Why $(\operatorname{Im}(M^{1/2}), (\cdot,\cdot))$ is a Hilbert space?

Question

Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on an infinite dimensional complex Hilbert space $(F,\langle\cdot,\cdot\rangle)$. Let $M\in \mathcal{B}(F)^+$ (i.e. $\langle Mx\;, \;x\rangle\geq 0$ for all $x\in F$).

Let $P$ denotes the orthogonal projection of $F$ onto the closure of $\operatorname{Im}(M)$.

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

Answer 1

Suppose that $\{M^{1/2}x_n\}$ is Cauchy in $\text{Im}\,M^{1/2}$. Given $\varepsilon>0$, there exists $n_0$ such that whenever $m,n\geq n_0$, $$ \|Px_m-Px_n\|^2=\|M^{1/2}x_m-M^{1/2}x_n\|^2_{\text{Im}\,M^{1/2}}<\varepsilon, $$ so $\{Px_n\}$ is Cauchy in $F$, and there exists $x\in F$ with $Px_n\to x$. It is clear that $Px=x$, and then $$ \|M^{1/2}x_n-M^{1/2}x\|_{\text{Im}\,M^{1/2}}=\|Px_n-x\|\to0. $$ So $\text{Im}\,M^{1/2}$ is complete with the twisted product.