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Proper proof of a delta function

  1. Apr 1, 2014 #1
    Prove:
    tδ(t) = 0

    The answer our TA has given isn't doing it for me:
    [itex]\int dt \delta(t)f(t) = (0)f(0) = 0[/itex]

    I'm wanting to write:
    [itex] t \frac{d}{dt}\int \delta(t) dt = t \frac{d}{dt}(1) = t * 0 = 0 [/itex]

    Am I right here? This doesn't make use of a test function. I'm very sloppy with proofs!

    Thanks for your time!
     
  2. jcsd
  3. Apr 1, 2014 #2

    joshmccraney

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    do you mean ##\delta (t)## is the dirac delta function?

    also, you are incorrect in stating this:

    note that ##\frac{d}{dt}\int \delta (t) dt= const. ##, thus and arbitrary ##\delta (t)## would work in the above "proof" (assuming it fits the criteria for the fundamental theorem). first, you misuse the fundamental theorem (no bounds). secondly, you assume a function of ##t## is not in fact a function of ##t##. and lastly, if this question is a hw question, i think it belongs in that section.
     
    Last edited: Apr 1, 2014
  4. Apr 1, 2014 #3

    micromass

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    Both ##t\delta(t)## and ##0## are distributions here. So you have to prove an equality of distributions. By definition, two distributions ##F## and ##G## are equal if

    [tex]\int \varphi(t)F(t)dt = \int \varphi(t)G(t)dt[/tex]

    for all test functions ##\varphi##. So this is what you need to check.
     
  5. Apr 2, 2014 #4

    pwsnafu

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    What don't you like about it? It's exactly the Schwartz product, that is if ##f## is a smooth function and ##\mu## a distribution and ##g## a test function then
    ##\langle f \, \mu , g \rangle := \langle \mu , fg \rangle##.
     
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