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I would like to ask the following question.

Let $\phi :\left[ 0,\infty \right) \rightarrow \left[ 0,\infty \right) $ be an increasing homeomorphism, and let $t>0$. Suppose $\underline{\lim }% _{x\rightarrow \infty }\frac{\phi \left( tx\right) }{\phi \left( x\right) }>0 $ and $\overline{\lim }_{x\rightarrow \infty }\frac{\phi \left( tx\right) }{% \phi \left( x\right) }<\infty $.

Is it true that $\lim_{x\rightarrow \infty }\frac{\phi \left( tx\right) }{% \phi \left( x\right) }$ exists ?

I have failed both trying to prove it or providing a counterexample.

Thanks in advance.

Uriel

1 Answers 1

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This might be easier to think about if you define a new increasing homeomorphism $\tilde{\phi} : \mathbb R \to \mathbb R$ by $\tilde{\phi}(y) = \log \phi(e^y)$. In terms of $\tilde{\phi}$, you know that $\lim \inf_{y \to \infty} (\tilde \phi(y + \delta) - \tilde \phi(y)) > -\infty$ and $\lim \sup_{y \to \infty} (\tilde \phi(y + \delta) - \tilde \phi(y)) < +\infty$, where $\delta = \log t$.

Then consider a $\tilde \phi$ that looks like a wavy line, e.g. $\tilde \phi(y) = 2y + \sin y$. Does this help?

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    Hi, thanks. I understand your idea, but I don´t see how this may help in providing explictly one $\phi$. Uriel – 2017-02-24
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    Hi again. I think your idea works. Solving for $\phi $ in the last equation you wrote you get $\phi \left( x\right) =x^{2}e^{\sin \left( \ln x\right) }$ . This satisfies the hypothesis, and taking different sequences $ x_{n}\rightarrow \infty $ you see that $\lim_{x\rightarrow \infty }\frac{ \phi \left( tx\right) }{\phi \left( x\right) }$ does not exist. Do you agree? – 2017-02-24
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    Yes, that is exactly my idea! – 2017-02-24