Let
- $U$ be a $\mathbb R$-Hilbert space
- $Q\in\mathfrak L(U)$ be nonnegative and self-adjoint
- $Q^{-1/2}$ denote the pseudoinverse of $Q^{1/2}$, i.e. $$Q^{-1/2}:=\left(\left.Q^{1/2}\right|_{\left(\ker Q^{1/2}\right)^\perp}\right)^{-1}$$
- $U_0:=Q^{1/2}U$ be equipped with $$\left\langle u_0,v_0\right\rangle_{U_0}:=\left\langle Q^{-1/2}u_0,Q^{-1/2}v_0\right\rangle_U\;\;\;\text{for all }u_0,v_0\in U_0$$
Are we able to show that $Q\in\mathfrak L(U_0)$?
Since $$Q^{1/2}U\subseteq\left(\ker Q^{1/2}\right)^\perp\tag1,$$ we should have $$\left\|Qu_0\right\|_{U_0}=\left\|Q^{-1/2}Q^{1/2}Q^{1/2}u_0\right\|_U=\left\|Q^{1/2}u_0\right\|_U=\left\|Q^{-1/2}Q^{1/2}u_0\right\|_{U_0}=\left\|u_0\right\|_{U_0}$$ for all $u_0\in U_0$ and hence $Q$ should even be an isometry on $U_0$. Am I overlooking something?