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Let $(E,\|.\|)$ is a normed vector space, $A\in E$ such that $A$ is compact. Let $f:A\to A$ such that $$\forall(x,y)\in A, x\neq y \Rightarrow \|f(x)-f(y)\| <\|x-y\|$$ Is $f$ contractant?

As I know, a function $g$ is contractant iff there exist $k\in (o,1)$ such that $\|g(x)-g(y)\| \le k\|x-y\|$. But I can't find a such $k$ that makes $f$ contractant. Can we conclude that $f$ is not contractant?

liverpool29
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  • Of course it is a contraction if $|f(x)-f(y)| <|x-y|$ consider $k=1-\varepsilon$ – Raffaele Oct 14 '17 at 21:10
  • Your reasoning is false. The constant $k$ does not depend on the choice of $x$ and $y$ in the definition of a contraction, while your reasoning only gives a $k$ depending on $x$ and $y$. In this exercise, you have to use the compactness of $K$. – Junkyards Oct 14 '17 at 21:19

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HINT take $f(x)=x^2/2$ on $[0,1]$ not a contraction

orangeskid
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As always in mathematics, if you think a statement is false, you have to provide a counter-example to conclude anything. Not being able to prove a statement is not a proof of the statement being false.

But in this special case, $f$ is indeed a contraction. As a hint : $K \times K$ is compact and the function $g : K \times K \to :\mathbb{R}, (x,y) \mapsto ||f(x) - f(y)||$ is continuous. You can go on with the proof.

Edit : I got too fast, the set you should consider is $\{(x,y) \in K \times K, x \not = y\}$, which is not always compact. Indeed, you can find counter-examples based on this idea.

Junkyards
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  • Can you please explain in more detail? Since I've found another answer which includes a counter-example here: https://math.stackexchange.com/questions/310786/is-every-weak-contraction-a-contraction?rq=1 – liverpool29 Oct 14 '17 at 21:21
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    I have indeed made a mistake, I went a little too fast. I will explain what my initial reasoning. Since $K$ is compact, I wanted to use the fact that a certain function would attain its maximum. But the problem here is that the function you need to consider is $h : {(x,y) \in K \times K, x \not = y}, (x,y) \mapsto \frac{||f(x) - f(y)||}{||x-y||}$. But this set is not always a compact : indeed, in the counter-example you provided, you have that${(x,y) \in K \times K, x \not = y}$ is a small square without its diagonal : it is not a compact. My bad, I will edit my post. – Junkyards Oct 14 '17 at 21:26