1. Jun 2, 2009

creepypasta13

1. The problem statement, all variables and given/known data
let S' be the unit circle, C the set of complex numbers, R the set of real numbers, ||f|| = sqrt[integral(f^2) from -pi to pi] (the length or norm of f)

find a function f: S'->C (so f is 2-pi periodic) and a sequence of functions {f_n}:R->C so that
||f-f_n|| converges to 0 but we dont have f_n(x) converging to f(x) for ANY x in S'

2. Relevant equations

3. The attempt at a solution
i was thinking of (-1)^n for f_n, and 0=f(x), but then ||f-f_n|| converges to 1, not 0

Last edited: Jun 2, 2009
2. Jun 2, 2009

Dick

You are right. (-1)^n doesn't work. You want a step type function whose center keeps oscillating over the interval [-pi,pi] while it's diameter keeps shrinking.

3. Jun 3, 2009

creepypasta13

if its diameter keeps shrinking, then its limit will approach 0, hence not satisfying the 2nd part of the problem where it can't approach 0 for ANY x

4. Jun 3, 2009

Dick

Let c(I)(x) be the characteristic function for an interval I (i.e. c(I)(x)=1 for x in I, 0 otherwise). Suppose I want such a sequence of functions on the interval [0,1] instead of S'. Pick f_1=c([0,1/2]), f_2=c([1/2,1]), f_3=c([0,1/3]), f_4=c([1/3,2/3]), f_5=c([2/3,1]), f_6=c([0,1/4])... Let f=0.

5. Jun 3, 2009

thanks