Analysis Problem: Finding f & f_n for ||f-f_n|| Convergence

[creepypasta13]

Homework Statement

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 don't have f_n(x) converging to f(x) for ANY x in S'

Homework Equations

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

 

[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.

 

[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

 

[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.

 

[creepypasta13]

thanks