prove or disprove: Let $f:[0,1]\rightarrow[0,1]$ be a monotone (need not be strict) function then f has a fixed point.

Can I have a hint?

  • $\begingroup$ But no continuity is given $\endgroup$ – Selvakumar A Mar 20 '18 at 11:44
  • $\begingroup$ I tried it but not succeeded. $\endgroup$ – Selvakumar A Mar 20 '18 at 11:46
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    $\begingroup$ Well, now you changed the question, yes it can. Take $f(x)=x$. But fixed points do not always exist for monotone functions in general. $\endgroup$ – The Phenotype Mar 20 '18 at 11:47
  • $\begingroup$ But in this case, I think it will that's why I asked proof. $\endgroup$ – Selvakumar A Mar 20 '18 at 11:49
  • $\begingroup$ I'll give you a way to visualize the problem: draw the plane $[0,1]\times [0,1]$ and $y=x$. Now construct any (discontinuous) function that does not intersect $y=x$. $\endgroup$ – The Phenotype Mar 20 '18 at 11:50

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).

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    $\begingroup$ You said you assume any monotone function, but infact you assumed a non-decreasing one. And a decreasing function may miss a fixed point, e.g. $f(x)=1-x/2$ for $0\le x\le 0.5$ and $f(x)=(1-x)/2$ for $0.5 < x\le 1$. $\endgroup$ – CiaPan Mar 20 '18 at 12:31
  • $\begingroup$ Actually, he assumes increasing $\endgroup$ – Selvakumar A Mar 20 '18 at 15:12
  • $\begingroup$ @SelvakumarA: the function $f(x)=\frac12$ meets the constraints of the answer, so $f$ is non-decreasing, not necessarily increasing. $\endgroup$ – robjohn Mar 20 '18 at 22:31
  • $\begingroup$ Your last paragraph is obscure, because you neglected to state a conclusion. It sounds as if you are saying that a function $f$ such that $x\le y\to f(x)\ge f(y)$ has a fixed point, which is of course false. $\endgroup$ – bof Jan 24 at 10:25

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.


Let $g(x):[0,1]$ be a monotone function. Suppose that $g$ has no fixed points. Then $g(x)\neq x$ for all $x\in [0,1]$, so that we can decompose $[0,1]$ into two (non empty) disjoint sets: $$ [0,1] = \{x: g(x)>x\} \cup \{x:g(x)<x\} = G^+ \cup G^-, \quad G^+\cap G^- = \emptyset, \quad 0 \in G^+,\ 1\in G^-. $$ Now let $\delta = \sup G^+$. I claim that $\delta \in G^+$. Indeed, by definition of the sup, there is a sequence $x_n \uparrow \delta, x_n\in G^+$ for all $n\in \mathbb{N}$. Then $$ g(\delta)\geq g(x_n) >x_n, \quad g(\delta)\geq \lim_n x_n = \delta. $$ Since $g(\delta)\neq \delta$ we conclude that $g(\delta) > \delta$ and $\delta \in G^+$. Now, since $\delta = \sup G^+$ and $\delta<1$, the interval $(\delta, 1]\subset G^-$. Now let $x_n \in G^-, x_n \downarrow \delta$. We obtain $$ g(\delta)\leq g(x_n)<x_n, \quad g(\delta)\leq \lim_n x_n = \delta, $$ and again, since $g(\delta)\neq \delta$, $g(\delta)<\delta$. So $\delta \in G^-$, but this is contradiction since $G^-\cap G^+=\emptyset$.

This shows that $g$ has a fixed point. In particular we showed that, if $g(0)>0$ and $g(1)<1$, then $g(\delta)=\delta$.


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 . . .


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


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.


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