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Derivative of Integral = Integral of Derivative

  1. Apr 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Let f : [0,1] x [0,1] -> R be a bounded function such that for each fixed y, the function g given by g(x) = f(x,y) is measurable. Suppose [itex]\partial f / \partial y[/itex] exists and is bounded. Prove that

    [tex]\frac{d}{dy} \int_0^1 f(x,y) \, dx = \int_0^1 \frac{\partial f}{\partial y} \, dx[/tex]


    2. Relevant equations
    This is from the chapter discussing the Lebesgue convergence theorem: Let g be an integrable function (integrable meaning the Lebesgue integral is finite) over E and let (fn) be a sequence of measurable functions that converge almost everywhere to f and |fn| < g. Then

    [tex]\int_E f = \lim \int_E f_n[/tex]

    Corollary: Let (gn) be a seq. of integrable functions which converge a.e. to an integrable function g. Let (fn) be a seq. of measurable functions such that [itex]|f_n| \le g_n[/itex] and [itex]f_n \to f[/itex] a.e. If [itex]\int g = \lim \int g_n[/itex], then [itex]\int f = \lim \int f_n[/itex].


    3. The attempt at a solution
    I need a sequence [itex](h_n)[/itex] such that [itex]h_n \to \partial f / \partial y[/itex] and such that [itex]\lim \int h_n = d/dt \int_0^1 f(x,y) \, dx[/itex]. I have no idea about the former. For the latter, since the derivative is a limit, I figure that we can use

    [tex]h_n(y) = \int_0^1 \frac{f(x, y+a_n) - f(x,y)}{a_n} \, dx[/tex]

    where [itex](a_n)[/itex] is any sequence that converges to 0. I can't think of anything else at the moment. Any tips?
     
  2. jcsd
  3. Apr 16, 2009 #2
    You're on the right track...hit this with the dominated convergence theorem!
     
  4. Apr 16, 2009 #3
    To use dominated convergence, I need to prove that [itex](h_n)[/itex] is point-wise bounded. Since f is a bounded function, [itex]f(x,y + a_n) - f(x,y)[/itex] is bounded. The problem is the [itex]1/a_n[/itex] factor of [itex]h_n[/itex], which is unbounded.
     
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