Functions and Analysis with a fixed-point

[Mr Davis 97]

Homework Statement

A fixed-point of a function f : A → A is a point a ∈ A such that f(a) = a.
The diagonal of A × A is the set of all pairs (a, a) in A × A.
(a) Show that f : A → A has a fixed-point if and only if the graph of f
intersects the diagonal.
(b) Prove that every continuous function f : [0, 1] → [0, 1] has at least one
fixed-point.
(c) Is the same true for continuous functions f : (0, 1) → (0, 1)?†
(d) Is the same true for discontinuous functions?

Homework Equations

The Attempt at a Solution

a) ---> Suppose that $f: A \rightarrow A$ has a fixed-point, call it $a$. Then $f(a) = a$, which, by virtue of how functions are defined, means that $(a,a) \in A \times A$. This is clearly a point on the diagonal of $A \times A$.
<--- Suppose that the graph of $f: A \rightarrow A$ intersects the diagonal. This means that that the graph of $f$ and the diagonal of $A \times A$ share a point, and that point must be of the form $(a,a)$, for some a. Thus, $f(a) = a$ is true for some $a$, and so $f$ has at least one fixed point.

b) We just need to show that every continuous function on $[0,1] \rightarrow [0,1]$ intersects the graph of $y=x$ at at least one point. So, for each function on this interval, we must show that $f(x) = x$ at at least one point, or, alternatively, we need to show that $g(x) = f(x) - x$ must have at least one zero. We see that it must be the case that $0 \le g(0) \le 1$ and $-1 \le g(1) \le 0$. So there are four cases: 1) $g(0)>0$ and $g(1)<0$. Then by the Bolzano's theorem $g(x)$ must have a zero on the inteval $[0,1]$. 2) $g(0)>0$ and $g(1) = 0$, then obviously we have a zero. 3) $g(0) = 0$ and $g(1)<0$, then obviously we have a zero. 4) $g(0) = 0$ and $g(1) = 0$, in which case we have at least two zeros. Hence, $g(x)$ must have at least one zero on the interval $[0,1]$, and the $f(x) -x = 0$ must have at least one zero on the interval. Or in other words, all continuous functions on the interval [0,1] must have at least one fixed point.

c) No. $f(x) = 1$ is a counterexample.

d) No. Consider
$f(x) = \begin{cases} -(x-\frac{1}{2})^2 + \frac{1}{2} & \text{if x \neq \frac{\sqrt{3}}{2}}\\ 0 & \text{if x = \frac{\sqrt{3}}{2}}\\ \end{cases}$

Should be b) be shortened down?

 

[Dick]

For c) $f(x)=1$ is NOT a function mapping $(0,1) \rightarrow (0,1)$.

 

[Orodruin]

For c) $f(x)=1$ is NOT a function mapping $(0,1) \rightarrow (0,1)$.

However, constructing an actual counter example is not very difficult.