- #1

Mr Davis 97

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## 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?