Let $T$ be a non-degenerate compact interval in $\mathbb R$ and $f:\mathbb R^2\to\mathbb R$ a strictly concave function such that (a) $f(0,0)=0$, (b) $f$ strictly increases in the first argument, and (c) it strictly decreases in the second. Consider the following set of functions: \begin{align*} C\equiv\Big\{(a,b):T\to\mathbb R^2\,\Big|\,&\text{(i) $a$ and $b$ are bounded and Borel measurable functions;}\\ &\text{(ii) $\int_{t\in T}a(t)\,\mathrm d t=\int_{t\in T}b(t)\,\mathrm d t$;}\\ &\text{(iii) $\int_{t\in T}f(a(t),b(t))\,\mathrm dt\geq0$}\Big\}. \end{align*} Note that $C$ is not empty, as it contains the identically zero function $(0,0)$.
My question is: envisioned as a subset of the Banach space of bounded functions from $T$ to $\mathbb R^2$ endowed with the (product) uniform norm, is $C$ compact? If not, would adding the following extra condition [which would imply also (i)]: “(iv) $a$ and $b$ are non-decreasing” make it compact? What about other possible extra conditions?
Note that Arzelà—Ascoli (a famous theorem on the compactness of certain subsets of function spaces) is off the table with no assumption on continuity. I guess the norm-closedness (and hence the completeness) of $C$ is not difficult to see (I'm not 100% percent sure, though, whether it's true in the first place), but I'm stuck with total boundedness. My gut feeling tells me that $C$ “should” be compact given the assumptions on $f$, but I'm stuck.
Any hints would be appreciated. (I don't need a complete solution just a direction to start.)