# Volterar operator is Vf(x) = int (f(s)ds,0,x)

1. Feb 6, 2009

### Nusc

Given that the volterar operator is Vf(x) = int (f(s)ds,0,x)

The adjoint of this operator is

V*f(x) = int(f(s)ds,x,1)

How do I show this ?

2. Feb 6, 2009

### HallsofIvy

Re: Operator

Well, I would think by using the definition of "adjoint operator"! Which is?

3. Feb 6, 2009

### Nusc

Re: Operator

If A in B(H,K), then the unique operator B in B(K,H) satisfying u(h,k) = <Ah,k>=<h,Bk> is called the adjoint of A.

How does this definition tell you that you integrate from x to 1?

4. Feb 7, 2009

### HallsofIvy

Re: Operator

Well, I guess now the question is "How is <u, v> defined?"

5. Feb 7, 2009

### Nusc

Re: Operator

$$<f,g> = \int f(x) \overline{g(x)} dx dy$$

and given that

$$Vf(x) = \int^{x}_{0} f(s) ds$$

$$<Vf,g> = \int^{x}_{0} f(s) \overline{g(s)}$$

Here we have to integrate by parts.
$$<Vf,g> = f(s)g(s)|^{x}_{0} - \int^{x}_0 f(s)g(s) ds$$

I'm still don't see where the integral from 1 to x arises or how I get rid of g

Last edited: Feb 7, 2009
6. Feb 7, 2009

### Nusc

Re: Operator

Let k:[0,1]X[0,1]-> R be the characteristic function of {(x,y);y<x>}. The corresponding operator V: L^2(0,1)->L^2(0,1) defined by

Vf(x) = int k(x,y)f(y)dy on [0,1] is called the volterra operator.

That was the textbook definition. Then they note Vf(x) = int f(y) dy on [0,x]

as with most online sources.

What's the later tell you?

7. Feb 7, 2009

### Nusc

Re: Operator

I"m sure I just use the later.

When integrating by parts when do you switch the order of integration?

8. Feb 7, 2009

### Unco

Re: Operator

First let's make sure we've answered Halls' questions, Nusc.

You have: $$\langle f, g \rangle = \int_0^1 f(x) \overline{g(x)} \, dx$$
and: $$Vf = \int_0^x f(s) \, ds$$

Now, you want: $$\langle Vf, g \rangle = \langle f, V^\ast g \rangle$$. The task being to find $$V^\ast$$. You have actually been given a $$V^\ast$$, so are only required to check it.

The left-hand side of this equality translates to
$$\int_0^1 (Vf)(x) \overline{g(x)} dx = \int_0^1 \left( \int_0^x f(s) ds\right) \overline{g(x)} dx$$

We may write this as a double integral: $$\int_0^1 \int_0^x f(s) \overline{g(x)} \, ds \, dx \; \; \; (\ast)$$

Noting that the required right-hand side is of the form
$$\langle f, V^\ast g \rangle = \int_0^1 f(t) \overline{(V^\ast g)(t)} dt$$
(where t is an arbitrary parameter: above we used 'x')

we see from (*) that finding $$V^\ast$$ just requires one to switch the order of integration: take t=s; we want ds dx to become dx ds.

There is no "integration by parts" required. This work comes solely under "double integrals".