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!

Distributional convergence

  1. Oct 25, 2008 #1
    This class is making my head hurt. I could use some help.

    1. The problem statement, all variables and given/known data
    Show that if f(x) is a function of slow growth on the real line,
    [tex] \lim}_{\substack \varepsilon \rightarrow0^+} \langle f(x)e^{- \varepsilon |x|}, \phi (x) \rangle = \langle f, \phi \rangle [/tex]
    where [itex] \phi (x) [/itex] is a test function.

    2. Relevant equations
    Definition of distribution:
    [tex] \langle f , \phi \rangle = \int_{R_n} f(x) \phi(x) dx [/tex]

    3. The attempt at a solution
    If I look at the function [itex] f_\varepsilon (x) = f(x)e^{-\varepsilon | x |} [/itex], I can say that the function is locally integrable, (actually [itex] L_1 (-\infty, \infty) [/itex]). Can't I just invoke the Lebesgue Dominated Convergence theorem here? That is, since [itex] f_\varepsilon [/itex] is locally integrable, and [itex] f_\varepsilon \rightarrow f [/itex] pointwise, that [itex] f_\varepsilon \rightarrow f [/itex] in a distributional sense.

    Or am I missing something here. Are there other considerations to take into account?
  2. jcsd
  3. Oct 25, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    I think dominated convergence theorem works, doesn't it? You know f*phi is integrable. e^(-epsilon*|x|)*f(x)*phi(x) converges pointwise to f*phi and is dominated by |f*phi|. It all sounds kosher to me.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?