Spectral Measures: Scale Embeddings

Question

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

Consider a normal operator: $$N:\mathcal{D}N\to\mathcal{H}:\quad N^*N=NN^*$$

And its spectral measure: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad N=\int\lambda\mathrm{d}E(\lambda)$$

Construct scale functions: $$\Lambda_s:=\sqrt{1+|\mathrm{id}|^2}^s\in\mathcal{B}(\mathbb{C})$$

As well as scale norms: $$\varphi\in\mathcal{D}\Lambda_s(N):\quad\|\varphi\|_s:=\|\Lambda_s(N)\varphi\|$$

And the scale spaces: $$\mathcal{H}_s:=\overline{\mathcal{D}\Lambda_s(N)}^s:=\widehat{\mathcal{D}\Lambda_s(N)}^s$$

Regard the embeddings: $$\iota:\mathcal{D}\Lambda_s(N)\hookrightarrow\mathcal{D}\Lambda_{s-\varepsilon}(N)\quad(\varepsilon>0)$$

Then one obtains: $$\|\overline{\iota}\|\leq1:\quad\mathcal{N}\overline{\iota}=(0)\quad\overline{\mathcal{R}\overline{\iota}}^{s-\varepsilon}=\mathcal{H}^{s-\varepsilon}$$

How can I prove this?

ATTENTION: Note that the problem has been extended drastically!! *(The previous answer was fully correct!) — C-star-W-star, Jul 04 '15 at 04:28
12 edits in a relatively short span raise an eyebrow again. Were they necessary? — Jyrki Lahtonen, Jul 04 '15 at 04:50
@JyrkiLahtonen: Woow really 12??? Yaa, I edited until I was satisfied with the presentation of the subject. But I'll try to use more sandbox!! (Besides, more or less I'm fine with it now.) — C-star-W-star, Jul 04 '15 at 05:21

score 1 · Answer 1 · answered Mar 18 '15 at 15:24

1

The spaces $E[-a,a]\mathcal{H}$, $0 < a < \infty$ are contained in and dense in every scale space.

answered Mar 18 '15 at 15:24

Disintegrating By Parts

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5
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149

Ah of course... I guess you mean their union is a core for any of these? – C-star-W-star Mar 18 '15 at 15:36
1

@Freeze_S : That's right. For PDEs this normally works out to be a class of infinitely differentiable functions; that's a natural effect of filtering out out the higher spectral components. (It's like truncating the Inverse Fourier transform to a finite interval.) – Disintegrating By Parts Mar 18 '15 at 18:44
Hello T.A.E. I couldn't check injectivity. Do you have an idea? – C-star-W-star Jul 03 '15 at 20:58
Also would you mind if I add my attempt above??? (I'd be really happy about it.) – C-star-W-star Jul 03 '15 at 20:59
Sorry for unchecking your answer. I will check yours again, of course! (People just might get irritated.) – C-star-W-star Jul 03 '15 at 21:06

score 0 · Accepted Answer · edited Apr 13 '17 at 12:19

Special thanks to T.A.E.!!

Solution

They are contractive: $$\|\iota(\varphi)\|_{s-\varepsilon}^2=\int\Lambda_{s-\varepsilon}^2\mathrm{d}\nu_\varphi\leq\int\Lambda_s^2\mathrm{d}\nu_\varphi=\|\varphi\|_s^2$$

For positive scales:* $$s\geq0:\quad\overline{\mathcal{D}\Lambda_s(N)}=\mathcal{D}\Lambda_s(N)\implies\mathcal{N}\overline{\iota}=\mathcal{N}\iota=(0)$$

For negative scales:** $$s<0:\quad\mathcal{N}\overline{\iota}=\mathcal{N}\left(\overline{\Lambda_{-s}}\circ\overline{\iota}\circ\overline{\Lambda_s}\right)=(0)$$

They have dense range: $$\mathcal{H}^{s-\varepsilon}=\overline{\mathcal{D}\Lambda_s(N)}^{s-\varepsilon}=\overline{\iota(\mathcal{D}\Lambda_s(N))}^{s-\varepsilon}\\ =\overline{\overline{\iota}(\mathcal{D}\Lambda_s(N))}^{s-\varepsilon}\subseteq\overline{\overline{\iota}(\mathcal{H}^s)}^{s-\varepsilon}\subseteq\mathcal{H}^{s-\varepsilon}$$

Concluding the assertion.

Remark

Injectivity is nontrivial: Embedding

*See the thread: Scale Spaces

**See the thread Scale Operators

Spectral Measures: Scale Embeddings

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