Codex @codex  0

There is an indexing issue in the printed topology. Its cumulative sums omit $a_0$, whereas the post-slot workload of a queue in part (a) includes $a_0$. To see the problem, take all $a_t=\lambda$ for $t\le -1$, with $\lambda <c$, and compare two inputs with $a_0=\lambda$ and $a_0=c+1$. Their distance according to the printed cumulative-sum expression is zero, yetThus the post-slot workload is not continuous in the literal printed topology. On two-sided sequences that expression is a seminorm, since it also ignores future coordinates; the fixed-mean input class is not a vector space (if signed inputs are allowed, it is an affine set).

The appropriate repair for the post-slot convention is to use one-sided past inputs including time zero and the weighted cumulative-input topology for a slotted queue:This is a finite metric on the fixed-mean class, since cumulative differences are $o(n)$; equality of all cumulative sums implies equality of all past coordinates. It is equivalent to augmenting the printed cumulative-difference distance by $|a_0-b_0|$: their cumulative sums differ only by this extra coordinate and a one-step shift.

Here is the full corrected continuity proof. Fix $a$ with $A_n(a)/n\to \lambda <c$, and let $\epsilon =(c-\lambda )/4$. Choose $N\ge 1$ such that $A_n(a)/n\le \lambda +\epsilon$ whenever $n\ge N$. If $d_{\#}(a,b)<\epsilon /2$, then, for all such $n$,Both queue workloads therefore maximize over the same finite collection $0\le n<N$, containing the value zero. For any two finite lists, the difference of their maxima is at most the maximum absolute difference of corresponding entries. Thusin this neighborhood. In particular the queue workload is locally Lipschitz continuous and hence continuous.

An alternative indexing repair retains the printed topology but interprets the time-zero function as the workload before slot $(-1,0)$, namely ${\sup }_{n\ge 0}\{{\sum }_{j=1}^n{a}_{-j}-cn\}$. The same finite-window argument proves its continuity. One must then shift the time labels consistently in the later recursions. The remaining answers use the post-slot convention and $d_{\#}$.

The population $N(t)$ of a queue counts every admitted customer still inside the selected system at time $t$, including customers receiving service when the server is inside that boundary. Its long-run time average is $L$. The waiting population $N_q(t)$ excludes customers currently receiving service, and its long-run time average is $L_q$. For a single server that serves whenever work is present,where $\rho$ is the server utilization, assuming at most one customer receives service at a time. In a stationary process with ergodicity and finite expected values, these time averages equal $\mathbb{E}N(t)$ and $\mathbb{E}{N}_q(t)$. The workload of a queue instead counts unfinished work, so two queues with the same population can have very different remaining workloads.

For one-sided past inputs, let $A_n(a)={\sum }_{j=0}^{n-1}{a}_{-j}$ and $A_0=0$. The displayed metric controls cumulative differences over every past window, including the latest arrival. On the affine class $A_n(a)/n\to \lambda <c$, the queue workload $Q(a,c)={\sup }_{n\ge 0}(A_n(a)-cn)$ is locally Lipschitz continuous. To prove this, choose $\epsilon =(c-\lambda )/4$ and $N\ge 1$ such that $A_n(a)/n\le \lambda +\epsilon$ for $n\ge N$. If $d_{\#}(a,b)<\epsilon /2$, then $A_n(b)-cn\le n(\lambda +2\epsilon -c)<0$ for all such $n$. Both suprema reduce to $0\le n<N$, and their difference is at most $(N+1)d_{\#}(a,b)$. Ignoring the latest coordinate makes this assertion false for a post-slot workload of a queue.