why is C([0,1]) banach? isnt the sequence x^n cauchy, but doesnt converge. (with sup norm)

2/12/2007 11:56:03 PM

For the sequence x^n, the limiting function is discontinuous (0 for 0<=x<1, 1 for x=1), so the limit isn't in C([0,1]); you're right on that part.

However, the sequence isn't Cauchy; without full detail, here's why not: for large values of n, the value of the function x^n stays near zero until x gets close to 1, then increases very quickly as x nears 1. Take epsilon = 1/4, and let n be as large as you want. Then there exists some x_0 such that (x_0)^n = 1/2. Consider m=2n. Then (x_0)^m = (x_0)^(2n) = ((x_0)^n)^2 = (1/2) ^2 = 1/4. So at the point x_0, the sequences x^n and x^m are separated by 1/4, so ||x_n - x_m|| >= 1/4 in the sup norm, and the sequence isn't Cauchy.

2/13/2007 9:28:33 AM

^what he said. Convergent implies Cauchy.

2/13/2007 10:47:57 AM

yeah, not really sure why you thought that sequence was Cauchy.

2/13/2007 1:15:29 PM

Yeah, not really sure why I thought it was cauchy either. I suppose I should have checked that out before posting....

2/13/2007 1:38:18 PM

It's OK. We all have our "dur" moments.

2/13/2007 4:46:24 PM