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How do I prove that $f(x) = x^2 : (0; 1/2) \to (0; 1/2)$ is not contraction mapping?

I'd like to prove this in $\mathbb{R}$ with the Euclidean metric.

Clayton
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Alex
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1 Answers1

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If it was a contraction map, then there would exist some $\lambda<1$ such that $|f(x)-f(y)| \le \lambda|x-y|$ for all $x,y \in I = (0,\frac{1}{2})$. If $f$ is differentiable, then this implies $\lim_{x \to x_0} \left| \frac{f(x)-f(x_0)}{x-x_0} \right| \le \lambda < 1$.

Note that $f'(x) = 2x$, so we see that $f'(x) \to 1$ as $x \to \frac{1}{2}$. In particular, we can choose $x_0$ so that $f'(x_0) > \lambda$.

Then $\lim_{x \uparrow x_0} \frac{f(x)-f(x_0)}{x-x_0} = f'(x_0)$, and so $\lim_{x \uparrow x_0} |\frac{f(x)-f(x_0)}{x-x_0}| = f'(x_0) > \lambda$, which means that $f$ is not a contraction.

copper.hat
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  • @Cameron: all those $f(1)$ and $\lim \uparrow 1$ are typos. Just having points arbitrarily close to $\frac12$ is enough to stop this being a contraction. It is a weak contraction on $(0,\frac12)$, but one must restrict to $(0,\frac12-\epsilon)$ to get a strong contraction. – Anthony Carapetis Nov 04 '13 at 16:18
  • @AnthonyCarapetis: Thanks! I modified the proof accordingly. – copper.hat Nov 04 '13 at 16:50
  • @Anthony: Thanks! I had not realized there was another notion of contractive. – Cameron Buie Nov 04 '13 at 17:10