Proof

Proof. Inclusion gives the implication by the adjoint identity. For necessity, let $H=H^{†}\in \text{cB}$. Any Hermitian $\Delta$ orthogonal to $\mathcal{A}$ is traceless because $I\in \mathcal{A}$. For sufficiently small $ϵ>0$, $I/d\pm ϵ\Delta$ are states agreeing on $\mathcal{A}$. The hypothesis gives $\mathrm{Tr}(\Delta {\Lambda }^{\ast }(H))=0$. Finite-dimensional Hilbert–Schmidt orthogonality places ${\Lambda }^{\ast }(H)$ in the self-adjoint part of $\mathcal{A}$. Extend by complex linearity.

For classical inputs use only block-diagonal $\Delta$; their pairings depend exactly on the pinched adjoint.