Fixed point of a monotone on [0,1].

This is a special case of the Knaster-Tarski fixed point theorem.

Suppose $f:[0,1] \to [0,1]$ is any monotonous function, i.e. whenever we have $x \le y$ in $[0,1]$ we have $f(x) \le f(y)$. (no continuity assumptions).

Define $A = \{x \in [0,1]: x \le f(x)\}$. The set $A$ is non-empty, as $0 \in A$.

So by completeness of $\mathbb{R}$ (every bounded above non-empty set has a supremum, and $1$ surely is an upperbound) $s:= \sup(A) \le 1$ exists. I claim that $f(s) = s$.

To see this: if $x \in A$ then $x \le s$, so $x\le f(x) \le f(s)$, and as $x \in A$ is arbitary, $f(s)$ is an upperbound for $A$ as well and $s$ is the least upper bound for $A$, so $s \le f(s)$.

$s \le f(s)$ shows that in fact $s \in A$ itself, and also implies that $f(s) \le f(f(s))$, which shows that $f(s) \in A$ as well. This implies $f(s) \le s$ (as $s$ is an upperbound for $A$) and so $f(s) = s$ from both inequalities.

So $f$ has a fixed point. If $f$ is monotonous the other way round ($x \le y \rightarrow f(x) \ge f(y)$) adapt the argument using $\inf$ e.g. (Or compose with an order reversing bijection of $[0,1]$, like $h(x) = 1-x$ and apply the above to the composed map first).

This might be a somewhat simpler argument.

Suppose that $f:[0,1]\mapsto[0,1]$ is non-decreasing and assume that $f$ has no fixed point.

Define $$ s=\sup\{x:f(x)\gt x\} $$ Note that $0\in\{x:f(x)\gt x\}$, so $s\in[0,1]$.

If $f(s)\gt s$, then for all $x\in(s,f(s))$, we must have $\ f(x)\lt x$ by the definition of $s$. But then $f(x)\lt x\lt f(s)$ contradicts that $f$ is non-decreasing.

If $f(s)\lt s$, then, by the definition of $s$, there must be some $x\in(f(s),s)$ so that $f(x)\gt x$. But then $f(s)\lt x\lt f(x)$ contradicts that $f$ is non-decreasing.

Thus, the assumption that $f$ has no fixed point must be false.

If you allow decreasing functions, counterexamples are easy . . .

For a counterexample where $f$ is non-strictly decreasing, let $f:[0,1]\to [0,1]$ be defined by $$ f(x)= \begin{cases} 1&\text{if}\;x=0 \qquad\;\;\;\;\;\,\\[4pt] 0&\text{otherwise} \end{cases} $$ For a counterexample where $f$ is strictly decreasing, let $f:[0,1]\to [0,1]$ be defined by $$ f(x)= \begin{cases} 1-\frac{x}{2}&\text{if}\;x < \frac{1}{2}\\[4pt] \frac{1}{2}-\frac{x}{2}&\text{otherwise} \end{cases} $$ On the other hand . . .

Claim:

If $f:[0,1]\to [0,1]$ is monotonically (not necessarily strictly) increasing, then $f$ has a fixed point.

Proof:

Suppose $f:[0,1]\to [0,1]$ is a monotonically (not necessarily strictly) increasing function such that $f$ does not have a fixed point.

Our goal is to derive a contradiction.

Let $A=\{x\in [0,1]\mid f(x) > x\}$, and let $B=\{x\in [0,1]\mid f(x) < x\}$.

Since $f$ has no fixed point, we get $f(0) > 0$, and $f(1) < 1$, so we have $0 \in A$, and $1\in B$.

Then $A,B$ are nonempty, disjoint, and $A \cup B = [0,1]$.

Let $c=\text{glb}(B)$, and let $d=f(c)$.

Consider two cases . . .

Case $(1)$:$\;c\in A$.

Then $f(c) > c$, hence $f(f(c)) \ge f(c)$, so $f(d) \ge d$.

Since $f$ has no fixed point, we get $f(d) > d$.

Thus, $c < f(c) = d < f(d)$.

Since $c=\text{glb}(B)$, and $c \notin B$, there exists $b\in B$, with $c < b < d$. \begin{align*} \text{Then}\;\;&c < b\\[4pt] \implies\;&f(c) \le f(b)&&\text{[by monotonicity of $f$]}\\[4pt] \implies\;&d \le f(b)&&\text{[since $d=f(c)$]}\\[4pt] \implies\;&b < f(b)&&\text{[since $b < d$]}\\[4pt] \end{align*} contrary to $b\in B$.

Case $(2)$:$\;c\in B$. \begin{align*} \text{Then}\;\;&d = f(c)\\[4pt] \implies\;&d < c&&\text{[since $f(c) < c$]}\\[4pt] \implies\;&f(d) \le f(c)&&\text{[by monotonicity of $f$]}\\[4pt] \implies\;&f(d) \le d&&\text{[since $d=f(c)$]}\\[4pt] \implies\;&f(d) < d&&\text{[since $f$ has no fixed points]}\\[4pt] \implies\;&d \in B\\[4pt] \end{align*} contradiction, since $c=\text{glb(B)}$, and $d < c$.

Thus, both cases yield a contradiction, which completes the proof.