Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on an infinite dimensionel complex Hilbert space $(F,\langle\cdot,\cdot\rangle)$. Let $M\in \mathcal{B}(F)^+$ (i.e. $\langle Mx\;, \;x\rangle\geq 0$ for all $x\in F$). Then $$\langle\cdot,\cdot\rangle_{M}:F\times F\longrightarrow\mathbb{C},\;(x,y)\longmapsto\langle Mx, y\rangle,$$ is a semi-inner product.

$\langle\cdot,\cdot\rangle_M$ induces an inner product on the quotient space $F/\text{Ker}(M)$ defined as: $$\langle[x],[y]\rangle = \langle Mx,y\rangle$$ ($[\cdot]$ denotes the class of equivalence).

I see in a paper the following result and the authors refer the reader to (1) but I don't find a proof in this reference.

Theorem: The completion of $F/\text{Ker}(M)$ under the above inner product is isometrically isomorphic to the Hilbert space $\text{Im}(M^{1/2})$ with the inner product $$(M^{1/2}x,M^{1/2}y)_{\text{Im}(M^{1/2})}:=\langle Px, Py\rangle,\;\forall\, x,y \in F,$$ where $P$ denotes the orthogonal projection of $F$ onto the closure of $\text{Im}(M)$.

I have two questions

• Why $\text{Im}(M^{1/2})$ endow with the inner product $$(M^{1/2}x,M^{1/2}y)_{\text{Im}(M^{1/2})}:=\langle Px, Py\rangle,\;\forall\, x,y \in F$$ is a Hilbert space?

• How to show that the completion of $F/\text{Ker}(M)$ is isomorphic to $\text{Im}(M^{1/2})$?

The following picture is from this paper (2):