A shrinking map that is not a contraction no fixed point.

Question

Working from Munkres.

Show that $f: \mathbb{R}\to \mathbb{R}$ given by $f(x) = [x+\sqrt{(x^2+1)}]/2$ is a shrinking map that is not a contraction that has no fixed point.

I figured out the fixed point part but have no clue how to show the other part.

For the fixed point part $$f(x) = x \Leftrightarrow x = [x+\sqrt{(x^2+1)}]/2$$

then $$2x = x + \sqrt{(x^2+1)}$$ $$x = \sqrt{(x^2+1)}$$ $$ x^2 = x^2+1$$

Which is a contradiction.

Thanks in advance.

What's "the other part"? Are you having trouble showing that the map is not a contraction? — Sammy Black, Mar 17 '14 at 00:15
What does the derivative say about a map being a shrinking map? — EgoKilla, Mar 17 '14 at 00:22

score 6 · Accepted Answer · answered Mar 17 '14 at 00:21

6

We have

$$\left\lvert f'(x)\right\rvert = \frac{1}{2} \left\lvert 1 + \frac{x}{\sqrt{x^2+1}}\right\rvert \leqslant \frac{1}{2}\left(1 + \frac{\lvert x\rvert}{\sqrt{x^2+1}}\right) < 1,$$

hence by the mean value theorem, $f$ is a shrinking map.

$f$ has no fixed point, $\mathbb{R}$ is complete, hence $f$ is not a contraction by Banach's fixed point theorem. (Alternative proof: $\lim\limits_{x\to+\infty} f'(x) = 1$.)

answered Mar 17 '14 at 00:21

Daniel Fischer

206,697

How does the mean value theorem show that f is a shrinking map? – EgoKilla Mar 17 '14 at 00:30
Oh I see it now, nevermind thank you! – EgoKilla Mar 17 '14 at 00:31

A shrinking map that is not a contraction no fixed point.

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