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Uniform convergence and improper integration

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose g and f_n are defined on [1,+infinity), are Riamann-integrable on [t,T] whenever 1<=t<T<+infinity. |f_n|<=g, f_n->f uniformly on every compact subset of [1,+infinity), and
    [tex]\int^{\infty}_{1} g(x)dx<\infty[/tex].
    Prove that
    [tex]lim_{n->\infty} \int^{\infty}_{1} f_{n}(x)dx =\int^{\infty}_{1} f(x)dx[/tex]


    2. Relevant equations



    3. The attempt at a solution
    If I let [tex]h_{n}(u)=\int^{u}_{1} f_{n}(x)dx[/tex], then [tex]lim_{n->\infty} h_{n}(u)=\int^{u}_{1} f(x)dx=h(u)[/tex] for each u in [1,+infinity). it is equivalent to prove that [tex]lim_{n->\infty}lim_{u->\infty} h_{n}(u)=lim_{u->\infty}lim_{n->\infty} h_n(u)[/tex]. This is true if h_n converges uniformly to h on [1,+infinity). This is where I got stuck. Actually, I'm not sure if h_n indeed converges uniformly...Or, is there any other way to prove it? Any hint? Thanks a lot!
     
  2. jcsd
  3. Nov 17, 2008 #2
    Well. I think you can prove uniform convergence of [itex]h_n[/itex] directly.
    You want to show that [itex]\forall \varepsilon >0: \exist N>0[/itex] such that whenever n>N [itex]|h_n(u)-h(u)|\leq\varepsilon\quad \forall u\in[1,\infty)[/itex], right?
    First pick [itex]u_0\in[1,\infty)[/itex] such that [itex]\int_{u_0}^\infty{dxg(x)}\leq \varepsilon/4[/itex]. Then pick an N with [itex]|f_n(x)-f(x)|\leq\frac{\varepsilon}{2(u_0-1)}[/itex] for all n>N and [itex]x\in[1,u_0][/itex] This N should do the job.

    You can write for any u
    [tex]
    |h_n(u)-h(u)|=\left|\int_1^u{dx[f_n(x)-f(x)]}\right|\leq\left|\int_1^{u_0\wedge u}{dx[f_n(x)-f(x)]}\right|+\chi_{\{u_0<u\}}\left|\int_{u_0}^u{dx[f_n(x)-f(x)]}\right|
    [/tex]

    For n>N the first term is clearly bounded by [itex]\varepsilon/2[/itex]. The second term is only there if [itex]u>u_0[/itex]. In this case you can use [itex]|f_n(x)|<=g(x)[/itex] for all x (which implies [itex]|f(x)|<=g(x)[/itex] so
    [tex]
    \left|\int_{u_0}^u{dx[f_n(x)-f(x)]}\right|\leq 2\int_{u_0}^u{dxg(x)}\leq 2\int_{u_0}^\infty{dxg(x)}\leq\varepsilon/2
    [/tex]
    where the last inequality holds by the choice of [itex]u_0[/itex].
     
    Last edited: Nov 17, 2008
  4. Nov 17, 2008 #3
    This is brilliant, thanks!
    u_0 is exactly what I didn't get! Thanks again!:)
     
  5. Nov 17, 2008 #4
    Don't be exaggerating:smile:
     
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