# Find all increasing idempotent functions on $\mathbb{R}$

Let $f: \Bbb R \to \Bbb R$ be an increasing function such that $f \circ f=f$, where $\circ$ denotes the composition of two functions. Find $f$.

• Do you mean "strictly increasing" with "increasing"? I.e., do you mean $x>y\implies f(x)>f(y)$ or $x>y\implies f(x)\ge f(y)$? – Hagen von Eitzen Apr 14 '17 at 13:37

Let's suppose you mean non-strictly increasing, since the other two answers have the strict case sewn up.

This permits the example of constant functions, since after all they are non-strictly monotonic, and $f \circ g = f$, regardless of $g$.

Does it permit anything else?

Well, suppose we're only piecewise constant. Then we require that each of $f$'s constant values are a fixed point of $f$, by idempotence, so we might imagine the class of functions which are piecewise constant and such that every "piece" intersects the line $f(x) = x$ on the graph.

Indeed, let's suppose, without loss of generality, that $x \le f(x)$. Then for all $y$ between the two, we have: $\begin{array}{rrrcll} & x &\le& y &\le& f(x) & \\ \implies & f(x) &\le& f(y) &\le& f(f(x)) & \text{by monotonicity} \\ \implies & f(x) &\le& f(y) &\le& f(x) & \text{by idempotence} \\ \implies & f(x) &=& f(y) \end{array}$

Thus, every $x$ with $x \not= f(x)$ has $f$ constant on the closed interval between them, and so:

Any such $f$ is the identity function, interrupted by up to countably many (open or closed or half-open) intervals on which it is instead constant, taking a value which falls inside that interval.

(Countably many because you can't interrupt a function by uncountably many disjoint intervals, since they each contain a rational).

A perhaps more familiar example of such a thing is the integer part (or floor) function.

Supposing that you mean strictly increasing :

$f \circ f=f \Leftrightarrow f(f(x)) = f(x)$

But $f: \Bbb R \to \Bbb R$ is a strictly increasing function, which means it is $"1-1"$ as well. So, we got :

$f(f(x)) = f(x) \Leftrightarrow f(x) = x$ which is an increasing function $\Bbb R \to \Bbb R$

Because $f$ is increasing it is also an injection, we have $f(f(x)) = f(x)$ for all $x$ which implies $f(x) = x$ for all $x$ therefore the only solution is the identity and it is easy to see that the identity is indeed a solution.

• Immediately, $f(x)=x$ only for $x$ in the image of $f$ – Hagen von Eitzen Apr 14 '17 at 13:39
• Not really. We know that $\forall x: f(f(x)) = f(x)$. And then from the definition of injection, $\forall x: f(x) = x$. – DHMO Apr 14 '17 at 13:41
• @HagenvonEitzen Maybe you meant only for $x$ in the domain of $f$ but the latter happens to be $\mathbb{R}$. – Errol.Y Apr 14 '17 at 13:43
• @DHMO I reapeat, immediately we only get $\forall x\in f(\Bbb R)\colon f(x)=x$. – Hagen von Eitzen Apr 14 '17 at 13:44
• I don't see the point of this remark. Do you mean to say that the answer is wrong? – DHMO Apr 14 '17 at 13:45

Some solutions :

• $x\mapsto x$
• $x\mapsto 4$
• $x\mapsto x$ if $x\in \mathbb{R}_+$, $x\mapsto -1$ otherwise

Consider some $a\in\mathbb{R}$. You have three cases :

• $f(a)=a$
• $f(a)<a$
• $f(a)>a$

If $f(a)>a$ then for all $x \in [a;f(a)]$ : $f(a)\leq f(x)\leq f(f(a))$. That is $f(a)\leq f(x)\leq f(a)$. Thus $f$ is a constant on $[a;f(a)]$. Same when $f(a)<a$ : $f$ is constant on $[f(a);a]$.

As a result every point $a$ is such that $f(a)=a$ or $a$ is in an interval with non zero width where $f$ is a constant.

$\mathbb{R}=A\cup \bigcup_{c\in C} f^{-1}(c)$

where $A$ is the set of points $a$ such as $f(a)=a$ (fixed points) and for all $c\in C$, $f^{-1}(c)$ is an interval with non zero width. Since these intervals are obviously disjoint, C is at most countable. Finally, the argument above proves that $c\in f^{-1}(c)$.

Solutions are the functions such as:

• $f$ is the identity function
• except on the union of (at most) countably many disjoint intervals $I$ with non zero width where $f$ is a constant on $I$, this constant belonging to $I$.