Finding adjoint of an operator

  • #1
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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
 

Answers and Replies

  • #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.
 
  • #3
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so inner product is <f,g>=integral f(x)g(x) dx
 
  • #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
 
  • #5
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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?
 
  • #6
HallsofIvy
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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) ?
No, du= f(x)dx. The t is a dummy variable.

but what happens to the limits of integration, do they just carry along.
Remember the formula for integration by parts?
[tex]\int_a^b udv= uv|_a^b- \int_a^b v du[/tex]

and for dv = g(x) dx,
dv/dx = g(x)
v = integ (g(x) dx)
is that correct?
It might be better to write
[tex]v= \int_0^x g(t)dt[/tex]

but what has changed, since there is still a double integral on LHS, and what should i do with the RHS?
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.
 
  • #7
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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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