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Dirac and weak convergence

  1. Dec 2, 2007 #1
    1. The problem statement, all variables and given/known data
    Show that if {x_k} is any sequence of points in space [tex] R^n [/tex] with [tex] |{x_k}| \rightarrow \infty [/tex], then [tex] \delta(x-x_k) \rightarrow 0 [/tex] weakly


    2. Relevant equations



    3. The attempt at a solution
    I'm still trying to grasp the concept of weak convergence for distributions. It would appear that this function doesn't converge pointwise. The distribution on a test function is
    [tex] \int \delta(x-x_k)\theta(x)dx = \theta(x_k)[/tex] Does the function converge weakly to zero because [tex] x_k [/tex] approaches infinity, and therefore would be outside of the region of support of any locally integrable test function?
     
  2. jcsd
  3. Dec 2, 2007 #2

    Dick

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    I think that's correct. But what exactly is your set of test functions?
     
  4. Dec 2, 2007 #3
    My set of test functions meet the following criteria:
    * function has a finite region of support, inside of which [tex] \theta(x) \neq 0 [/tex], outside of which [tex] \theta(x)=0 [/tex]
    * [tex] \theta(x) [/tex] has derivatives of all orders.
     
    Last edited: Dec 2, 2007
  5. Dec 2, 2007 #4

    Dick

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    Then you are right. Finite support is enough.
     
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