Proposition: Continuity of strengths, not of exact edge membership

For $W,W^{\prime }\in {\mathsf{W}}_3$,

Nevertheless, trace-norm convergence of processes does not generally stabilize the exact predicate ${\Delta }_W(a,b)>0$.

Proof. Each positive tester element has operator norm at most its trace, hence at most two. Moreover $0\le S,T\le 2I$ implies $\Vert S-T{\Vert }_{\infty }\le 2$. Consequently

Trace duality bounds the change of each objective in (eq:strength-maximum) by $8\Vert W-W^{\prime }{\Vert }_1$. Taking maxima gives (eq:strength-stability).

For the last assertion, let $W_{\to }$ be the definite-order process that initializes a qubit at $A$, sends $A$'s output through an identity channel to $B$, and discards $B$'s output. Party $C$ receives an independent initialized qubit and has its output discarded. Preparing $|0⟩$ or $|1⟩$ at $A$ and measuring at $B$ shows that

Such definite-order circuits are valid processes OCB. The process $W_0$ in (eq:base-process) has zero signalling strengths, as also checked in Proposition prop:empty-query.

For $0<\delta \le 1$, convexity of the defining conditions gives

The $W_0$-contribution to every signalling difference vanishes, so (eq:strength-maximum) gives

Thus $W_{\delta }\to W_0$ in trace norm as $\delta \downarrow 0$, but the exact edge $A{R}_{\inf }B$ is present for every $\delta >0$ and absent at the limit.