\|f(x)-f(y)\|$ with $K_{\delta} <1$ because of the condition on the derivative (Think about the one dimensional case to motivate the argument) and then use these suitable neighborhoods to cover $[0,1]^n$, and use compactness.

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  • $\begingroup$ Please check all your inequalities :P $\endgroup$ – Niklas Sep 21 '17 at 14:10
  • $\begingroup$ @NiklasHebestreit I have no excuse for myself hahaha $\endgroup$ – Andres Mejia Sep 21 '17 at 14:11
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    $\begingroup$ Thank you: does it matter whether I show $<1$ versus $\leq 1$ in $(\star)$? My intuition is that the inequality should be STRICT to have $c<1$ in the contract definition, correct? $\endgroup$ – STF Sep 21 '17 at 14:13
  • $\begingroup$ @AndresMejia I am glad, I am not the only one who is making mistakes :D $\endgroup$ – Niklas Sep 21 '17 at 14:15
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    $\begingroup$ Oh I am sorry, atm I am working with a Browder theorem :D Yes, $k=1$ is enough and guarantess the continuity of your map. Sorry for confusing you! $\endgroup$ – Niklas Sep 21 '17 at 14:32

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