# Integrated Information in Relational Quantum Dynamics (RQD)
## Appendix C Proof of Proposition 1 (Canonical Optimal Witness)
### Strict Convexity and Uniqueness of the Minimizer  
We first establish that the minimizer of \(D_{JS}^{Q}(\rho\|\sigma)\) over \(\sigma\in\mathcal{P}*A|B\) is unique. This follows from the strict convexity of the Jensen-Shannon divergence as a function of \(\sigma\). Intuitively, because \(D*JS^{Q}\) behaves like a squared distance in a suitable feature space, it has a single well-defined projection point on any \"convex\" region of states. In particular, even though \(\mathcal{P}*A|B\) itself is not convex, one can argue (using the joint convexity of quantum relative entropy or properties of negative-definite kernels) that \(D*JS^{Q}(\rho\|\sigma)\) is a jointly convex function in \((\rho,\sigma)\) and in each argument separately. We sketch the reasoning here for completeness. **Convexity in \(\sigma\). ** Fix \(\rho\) and consider two product states \(\sigma^{(1)}=\sigma_{A}^{(1)}\otimes\sigma_{B}^{(1)}\) and \(\sigma^{(2)}=\sigma_{A}^{(2)}\otimes\sigma_{B}^{(2)}\) in \(\mathcal{P}_{A|B}\). Let \(\sigma^{(t)}:=\sigma_{A}^{(t)}\otimes\sigma_{B}^{(t)}\) be any product state on the same cut (not necessarily a convex combination of \(\sigma^{(1)},\sigma^{(2)}\), since the manifold is not closed under mixing unless \(\sigma_{A}^{(1)}=\sigma_{A}^{(2)}\) or \(\sigma_{B}^{(1)}=\sigma_{B}^{(2)}\)). We can nevertheless compare the divergence at \(\sigma^{(t)}\) to that at \(\sigma^{(1)},\sigma^{(2)}\) by considering their entropy components. Using Equation (A3) and the concavity of the entropy, one finds the following:  
\[ \begin{array}{ll}\bullet&S!\left(\frac{\mu+\sigma^{(t)}}{2}\right);\leq;\frac {1}{2},S!\left(\frac{\mu+\sigma^{(1)}}{2}\right)+\frac{1}{2},S!\left(\frac{\mu +\sigma^{(2)}}{2}\right),\\ \bullet&S(\sigma^{(t)});\geq;\frac{1}{2},S(\sigma^{(1)})+\frac{1}{2},S(\sigma ^{(2)}),\end{array} \]  
Since \(\sigma^{(t)}\) is a \(50/50\) mixture of \(\sigma^{(1)}\) and \(\sigma^{(2)}\) in each local subsystem (by construction of \(\sigma_{A}^{(t)},\sigma_{B}^{(t)}\)), even though it may not equal their mixture on \(AB\). Combining these inequalities, the Jensen-Shannon divergence obeys  
\[D_{IS}^{Q}(\rho\|\sigma^{(t)})\ \leq\ \frac{1}{2}\,D_{IS}^{Q}(\rho\|\sigma^{(1)}) \ +\ \frac{1}{2}\,D_{IS}^{Q}(\rho\|\sigma^{(2)})\,,\]  
showing convexity in this context. In fact, except in the trivial case where \(\rho\) itself is a product state (\(\Phi(\rho)=0\)), this convexity is strict. Any significant deviation of \(\sigma\) from the true minimizer will increase \(D_{IS}^{Q}\). Thus, there cannot be two distinct product states that both minimize the divergence. If \(\sigma_{1}^{*}\) and \(\sigma_{2}^{*}\) were two different minimizers of \(\Phi(\rho)\), then any product state \"between\" them would (by the above) have divergence less than or equal to the same minimum value, contradicting the assumption that \(\sigma_{1}^{*}\) and \(\sigma_{2}^{*}\) are strict local minima. Therefore, the closest product state \(\sigma^{*}(\rho)\) achieving \(\Phi(\rho)\) is unique. **Closest product state shares marginals with \(\rho\)**. A further property of the minimizer can be deduced by symmetry: the optimal product state \(\sigma^{*}=\rho_{A}^{*}\otimes\rho_{B}^{*}\) must reproduce the correct reduced states on each subsystem. In other words, \(\rho_{A}^{*}=\mathrm{Tr}_{B}[\sigma^{*}]\) should equal \(\mathrm{Tr}_{B}[\rho]\) (the actual reduced state of \(\rho\) on \(A\)), and similarly \(\rho_{B}^{*}=\mathrm{Tr}_{A}[\rho]\). If this were not the case--say \(\rho_{A}^{*}\) differed from \(\rho_{A}:=\mathrm{Tr}_{B}[\rho]\)--then one could _decrease_ the divergence by adjusting \(\rho_{A}^{*}\) closer to \(\rho_{A}\). Intuitively, any mismatch in local marginals adds extra divergence (since \(D_{IS}^{Q}\) penalizes even classical distribution differences).