I have that $\Phi:\mathbb{R}\rightarrow [0,+\infty)$ is a continuous, increasing and convex function such that $$t^l\Phi(1)\leq \Phi(t)\leq t^m \Phi(1), \;\;\forall t\geq1$$ where $1<l<m$.

and $\displaystyle\lim_{t\rightarrow0}\frac{\Phi(t)}{t}=0~\text{and}~\lim_{t\rightarrow+\infty}\frac{\Phi(t)}{t}=+\infty.$

I want to prove that $$t^l\leq \Phi(t)+1,\;\;\forall t\geq1$$

Can someone help me?


This is not true. Take $l = 2$, $m = 3$ and $\Phi(t) = \frac12 \, t^2$. It is clear that your assumptions are satisfied. However, with $t = 2$ you get $$4 = t^2 \not\le \Phi(t) + 1 = \frac12 \, 2^2 + 1 = 3.$$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.