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A limit of an indexed integral at closed interval

  1. Nov 4, 2012 #1
    Let ρ(x) be a continuous function on ℝ, which evaluates ρ(x)=0 when |x|≥1 and that meets following.

    ∫[-1,1]ρ(x)dx=1

    And let ψ(x) be a continuous function on interval [-1,1], prove

    lim[n→∞] n∫[-1,1]ρ(nx)ψ(x)dx = ψ(0).

    is denoted.


    This is NOT a homework but a past exam problem of a college that has no answer.
    I've been thinking for a whole day now.

    What I have found so far:

    First I defined sequencing functions as following:

    F_n(x)=∫[-1,x]ρ(ny)ψ(y)dy
    G_n(x)=∫[x,1]ρ(ny)ψ(y)dy

    Now given a properly large natural number of n_0 which completes n_0≥1/y (where y≠0),
    for all n∈N of n≥n_0, ρ(ny) evaluates 0. In this sense, taken a small positive number ε>0 instead of y, we only have to concern that of a small closed interval of [-ε,ε] as to nF_n(-ε) and nG_n(ε) gives 0 (for all n≥n_0≥1/ε).

    So for all n same above,

    n∫[-1,1]ρ(nx)ψ(x)dx = n∫[-ε,ε]ρ(nx)ψ(x)dx


    And I have a stuck.
    Please help ;)
     
  2. jcsd
  3. Nov 4, 2012 #2

    lurflurf

    User Avatar
    Homework Helper

    First change variables (just for simplicity)
    n∫[-1,1]ρ(nx)ψ(x)dx =∫[-1,1]ρ(x)ψ(x/n)dx
    now by the mean value theorem for integrals
    ∫[-1,1]ρ(x)ψ(x/n)dx =ψ(t/n)∫[-1,1]ρ(x)dx =ψ(t/n)
    where |t|<1, but may depend upon ψ and n
    consider
    lim[n→∞] ψ(t/n)
     
  4. Nov 5, 2012 #3
    >lurflurf

    I got it. Changing variable was out of my thought.
    Thank you:)
     
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