Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function. We say that f has the run-away property if for every compact subset $K\subseteq \mathbb{R}$ there is some positive integer N such that for every $n \geq N$
$$
f^n(K) \cap K = \emptyset.
$$


Some toy examples include:


1. $f(x)=x+b$ for non-zero b.
2. $f(x)=\exp(x)$.


Some non-examples are:


1. $f(x)=x^2 - b$ for positive b *(since iterates of any compact neighborhood of the root of $x^2 -x -b$ always contain that fixed point).*
2. $f(x)=\sin(x)$ *(since the compact $[-1,1]$ is never escaped)*.


In general is there a known *sufficient* condition on f for it to be run-away?


**Thoughts:**


It seems that unbounded range (not necessarily surjective), and no periodic points are necessary...



Run-away functions
As noted in the question's comments by [Aleksei Kulikov](https://mathoverflow.net/questions/364244/run-away-functions#comment919269_364244), a necessary and sufficient condition is given by the following:


**Theorem 1**


A real continuous function f is a runaway function iff $f(x)=x$ has no solution for $x\in \mathbb{R}$.


To prove this we need the following lemma:


**Lemma**


Let $f$ be continuous on $\mathbb{R}$ and $f(x)>x$ for all $x \in \mathbb{R}$.


Then for any real values $x$ and $u$ with $x<u$ $$\exists\_{N \in \mathbb{N}} \mid \forall\_{n\geq N} f^{(n)}(x) > u,$$ and $N$ can be chosen to be less than $1+(u-x)/G$ where $G=\min\_{s \in [x,u]} (f(s)-s)$.


**Proof**


Since $g(x)=f(x)-x$ is continuous, by the [extreme value theorem](https://en.wikipedia.org/wiki/Extreme_value_theorem) it attains its bounds on $[x,u]$ and in particular there exists $\theta \in [x,u]$ s.t. $g(\theta)\leq g(t)$ for all $t \in [x,u]$. Since $f(x)>x$ for all $x$, $g(\theta)>0$. Hence there exists real $G=g(\theta)>0$ s.t. $f(s)\geq s+G$, for all $s \in [x,u].$


We know that $f(x)\geq x+G$. If $f(x)>u$ we are done so assume that $f(x)\in [x,u]$. Then by the above we have $f^{(2)}(x)=f(f(x))\geq f(x)+G\geq x+2G$. Clearly by induction we can prove $$\forall\_{n \in \mathbb{N}}f^{(n)}(x)\leq u \implies \forall\_{n \in \mathbb{N}}f^{(n)}(x)\geq x+nG.$$


However if we choose $n>(u-x)/G$ then $f^{(n)}(x)\geq x+nG > x+(u-x)=u$.


This is a contradiction. Hence there must exist an $N \in \mathbb{N}$ s.t. $f^{(N)}(x)>u$ and then clearly since $f(t)>t$ for all $t \in \mathbb{R}$, $f^{(n)}(x)>f^{(n-1)}(x)>\dotsb>f^{(N)}(x)>u$ for all $n\geq N$. Clearly $N$ can be chosen to not exceed $1+(u-x)/G$ and we are done. $\blacksquare$


**Proof of Theorem 1**


If $f(x)=x$ for some $x\in \mathbb{R}$ then the non-empty compact set $X=\{x\}$ is fixed by $f$ and hence $f^{(n)}(X)\cap X = X \cap X = X\neq \emptyset$ for all $n \in \mathbb{N}$. Thus $f$ is not a runaway function.


If $f(x)\neq x$ for any $x\in \mathbb{R}$ then since $f$ is continuous either $f(x)>x$ or $f(x)<x$ for all $x\in \mathbb{R}$. This is because if the continuous function $g(x)=f(x)-x$ takes both strictly positive and strictly negative values then by the intermediate value theorem it has a real root $a$ which satisfies $f(a)=a$.


Assume wlog $f(x)>x$ for all $x$ (for the other case take continuous $f\_1(x)=-f(-x)>x$).


Take any interval $[a,b]$. Lemma 1 then says that for any $x\in[a,b]$ we can find $h(x) \in \mathbb{Z}\_{>0}$ s.t. $\forall\_{n\geq h(x)} f^{(n)}(x) > b$ and $h(x)$ can be chosen to not exceed $1+(b-x)/G$ where $G=\min\_{s \in [a,b]} (f(s)-s)$.


However $1+(b-x)/G<1+(b-a)/G$ and $G'=\min\_{s \in [a,b]} (f(s)-s)\leq G = \min\_{s \in [x,b]} (f(s)-s)$, and since $f(x)-x$ is continuous, by the [extreme value theorem](https://en.wikipedia.org/wiki/Extreme_value_theorem), the minimum is attained at some point $\mu \in [a,b]$. Thus $G'=f(\mu)-\mu>0$.


Hence $h(x)$ can be chosen not to exceed $h=1+(b-a)/G'$ where $G'=\min\_{s \in [a,b]} (f(s)-s)>0$. Thus for any $x\in[a,b]$ we can find $h \in \mathbb{Z}\_{>0}$ s.t. $\forall\_{n\geq h} f^{(n)}(x) > b$. Clearly this implies that $\forall\_{n\geq h} f^{(n)}([a,b]) \cap [a,b] = \emptyset$.


Now any compact set $S$ in $\mathbb{R}$ is bounded hence we can find a closed interval $[a,b]$ which contains it. By the above we can find $h \in \mathbb{Z}\_{>0}$ s.t. $\forall\_{n\geq h} f^{(n)}([a,b]) \cap [a,b] = \emptyset $ which implies that $$\forall\_{n\geq h}( f^{(n)}(S)\cap S \subset f^{(n)}([a,b]) \cap [a,b] = \emptyset ).$$


Hence we have proved that $f$ is a runaway function if there does not exist $x\in \mathbb{R}$ s.t. $f(x)=x$. This, combined with the first implication, proves the result. $\blacksquare$


