Spectral Measures: Domain Criterion

Question

Problem

Given a Hilbert space $\mathcal{H}$.

Consider a spectral measure: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$$

Then the criterion holds: $$\mathcal{D}f(E)\subseteq\mathcal{D}g(E)\iff|g|^2\leq R^2(|f|^2+1)$$

How can I prove this?

Application

See the thread: Dominated Convergence

Ok, back to the problem... I couldn't read anymore your last comment but I still saw it was something along the lines linear map on closed subspaces the rest I'm missing. But I guess you wanna do something similar to the proof of closed graph theorem from bounded inverse? — C-star-W-star, Oct 30 '14 at 19:26
I mean something not precisely but similar to: $\Gamma:\mathcal{G}(T)\to\mathcal{D}(T):(x,Tx)\mapsto x$ That one was a continuous bijection and for the graph and domain being closed the bounded inverse theorem applies and one gets the estimate: $|Tx|+|x|\leq C|x|$. Was this kind of your idea but for the context herein? — C-star-W-star, Oct 30 '14 at 19:32
The map from $(x,T_{f}x)\mapsto (x,T_{g}x)$ is a well-defined linear map from the Hilbert space $\mathcal{G}(T_{f})\subseteq H\times H$ into the Hilbert space $\mathcal{G}(T_{g})\subseteq H\times H$. (These are Hilbert spaces because $T_{f}$, $T_{g}$ are closed linear operators.) The map is well-defined because $(x,T_{f}x)\mapsto x \in H \mapsto (x,T_{g}x)$ is linear, and because $\mathcal{D}(T_{f})\subseteq \mathcal{D}(T_{g})$. — Disintegrating By Parts, Oct 30 '14 at 20:17
Moreover, the map is closed as well as its domain that is the graph of one of the operators. That together implies that it was already continuous by the closed graph theorem yeii :D ...besides was the closeness of the other operator's graph really needed? — C-star-W-star, Oct 30 '14 at 20:27
The graph of this map is in $\mathcal{G}(T_{f})\times \mathcal{G}(T_{g}) \subset (H\times H)\times(H\times H)$. You start by assuming a sequence converges in $\mathcal{G}(T_{f})$ and its image under this map converges in $\mathcal{G}(T_{g})$. I think you're going to need both to be closed, but you can check. — Disintegrating By Parts, Oct 30 '14 at 20:39
As far as I know the codomain has no effect at all for this special direction of the closed graph theorem one could just choose any codomain the boundedness of the operator won't be affected by a change as it is already bounded a fortiori... or? — C-star-W-star, Oct 30 '14 at 20:46
But I think you'll need $\mathcal{G}(T_{g})$ to be closed in order to show that the map from $\mathcal{G}(T_{f})$ to $\mathcal{G}(T_{g})$ is closed anyway. Please check on that. You can post an answer. I just wanted to give hints this time, as you originally requested. — Disintegrating By Parts, Oct 30 '14 at 20:53

C-star-W-star · Accepted Answer · 2015-06-24T09:38:17.010

0

Criterion

Denote for shorthand: $$S:=f(E)\quad T:=g(E)$$

Consider the operator: $$\Gamma:\mathcal{G}(S)\to\mathcal{G}(T):(x,Sx)\mapsto(x,Tx)$$

It is well defined since: $$\mathcal{D}(S)\subseteq\mathcal{D}(T)$$

And it is closed since: $$\varphi_n\to\varphi,S\varphi_n\to\psi,\,T\varphi_n\to\chi\implies S\varphi=\psi,T\psi=\chi$$

Also its domain is complete: $$\mathcal{D}(\Gamma)=\mathcal{G}(S):\quad\mathcal{D}(\Gamma)=\overline{\mathcal{D}(\Gamma)}$$

By closed graph theorem: $$\varphi\in\mathcal{D}(S):\quad\|T\varphi\|^2\leq\|T\varphi\|^2+\|\varphi\|^2\leq\|\Gamma\|^2\left(\|S\varphi\|^2+\|\varphi\|^2\right)$$

By mean value theorem: $$\int_A|g|^2\mathrm{d}\nu_\varphi\leq\int_A\|\Gamma\|^2\left(|f|^2+1\right)\mathrm{d}\nu_\varphi\implies|g|^2\leq\|\Gamma\|^2\left(|f|^2+1\right)\mod\nu_\varphi$$

So one obtains that: $$|g|^2\leq\|\Gamma\|^2\left(|f|^2+1\right)\mod E$$

Conversely one has: $$\int|g|^2\mathrm{d}\nu_\varphi\leq R^2\int|f|^2\mathrm{d}\nu_\varphi+R^2\|\varphi\|^2$$

Concluding the assertion.

Null Set

Remind that it holds: $$\nu_\varphi(A)=0\iff E(A)\varphi=0$$

Regard the null set:* $$N_0:=\bigcap_{\varphi\in\mathcal{H}}N_\varphi:\quad\nu_\varphi(N_\varphi)=0$$

Indeed one has: $$E(N_0)\varphi\quad(\varphi\in\mathcal{H})\implies E(N_0)=0$$

That was desired!

Separable Spaces

For separable spaces: $$N_0:=\bigcup_{\alpha\in\mathcal{A}}N_\alpha:\quad N_\alpha\in\mathcal{B}(\mathbb{C})\implies N_0\in\mathcal{B}(\mathbb{C})$$

By density one has: $$E(N_0)\alpha=0\quad(\alpha\in\mathcal{A})\implies E(N_0)=0$$

That was desired.

*There's a flaw: It may fail to be Borel!

edited Jun 24 '15 at 09:38

answered Oct 30 '14 at 21:32

C-star-W-star

16,275

Completeness corrected. What do you mean by your hint, I did Lebesgue mean value now, is it right this way? – C-star-W-star Oct 30 '14 at 22:50
If $\int_{E}f,d\mu \ge 0$ for a positive measure $\mu$ for all measurable subsets $E$, set $E_{n}={ \lambda : f(\lambda) \le -1/n }$ to conclude $\mu E_{n}=0$ so that $f \ge 0$ a.e. $[d\mu]$. – Disintegrating By Parts Oct 30 '14 at 22:53
That's a variant of Lebesgue mean value ;) – C-star-W-star Oct 30 '14 at 22:58
Yeeii :) I guess another lemma, check. – C-star-W-star Oct 30 '14 at 23:02
If by mod E you mean an exception set $S$ with $E(S)=0$, then you may need separability of $H$. – Disintegrating By Parts Oct 31 '14 at 00:02
You may want to assume $\Omega=\mathbb{C}$. Otherwise, you probably need some kind of regularity on the measure in order to effectively study the operator $T_{f}$. All the results I'm used to are for $\Omega=\mathbb{C}$. – Disintegrating By Parts Oct 31 '14 at 00:24
For every $\varphi\in\mathcal{H}$ there exists $N_\varphi$ such that $|g|^2\leq\ldots$ on $N_\varphi^c$. Take $N:=\bigcap_{\varphi\in\mathcal{H}}$. Then $|g|^2\leq\ldots$ on $N^c$ but on the same time $N\subseteq N_\varphi$ so by completeness of the probability measures it is measurable and $\nu_\varphi(N)=0$. As this holds for any vector necessarily $E(N)=0$. – C-star-W-star Oct 31 '14 at 00:31
I guess in most cases separability is just a historical obstacle... – C-star-W-star Oct 31 '14 at 00:33
Borel measures are not necessarily complete. – Disintegrating By Parts Oct 31 '14 at 00:37
By the way does it also show in the second last line of the answer '[Math Processing Error]' on your screen? On some devices of mine it fails and on some it works... :/ – C-star-W-star Oct 31 '14 at 00:37
But can be completed - also Borel spectral measures - if I'm not mistaken I read it in Birman-Solomyak. – C-star-W-star Oct 31 '14 at 00:40
Hmm not sure about it... – C-star-W-star Oct 31 '14 at 00:51
I gave you (+) ; no sure why someone else had (-). – Disintegrating By Parts Nov 01 '14 at 17:03

Spectral Measures: Domain Criterion

1 Answers1

Linked