If $Q$ is an operator on a Hilbert space $U$, $(e_n)$ is an ONB of $U$ consisting of eigenvectors of $Q$, then $(Q^{1/2}e_n)$ is an ONB of $Q^{1/2}U$

Question

Let

• $(U,\langle\;\cdot\;,\;\cdot\;\rangle)$ be a separable Hilbert space
• $Q$ be a bounded, linear, nonnegative and symmetric operator on $U$
• $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $U$ with $$Qe_n=\lambda_ne_n\;\;\;\text{for all }n\in\mathbb N$$ for some $(\lambda_n)_{n\ge 0}\subseteq[0,\infty)$
• $U_0:=Q^{\frac 12}(U)$ and $$\langle u,v\rangle_0:=\langle Q^{-\frac 12}u,Q^{-\frac 12}v\rangle\;\;\;\text{for }u,v\in U_0$$ where $Q^{-\frac 12}$ is the pseudo inverse of $Q^{\frac 12}$

We can prove that $(U_0,\langle\;\cdot\;,\;\cdot\;\rangle_0)$ is a separable Hilbert space. Let $$e^{(0)}_n:=Q^{\frac 12}e_n\;\;\;\text{for }n\in\mathbb N\;.$$ How can we prove that $\left(e^{(0)}_n\right)_{n\in\mathbb N}$ is an orthonormal basis of $U_0$?

I fail even to prove that $\left(e^{(0)}_n\right)_{n\in\mathbb N}$ is an orthonormal system, cause I don't know how I need to deal with $Q^{-\frac 12}$.

Answer 1

I am little bit rusty at operator theory, so bear with me.

1. $Q: D(Q)\to U$ is bounded, $E:=(e_n)_{n\in\mathbb N}$ is an ONB, and $E\subseteq D(Q)$. So, $D(Q) = U$
2. Since $Q$ is symmetric and defined on $U$, $Q$ is self-adjoint.
3. From $Qe_n = \lambda_n e_n$ for every $n\in\mathbb N$, it follows that $E$ is orthonormal eigenbasis. Since $Q$ is nonnegative, for $L:=\{l\in\mathbb N \mid \lambda_l > 0\}$ we have \begin{align} Q &= \sum_{l\in L} \lambda_l e_l e_l^*, & R:= Q^{\frac12} &= \sum_{l\in L} \lambda_l^{\frac12} e_l e_l^*, \end{align} where $e_l^*$ denotes linear functional $x\mapsto \langle e_l, x\rangle$.
4. Notice that $$ U_0 = R(U) = \left\{ \sum_{l\in L} x_l e_l \mid \sum_{l\in L} \frac{x_l^2}{\lambda_l} < \infty \right\}. $$ Thus, on $U_0$ the pseudo inverse $R^+$ of $R$ is given by $$ R^+ = \sum_{l\in L} \lambda_l^{-\frac12} e_l e_l^*. $$ In particular, for $l\in L$ we have $$ R^+ R e_l = e_l. $$