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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?

Daniel Robert-Nicoud
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Star
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    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. – Daniel Robert-Nicoud Sep 21 '17 at 13:04
  • Yes, of course, I have a specific $H_n$ to consider. My questions posted here are just about the general interpretation of the exercise. – Star Sep 21 '17 at 13:05
  • @DanielRobert-Nicoud Could you give me some help with question 2? – Star 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.

Andres Mejia
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  • Please check all your inequalities :P –  Sep 21 '17 at 14:10
  • @NiklasHebestreit I have no excuse for myself hahaha – Andres Mejia Sep 21 '17 at 14:11
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    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? – Star Sep 21 '17 at 14:13
  • @AndresMejia I am glad, I am not the only one who is making mistakes :D –  Sep 21 '17 at 14:15
  • It matters a lot. I should be less sloppy. – Andres Mejia Sep 21 '17 at 14:16
  • @STF Yes, I agree (but this might depend on the definition of contraction). If you want to proofe that $H_n$ has fixed points, you might want to use Browders fixed point theorem since $[0,1]^n$ is compact and $H_n$ (lipschitz) continuous. –  Sep 21 '17 at 14:16
  • @NiklasHebestreit is it correct to say that in order to apply B. fixed point theorem I would need to show 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 $k=1$? – Star Sep 21 '17 at 14:19
  • on the other hand, B. fixed point theorem does not allow to show that the fixed point is unique (as instead the contraction mapping theorem tells) – Star Sep 21 '17 at 14:20
  • Ah, OK, thank you. you meant BROWER and not BROWDER.. these are two different fixed point theorems https://en.wikipedia.org/wiki/Brouwer_fixed-point_theorem and https://en.wikipedia.org/wiki/Browder_fixed-point_theorem – Star Sep 21 '17 at 14:26
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    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! –  Sep 21 '17 at 14:32