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]): 

[![enter image description here][3]][3]


  [1]: https://www.math.purdue.edu/~branges/square-summable.pdf
  [2]: http://dx.doi.org/10.1080/03081087.2012.667094
  [3]: https://i.sstatic.net/DLOHY.png