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Isometric Equality

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

Consider a closed operator: $$A:\mathcal{D}(A)\to\mathcal{H}:\quad A=A^{**}$$

Denote for shorthand: $$H:=A^*A:\quad H=H^*$$

Regard elements: $$\varphi\in\mathcal{D}(H)\subseteq\mathcal{D}(A)\cap\mathcal{D}(\sqrt{H})$$

A calculation gives: $$\|\sqrt{H}\varphi\|^2=\langle\sqrt{H}\varphi,\sqrt{H}\varphi\rangle=\langle H\varphi,\varphi\rangle=\langle A\varphi,A\varphi\rangle=\|A\varphi\|^2$$

But for both it is a core: $$\overline{A_{\mathcal{D}(H)}}=A\quad\overline{\sqrt{H}_{\mathcal{D}(H)}}=\sqrt{H}$$

How can I prove the latter?

Partial Isometry

By the check above: $$\varphi\in\mathcal{D}(A)=\mathcal{D}(|A|):\quad\||A|\varphi\|=\|A\varphi\|\quad$$

Construct the isometry: $$U_0:\mathcal{R}(|A|)\to\mathcal{R}(A):\quad U_0(|A|\varphi):=A\varphi$$

Extend this uniformly: $$U_E:\overline{\mathcal{R}(|A|)}\to\overline{\mathcal{R}(A)}:\quad U_E(\varphi):=\lim_nU_0(\varphi_n)$$

Lift to partial isometry: $$U:\overline{\mathcal{R}(|A|)}\oplus\mathcal{R}(|A|)^\perp\to\mathcal{H}:\quad U(\varphi+\varphi^\perp):=U_E(\varphi)$$

The polar decomposition.

freishahiri
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Let $S$ be any densely-defined selfadjint linear operator on a Hilbert Space with spectral measure $E$. The range of $E[-\lambda,\lambda]$ is in the domain of every positive power of $S$. In fact, $$ \mathscr{C}=\bigcap_{n=1}^{\infty}\mathcal{D}(S^{n}) $$ is a core for $S$. This is because the range of $E[-\lambda,\lambda]$ is in $\mathscr{C}$, and $x\in\mathcal{D}(S)$ implies $$ \lim_{\lambda\uparrow\infty}E[-\lambda,\lambda]x = x,\\ \lim_{\lambda\uparrow\infty}SE[-\lambda,\lambda]x = Sx. $$ The last limit holds because of the Spectral Theorem characterization of the domain of $S$ as the set of $x$ for which $\int_{\mathbb{R}}\lambda^{2}\,d\|E(\lambda)x\|^{2} < \infty$.

In your case: Based on the phrasing of your comment it appears that you know $\mathcal{D}(A^{\star}A)$ is a core for $A$, and you want to know if $\mathcal{D}(A^{\star}A)$ is a core for $|A|$. So I'll only address the statement that $\mathcal{D}(A^{\star}A)$ is a core for $|A|$.

If $A$ is as you state, then $|A|=(A^{\star}A)^{1/2}$ has $\mathcal{D}(|A|^{2})=\mathcal{D}(A^{\star}A)$ as a core, based on the analysis for the case where $S=|A|$.

freishahiri
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Disintegrating By Parts
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  • Cool thanks!!!! ...I start getting comfy with spectral measures thanks to you yeii :D – freishahiri Oct 29 '14 at 21:35
  • In fact, it suffices to know: that $\mathcal{D}0:=\bigcup{n=0}^\infty\mathcal{R}E(B_n)$ is a core simultaneously for every $f(E)$. – freishahiri Oct 29 '14 at 21:42
  • Oh only for the positive powers, you're right! – freishahiri Oct 29 '14 at 22:00
  • @Freeze_S : Once you get accustomed to a handful of core ideas, the whole thing seems to come into focus. So I keep posting using the same basic core ideas, with few tricks. Tricks can obscure the big picture. – Disintegrating By Parts Oct 29 '14 at 22:09
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    @Freeze_S : I heard one Mathematician claim that unbounded operators were easier to study because there were only a handful of ideas that worked, and almost all of those ideas were due to John von Neumann. So, after you tried all of those ideas, you could give up if nothing worked. :) There was a lot of truth in what he said. – Disintegrating By Parts Oct 29 '14 at 22:26
  • I rearranged the layout. But I don't want others to think you answer part of my question only. :/ As it was earlier noting that the phrasing of my question appears I know the first core already seemed enough. Now I made a special comment below my question remarking I only need a proof for the second core. ... – freishahiri Jul 01 '15 at 10:21
  • ... My suggestion so your answer won't look weird in the new context: Instead of "Based on the phrasing of your question..." rather "Base on the phrasing of your comment..." I'll profilactically edit that in your answer and only that in your answer in case you want find time soon. I hope this is fine, is it? Ah and by the way thanks for so much support of you. You are really a great teacher!!!!!!!!!!!!! :) – freishahiri Jul 01 '15 at 10:21