# 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.

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It would help to link to the paper you speak of... –  Ｊ. Ｍ. 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