It is proven in the very helpful answer here that using the optimal POVM for unambiguous discrimination of the equally-likely non-orthogonal states $\lvert 0\rangle$ and $\lvert +\rangle$, we can discriminate between the states with a success probability of $\tfrac{2-\sqrt 2}{4}$. I would like to prove (or disprove?) that this method of discrimination is optimal in a broader sense, namely optimal among all quantum channels.
Specifically, I would like to prove (disprove?) that if $T$ is a quantum channel (i.e. a completely positive trace-preserving superoperator) on $\mathbb C^2\otimes\mathbb C^3$ satisfying the equalities \begin{align*} T\big(\lvert 0\rangle\langle 0\rvert\otimes\lvert0\rangle\langle 0\rvert\big) &= p ~ \rho_0\otimes \lvert0\rangle\langle 0\rvert + (1-p) ~ \rho_1 \lvert 2\rangle\langle 2\rvert \\ T\big(\lvert +\rangle\langle +\rvert\otimes\lvert0\rangle\langle 0\rvert\big) &= q~ \rho_2\otimes \lvert 1\rangle\langle 1\rvert + (1-q) ~ \rho_3 \lvert 2\rangle\langle 2\rvert \\ \end{align*} for some density matrices $\rho_i$ and probabilities $p,q\in [0,1]$, then it is necessarily the case that $$\frac{p+q}{2}\leq \frac{2-\sqrt 2}{4}$$ Intuitively, I'm treating $T$ as some sort of quantum black-box in which the first input qubit is the state to be identified as either $\lvert 0\rangle$ or $\lvert +\rangle$, and the second input qutrit is a register that will either read $\lvert 0\rangle, \lvert 1\rangle$ or $\lvert 2\rangle$ after applying $T$, indicating an outcome of either a successful identification of $\lvert 0\rangle$, a successful identification of $\lvert +\rangle$ or an inconclusive result, respectively.
Does anyone know how to prove (disprove?) this more general claim? Can it be reduced to the optimality of the optimal unambiguous POVM?
