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For a stable input with path action $I(a)={\int }_{-\infty }^{0}g({\dot{a}}_s)ds$, a workload x requires some interval of length t with average input near $C+x/t$. Jensen inequality bounds its cost below by $tg(C+x/t)$; taking the supremum workload and then the infimum over t yields the displayed lower bound. Constant-rate burst paths give the matching upper bound. At zero the typical-rate path has zero cost; negative workloads have infinite rate. If no finite-cost rate exceeds C, every positive workload has infinite rate.

A local path action integrates a nonnegative convex cost g of the path's instantaneous input rate, assigning infinity to paths that are not absolutely continuous. When $g(\mu )=0$, the typical-rate path has zero cost. On any interval with prescribed total input, Jensen inequality makes a constant derivative minimize its cost. With a queue workload constraint this leads to constant-rate burst paths and the linear workload rate for a local convex action. Such a formula alone does not establish goodness in an unspecified path topology; goodness must be checked or assumed as part of the input large-deviation principle.

Put $g={\Lambda }^{\ast }$ and $\gamma ={\inf }_{v>C}g(v)/(v-C)$, allowing gamma to be infinity if no rate above C has finite cost. We prove the linear workload rate for a local convex action, which in particular proves monotonicity on the physical domain $[0,\infty )$.

For an absolutely continuous finite-cost path and any interval of length t with positive net work $z(t)$, its average arrival rate is $C+z(t)/t$. Nonnegativity of g and Jensen inequality giveIf $q(a)=x>0$, choose intervals with $z(t)$ approaching its supremum x, obtaining $I(a)\ge \gamma x$. If gamma is infinite, the existence of any positive net interval instead forces infinite cost, giving the same extended-value conclusion.

Conversely, choose any $v>C$ of finite cost, use arrival rate v during the last $t=x/(v-C)$ time units and the typical rate mu before that. This constant-rate burst path has queue workload exactly x: the lookback net input increases to x over the burst and then decreases with slope $\mu -C<0$. Its action is $tg(v)=xg(v)/(v-C)$. Taking the infimum over v proves the matching upper bound. The typical-rate path has zero queue workload and zero cost. ThereforeStrict convexity and $g(\mu )=0$ ensure $\gamma >0$ whenever it is finite. Indeed, choose a finite-cost $v>C$ and some $u\in (\mu ,C)$; then $g(u)>0$, and convexity gives $g(v)/(v-C)\ge g(u)/(u-\mu )$ for every finite-cost $v>C$. Thus J is increasing on nonnegative queue workloads, and the preceding fiber bound yields $\boxed{\bar{J}(x)\ge J(x)\text{ for }x\ge 0}$. Negative values are impossible for both queue workloads and have infinite rate. No monotonicity is asserted across that impossible negative domain.

Take $0<x\le B$ and a finite-cost burst rate $v>C$. For the constant-rate burst path just constructed, let $t_0=x/(v-C)$. On $0\le s\le t_0$ its net input is $z(s)=(v-C)s$, so the supremum is x and the infimum is zero. The finite-buffer formula evaluated at $t=t_0$ is therefore $\min (x,B)=x$. Since $\bar{q}\le q=x$, we conclude $\bar{q}=x$ exactly: the burst never needs more space than the permitted capacity.

That path's action bounds the contracted finite-buffer rate, and optimizing givesAt zero use the typical-rate path to obtain $\bar{J}(0)=J(0)=0$. If the positive infinite-buffer rate is infinite, the asserted upper inequality is automatic. Approximating burst rates suffice, so the argument does not rely on an unproved existence of an optimal duration. Consequently $\boxed{\bar{J}(x)\le J(x)\text{ for }0\le x\le B}$.