1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Real analysis

  1. Nov 15, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that the closed ball in ##C([0,1])## of center ##0## and radius ##1## is not compact.


    2. The attempt at a solution

    I was given a hint, to look at the sequence of continuous functions ##f_n(x) = x^n## on the closed ball in ##C([0,1])##. Why is that sequence continuous? Isn't it discontinuous in the closed ball ##[0,1]##?
     
    Last edited: Nov 15, 2013
  2. jcsd
  3. Nov 15, 2013 #2

    pasmith

    User Avatar
    Homework Helper

    Continuity is not a notion that applies to sequences. Convergence is.

    I assume you are using the sup metric, [itex]d(f,g) = \sup \{|f(x) - g(x)| : x \in [0,1]\}[/itex].

    The function [itex]f_n : x \mapsto x^n[/itex] is continuous (indeed, differentiable) on [itex][0,1][/itex] for any [itex]n \in \mathbb{N}[/itex]. Hence [itex]f_n \in C([0,1])[/itex]. It is straightforward to show that [itex]d(f_n,0) = 1[/itex], so [itex]f_n[/itex] is in the closed unit ball for all [itex]n[/itex].

    The question is: Does this sequence converge to a limit in [itex]C([0,1])[/itex] or not? Does it have any subsequences which do? And what does this tell you about the compactness of the closed unit ball in [itex]C([0,1])[/itex]?
     
  4. Nov 15, 2013 #3
    But I thought ##f_n : x \mapsto x^n## is not continuous on ##[0,1]## since ##
    f(x) = \begin{cases} 1 & \quad \text{if }x= 1 \\ 0 & \quad \text{if }x\in [0,1) \\ \end{cases}##
     
  5. Nov 15, 2013 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    ##f_n## is a polynomial and is continuous for all ##x##. Surely you know that. All polynomials are.
     
  6. Nov 15, 2013 #5
    LCKurtz - Wow, you're right! I should go run outside and clear my mind. Very dumb mistake by me. Thank you both!
     
  7. Nov 15, 2013 #6

    WannabeNewton

    User Avatar
    Science Advisor

    What you wrote down is the pointwise limit of ##(f_n)## and indeed the pointwise limit is discontinuous. However ##f_n## is continuous for all ##n##.

    Regardless, does the uniform limit of ##(f_n)## exist? Furthermore, does any subsequence ##(f_{n_k})## of ##(f_n)## have a uniform limit? Recall that convergence in ##C([0,1])## is uniform convergence in ##[0,1]##. And if no ##(f_{n_k})## has a uniform limit then what does that tell you about sequential compactness of the closed unit ball in ##C([0,1])##?
     
  8. Nov 15, 2013 #7
    Thank you, WannabeNewton! I know what to do now. Since it is not sequentially compact, therefore it is not compact.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Real analysis
  1. Real analysis (Replies: 3)

  2. Real analysis (Replies: 0)

  3. Real analysis (Replies: 16)

  4. Real Analysis (Replies: 1)

  5. Real Analysis (Replies: 5)

Loading...