• $U$ be a separable $\mathbb R$-Hilbert space
  • $Q\in\mathfrak L(U)$ be self-adjoint and nonnegative with $\operatorname{tr}Q=1$
  • $Q^{-\frac12}$ denote the pseudoinverse of $Q^{\frac12}$
  • $U_0:=Q^{\frac12}U$ be equipped with $$\langle u_0,v_0\rangle_{U_0}:=\left\langle Q^{-\frac12}u_0,Q^{-\frac12}v_0\right\rangle_U\;\;\;\text{for }u_0,v_0\in U_0$$
  • $H$ be a separable $\mathbb R$-Hilbert space

Note that $$\left\|u_0\right\|_U^2\le\left\|Q\right\|_{\mathfrak L(U)}\left\|u_0\right\|_{U_0}^2\le\left\|u_0\right\|_{U_0}^2\tag1\;\;\;\text{for all }u_0\in U_0$$ and hence $$\left\|\Phi^\ast\right\|_{\mathfrak L(H,\:U)}^2\le\left\|Q\right\|_{\mathfrak L(U)}\left\|\Phi\right\|_{\mathfrak L(U_0,\:H)}^2\le\left\|\Phi\right\|_{\mathfrak L(U_0,\:H)}^2\tag2$$ (the crucial point is that $\Phi^\ast$ is bounded from $H$ to $U$, while by definition this operator is only bounded from $H$ to $U_0$).

Let $u\in U$ and $h\in H$. Are we able to write $\langle u,\Phi^\ast h\rangle_U$ in the form $\langle f(\Phi,u),h\rangle_H$?

To be clear, I know that $$\langle u_0,\Phi^\ast h\rangle_{U_0}=\langle \Phi u_0,h\rangle_H\;\;\;\text{for all }u_0\in U_0\text{ and }h\in H;\tag3$$ that's not what I'm looking for.

In order to find the desired expression, I thought we should consider the inclusion $\iota_Q$ from $U_0$ into $U$. By $(1)$, $\iota_Q\in\mathfrak L(U_0,U)$ with $\left\|\iota_Q\right\|_{\mathfrak L(U_0,\:U)}\le1$.

Now, we should consider the adjoint of $\iota_Q$: Let $u_0\in U_0$ and $v\in U$. There is a unique $u\in(\ker Q)^\perp$ with $u_0=Q^{\frac12}u$ and hence (since $Q^{-\frac12}Q^{\frac12}=\operatorname P_{(\ker Q)^\perp}$ is the orthogonal projection from $U$ onto $(\ker Q)^\perp$) $$\langle\iota_Qu_0,v\rangle_U=\left\langle u,Q^{\frac12}v\right\rangle_U=\left\langle\operatorname P_{(\ker Q)^\perp}u,\operatorname P_{(\ker Q)^\perp}Q^{\frac12}v\right\rangle_U=\left\langle u_0,Qv\right\rangle_{U_0}.\tag4$$ So, we should obtain $\iota_Q^\ast=Q$. (Note that $Q$ is actually bounded from $U$ into $U_0$.)

Thus, we should be able to conclude $$\langle u,\Phi^\ast h\rangle_U=\langle u,\iota_Q\Phi^\ast h\rangle_U=\langle Qu,\Phi^\ast h\rangle_{U_0}=\langle \Phi Qu,h\rangle_H.\tag5$$

Since $(5)$ is not what I expected, I'm unsure whether I made any mistake. So, is there anything wrong in my reasoning?


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