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Suppose $f:(a,b)\to\mathbb R$ that $ f $ satisfies:

$$f\in C^1$$

$$\lim_{x\to a ^ +}f^2(x)=0$$ $$\lim_{x\to b ^ -}f^2(x)=e-1$$

if $\forall x \in(a,b) : 2f(x)f '(x)-f^2(x)\ge1 $, then how to prove $0\le b-a\le1$

Thanks in advance

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It must have something to do with the fact that $2ff'=(f^2)'$. – xavierm02 Mar 20 '13 at 19:29
I'm not sure what you mean but I'm not the one who downvoted. – xavierm02 Mar 20 '13 at 19:32
up vote 2 down vote accepted

Let $g(x)=e^{-x}f^2(x),$ then $g'(x)=e^{-x}(2f(x)f'(x)-f^2(x)) \geq e^{-x}.$ Take integral of $g'$ over $(a,b),$ to get $e^{-b}(e-1) \geq -e^{-b}+e^{-a}$ which leads you to $b-a \leq 1.$

I'll let you to fill out some little gaps!

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It's indeed more than a hint! – Ehsan M. Kermani Mar 20 '13 at 20:04

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