= Solution
Put $g=\Lambda^*$ 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> give
$$
I(a)\geq\int_{-t}^0g(\dot a_s)\,ds\geq t\,g\left(C+\frac{z(t)}t\right)\geq\gamma z(t).
$$
If $q(a)=x>0$, choose intervals with $z(t)$ approaching its <supremum> x, obtaining $I(a)\geq\gamma x$. 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. Therefore
$$
\boxed{J(0)=0,\qquad J(x)=\gamma x\ (x>0),\qquad J(x)=\infty\ (x<0).}
$$
Strict <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 $g(v)/(v-C)\geq g(u)/(u-\mu)$ for every finite-cost $v>C$. Thus J is increasing on nonnegative <queue workloads>, and the preceding fiber bound yields $\boxed{\bar J(x)\geq J(x)\text{ for }x\geq0}$. Negative values are impossible for both <queue workloads> and have infinite rate. No monotonicity is asserted across that impossible negative domain.
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