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May 1st, 2007, 02:16 AM   #1
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Bounded derivatives

Here's a cute problem. Let f: R->R be a mapping which is N+1 times differentiable, and such that both f and f^(N+1) (the N+1-th derivative of f) are bounded on R. Show that all derivatives f^(k) of f are bounded on R.

R designates the field of all real numbers.
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October 30th, 2007, 09:57 AM   #2
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You're assuming f^k exists on R for every k, right? This is not part of the conclusion, right? If, for example, n =1, then the fact that f is twice differentiable and f and f'' are bounded on R does not impliy f has derivatives of all orders on R.

I'm confused here.

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October 30th, 2007, 10:47 AM   #3
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No, what I meant is to prove that given the hypotheses, the result holds for every k<=N+1. Of course, f does not end up being infinitely differentiable.
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October 30th, 2007, 11:48 AM   #4
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Can we suppose that n=N?
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October 30th, 2007, 01:47 PM   #5
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Sorry, I edited the original message. Actually there's only N.
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November 1st, 2007, 09:35 AM   #6
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Well it is true for N=1... thats all i can show for now, dunno how to expand it to N=n.

First if f is N+1 times differentialable on R f^k is continous everywhere. so if any derivate is unbounded its at x>>1

okay... if f is bounded then f(x)<M for all x in R
thus given any 2 elements in R: a and b, where a<b
i) 2M>f(a)-f(b)=f'(c)(b-a) for some c contained in (a,b) {meanvaluetheorem continous functions}
if f' is unbounded, then given any £ > 0 there exists a real number d in R such that f'(d) > 1/£^2
Now consider the interval (d-£,d+£) by i) there exists a q contained in the interval where f'(q)<2M*2£
Meanvalue theorem applied again yields f'(d)-f(q)=f''(k)*(q-d) for some k contained in (q,d) or (d,q) who have length < £
so f''(k)=(f'(d)-f(q))/(q-d)> (1/£^2 - 2M*2£) / £ which gets arbitrary large ie we have shown the existance of a such k for every chosen £, thus f''(k) must be unbounded (contradiction, so f' is unbounded is false)
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