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Let $M$ be a positive semidefinite operator on a complex Hilbert space $(F,(\cdot,\cdot))$.

On the quotient space $F/\text{Ker}(M)$ we have the following inner product $$\langle \overline{x},\overline{y}\rangle = (Mx,y),$$ for all $\overline{x},\overline{y}\in F/\text{Ker}(M)$.

According to some papers, the following theorem figures in this book. However, I don't find it and I hope to get its proof.

Theorem: The completion of $F/\text{Ker}(M)$ denoted $\overline{F/\text{Ker}(M)}$ is isometrically isomorphic to the Hilbert space $\text{Im}(M^{1/2})$ with the inner product $$( M^{1/2}x,M^{1/2}y)_0=(P_{\overline{\text{Im}(M)}}x, P_{\overline{\text{Im}(M)}}y),\;\forall\, x,y \in F.$$ $P_{\overline{\text{Im}(M)}}$ is the orthogonal projection onto $\overline{\text{Im}(M)}$.

I think that in order to prove the theorem it suffices to show that

The completion of $F/\text{Ker}(M)$ is isomorphic to $(\overline{\text{Im}(M)}, (\cdot,\cdot)_0)$. Is this claim true?

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  • $\begingroup$ Why do you expect this to be true? $\endgroup$
    – YCor
    May 17, 2018 at 12:11
  • $\begingroup$ Oh, actually these are two Hilbert spaces, so the whole point is to check that they have the same dimension... $\endgroup$
    – YCor
    May 17, 2018 at 12:56
  • $\begingroup$ @YCor But we work in infinite dimensionel Hilbert spaces $\endgroup$
    – Schüler
    May 17, 2018 at 13:00
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    $\begingroup$ Yes, a Hilbert space is determined up to linear isometry by its dimension (in the Hilbert sense: cardinal of an orthogonal basis). This is one of the very first things to know about a Hilbert space. $\endgroup$
    – YCor
    May 17, 2018 at 13:36
  • $\begingroup$ Completion of $F/\operatorname{Ker}(M)$ with respect to the $\langle\cdot,\cdot\rangle_M$ norm? $\endgroup$
    – Hannes
    May 17, 2018 at 14:29

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I think what is going on is the following.

As Meisam Soleimani Malekan says, for any Hilbert space $H$ with inner product $(\cdot|\cdot)$, and $\newcommand{\mc}{\mathcal}T\in\mc B(H)$, we have that $\ker T^*T=\ker T$, because $T^*T\xi=0\implies (T^*T\xi|\xi)=0 \implies \|T\xi\|^2=0\implies T\xi=0$. Also, $\newcommand{\im}{\operatorname{Im}}(\im T)^\perp = \ker T^*$ by an easy calculation.

For your $F$ and $M\in\mc B(F)$ positive we hence have that $\ker M = \ker M^{1/2}$ and so on $F / \ker M$ we may define an inner product $$ \newcommand{\ip}[2]{\langle{#1},{#2}\rangle} \ip{\overline{x}}{\overline{y}} = (Mx|y) = (M^{1/2}x|M^{1/2}y). $$ Here $\overline{x} = x+\ker M = x+\ker M^{1/2}$ is the equivalence class of $x\in F$. It is evident that $\overline{x} = \overline{x'} \implies x-x'\in\ker M^{1/2} \implies M^{1/2}x = M^{1/2}x'$, so the definition is well-defined. Notice that we have not established that $F/\ker M$ is complete for this inner-product.


Alternatively, in any Hilbert space $H$, if $X\subseteq H$ is a closed subspace, then we can identify $H/X$ with $X^\perp$. Indeed, let $P:H\rightarrow X^\perp$ be the orthogonal projection, and define $\theta:H/X\rightarrow X^\perp; \overline{\xi}\mapsto P(\xi)$. This is well-defined, for $\overline{\xi}=\overline{\eta}\implies \xi-\eta\in X \implies P(\xi-\eta)=0$, and similarly $\theta$ is injective, as $P(\xi)=0$ exactly when $\xi\in X$. By construction $\theta$ is onto, and so $\theta$ is an isomorphism.

However this construction, applied to your setting, identifies $F/\ker M$ with $(\ker M)^\perp = (\im M)^{\perp\perp} = \overline{\im M}$ which is already a Hilbert space, so complete. The inner product we get on $F/\ker M$ is $(\overline x|\overline y) = (P(x)|P(y))$. Suppose we choose $x$ and $y$ already in $\overline{\im M}$. Then $(\overline x|\overline y) = (P(x)|P(y)) = (x|y)$. This is not the inner product you want.


Instead let's define $U:\im(M^{1/2})\rightarrow F/\ker M$ as follows. Define $U(\xi) = \overline{x}$ if $\xi = M^{1/2}x$ for some $x\in F$. This is well-defined, for if also $\xi=M^{1/2}y$ then $x-y\in\ker M^{1/2}=\ker M$ so $\overline{x}=\overline{y}$. Furthermore, for $\xi=M^{1/2}x$ and $\eta=M^{1/2}y$, $$ \ip{U(\xi)}{U(\eta)} = \ip{\overline x}{\overline y} = (Mx|y) = (M^{1/2}x|M^{1/2}y) = (\xi|\eta). $$ Hence $U$ is an isometry. Clearly $U$ is surjective. So $$ U^{-1} : \big( F/\ker M, \ip{\cdot}{\cdot} \big) \rightarrow \big( \im(M^{1/2}), (\cdot|\cdot) \big) $$ is an isometric linear isomorphism. This identifies $F/\ker M$, given the inner-product you have defined, with the (in general not closed) subspace $\im(M^{1/2})$ of $F$. In particular, $F/\ker M$ might fail to be a Hilbert space.

However, if we take the completion of $F/\ker M$ then $U^{-1}$ extends to a unitary showing that $\overline{F/\ker M}$ is isomorphic to $\overline{\im(M^{1/2})} = \overline{\im(M)}$, the latter viewed as a subspace of $F$.


Your theorem suggests that we give $\im(M^{1/2})$ a different inner-product, $$ (M^{1/2}x|M^{1/2}y)_0 = (Px|Py) $$ where I have written $(\cdot|\cdot)_0$ to avoid confusion with $(\cdot|\cdot)$ which is the given inner-product on $F$. Define a different map $V:F/\ker M \rightarrow \im M^{1/2}$ by $$ V(\overline x) = Mx = M^{1/2} M^{1/2} x \in \im M^{1/2}. $$ Again, this is well-defined. As $\overline{\im M} = (\ker M)^\perp = (\ker M^{1/2})^\perp = \overline{\im M^{1/2}}$, we have that $PM^{1/2}x=M^{1/2}x$, and so $$ (Mx|My)_0 = (M^{1/2} M^{1/2} x|M^{1/2} M^{1/2} y)_0 = (PM^{1/2}x|PM^{1/2}y) = (M^{1/2}x|M^{1/2}y) = \ip{\overline x}{\overline y}. $$ Thus $$ V:\big( F/\ker M, \ip{\cdot}{\cdot} \big) \rightarrow \big( \im M^{1/2}, (\cdot|\cdot)_0 \big) $$ is an isometry, as you want. Notice that $V$ is not (in general) onto. Finally, we can extend $V$ to the completion of $F/\ker M$, and then we will obtain a unitary transformation onto $\overline{\im M^{1/2}}$, completion with respect to $(\cdot|\cdot)_0$.

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  • $\begingroup$ I don't fully understand. Firstly, $\operatorname{Im}(M^{1/2})$ is not closed, so it is not a Hilbert space. Secondly, in what sense does $M^{1/2}$ map between $\overline{\operatorname{Im}(M)}$ and $\operatorname{Im}(M^{1/2})$? Do you closure here? But then $\overline{\operatorname{Im}(M)} = \overline{\operatorname{Im}(M^{1/2})}$ so $M^{1/2}$ does not map between them bijectively. Thirdly, you keep writing $\langle\cdot,\cdot\rangle$ but this is used in lots of different ways in your question, so I don't really know what inner product is meant. $\endgroup$ Mar 21, 2019 at 12:15
  • $\begingroup$ $\text{Im}(M^{1/2})$ is a Hilbert space with respect to the following inner product $$( M^{1/2}x,M^{1/2}y)=\langle P_{\overline{\text{Im}(M)}}x, P_{\overline{\text{Im}(M)}}y\rangle,\;\forall\, x,y \in F.$$ $P_{\overline{\text{Im}(M)}}$ is the orthogonal projection onto $\overline{\text{Im}(M)}$. $\endgroup$
    – Schüler
    Mar 21, 2019 at 12:18
  • $\begingroup$ @Schüler: I think I understand. I have edited my answer to correct a mistake, and to directly address your question, I hope. $\endgroup$ Mar 21, 2019 at 16:30
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    $\begingroup$ @Schüler: Ah, I think I see. The hint is actually at the very bottom of page 4 of the paper you link to. I have yet again altered by answer. $\endgroup$ Mar 21, 2019 at 21:58
  • $\begingroup$ Yes, I think that is right. If you want, extend to the closures on both sides. $\endgroup$ Mar 22, 2019 at 9:55
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Firstly, with regard to the above comment, I imagine that the claim is that the spaces are isometric in a natural way and this is indeed the case. For the sake of simplicity, I will assume that $M$ is injective and is multiplication by the positive sequence $(\lambda_n)$ on $\ell^2$. Then a simple calculation shows that both spaces consist of the weighted $\ell^2$-spaces of sequences $(\xi_n)$ for which $\sum \lambda_n \xi_n^2 <\infty$. One can use the spectral theorem to reduce to this case, at least for operators with discrete spectrum. The general case is much the same except that,rather than sequences, one has to consider multiplication operators on $L^2$-spaces.

I should perhaps add that one can prove this directly by con sidering the map, $x \mapsto M^{(1/2)} (x)$, factoring over the quotient by its kernel, and extending it to the completions. This dispenses with the recourse to the spectral theorem, but only apparently since it is there implicitly in the use of the square root. I think the above version is more transparent but that is a matter of taste.

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For $T\in B(F)$, we have $\ker T=\ker T^*T$ and $\ker T=(\text{Im} T^*)^\perp$, so for a positive operator $M$ we obtain $\ker M^{1/2}=\ker M$, whence $$F=\ker M\oplus\overline{\text{Im} M}=\ker M^{1/2}\oplus\overline{\text{Im} M^{1/2}}$$ If $P$ denote the projection on $\overline{\text{Im} M}=\overline{\text{Im} M^{1/2}}$,then $$\langle M^{1/2}x,M^{1/2}y\rangle=\langle M^{1/2}Px,M^{1/2}Py\rangle=\langle MPx,Py\rangle=\langle Px,Py\rangle_M$$ so $F/\ker M=\overline{\text{Im} M^{1/2}}$ and the product defined on $F/\ker M$ is just the one defined on $\overline{\text{Im} M^{1/2}}$.

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