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

[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 ?

 

[HallsofIvy]

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

 

[Nusc]

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?

 

[HallsofIvy]

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

 

[Nusc]

&lt;f,g&gt; = \int f(x) \overline{g(x)} dx dy

and given that

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

&lt;Vf,g&gt; = \int^{x}_{0} f(s) \overline{g(s)}

Here we have to integrate by parts.
&lt;Vf,g&gt; = 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

 

[Nusc]

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?

 

[Nusc]

I"m sure I just use the later.

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

 

[Unco]

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".