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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.

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  • $\begingroup$ 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. $\endgroup$
    – 52heartz
    Dec 22, 2017 at 9:47
  • $\begingroup$ @52heartz From the topological point of view it is pretty natural definition. $\endgroup$ Dec 22, 2017 at 9:49
  • $\begingroup$ @52heartz it just another way to say that $\sup f(x)$ and $\inf f(x)$ are equal when taking the limit $\endgroup$
    – ℋolo
    Dec 22, 2017 at 9:50

3 Answers 3

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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.

Plot of function

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  • $\begingroup$ 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 $\endgroup$
    – Atbey
    Dec 23, 2017 at 8:21
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    $\begingroup$ Why the downvote? This is a counterexample to the claim, hence a valid answer. $\endgroup$
    – Ennar
    Dec 23, 2017 at 8:39
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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 $x>11$

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  • $\begingroup$ Is this a counter-example? If so, why? $\endgroup$ Dec 22, 2017 at 20:50
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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 $[x]=x-\lfloor x\rfloor$ $$f(x)=\begin{cases}x&\text{if}&[x]=0\\ f(\lfloor x\rfloor)-2[x]&\text{if}&[x]\in(0,0.5)\\ f(\lfloor x\rfloor)+2[x]&\text{if}&[x]\in[0.5,1)\end{cases}$$

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

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