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 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?
$t^l\leq \Phi(t)+1,\forall t\geq1$
3
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real-analysis
functional-analysis
functions
1 Answers
1
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 $$4 = t^2 \not\le \Phi(t) + 1 = \frac12 \, 2^2 + 1 = 3.$$