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 $z \in \mathbb{R}$.
Analysis of convergence 2
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1Please don't ask question in imperative mode. $A$lso, it would be $n$ice if you included your work on the problem in question. – 2011-03-30
1 Answers
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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0Can you also explain how to use the IVT to find at least 1 fixed point of f(z)? – 2011-03-30