Thread: Banach space
View Single Post
dirk_mec1
#3
Sep28-08, 01:23 PM
P: 682
Quote Quote by Dick View Post
You have that backwards. ||fn-fm||_E<epsilon implies ||f'n-f'm||_infinity<epsilon.
Really? I don't see why this is so.

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?
But what good will that do?



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_n-f||_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?