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Integral equation with bounded unknown kernel

  1. Aug 2, 2013 #1
    I need to solve an integral equation of the form

    $$\forall \omega \in [0,1], ~ \int_{\mathbb{R}} K(\omega,y)f(y)dy = \omega$$

    where

    - f is known and positive with $$\int_{\mathbb{R}} f(y)dy = 1$$

    - K: [0,1] x R -> [0,1] is the unknown kernel

    I am looking for a solution other than K(omega,y) = omega. I do not know if such a solution exists, so I am looking either for a solution, or for a proof that K(omega,y) = \omega is the only solution.

    ----------------------

    Let me tell you more about my attempts :

    - looking for K(omega,y) = a(omega)b(y) fails because it leads to a(omega) = \omega and b(y) = 1

    - looking for K(omega,y) = a(omega-y) and trying to solve the convolution equation $$\forall \omega \in [0,1], ~ (a \star f)(\omega) = \omega$$ fails (I proved it with Fourier transforms, although I'm not 100% confident about my proof)

    - setting K(omega,y) = omega + H(omega,y) and trying to solve $$\int_{\mathbb{R}} H(\omega,y) f(y) = 0$$ with either H(omega,y) = a(omega)b(y) or H(omega,y) = a(y-omega) fails as well

    There may not be a solution but I hope you can help me find one or prove that it's impossible ! Thank you
     
  2. jcsd
  3. Aug 3, 2013 #2

    Stephen Tashi

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    Science Advisor

    One way of thinking about it is to assume we can differentiate inside the integral sign with respect to [itex] \omega [/itex]

    [tex] \int_{\mathbb{R}} K'(\omega,y)f(y)dy = 1 [/tex]
    [tex] \int_{\mathbb{R}} K"(\omega,y)f(y)dy = 0 [/tex]

    We can consider [itex] K" [/itex] to be a member of a family of functions of [itex] y [/itex] parameterized by [itex] \omega [/itex] such that each member of the family is orthogonal to [itex] f(y) [/itex] on [itex] \mathbb{R} [/itex].

    Suppose we have an orthogonal basis [itex]\{f_1(y), f_2(y),f_3(y)...\}[/itex] for some space of functions and that [itex] f = f_1 [/itex] Then [itex] \omega f_2 + (1 - \omega) f_3 [/itex] is a family of functions that is orthogonal to [itex] f [/itex].

    I'm just making these suggestion off the top of my head. I haven't tried to work this out with any concrete example.
     
  4. Aug 3, 2013 #3
    Thank you for your answer, I will also try your approach. Working from K'' may still be difficult since we have a boundedness condition on K.
     
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