Nondecreasingness of $\frac{\Phi(t)}{t}$ implies the nondecreasingness of $\frac{\Phi(t)}{t^{1+\varepsilon}}$?

Question

Let $\Phi:[0,\infty)\rightarrow[0,\infty)$ be a convex function, i.e. $\Phi(\lambda t_1+(1-\lambda)t_2)\leq \lambda\Phi(t_1)+(1-\lambda)\Phi(t_2)$ for all $\lambda\in[0,1]$ and $t_1,t_2\in[0,\infty)$. Let also $\Phi(0)=0$. Therefore, $t\in(0,\infty)\mapsto\frac{\Phi(t)}{t}$ is a nondecreasing function. Indeed, $t_1=\frac{t_1}{t_2}t_2+\big(1-\frac{t_1}{t_2}\big)0$, then $$ t_1\le t_2\Rightarrow\Phi(t_1)\leq\dfrac{t_1}{t_2}\Phi(t_2). $$ My question is can we say that "nondecreasingness of $\frac{\Phi(t)}{t}$ implies the nondecreasingness of $\frac{\Phi(t)}{t^{1+\varepsilon}}$ for some $\varepsilon>0$."

Answer 1

Edit: Here's a very simple example, in fact a bit trivial. Let $$\Phi(t)=t.$$ This is (non-strictly) convex on $[0,\infty)$ and $\Phi(0)=0$. The function $$\frac{\Phi(t)}{t}=1$$ is nondecreasing in $(0,\infty)$, but $$\frac{\Phi(t)}{t^{1+\epsilon}}=t^{-\epsilon}$$ is not, because it is a negative power.