# Continuity of actions: scalar tests, preduals, and bounded nets ## OA-FLOW.TOP.JOINT — Simultaneously varying the automorphism and the element **Theorem.** Let $\beta_i\to\beta$ in the $u$ topology. Let $(x_i)$ be a uniformly norm-bounded net with $x_i\to x$ strong-star in a faithful normal representation. Then $$\beta_i(x_i)\longrightarrow\beta(x)\quad\text{strong-star}.\tag{11}$$ Equivalently, on each norm-bounded part of $M$, evaluation is jointly continuous from the $u$ topology and the strong-star topology to the strong-star topology. In particular, for a point-ultraweakly continuous action of a locally compact group, $$t_i\to t,\quad x_i\to x\text{ strong-star},\quad\sup_i\|x_i\|<\infty \quad\Longrightarrow\quad\alpha_{t_i}(x_i)\to\alpha_t(x)\text{ strong-star}.\tag{12}$$ **Proof.** Put $y_i=x_i-x$ and choose $R$ with $\|y_i\|\leq R$ for every $i$. For any $\varphi\in M_*^+$, positivity gives $$ \begin{aligned} \varphi(\beta_i(y_i)^*\beta_i(y_i)) &=(\varphi\circ\beta_i)(y_i^*y_i)\\ &\leq(\varphi\circ\beta)(y_i^*y_i) +R^2\|\varphi\circ\beta_i-\varphi\circ\beta\|. \end{aligned} \tag{13} $$ The first term tends to zero because $\varphi\circ\beta$ is positive normal and $y_i$ tends $\sigma$-strong-star to zero by the bounded-set comparison. The second tends to zero by $u$ convergence. Replacing $y_i^*y_i$ by $y_i y_i^*$ proves the other half of $p_\varphi(\beta_i(y_i))\to0$. By the preceding lemma, $p_\varphi(\beta_i(x)-\beta(x))\to0$. The triangle inequality now gives $p_\varphi(\beta_i(x_i)-\beta(x))\to0$. The output net is norm bounded because automorphisms are isometric. The bounded-set comparison converts this intrinsic convergence back to concrete strong-star convergence, proving (11). Apply OA-FLOW.TOP.PREDUAL to obtain (12). $\square$ The uniform norm bound is used explicitly in (13) and in the representation-independent topology comparison. It is part of the theorem. The statement does not assert joint continuity on an arbitrary unbounded set with the concrete strong-star topology.