If $\Phi$ is bounded from $\mathcal RQ^{1/2}$ to $H$, are we able to show that $Q^{1/2}\Phi^\ast=(\Phi Q^{1/2})^\ast$?

Question

Let

• $U,H$ be separable $\mathbb R$,Hilbert spaces
• $Q\in\mathfrak L(U)$ be injective, nonnegative and self-adjoint
• $U_0:=Q^{1/2}U$ be equipped with $$\langle u_0,v_0\rangle_{U_0}:=\langle Q^{-1/2}u_0,Q^{-1/2}v_0\rangle_U\;\;\;\text{for }u_0,v_0\in U_0$$
• $\Phi\in\mathfrak L(U_0,H)$

Are we able to show that $Q^{1/2}\Phi^\ast=(\Phi Q^{1/2})^\ast$?

Note that, by definition of the adjoint, $\Phi^\ast\in\mathfrak L(H,U_0)$, but since $$\left\|\Phi^\ast h\right\|_U^2\le\left\|Q\right\|_{\mathfrak L(U)}\left\|\Phi^\ast h\right\|_{U_0}^2\le\left\|Q\right\|_{\mathfrak L(U)}\left\|\Phi^\ast\right\|_{\mathfrak L(H,\:U_0)}^2\left\|h\right\|_H^2\;\;\;\text{for all }h\in H,\tag1$$ $\Phi^\ast$ is even bounded when it is considered as being an operator from $H$ to $U$. Hence, the desired result is at least formally possible.

Answer 1

If I understand correctly, your question arises from the diference between the adjoint in $\mathfrak{L}(H,U_0)$ and $\mathfrak{L}(H,U)$. But consider the following: let $u_0, v_0 \in U_0$ \begin{align*} \langle Q^{1/2}\Phi^* u_0, v_0\rangle_{U_0} &= \langle Q^{-1/2}Q^{1/2}\Phi^* u_0, Q^{-1/2}v_0\rangle_{U} \\ &= \langle Q^{1/2}Q^{-1/2}\Phi^* u_0, Q^{-1/2}v_0\rangle_{U} \\ &= \langle Q^{-1/2}\Phi^* u_0, Q^{1/2}Q^{-1/2}v_0\rangle_{U} \\ &= \langle Q^{-1/2}\Phi^* u_0, Q^{-1/2} Q^{1/2}v_0\rangle_{U} \\ &= \langle \Phi^* u_0, Q^{1/2}v_0\rangle_{U_0} \\ &= \langle u_0, \Phi Q^{1/2}v_0\rangle_{U_0} \\ \end{align*}

Showing that $(\Phi Q^{1/2})^* = Q^{1/2} \Phi^*$ (and also that $(Q^{1/2})^* = Q^{1/2}$).