Example of function satisfying the growth condition: $\phi\big(\theta \frac{s}{t}\big) \leq \frac{\phi(s)}{\phi(t)}$ I am barely looking for example(s) of invertible convex functions $\phi: [0,\infty)\to [0, \infty)$ such that $\phi(0)=0$ and there exists $\theta>0$ and for all $s\leq t$ we have
\begin{align}\label{EqI}\tag{I}
\phi\big(\theta \frac{s}{t}\big) \leq \frac{\phi(s)}{\phi(t)} \qquad\text{or equaly}  \qquad  \theta  \leq \phi^{-1}\big(\frac{s}{t}\big)\frac{\phi^{-1}(t)}{\phi^{-1}(s)}
\end{align}
The most simple class consists of polynomial functions of the form $\phi(t)= ct^p$ with $c>0$ and $p>0$.
Question: Are there other possible non-polynomial examples satisfying $\eqref{EqI}$?
I have tried without success with $\phi(t)= e^{t^\alpha}-1$, $\alpha>0$.
 A: I think you can basically just glue two such polynomial functions together:
$$
\phi(x) = 
\begin{cases}
 x^2 &:& 0\leq x\leq1,
 \\
 2x-1 &:& x>1.
\end{cases}
$$
verification for this particular function:
One can check that this function is continuous and convex.
Let us check that (I) is satisfied.
We assume that $\theta$ satisfies $\theta^2\leq 1/2$.
Due to $s\leq t$ we have to consider three cases:
first case: $s \leq t\leq 1$. We have
$$
\phi(\theta \frac st) 
= \theta^2 (s/t)^2
\leq (s/t)^2
= \phi(s)/\phi(t).
$$
second case: $s\leq 1\leq t$. We have
$t^2\geq 2t-1$ and therefore
$$
\phi(\theta \frac st) 
= \theta^2 (s/t)^2
\leq \theta^2 s^2/(2t-1)
\leq s^2/(2t-1)
= \phi(s)/\phi(t).
$$
third case: $1\leq s \leq t$. We have
$t^2\geq 2t-1$ and therefore
$$
\phi(\theta \frac st) 
= \theta^2 (s/t)^2
\leq \theta^2 s/t
\leq \frac12\cdot \frac{2s-1}{t}
\leq \frac{2s-1}{2t-1}
= \phi(s)/\phi(t).
$$
Thus, (I) is satisfied.
general remarks:
I think any convex function with
$$
\phi(t) \leq c_1 t^{p_1}
$$
for large $t$ and
$$
c_2t^{p_2} \leq \phi(t) \leq c_3 t^{p_1}
$$
for small $t$,
where $p_2 \geq p_1 \geq 1$ and $c_1,c_2,c_3>0$
are constants,
should satisfy (I).
I think this is even an if and only if condition, but a full proof would be complicated. I can provide some ideas/justifications for that if requested.
With this condition, it should be easy to verify that the above function satisfies (I), and also that the function $\ln(\frac{k+e^x}{k+1})$ from the comments is a valid solution and explains why $e^{x^\alpha}-1$ or anything else that grows exponentially cannot work.
A: Let $1\leq p<q$, the function $\phi_0:t\mapsto \min(t^p,t^q)$ isn't convex but one can check that
$$\phi(t)= \int_0^t\frac{\min(s^p,s^q)}{s}d s$$ is convex since $\phi'$ is nondecreasing or $\phi''\geq0$. Moreover, for all $t\geq 0$, we have
$\phi(t/2)\leq \phi_0(t)\leq\phi(2t)$
that is
$$\int_0^{t/2}\frac{\min(s^{p}, s^{q})}{s}d s\leq \min(t^{p}, t^{q})\leq \int_0^{2t} \frac{\min(s^{p}, s^{q})}{s}d s$$
Indeed, for $s\leq t/2\leq t$ we have $s^{p-1}\leq t^{p-1}$ and $s^{q-1}\leq t^{q-1}$ so that
$$\int_0^{t/2}\frac{\min(s^{p}, s^{q})}{s}d s \leq\int_0^{t/2}\min(t^{p-1}, t^{q-1})d s=\frac12 \min(t^{p}, t^{q}).$$
Similarly $2t\geq s\geq t$ implies $s^{p-1}\geq t^{p-1}$ and $s^{q-1}\geq t^{q-1}$ so that
$$\int_0^{2t}\frac{\min(s^{p}, s^{q})}{s}d s \geq\int_t^{2t}\min(t^{p-1}, t^{q-1})d s=\min(t^{p},t^{q}).$$
We can easily check that $\phi_0$ satisfies
$$\phi_0\big(\frac{s}{t}\big) \leq \frac{\phi_0(s)}{\phi_0(t)} \qquad\text{for $s\leq t$.} 
$$
So that, taking $\theta=\frac{1}{8}$ we get
$$\phi\big(\frac{s}{8t}\big)\leq \phi_0\big(\frac{s}{4t}\big) \leq \frac{\phi_0(s/2)}{\phi_0(2t)}\leq \frac{\phi(s)}{\phi(t)}  \leq \qquad\text{for $s\leq t$.} 
$$
