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Finding adjoint of an operator

  1. Feb 5, 2008 #1
    Find the adjoint of the operator A:L^2[0,1] -> L^2[0,1] defined by (Af)(x) = integ from (0 to x) f(t)dt

    so from my notes it says: the operator A* is called the adjoint of A if:
    <Ax, y> = <x, A*y> for all x, vE H

    i am not sure how to do this, and need to know how to do it for a test
     
  2. jcsd
  3. Feb 5, 2008 #2

    morphism

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    Think about what the inner product in this case is, and keep in mind that the operator A* that satisfies <Ax,y>=<x,A*y> for all x and y is unique.
     
  4. Feb 5, 2008 #3
    so inner product is <f,g>=integral f(x)g(x) dx
     
  5. Feb 5, 2008 #4

    HallsofIvy

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    Okay, so <Af, g>= <f, A*g> is [itex]\int (Af)g dx= \int f(Ag)dx[/itex]

    Since [itex]Af= \int_0^x f(t)dt[/itex] that becomes
    [tex]\int\left[\int_0^x f(t)dt\right]g(x)dx= \int f(x)\left[A*g(x)\right]dx[/tex]

    You might try doing the left side by "integration by parts" letting [itex]u= \int_0^x f(t)dt[/itex] and dv= g(x)dx
     
  6. Feb 5, 2008 #5
    i am trying to do the lhs as you said
    but i am getting stuck, i have not done integration by parts like this before.
    so for u = integ (0 to x) f(t) dt
    u' = f(t) ? but what happens to the limits of integration, do they just carry along.
    and for dv = g(x) dx,
    dv/dx = g(x)
    v = integ (g(x) dx)
    is that correct?
    but what has changed, since there is still a double integral on LHS, and what should i do with the RHS?
     
  7. Feb 6, 2008 #6

    HallsofIvy

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    No, du= f(x)dx. The t is a dummy variable.

    Remember the formula for integration by parts?
    [tex]\int_a^b udv= uv|_a^b- \int_a^b v du[/tex]

    It might be better to write
    [tex]v= \int_0^x g(t)dt[/tex]

    What has changed is the the "inner integral" on the left is now an integral of g rather than f. Compare the right and left sides and you should conclude that A*g must also be an integral.
     
  8. Feb 6, 2008 #7
    ok, after the udv part i got
    integ (0-x) f(t)dt integ (0-x) g(t)dt - integ (integ (0-x) g(t)dt ) f(x) dt
    I see what you are talking about the inner integral, but can i just conclude after this step that A*g must also be an integral?
     
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