Example of function satisfying the growth condition: $\phi\big(\theta \frac{s}{t}\big) \leq \frac{\phi(s)}{\phi(t)}$

Question

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 $\phi(t)= e^{t^\alpha}-1$, $\alpha>0$.

Answer 1

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.

Answer 2

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

$$\phi\big(\frac{s}{8t}\big)\leq \phi_0\big(\frac{s}{4t}\big) \leq \frac{\phi_0(s/2)}{\phi_0(2t)}\leq \frac{\phi(s)}{\phi(t)} \leq \qquad\text{for $s\leq t$.} $$