Examples of increasing homeomorphism from $\mathbb{R}_+$ onto $\mathbb{R}_+$ satisfying some inequalities Let $\varphi : \mathbb{R}_+ \to \mathbb{R}_+$ be an increasing homeomorphism satisfying $\varphi(0)=0,$ where $ \mathbb{R}_+:=[0,\infty).$
For example,  $\varphi(s)=\frac{s^3}{1+s^2}$ for $s \in \mathbb{R}_+.$ Then $\varphi'(s)=\frac{s^4+3s^2}{(1+s^2)^2}>0$ for $s \neq 0,$ so that  $\varphi:\mathbb{R}_+ \to \mathbb{R}_+$ is an increasing homeomorphism. In this case, there are increasing homeomorphisms $\psi_1,\psi_2: \mathbb{R}_+\to \mathbb{R}_+$ satisfying
    $$ \varphi(s)\psi_1(t)\le \varphi(st) \le \varphi(s) \psi_2(t)~\hbox{for all }~s,t \in \mathbb{R}_+.\label{f1}\tag{F1}$$
$\psi_1(t)=\min\{t^3,t\}$  and $\psi_2(t)=\max\{t^3,t\}$  are the examples (see [Examples for increasing homeomorphisms related to $\varphi$-laplacian).
Seeing this example, I have a few question as follows:

(1) Are there any examples of $\varphi$ such that $\not \exists$ $\psi_1$ and $\not \exists$ $\psi_2$ satisfying \eqref{f1}? Here $\psi_1, \psi_2: \mathbb{R}_+\to \mathbb{R}_+$ are homeomorphisms.
(2) Are there any examples of $\varphi$ such that $\exists$ $\psi_1$, but $\not \exists \psi_2$ satisfying \eqref{f1}? Here $\psi_1, \psi_2: \mathbb{R}_+\to \mathbb{R}_+$ are homeomorphisms.
(3) Are there any examples of $\varphi$ such that $\exists$ $\psi_1$ and $\exists$ $\psi_2$ satisfying \eqref{f1}? Here $\psi_1: \mathbb{R}_+\to \mathbb{R}_+$ is a homeomorphism and $\psi_2: \mathbb{R}_+\to \mathbb{R}_+$ is a function, not a homeomorphism.
(4)  Are there any examples of $\varphi$ such that we don't know if $\exists$ $\psi_1$ and $ \exists$ $\psi_2$ satisfying \eqref{f1} or not, but it is hard to find them? Here $\psi_1, \psi_2: \mathbb{R}_+\to \mathbb{R}_+$ are homeomorphisms.

It seems that the questions (1) and (2) are hard to answer, but (3) or (4) may be possible.
Please let me know if you have any idea or comment for my questions. Thanks in advance.
 A: Given a homeomorphism $\varphi$ of $\Bbb R_+=[0,\infty)$ (which is necessarily increasing and $\varphi(0)=0$), for each $t\in [0,\infty)$ put $\underline{\varphi}(t)=\inf_{s>0} \varphi(st)/ \varphi(s)$ and $\overline{\varphi}(t)=\sup_{s>0} \varphi(st)/ \varphi(s)$. In particular, for each $t\in\Bbb R_+$ we have 
$$\underline{\varphi}(t)\le \frac{{\varphi}(t)}{\varphi(1)}\le \overline{\varphi}(t).$$
Remark that it is possible that $\overline{\varphi}(t)=\infty$ for some $t>0$. Since the map $\varphi$ is increasing and $\varphi(0)=0$, the maps $\underline{\varphi}$ and $\overline{\varphi}$ are non-decreasing and $\underline{\varphi}(0)= \overline{\varphi}(0)=0$. Also we have $\underline{\varphi}(1)= \overline{\varphi}(1)=1$. Moreover, it is easy to show the following Multiplicative Inequality: for each $t,t’>0$ we have $\underline{\varphi}(tt’)\ge \underline{\varphi}(t) \underline{\varphi}(t’)$ and $\overline{\varphi}(tt’)\le \overline{\varphi}(t) \overline{\varphi}(t’)$. 
If $\psi_1$ and $\psi_2$ are functions from $\Bbb R_+$ to $\Bbb R_+$ satisfying F1 then $\psi_1(t)\le \underline{\varphi}(t)$ and $\psi_2(t)\ge \overline{\varphi}(t)$ for each $t>0$. 
So if such a function $\psi_2$ exists then the function $\overline{\varphi}$ is finite, that is $\overline{\varphi}(t)<\infty$ for each $t>0$. Multiplicative Inequality implies that the function $\overline{\varphi}$ is finite iff there exists $\overline{t}_+>1$ such that $\overline{\varphi}(\overline{t}_+)$ is bounded. 
Nevertheless, if $\varphi(s)=\ln (1+s)$ for each $s\in\Bbb R_+$ then there is no homeomorphism $\psi_2$ of $\Bbb R_+$, satisfying F1, because $\overline{\varphi}(t)=\max\{1,t\}$ for each $t>0$. Indeed, $\varphi(st)/ \varphi(s)=\log_{1+s} (1+st)$ for each $s>0$. 
If $t\ge 1$ then $(1+s)^t\ge 1+st$ by Bernoulli's inequality, so $\log_{1+s} (1+st)\le t$. On the other hand, by L'Hôpital's rule,
$$\lim_{s\to +0} \log_{1+s} (1+st)= \lim_{s\to +0} \frac {\ln (1+st)}{\ln(1+s)}= \lim_{s\to +0} \frac {\frac{t}{1+st}}{\frac 1{1+s}}=t
.$$
If $0<t\le 1$ then $1+s\ge 1+st$, so $\log_{1+s} (1+st)\le 1$. On the other hand,
$$\frac {\ln (1+st)}{\ln(1+s)}\ge \frac {\ln (t+st)}{\ln(1+s)}= \frac {\ln t+\ln s}{\ln(1+s)},$$
and the last value tends to $1$ when $s$ tends to infinity. 
So in order to provide that there exists a homeomorphism $\psi_2$ of $\Bbb R_+$, satisfying F1, we need an addtional condition. A necessarily condition is there exists $0<\overline{t}_-<1$ such that $\overline{\varphi}(\overline{t}_-)<1$. 
We claim that if the points $\overline{t}_-$ and $\overline{t}_+$ exist then there exists such a homeomorphism $\psi_2$. Indeed, by Multiplicative Inequality for each non-negative integer $n$ we have $\overline{\varphi}(t)\le \overline{\varphi}(\overline{t}_-)^n$ if $t\le \overline{t}_-^n$ and $\overline{\varphi}(t)\le \overline{\varphi}(\overline{t}_+)^n$, if $t\le \overline{t}_+^n$. In particular, $\overline{\varphi}(0)=0$. So we can put $\psi_2(0)=0$ and for each $t>0,$
$$\psi_2(t)=\max\left\{\overline{\varphi}(\overline{t}_+)^{1+\tfrac{\ln t}{\ln \overline{t}_+}},\overline{\varphi}(\overline{t}_-)^{\tfrac {\ln t}{\ln \overline{t}_-}-1}\right\}.$$
Indeed, it is easy to check that $\psi_2(0)=0$, $\lim_{t\to\infty} \psi_2(t)=+\infty$ and a function $\psi_2$ is continuous and increasing. So it is open that is it is a homeomorphism of $\Bbb R_+$, that is there exists a continuous inverse bijection $\psi_2^{-1}:\Bbb R_+\to \Bbb R_+$. 
Moreover,
$$\psi_2(1)=\max\left\{\overline{\varphi}(\overline{t}_+),\overline{\varphi}(\overline{t}_-)^{-1}\right\}\ge 1=\overline{\varphi}(1).$$
If $t>1$ then let $n$ be a positive integer such that $\overline{t}_+^{n-1}<t\le \overline{t}_+^n$. Then $\overline{\varphi}(t)\le \overline{\varphi}(\overline{t}_+)^n\le \psi_2(t)$, because $1+\log_{\overline{t}_+} t\ge n$.
If $t<1$ then let $n$ be a positive integer such that $\overline{t}_-^{n-1}>t\ge \overline{t}_-^n$. Then $\overline{\varphi}(t)\le \overline{\varphi}(\overline{t}_-)^{n-1}\le \psi_2(t)$, because $\log_{\overline{t}_-} t\le n$.
Also the following properties of the function $\overline{\varphi}$ can be useful. Since the function $\overline{\varphi}$ is non-decreasing, by Multiplicative Inequality for each natural number $n$ we have 
$$\overline{\varphi}(0)\le \lim_{t\to 0+} \overline{\varphi}(t)\le \overline{\varphi}(\overline{t}_-^n)\le \overline{\varphi}(\overline{t}_-)^n\to 0,$$
and 
$$\lim_{t\to\infty} \overline{\varphi}(t)\ge \overline{\varphi}(1)/\overline{\varphi}(\overline{t}_-^n) \to+\infty.$$
So $\overline{\varphi}(0)=0$ and $\lim_{t\to\infty} \overline{\varphi}(t)=+\infty$.
We claim that the function $\overline{\varphi}$ is increasing. Indeed, suppose to the contrary that there exist real numbers $t<t’$ such that $\overline{\varphi}(t)=\overline{\varphi}(t’)$.
If $t=0$ then for each $t^*>0$ there exists a natural number $n$ such that $t^*\le t\overline{t}_+^n$ so, since the function $\overline{\varphi}$ is non-decreasing, by Multiplicative Inequality 
$$\overline{\varphi}(t^*)\le \overline{\varphi}(t) \overline{\varphi}(\overline{t}_+^n)=0,$$
a contradiction with $\lim_{t\to\infty} \overline{\varphi}(t)=+\infty$.
If $t>0$ then by the above $\overline{\varphi}(t)>0$ and, since the function $\overline{\varphi}$ is non-decreasing, by Multiplicative Inequality
$$1=\overline{\varphi}(1)\le \overline{\varphi}(t’/t)\le \overline{\varphi}(t’)/ \overline{\varphi}(t)=1.$$
So $\overline{\varphi}(t’/t)=1$.  Then for each $t^*>0$ there exists a natural number $n$ such that $t^*\le (t’/t)^n$ so, since the function $\overline{\varphi}$ is non-decreasing, by Multiplicative Inequality 
$$\overline{\varphi}(t^*)\le \overline{\varphi}(1) \overline{\varphi}((t’/t)^n)=1,$$
a contradiction with $\lim_{t\to\infty} \overline{\varphi}(t)=+\infty$.
The $\psi_1$ case is considered similarly. In order to provide that there exists a homeomorphism $\psi_1$ of $\Bbb R_+$, satisfying F1, we need that $\lim_{t\to+\infty} \underline{\varphi}(t)=+\infty$. Multiplicative Inequality implies that this condition holds iff there exists $\underline{t}_+>1$ such that $\underline{\varphi}(\underline{t}_+)>1$. This condition is not automatic. Indeed, if again $\varphi(s)= ln(1+s)$ for each $s\in\Bbb R_+$ then $\varphi(st)/ \varphi(s)=\log_{1+s} (1+st)$ for each $s>0$. If $s\ge t$ then $(1+s)^2=1+2s+s^2>1+s^2\ge 1+st$, so $\underline{\varphi}(t)\le 2$ for each $t$.  
An other necessary condition to provide that there exists a homeomorphism $\psi_1$ of $\Bbb R_+$, satisfying F1, is: there exists $0<\underline{t}_-<1$ such that $\underline{\varphi}(\underline{t}_-)>0$. This condition is not automatic too, as shows the following example. Let $\varphi(s)=e^s-1$ for each $s\in\Bbb R_+$. Then for each $s,t>0$,  $\varphi(st)/ \varphi(s)=\tfrac{e^{st}-1}{e^s-1}$. 
 If $t<1$ then $\underline{\varphi}(t)=0$ because 
$$\lim_{s\to +\infty}\frac{e^{st}-1}{e^s-1}=\lim_{s\to +\infty} \frac{e^{s{t-1}}-e^{-s}}{1-e^{-s}}=0.$$
Now assume that $t\ge 1$ and $s>0$. Put $x=e^s-1$. By Bernoulli's inequality, $(1+x)^t\ge 1+xt$. Thus $$\frac{e^{st}-1}{e^s-1}= \frac{(1+x)^t-1}{x}\ge t.$$
Then  $\underline{\varphi}(t)=t$ because by L'Hôpital's rule 
$$\lim_{s\to +0} \frac{e^{st}-1}{e^s-1}= \lim_{s\to +0} \frac{te^{s}}{e^s}=t.$$
We claim that if the points $\underline{t}_-$ and $\underline{t}_+$ exist then there exists such a homeomorphism $\psi_1$.  Indeed, pick arbitrary numbers $\alpha_+,\alpha_-, C>0$ such that $C\underline{t}_+\le 1$, $\underline{t}_+^{2\alpha_+}\le\underline{\varphi}(\underline{t}_+)$, $C\le \underline{\varphi}(\underline{t}_-)$, and $\underline{t}_-^{\alpha_-}\le  \underline{\varphi}(\underline{t}_-)$. It is easy to check that $C\underline{t}_+^{\alpha_+(n+1)}\le \underline{\varphi}(\underline{t}_+)^n$ and  $C\underline{t}_-^{\alpha_- n}\le \underline{\varphi}(\underline{t}_-)^{n+1}$ for each non-negative integer $n$. 
Put $\psi_1(0)=0$,  and $\psi_1(t)=Ct^{\alpha_-}$ for any $0\le t\le 1$, and $\psi_1(t)=Ct^{\alpha_+}$ for any $t\ge 1$.  Iit is easy to check that $\psi_1(0)=0$, $\lim_{t\to\infty} \psi_1(t)=+\infty$ and a function $\psi_1$ is continuous and increasing. So it is open that is it is a homeomorphism of $\Bbb R_+$, that is there exists a continuous inverse bijection $\psi_1^{-1}:\Bbb R_+\to \Bbb R_+$. 
If $t\ge 1$ then let $n$ be the largest integer such that $t\ge \underline{t}_+^n$. Then $t<\underline{t}_+^{n+1}$ so 
$$\psi_1(t)\le \psi_1(\underline{t}_+^{n+1})=C\underline{t}_+^{\alpha_+(n+1)}\le \underline{\varphi}(\underline{t}_+)^n\le \underline{\varphi}(\underline{t}_+^n)\le \underline{\varphi}(t).$$
If $t\le 1$ then let $n$ be the largest integer such that $t\le \underline{t}_-^n$. Then $t>\underline{t}_-^{n+1}$ so 
$$\psi_1(t)\le \psi_1(\underline{t}_-^n)=C\underline{t}_-^{\alpha_- n}\le \underline{\varphi}(\underline{t}_-)^{n+1}\le \underline{\varphi}(\underline{t}_-^{n+1})\le \underline{\varphi}(t).$$
A: This is an answer to (1) and (2), and I am not quite clear what exactly (3) and (4) are asking.
If $\phi(s) = e^s-1$, then any function $\phi_2$ satisfying (F1) would have
$$
\psi_2(t) \ge \frac{\phi(st)}{\phi(s)} = \frac{e^{st}-1}{e^{s}-1} = \frac{e^{s(t-1)}-e^{-s}}{1-e^{-s}}
$$
for all $s,t>0$. In particular, if you fix $t>1$ and let $s\to\infty$, you can see that $\psi_2(t)=\infty$ for all $t>1$.
Similarly, for the inverse function $\phi(s) = \ln(s+1)$ you get that one would have $\psi_1(t)=0$ for all $t \in (0,1)$.
