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Let $(X,d)$ be a metric space and let $F:A(\subset X)\to X$. We say $F$ is a contraction if there exists $\lambda$ where $0\leq\lambda<1$ such that

$$d(F(x),F(y))\leq\lambda d(x,y)$$

for all $x,y\in X$.

My question is:

I understand that the function $f(x)=x^2$ is a contraction on each interval on $[0,a], 0<a<0.5$.

But my doubt is why is it NOT a contraction on $[0,0.5]$?

enter image description here

As we can see any distance on any interval on the horizontal axis is less than those on the vertical axis, so there should be some $\lambda$ that satisfy the inequality.

I don't really know the reason. Maybe is there any counterexample that makes it not a contraction?

Many thanks in advance for the help.

user71346
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3 Answers3

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Suppose by contradiction that $x^2$ is a contraction of $[0, 0.5]$ with some constant $0<c<1$ satisfying for all $x \neq y \in [0,0.5]$ $$|x^2-y^2|< c|x-y|$$

As a simple consequence of the MVT, $c > 0.5$. But now, $$c|x-y|>|x^2-y^2|=|x+y| \cdot |x-y|$$ which implies $c > |x+y|=x+y$. Now, take simply $x=0.5$ and $y= c - 0.5$ to get a contradiction $c>c$.

The fact is that $x^2$ is a contraction of $[0,a]$ for all $a \in (0,0.5)$ because of the MVT. But as $a \to 0.5$, the best Lipschitz constant $c$ approaches to $1$, so that there is no constant $c<1$ working for the whole interval $[0,0.5]$. For this limit case, necessarily $c\ge 1$ so that $x^2$ is Lipschitz but not a contraction.

Crostul
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  • Alternatively, for any $\epsilon$, take $x=0.5$ and $y=0.5-\epsilon$ to get $c>1-\epsilon$. But that can't possibly be try for every $\epsilon$ (since that would contradict $0<c<1$). Right? – Akiva Weinberger Jan 21 '16 at 12:23
  • @AkivaWeinberger Yes, what you say is equivalent to my argument. – Crostul Jan 21 '16 at 12:25
  • May I please ask how can we conclude $c>0.5$? – Y.X. Apr 12 '17 at 10:22
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    @PropositionX Pick $x=0.3$, $y=0.4$. By the MVT there exists some point $z \in [0.3, 0.4]$ such that $|x^2-y^2|=(2z)|x-y|$. This implies $c \ge 2z \ge 0.6$. – Crostul Apr 12 '17 at 12:02
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A quick argument using derivatives: note that $f'(0.5) = 1$. So, for any $c < 1$, we have $$ \lim_{a \to 0.5^-} \frac{f(0.5) - f(a)}{0.5 - a}> c $$ So, for any such $c$, there exists an $a \in (0,0.5)$ such that $$ \frac{f(0.5) - f(a)}{0.5 - a} > c \implies\\ |f(0.5) - f(a)| > c|0.5 - a| $$ The conclusion follows.

Ben Grossmann
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Let me add a point that might help for the understanding. In the other answers, we already have seen that $$|x^2 - y^2| \le C \, |x - y| \quad\forall x,y\in [0,0.5]$$ implies $C \ge 1$.

Now, the OP correctly observed "As we can see any distance on any interval on the horizontal axis is less than those on the vertical axis" Indeed, we have $$|x^2 - y^2| < |x - y| \quad \forall x,y\in[0,0.5], x\ne y.$$

But now, he OP concludes "so there should be some $\lambda$ that satisfy the inequality." And this conclusion is not valid. In fact, for every $x,y \in [0,0.5]$, $x\ne y$, we find $\lambda(x,y) < 1$, such that $$|x^2 - y^2| \le \lambda(x,y) \, |x - y|$$ However, you cannot choose $\lambda(x,y)$ to be uniformly smaller than $1$.

gerw
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  • Is this an example that to have a Banach fix point (sequence of function iteratively applied converges to a unique fixpoint) it is sufficient but not necessary that the function is a contraction? – Elsa Jan 21 '16 at 15:11
  • @Elsa: I am not sure if I understood you correctly. If you iteratively apply $x \mapsto x^2$ to some $x_0 \in [0,0.5]$, you end up with $0$, although $x \mapsto x^2$ is not a contraction. But $x_0^2 \in [0,0.25]$ and on this interval $x \mapsto x^2$ is a contraction. – gerw Jan 21 '16 at 15:28
  • Yes, this is what I meant. The example in the question just "behaves" like a contraction (regarding iterations) but it is not a contraction. Shouldn't then the definition of a contraction been made more general? – Elsa Jan 21 '16 at 15:32
  • But then, the proof of the Banach fix point theorem will not work. There, the definition of a contraction is essential. – gerw Jan 21 '16 at 18:48