- #1

- 162

- 0

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter braindead101
- Start date

- #1

- 162

- 0

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

morphism

Science Advisor

Homework Helper

- 2,015

- 4

- #3

- 162

- 0

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

- #4

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 964

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

- 162

- 0

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

Science Advisor

Homework Helper

- 41,847

- 964

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

Remember the formula for integration by parts?but what happens to the limits of integration, do they just carry along.

[tex]\int_a^b udv= uv|_a^b- \int_a^b v du[/tex]

It might be better to writeand for dv = g(x) dx,

dv/dx = g(x)

v = integ (g(x) dx)

is that correct?

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

- #7

- 162

- 0

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?

Share: