
#1
Sep2708, 10:30 AM


#2
Sep2708, 12:34 PM

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P: 25,166

You have that backwards. fnfm_E<epsilon implies f'nf'm_infinity<epsilon. Can you use the fact the difference in derivatives of fn and fm is small to prove the difference between fn and fm is small? Hence that fn(x) is a cauchy sequence for each x?




#3
Sep2808, 01:23 PM

P: 664

So here is the interpretation of the assignment in my eyes: Given a Cauchy sequence [tex](f_n)_n \in\ E[/tex] prove that [tex]f_nf_E \rightarrow 0 [/tex] and that f is in E. So we have: [tex] f'_n f'_m_{\infty} < \epsilon\ \forall m,n \geq N [/tex] and we want: [tex] f'_n f'_{\infty} \rightarrow 0\ \forall n \geq N [/tex] Is this correct? 



#4
Sep2808, 02:18 PM

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P: 25,166

Banach space
Yes, that's it. Show f exists and has bounded derivative.




#5
Sep2908, 03:22 PM

P: 664

I'm sorry Dick, I have have been thinking about this but I can't seem to get f in E and converging.
Your posts imply that I should prove that f_n is a Cauchy sequence but what do I get from that? You also mention to use the deratives to prove that f_n is Cauchy: do you mean that I should use the definition of the derative? 



#6
Sep2908, 03:42 PM

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You know there is a limiting function f', since f'_n(x) is a cauchy sequence in R for every x. So that sequence has a limit, define f'(x) to be that limit. f' is also continuous since it's a uniform limit of continuous functions. Once you have f' just define f to be the integral from 0 to x of f'(t)dt.



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