# $f\colon(0,\infty)\to \mathbb R$ be continuous ; $f(x)\le f(nx) , \forall n \in \mathbb N , \forall x >0$ , then $\lim_{x\to \infty} f(x)$ exists? [closed]

Let $f\colon(0,\infty)\to \mathbb R$ be a continuous function such that $f(x)\le f(nx) , \forall n \in \mathbb N , \forall x >0$ , then is it true that $\lim_{x\to \infty} f(x)$ exists (may be infinite) ?

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– user378947
Dec 18, 2016 at 14:35

Let $\ell=\sup f$. First assume that $\ell\in\mathbb{R}$. Let's show that $\lim\limits_{+\infty}f=\ell$. Let $\varepsilon>0$. By definition of $\ell$, there exists $x_0\in(0,+\infty)$ such that $f(x_0)>\ell-\varepsilon/2$. Since $f$ is continuous, there exists $\delta>0$ (with, of course, $\delta<x_0$ too) such that $$\forall x\in(x_0-\delta,x_0+\delta),\ f(x_0)-\varepsilon/2<f(x)<f(x_0)+\varepsilon/2,$$ and in particular, $$\forall x\in(x_0-\delta,x_0+\delta),\ \ell-\varepsilon<f(x)\leq\ell<\ell+\varepsilon.$$ Since $\delta>0$ there exists $n_0\in\mathbb{N}$ such that $x_0<n_0\delta$. Let $M=n_0(x_0+\delta)$ and let $x\in(M,+\infty)$. Let $k=\lfloor x/x_0\rfloor$ ($k$ is the integer part of $x/x_0$). By definition, $$k\leq\frac x{x_0}<k+1$$ but since $x>n_0(x_0+\delta)$ we conclude that $x/x_0>n_0+n_0\delta/x_0>n_0+1$, and hence $k>n_0$. Finally, we conclude that $$-k\delta<0\leq x-k x_0<x_0<n_0\delta<k\delta,$$ i.e., $$x\in(k x_0-k\delta,k x_0+k\delta).$$ Hence $f(x)=f(k x/k)\geq f(x/k)>\ell-\varepsilon$ since $x/k\in(x_0-\delta,x_0+\delta)$, and hence $$\ell-\varepsilon<f(x)\leq\ell<\ell+\varepsilon.$$
We hence showed that: $$\forall\varepsilon>0,\ \exists M>0,\ \forall x\in(M,+\infty),\ \ell-\varepsilon<f(x)<\ell+\varepsilon,$$ that is, $\lim\limits_{x\to+\infty}f(x)=\ell$.
In the case where $\ell\not\in\mathbb{R}$, i.e., $\ell=+\infty$, either repeat the previous proof with some straightforward adaptations, or consider the function $g=\arctan\circ f$. Then, since $\arctan$ is increasing, $g$ also satisfies the given relation; moreover, $\ell=\sup g=\pi/2\in\mathbb{R}$ so we can apply the previous proof to obtain $\lim\limits_{x\to+\infty}g(x)=\pi/2$; since the values of $g$ lie in $(-\pi/2,\pi/2)$ we conclude that $\lim\limits_{x\to+\infty}f(x)=+\infty$.