 Solution

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.