Distributional convergence

  • Thread starter tom_rylex
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  • #1
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This class is making my head hurt. I could use some help.

Homework Statement


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.

Homework Equations


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


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?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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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.
 

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