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Proof involving sequences of functions and uniform convergence

  1. Apr 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Let [tex] \phi_n(x) [/tex] be positive valued and continuous for all x in [tex] [-1,1] [/tex] with:

    [tex] \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x) = 1 [/tex]

    Suppose further that [tex] \{\phi_n(x)\} [/tex] converges to 0 uniformly on the intervals [tex] [-1,-c] [/tex] and [tex] [-c,1] [/tex] for any [tex] c > 0 [/tex]

    Let g be any function which is continuous on [tex] [-1,1] [/tex].

    Show that:

    [tex] \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x) g(x)dx = g(0) [/tex]

    2. Relevant equations


    If the functions [tex] \{f_n\} [/tex] and F are integrable on a bounded closed set E, and [tex] \{f_n\} [/tex] converges to F pointwise on E, and if [tex] \Vert {f_n} \Vert < M [/tex] for some M and all n = 1,2,3,..., then:

    [tex] \lim_{n\rightarrow\infty} \int_{E} {f_n} = \lim_{n\rightarrow\infty} \int_{E} F [/tex]

    3. The attempt at a solution

    Alrighty, so the first thing I did was apply the theorem on:

    [tex] f_n = \phi_n [/tex] on the intervals [-1,-c] and [-c,1]

    and came up with the following integrals(since uniform convergence implies pointwise convergence):

    [tex] \lim_{n\rightarrow\infty} \int_{-1}^{-c} \phi_n(x) = 0 [/tex]
    [tex] \lim_{n\rightarrow\infty} \int_{-c}^{-1} \phi_n(x) = 0 [/tex]

    If we add these integrals together we know what happens at c=0.

    Looking at our new sequence:

    [tex] f_n = \phi_n g(x) [/tex]

    This sequence is uniformly bounded on [-1,-c] and [-c,1] and converges pointwise to 0.

    Thus I think because of the continuity of [tex] \phi_n [/tex] and g, and the uniform convergence of [tex] \phi_n [/tex]:

    [tex] \lim_{n\rightarrow\infty} \int_{-1}^{-c} \phi_n(x) g(x) = 0 [/tex]
    [tex] \lim_{n\rightarrow\infty} \int_{-c}^{-1} \phi_n(x) g(x) = 0 [/tex]

    Again if we add these integrals together we know what happens when c = 0. But I'm not entirely sure where to go from here or if I'm going in the right direction. Any help is appreciated.
  2. jcsd
  3. Apr 6, 2009 #2
    Second attempt at solution

    Okay so I've tried a little bit more now:

    Just adding and subtracting a term from the limit we want to prove:

    \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x) g(x)dx =
    \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x)(g(x)-g(0))dx + \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x) g(0)dx


    [tex] \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x) g(0)dx = g(0) [/tex]

    and by the mean value theorem for integrals we know there exists a t in [tex] [-1,1] [/tex] such that:

    [tex] \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x)(g(x)-g(0))dx = (g(t)-g(0))\lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x) = g(t) - g(0) [/tex]

    and now I want to make an argument that states g(t) must be g(0).

    So I started by breaking my integral up:
    \lim_{n\rightarrow\infty} \int_{-1}^{-c} \phi_n(x) g(x) + \lim_{n\rightarrow\infty} \int_{-c}^{1} \phi_n(x) g(x) = 0

    as c goes to zero.

    But then g(t) - g(0) = 0 and thus g(t) = g(0) and then we would have:

    \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x) g(x)dx = \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x)(g(x)-g(0))dx + \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x) g(0)dx = g(t) - g(0) + g(0) = g(0) [/tex]

    The very last step seems a little hand wavy, I hope someone can help me.

    Edit: (I also realize some of the stuff in my first post is wrong, but it won't let me edit it)
    Last edited: Apr 6, 2009
  4. Apr 6, 2009 #3


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    You MUST mean phi_n converges uniformly to zero on [-1,-c] and [c,1] for any c>0. Not [-1,-c] and [-c,1]. If that were the case then the integral of phi_n would go to zero as n->infinity.
  5. Apr 6, 2009 #4
    Right, thanks for pointing that out.

    Then my integrals at the bottom of my second post change to:


    \lim_{n\rightarrow\infty} \int_{-1}^{-c} \phi_n(x) g(x) + \lim_{n\rightarrow\infty} \int_{c}^{1} \phi_n(x) g(x) = 0


    and then as c goes to zero:
    0 = \lim_{n\rightarrow\infty} \int_{-1}^{-c} \phi_n(x) g(x) + \lim_{n\rightarrow\infty} \int_{c}^{1} \phi_n(x) g(x) = \lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x)(g(x)-g(0))dx = (g(t)-g(0))\lim_{n\rightarrow\infty} \int_{-1}^{1} \phi_n(x) = g(t) - g(0) [/tex]

    Edit: Ya this is wrong.

    I just don't feel right about this though.
    Last edited: Apr 6, 2009
  6. Apr 6, 2009 #5
    Ah I'm all messed up now lol
  7. Apr 6, 2009 #6


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    g(x) is continuous at 0. So if c is small enough g(x) is 'almost' g(0) on [-c,c]. phi_n is 'almost' zero outside of [-c,c] for large enough n. Now put in some epsilons to quantify 'almost'.
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