# Analysis of convergence 2

Let $f: \mathbb{R} \to \mathbb{R}$. Prove that if $f$ is bounded such that for all $x, y \in \mathbb{R}$ $x\neq y$ implies $|f(x) - f(y)| \lt |x -y|$ and for all $x \in \mathbb{R}$, $f$ is differentiable at $x$ with $|f'(x)| \lt 1$ then $f(z) = z$ for exactly one number $z \in \mathbb{R}$.

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mary = Omar? mathoverflow.net/questions/60054 – Did Mar 30 '11 at 11:00
Please don't ask question in imperative mode. Also, it would be nice if you included your work on the problem in question. – user5501 Mar 30 '11 at 23:49

Hints: Intermediate value theorem. If $x \neq y$ are fixed points of $f$, that is, $x = f(x)$ and $y = f(y)$ you get a contradiction to $|f(x) - f(y)| \lt |x - y|$ for all $x \neq y$. If $f$ is differentiable, the property $|f'(z)| \lt 1$ implies $|f(x) - f(y)| \lt |x - y|$ by the mean value theorem.

On mary's request I elaborate a little:

• Let $g(x) = f(x) - x$. Since $f$ is bounded by hypothesis (that is $|f(x)| \lt C$ for some $C$), we have $g(x) \gt 0$ for $x \leq -2C$ and $g(x) \lt 0$ for $x \geq 2C$. Since $f$ is continuous (by $|f(x) - f(y)| \lt |x - y|$), so is $g$. Therefore the intermediate value theorem tells us that there exists at least one $x_0 \in [-2C,2C]$ such that $g(x_0) = 0$. But $g(x_0) = 0$ means $f(x_0) = x_0$.

• I argued above why there is at most one fixed point.

• Assume $x \lt y$ and that $f$ is differentiable with $|f'(c)| \lt 1$ for all $c \in \mathbb{R}$. The mean value theorem tells us that there is some $c$ such that $f'(c) = \frac{f(y) - f(x)}{y - x}$ for some $c \in (x,y)$. But this is the same as saying $|f(y) - f(x)| = |f'(c)| \cdot |y - x| \lt |y - x|$ because $|f'(c)| \lt 1$.

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Hints: Intermediate value theorem. If $x \neq y$ are fixed points of $f$, that is, $x = f(x)$ and $f(y) = y$ you get a contradiction to $|f(x) - f(y)| \lt |x - y|$ for all $x \neq y$. The property $|f'(z)| \lt 1$ implies $|f(x) - f(y)| \lt |x - y|$ by the mean value theorem. – user8883 Mar 30 '11 at 11:04
@mary: ? ${}{}{}$ – t.b. Mar 30 '11 at 11:07
Is this the answer or a hint, I am not understanding it – user8883 Mar 30 '11 at 11:08
@mary: Well, I declared it as hints. Use the intermediate value theorem to find at least one fixed point $f(z) = z$ of $f$. (Consider $g(x) = f(x) - x$. For $x \ll 0$ you have $g(x) \gt 0$ and for $x \gg 0$ you have $g(x) \lt 0$. Now observe that $g$ is continuous). Then I argued that there cannot be two distinct fixed points. Finally, I told you how to apply the mean value theorem in order to reduce your second question to the first one. – t.b. Mar 30 '11 at 11:11
@mary: Actually I answered the following question: If $f$ is satisfies $|f(x) - f(y)| \lt |x - y|$ for all $x \neq y$ then $f$ has a unique fixed point. It is a good exercise to show that such an $f$ is continuous. – t.b. Mar 30 '11 at 11:17