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Let $K=[0,1]\times [0,1]$. Find a continuous mapping $F:K\rightarrow \mathbb{R}^2$ satisfying:

  • $\|F(x)-F(y)\|\leq \|x-y\| \quad\forall x,y\in K,$

  • There exists $\gamma>0$ such that for all $x,y\in K$ $$\left\geq 0\Longrightarrow\left\geq \gamma\|x-y\|^2,$$

  • There exist $u,v\in K$ such that $\left<0$.

Here, $\|.\|$ is the Euclidean norm and $\left<,.,\right>$ is the scalar product in $\mathbb{R}^2$.

Thank you for all comments and helping.

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    Could you tell us how this question arose, and what your progress is on it? For instance, have you ruled out certain classes of maps, like restriction of linear or affine maps?2012-07-23
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    @Olivier Bégassat: I have tried it. It is easily to construct a mapping satisfying all conditions in the case $\mathbb{R}$. It is very difficult to to construct such mapping in $\mathbb{R}^2$. It is interesting to construct an $\textbf{affine mapping}$ satisfying all above conditions. Thank you for your comments.2012-07-23
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    How did you come to consider this qestion? Can you motivate the two final conditions? It might help others to think about your problem to know its context and why the conditions are what they are. I don't understand your last comment fully, do you have a solution to this problem?2012-07-23
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    @OlivierBégassat: Two final conditions relate to some kind of monotonicity conception. The last condition means that $F$ is not monotone on $K$. Thank you for your consideration.2012-07-23
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    Could you expand on your comment, it is unfortunately too vague to be of any help for me ^^, also, please tell us your progress.2012-07-23
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    @OlivierBégassat: I give an example of my question in $\mathbb{R}$. Let $$K=[0,1] \text{ and } F(x)=\frac{1}{1+x}.$$ Note that $$|F(x)-F(y)|=\frac{|x-y|)}{(1+x)(1+y)}\leq |x-y|.$$ Suppose that $F(x)(y-x)=\frac{1}{1+x}(y-x)\geq 0$. Then $$F(y)(y-x)=\frac{1}{1+y}(y-x)\geq \frac{1}{4}(y-x)^2.$$ Moreover, we have $$(F(0)-F(1))(0-1)=(1-\frac{1}{2})(0-1)=-\frac{1}{2}<0.$$ I would like to expand my example in $\mathbb{R}^2$. Thank you for your disscusion.2012-07-23
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    What is the motivation behind this?2012-07-23
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    @OlivierBégassat: To expand from $\mathbb{R}$ to $\mathbb{R}^2$. That's enough. I have tried it and can't find a solution. In my opinion this is a difficult question.2012-07-23
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    It seems like the conditions you impose are completely random. Why are you considering them?2012-07-23
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    @Oliver Bégassat: I have tried but i found that it is difficult. Do you have any comment and hint for this question?2012-07-23
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    @Robert Israel: Dear Sir. I would like to ask your opinion about this problem.2012-07-24

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