Tag: Fixed Points

Prove a function with given properties has a fixed point

by RoRi

September 5, 2015

Given a real function $f$ continuous on $[a,b]$ , with $f(a) \leq a$ and $f(b) \geq b$ . Prove there is some $c \in [a,b]$ such that $f(c) = c$ (i.e., $f$ has a fixed point in $[a,b]$ ).

Proof. If $f(a) = a$ or $f(b) = b$ then we are done since these would be fixed points.

Assume then that $f(a) < a$ and $f(b) > b$ and let $g(x) = f(x) - x$ . We know that $g$ is continuous on $[a,b]$ and

$g(a) = f(a) - a < 0, \qquad \text{and} \qquad g(b) = f(b) - b > 0.$

Thus, by Bolzano’s theorem, there is some $c \in [a,b]$ such that $g(c) = 0$ ; hence, $f(c) - c = 0$ or $f(c) = c. \qquad \blacksquare$

Prove that a particular function has a fixed point on [0,1]

by RoRi

September 5, 2015

Let $f$ be a continuous, real function on the interval $[0,1]$ . Assume

$0 \leq f(x) \leq 1 \qquad \text{for all } x \in [0,1].$

Prove that there exists a real number $c \in [0,1]$ such that $f(c) = c$ .

Proof. Let $g(x) = f(x) - x$ . Then $g$ is continuous on $[0,1]$ since it is the difference of functions which are continuous on $[0,1]$ .

Then, $g(0) = f(0)$ and $g(1)= f(1) - 1$ . If $f(0) = 0$ , then $c = 0$ and we are done. Similarly, if $f(1) = 1$ , then $c = 1$ and we are done as well.

Assume then that $0 < f(0) \leq 1$ , and $0 \leq f(1) < 1$ . Then

$\begin{align*} f(0) > 0 \quad \implies \quad g(0) &>0 \\ f(1) < 1 \quad \implies \quad g(1) &= f(1) - 1 < 0. \end{align*}$

Hence, applying Bolzano’s theorem to $g$ , there is some $c \in [0,1]$ such that $g(c) = 0$ . This implies $f(c) - c = 0$ , or $f(c) = c. \qquad \blacksquare$