= Solution
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 $t\leq-1$, 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, yet
$$
Q_0(a,c)=0,\qquad Q_0(b,c)=1.
$$
Thus \b[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>:
$$
d_\#(a,b)=\sup_{n\geq1}\frac{|A_n(a)-A_n(b)|}{n+1}.
$$
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 $\varepsilon=(c-\lambda)/4$. Choose $N\geq1$ such that $A_n(a)/n\leq\lambda+\varepsilon$ whenever $n\geq N$. If $d_\#(a,b)<\varepsilon/2$, then, for all such $n$,
$$
A_n(b)-cn\leq n(\lambda+\varepsilon-c)+(n+1)d_\#(a,b)
\leq n(\lambda+2\varepsilon-c)<0.
$$
Both <queue workloads> therefore maximize over the same finite collection $0\leq n<N$, containing the value zero. For any two finite lists, the difference of their maxima is at most the maximum absolute difference of corresponding entries. Thus
$$
\boxed{|Q_0(a,c)-Q_0(b,c)|\leq(N+1)d_\#(a,b)}
$$
in 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 $\sup_{n\geq0}\{\sum_{j=1}^na_{-j}-cn\}$. 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 $d_\#$.
Back to article page
