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

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.

• The fact that infinite limit exists is a weird (i.e. mathematical) way to say: there is no limit. In other words, it means that a sequence not bound to any concrete value. Dec 22, 2017 at 9:47
• @52heartz From the topological point of view it is pretty natural definition. Dec 22, 2017 at 9:49
• @52heartz it just another way to say that $\sup f(x)$ and $\inf f(x)$ are equal when taking the limit
– ℋolo
Dec 22, 2017 at 9:50

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

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.

• Thank you for comment. Legend was not reversed but I was plotting $f(x/2)$ and $f(x/3)$, not $f(2x)$ and $f(3x)$. I fixed it. Function is not increasing on $[1,4/3]$. @uniquesolution Dec 23, 2017 at 8:21
• Why the downvote? This is a counterexample to the claim, hence a valid answer. Dec 23, 2017 at 8:39

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 $x>11$

• Is this a counter-example? If so, why? Dec 22, 2017 at 20:50

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

• Your answer for the previous question really needs editing or deleting. All you proof was based on the assumption, that $f$ is (weakly) increasing. Dec 22, 2017 at 10:46
• @PrzemysławScherwentke I edit it, I said it was wrong and link to here
– ℋolo
Dec 22, 2017 at 10:47