# A question on Hilbert space theory

1. Sep 25, 2006

### dextercioby

Let's say we have a self-adjoint, densly defined closed linear operator acting on a separable Hilbert space $H$

$$A:D_{A}\rightarrow H$$

Let $\lambda$ be an eigenvalue of A and let

$$\Delta_{A}\left(\lambda\right) = \{\left(A-\lambda \hat{1}_{H}\right)f, \ f\in D_{A}\}$$

How do i prove that

$$D_{A}\perp \Delta_{A}(\lambda) \Leftrightarrow \bar{\Delta_{A}(\lambda)} \neq H$$.

Daniel.

2. Sep 25, 2006

### matt grime

Unless you define D_A I don't think we can do the question.

3. Sep 25, 2006

### Hurkyl

Staff Emeritus
I presume D_A is the domain of A.

4. Sep 26, 2006

### matt grime

The domain of A is H.

5. Sep 26, 2006

### George Jones

Staff Emeritus
No, Hurkyl is right. By the Hellinger-Toeplitz theorem, any self-adjoint operator whose domain is all of H is bounded. So, the best that one can expect for an unbounded self-adjoint operator A is that its domain D_A is a dense subset of H, as Daniel says.

6. Sep 26, 2006

### matt grime

Then why did the OP say we had an operator acting on H, if H is not the domain? Must just be a convention thing, but seems odd to me.

7. Sep 26, 2006

### Hurkyl

Staff Emeritus
It's not that much different from talking about meromorphic functions on C, or rational functions on an an algebraic variety, is it?

8. Sep 26, 2006

### matt grime

As in they have poles? No, I think that is different, slightly, since we implicitly extend the codomain to have a point at infinity thus the functions are defined at all points of the domain. I get the impression here that there is no possible extension to an operator defined on the whole of H, though I may be mistaken in this assumption.

9. Sep 27, 2006

### dextercioby

George Jones nailed it on the head. It's assumed unbounded, so by Hellinger-Toeplitz theorem

$$D_{A}\neq H$$

but in order to be self-adjoint one must have that the closure of $D_{A}$ is H.

I guess the question still remains open. :uhh:

Daniel.

10. Sep 27, 2006

### Hurkyl

Staff Emeritus
That can't be right.

Let A be any projection operator. So $D_A = H$. Then, $\Delta_A(0)$ is the image of A. For most such operators, $D_A$ is clearly not orthogonal to $\Delta_A(0)$... and yet $\Delta_A(0)$ is not dense in H.

Or, did you mean to only consider unbounded operators?

11. Sep 29, 2006

### dextercioby

Yes, that operator is meant to be unbounded. Look at it this way, if it were bounded, why would one consider $D_{A}$ (the domain of A) dense in H but yet different from H ? It would have been simply H, as a bounded operator could be extended by continuity on all H...

Daniel.

Last edited: Sep 29, 2006
12. Sep 29, 2006

### matt grime

But the case Hurkyl gave is a special case of what you wrote and it breaks down in the special case so why should it be true if we exclude the special case above? I think you should taken that post as a hint as to what you need to do. (Not that I know what the proof is, before you ask.) There are a lot of hypotheses on A up there, and H, so how are you attempting to use them?

13. Sep 29, 2006

### dextercioby

Perhaps the problem that i stated is not quite correct. It would be fair to post the whole context. Here's a scanned version of page #89 of Ahiezer & Glazman's book. It's part of the 43-rd section which i'm trying to fully understand.

The thing i didn't understand is the argument he uses in the proof of theorem #2 namely that under the assumptions i stated in my first post (which appear on page 88).

He asserts that if $Af=\lambda f$ for an "f" both in the domain and in the range of A (closed, densly-defined self-adjoint linear operator) then

$$f\perp \Delta_{A}\left(\lambda\right) \Leftrightarrow \Delta_{A} \ \mbox{ is not everywhere dense in H}$$.

Huh?

Daniel.

Last edited: Nov 22, 2006
14. Sep 29, 2006

### dextercioby

So i was basically wrong in the first post, since "f" is not arbitrary in D_{A} so i can't assert that generally D_{A} \perp \Delta_{A}\left(\lambda\right)

Daniel.

15. Sep 29, 2006

### Hurkyl

Staff Emeritus
Let me write $D = D_A$ and $\Delta = \Delta_A(\lambda)$ for simplicity.

Incidentally, it's easy to see that you can't have both $D \perp \Delta$ and $\bar{\Delta} = H$: the inner product is continuous, and this implies that it's identically zero.

16. Sep 29, 2006

### dextercioby

So lemme see why you think it's trivial. If i assume

$$\bar{\Delta} =H$$

, then

$$\bar{\Delta}\supset H \supset D[/itex] But $D\perp \Delta$ , so considering a sequence $g_{n}$ from $\Delta$ convergent to "g" in $\bar{\Delta}$, then from [tex] \langle f,g_{n}\rangle =0$$

it follows that

$$\langle f,g\rangle =0$$

,using continuity of the scalar product.

So i proved that $D \perp \bar{\Delta}$. Since above i assumed that $D\subset \bar{\Delta}$, it follows that $D={0}_{H}$ which is false. Ergo, the initial assumption is false.

Daniel.

17. Sep 29, 2006

### George Jones

Staff Emeritus
Let my try and elaborate a bit on Hurky's last post.

Let $U$ be any any dense subset of $H$, and let $v \in H$ be such that $\left< v , u \right> = 0$ for every $u \in U$. Then, by continuity and density (insert small analysis argument), $v = 0$.

Now, in the first half of the proof, one has $\left< f , u \right> =0$ for every $u \in \Delta$. Therefore, if $\Delta$ were dense in $H$, then $f$ would have to be zero. But this can't be, since $f$ is an eigenvector.

Oops, I didn't see Daniel's last post.

Last edited: Sep 29, 2006
18. Sep 29, 2006

### dextercioby

I see, and there's no reverse of that statement. So i'm supposed to get that

$$\bar{\Delta}\neq H \nRightarrow D\perp \Delta$$

, but instead one can find a nonzero vector from D which is orthogonal to $\Delta$.

Daniel.