Is a continuous function $ f\colon(0,\infty)\to R$, such that $f(x)\leq f(nx)$ increasing?

Question

My question is related to: LeL $f: (0, \infty)\to R$ be continuous and $f(x)\leq f(nx)$ prove $\lim\limits_{x\to\infty} f(x)$ exists and $f\colon(0,\infty)\to \mathbb R$ be continuous ; $f(x)\le f(nx) , \forall n \in \mathbb N , \forall x >0$ , then $\lim_{x\to \infty} f(x)$ exists?

Let $f\colon (0, \infty)\to R$ be continuous such that $f(x)\leq f(nx)$ for all positive $x$ and natural $n$.

It was proved that the limit (finite or infinite) in the infinity exists. Do we know if such a function must be (weakly) increasing? I believe that there might be counterexamples.

Answer 1

Let \begin{equation} f(x)= \begin{cases} x \quad &\text{if} \quad x\leq 1\\ 2-x \quad &\text{if} \quad 1\leq x \leq 4/3\\ x- 2/3 \quad &\text{if} \quad x\geq 4/3 \end{cases} \end{equation}

In $[1,4/3]$, $f(x)$ has minimum $2/3$, and in $[1/2,2/3]$, it has maximum $2/3$. Hence satisfies the condition.
Other regions also satisfies the condition. It is also continuous.

Answer 2

You can take $f(x)$ such that: $f(x)=10^x$ for $x \in [0,10]$

$f(x)=10^{10}-(x-10)^{100}$ for $x\in [10,11]$

$f(x)=10^{10}-1+10^{10}\times (x-11)$ for $x>11$

Answer 3

I thought about this and the answer is no, I'll edit my answer in the other question, thank you for letting me know for my mistake. The function:

We will define $[x]=x-\lfloor x\rfloor$ $$f(x)=\begin{cases}x&\text{if}&[x]=0\\ f(\lfloor x\rfloor)-2[x]&\text{if}&[x]\in(0,0.5)\\ f(\lfloor x\rfloor)+2[x]&\text{if}&[x]\in[0.5,1)\end{cases}$$