Let
- $U,U_1$ be separable $\mathbb R$-Hilbert spaces
- $U_0:=Q^{1/2}U$ be equipped with $$\langle u_0,v_0\rangle_{U_0}:=\langle Q^{-1/2}u_0,Q^{-1/2}v_0\rangle_U\;\;\;\text{for }u_0,v_0\in U_0$$ where $$Q^{-1}:=\left(\left.Q\right|_{\ker(Q)^\perp}\right)^{-1}$$
- $Q_1:=\iota\iota^\ast$
- $Q_1^{1/2}U_1$ be in the same way equipped with an inner product as $U_0$
- $(e^n)_{n\in\mathbb N}$ be an orthonormal basis of $U$ and $$e_0^n:=Q^{1/2}e^n\;\;\;\text{for }n\in\mathbb N$$
Can we show that $(\iota e_0^n)_{n\in\mathbb N}$ is an orthogonal set in $U_1$?
Note that
- $(e_0^n)_{n\in\mathbb N}$ is an orthonormal basis of $U_0$
- $\iota U_0=Q_1^{1/2}U_1$ (in the sense of set equality)
- $\iota$ is an isometry between $U_0$ and $Q_1^{1/2}U_1$
The answer is no.
Let $(e_n)_{n \in \mathbb{N}}$ be an orthonormal basis for $U$ and let $(f_n)_{n \in \mathbb{N}}$ be an orthonormal basis for $U_1$. Then we can consider the projection $Q: U \to U$ onto the subspace generated by $e_1$ and $e_2$.
Thus we have $U_0 = \{ \lambda e_1 + \mu e_2 \mid \lambda,\mu\in \mathbb{R} \}$ with the normal inner product. since $Q^{-1} = Q|_{U_0}$, which is the identity, so we may say $U_0 \simeq \mathbb{R}^2$.
In particular, we have $(e_n^0)_{n \in \mathbb{N}} = (e_1,e_2,0,0,\cdots)$ which is orthogonal. Clearly $\iota 0$ is orthogonal for all $\iota$, thus we are only concered about $\iota e_1$ and $\iota e_2$.
Now, let $\iota:U_0 \to U_1$ be defined as $\iota e_1 = 2 f_1 +2 f_2$ and $\iota e_2 = 2 f_1 + f_2$. Then we have that $(\iota e_1 , \iota e_2)$ is no longer orthogonal.