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Homework Help: Cauchy sequence

  1. Mar 6, 2005 #1
    We consider the space [tex]C^0 ([-1,1])[/tex] of continuous functions from [tex][-1,1][/tex] to [tex]\mathbb{R}[/tex] supplied with the following norm:

    [tex]||f||_1 = \int_{-1}^{1} |f(x)| dx [/tex]

    a. Show that [tex]||.||_1[/tex] defines indeed a norm.

    b. Show that the sequence of functions [tex](f_n)[/tex], where

    [tex]
    \begin{align*}
    f_n(x) &= -1, \quad & -1 \leq{x} \leq{\frac{-1}{n}} \\
    \ &= nx, \quad & \frac{-1}{n} \leq{x} \leq{\frac{1}{n}} \\
    \ &= 1, \quad & \frac{1}{n} \leq{x} \leq{1}
    \end{align*}[/tex]

    is a Cauchy-sequence with respect to the given norm.

    c. Show that [tex]C^0 ([-1,1])[/tex] is not complete with respect to the given norm.

    I figured out a. myself, by showing this norm satisfies the properties of a norm, but I can't find out how to tackle b. and c.
     
  2. jcsd
  3. Mar 6, 2005 #2
    b) I assume you can compute [tex] ||f_m - f_n||_1 [/tex] for arbitrary m > n > 0. Now [tex]f_n[/tex] is called a Cauchy-sequence if for all [tex]\epsilon >0[/tex] there exists a N>0 such that: [tex] ||f_m - f_n||_1 < \epsilon [/tex] for all m>n>N. You can do the estimates yourself.

    c) Does the sequence converge to a continuous function? A set with a given norm is called closed if all Cauchy-sequences converge to an element in the same set.
     
  4. Mar 6, 2005 #3
    I knew the definition of a Cauchy sequence, but I still can't find the solution.

    I think I can't, given a certain n and [tex]f_n[/tex] I can find [tex]f_{n+1}[/tex] and I see that once n approaches infinity [tex]f_n[/tex] becomes either -1 or 1, but I don't know how to work from there. In fact this is all quite new to me.
     
  5. Mar 7, 2005 #4
    When we assume m > n:

    [tex] \begin{align*}|f_m(x) -f_n(x)| &= 0 & x > 1/n \\
    \ &= 1-nx & 1/m < x < 1/n \\
    \ &= (m-n)x & 0 \leq x < 1/m \end{align*}[/tex]

    and [tex] |f_m(-x) -f_n(-x)| = |f_m(x) -f_n(x)| [/tex]. So:

    [tex]||f_m - f_n||_1 = \int_{-1}^{1} |f_m(x)-f_n(x)| dx = 2 \int_{0}^{1} |f_m(x)-f_n(x)| dx = 2 ( \int_{0}^{1/m} (m-n)x dx + \int_{1/m}^{1/n} (1-n x) dx ) = [/tex]
    [tex]= 2 ( \frac{m-n}{2 m^2} + 1/n - 1/m - \frac{n}{2 n^2} + \frac{n}{2 m^2}) = 1/n-1/m < 1/n [/tex]

    Let [tex] \epsilon > 0 [/tex]. Take [tex] N > 1/\epsilon [/tex], then for all n,m > N, we have [tex]||f_m - f_n||_1 < 1/N < \epsilon [/tex].

    Now we have proven that [tex]f_n[/tex] is indeed a Cauchy-sequence.
     
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