Let $E$ be a complex Hilbert space.
If $S\in\mathcal{L}(E)^+$ such that $\mathcal{R}(S)=\mathcal{R}(S^{1/2})$, I want to show that $\mathcal{R}(S)$ is a closed subspace of $E$.
Since $S\geqslant 0$, we have $\mathcal{R}(S)^\perp = \mathcal{N}(S)$. Thus, $\mathcal{R}(S)$ is closed if and only if $\mathcal{N}(S)^\perp\subset \mathcal{R}(S)$. So pick $y\in \mathcal{N}(S)^\perp$. Since $\mathcal{R}(S)=\mathcal{R}(S^{1/2})$, we can pick $x\in \mathcal{N}(S)^\perp$ such that $$S^{1/2}S^{1/2} x = Sx = S^{1/2} y.$$ So, $S^{1/2}(S^{1/2} x - y)=0$, i.e. $S^{1/2} x - y\in \mathcal{N}(S^{1/2})$. Moreover, Clearly $S^{1/2}x\in \mathcal{R}(S^{1/2})\subset\mathcal{N}(S^{1/2})^\perp$ so $S^{1/2} x - y\in \mathcal{N}(S^{1/2})^\perp$. Thus, we have $y=S^{1/2}x\in\mathcal{R}(S^{1/2})=\mathcal{R}(S)$.
My question why for all $y\in \mathcal{N}(S)^\perp$ there exists $x\in \mathcal{N}(S)^\perp$ such that $$S^{1/2} y=Sx?$$
Notice that there is an answer here (1), but in this question we give a different proof.
Since $\mathcal{R}(S)=\mathcal{R}(S^{1/2})$, there exists $x_0\in E$ such that $S^{1/2}y=Sx_0$. Now let $x$ be the projection of $x_0$ onto $\mathcal{N}(S)^\perp$. Then $x-x_0\in\mathcal{N}(S)$, so $Sx=Sx_0=S^{1/2}y$.