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 November 2nd, 2008, 08:46 PM #1 Member   Joined: Sep 2008 Posts: 46 Thanks: 0 differentiable function Let f:R --> R be differentiable where f'(x) <= a*f(x) with some constant a. WTS that f(x) <= f(0) * e^(ax)for all non-negative x.
 November 2nd, 2008, 10:51 PM #2 Senior Member   Joined: Jul 2008 Posts: 144 Thanks: 0 Re: differentiable function just try to use Gronwall-Belman inequation. [a series of famous inequations,called Gromwall/Bellman inequation]
 November 3rd, 2008, 07:00 AM #3 Member   Joined: Sep 2008 Posts: 46 Thanks: 0 Re: differentiable function I tried looking this up but couldn't find much on it. Any resources. All I have are basic analysis texts (rudin,royden,etc.).
 November 4th, 2008, 01:18 AM #4 Site Founder     Joined: Nov 2006 From: France Posts: 824 Thanks: 7 Re: differentiable function Try looking for Gronwall's lemma on Google (this is under this name that this proposition is usually called; it is used in ODE theory in order to prove the uniqueness part of various existence theorems, for instance the Cauchy-Lipschitz theorem).
 November 6th, 2008, 11:39 PM #5 Member   Joined: Sep 2008 Posts: 46 Thanks: 0 Re: differentiable function Can you perhaps explain how to apply Gronwall's inequality to this problem. I read some references on the inequality but most of them include the integral-version, and I can't seem to apply it properly. I think I need to come up with some function g in terms of f such that g' < 0. still not sure.
 November 8th, 2008, 03:50 AM #6 Senior Member   Joined: Jul 2008 Posts: 144 Thanks: 0 Re: differentiable function f'(x) <= a*f(x) ==> f(x)<=a*int(f,[0,x])+f(0) then use the inequation. now,let me show a problem: complete lattics(L,<=) denote G={x:L|x<=f(x)},here f is a increase function on L->L prove:if fix(f)={m},then m=supG(or,m=supG=>f(m)=m:A)

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