# simple functional inequality [closed]

This comes up in a paper by Beskos and Roberts on exact simulation of diffusions. I'm sure it's easy, but I can't work out what's going on.

Given a differentiable function $\alpha$ satisfying

$\alpha(u) \geq k_1$

and

$\alpha'(u) \leq k_2$,

for some constants $k_1$ and $k_2$, show that there exist constants $c_1$ and $c_2$ such that

$c_1 \leq \alpha^2(u) + \alpha'(u) \leq c_2$.

Clearly one condition implies alpha is bounded below and one implies that alpha has at most linear growth, but I don't see how this is enough for the conclusion.

Edit - I'm a fool. They wrote $k_1 \leq \alpha(u), \alpha'(u) \leq k_2$, but there was a line break after the comma. I was reading them as two seperate conditions on $\alpha$ and $\alpha'$. Sorry everyone, thanks for the help. If it wasn't for the counterexamples it would have taken a lot longer to realise that.

Thanks.

## closed as too localized by Lord_Farin, Start wearing purple, Martin, Ittay Weiss, Julian KuelshammerJun 4 '13 at 11:10

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

• It would help to link to the paper you speak of... – J. M. is a poor mathematician Nov 5 '10 at 16:17
• $\alpha(u) = u$ for $u > 0$ seems to be a counterexample. – Aryabhata Nov 5 '10 at 16:19
• Sorry, I think they require $\alpha$ defined on the whole real line and everywhere differentiable. The relevant paper is here: arxiv.org/pdf/math/0602523. Page six, conditions 3 and 3' – Simon Nov 5 '10 at 16:24
• $\alpha'(u)=-1$ for $u\leq-1$, $\alpha'(u)=1$ for $u\geq1$, and $\alpha'(u)=u$ for $-1\leq u\leq1$ gives a family of counterexamples on $\mathbb{R}$ with $\alpha'\leq 1$ and $\alpha\geq \alpha(0)$. – Jonas Meyer Nov 5 '10 at 16:27
• That would a priori be a reasonable interpretation even without the line break, but with the line break it was especially ambiguous. No need to apologize. – Jonas Meyer Nov 5 '10 at 16:45