The completion of $F/\text{Ker}(M)$ is isomorphic to the closure of the range of $M$

Question

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?

Answer 1

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 $(\cdot|\cdot)_0$.

Answer 2

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.

Answer 3

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 $\overline{\text{Im} M^{1/2}}$.