# Finding adjoint of an operator

1. Feb 5, 2008

### braindead101

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. Feb 5, 2008

### morphism

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. Feb 5, 2008

### braindead101

so inner product is <f,g>=integral f(x)g(x) dx

4. Feb 5, 2008

### HallsofIvy

Okay, so <Af, g>= <f, A*g> is $\int (Af)g dx= \int f(Ag)dx$

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

You might try doing the left side by "integration by parts" letting $u= \int_0^x f(t)dt$ and dv= g(x)dx

5. Feb 5, 2008

### braindead101

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. Feb 6, 2008

### HallsofIvy

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

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

It might be better to write
$$v= \int_0^x g(t)dt$$

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. Feb 6, 2008

### braindead101

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?

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook