Run-away functions 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:

*

*$f(x)=x+b$ for non-zero b.

*$f(x)=\exp(x)$.

Some non-examples are:

*

*$f(x)=x^2 - b$ for positive b (since iterates of any compact neighborhood of the root of $x^2 -x -b$ always contain that fixed point).

*$f(x)=\sin(x)$ (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...
 A: As noted in the question's comments by Aleksei Kulikov, 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 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, 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. $\blacksquare$
