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

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

Following some papers, the following theorem figures in this book (chapter: Appendix on square summable power series).

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 $$( M^{1/2}x,M^{1/2}y)=\langle P_{\overline{\text{Im}(M)}}x, P_{\overline{\text{Im}(M)}}y\rangle,\;\forall\, x,y \in F.$$ $P_{\overline{\text{Im}(M)}}$ is the orthogonal projection onto $\overline{\text{Im}(M)}$.

I hope to get a proof or a sketch of proof of the above theorem.

Clearly $$M^{1/2}:(\overline{\text{Im}(M)},\langle\cdot,\cdot\rangle)\to \text{Im}(M^{1/2}),$$ is an isomorphism. So in order to prove the theorem it suffices to show that

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

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?

Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on an infinite-dimensional complex Hilbert space $(F,\langle\cdot,\cdot\rangle)$. Let $M$ be a positive semidefinite operator.

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

The following theorem figures without proof in this paper (page 5).

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 $$( M^{1/2}x,M^{1/2}y)=\langle P_{\overline{\text{Im}(M)}}x, P_{\overline{\text{Im}(M)}}y\rangle,\;\forall\, x,y \in F.$$ $P_{\overline{\text{Im}(M)}}$ is the orthogonal projection onto $\overline{\text{Im}(M)}$.

I hope to get a proof or a sketch of proof of the above theorem.

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

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

Following some papers, the following theorem figures in this book (chapter: Appendix on square summable power series).

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 $$( M^{1/2}x,M^{1/2}y)=\langle P_{\overline{\text{Im}(M)}}x, P_{\overline{\text{Im}(M)}}y\rangle,\;\forall\, x,y \in F.$$ $P_{\overline{\text{Im}(M)}}$ is the orthogonal projection onto $\overline{\text{Im}(M)}$.

I hope to get a proof or a sketch of proof of the above theorem.

Clearly $$M^{1/2}:(\overline{\text{Im}(M)},\langle\cdot,\cdot\rangle)\to \text{Im}(M^{1/2}),$$ is an isomorphism. So in order to prove the theorem it suffices to show that

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