Suppose $P$ is the set of functions where $p\in P: R^{+2}\to R^+$ and $p(t,s)$ is differentiable in $t$. $\forall t, p(t,\cdot)$ is a probability distribution on the positive axis $s\in [0,\infty)$, i.e. $s,t\ge 0$, $\int_0^\infty p(t,s)ds=1,\forall t$ and $p(t,s)\ge 0, \forall s,t$. Let $$f(p;t,k) := \int_0^\infty (s-k)_+\frac{\partial p(t,s)}{\partial t}ds,$$ where $x_+:=\max(x,0)$. For given $\forall t,k$,
solve for $\displaystyle\max_{p\in P} f(p;t,k)$;
solve for $\displaystyle\max_{p\in P} f(p;t,k)$, if in addition $p\in P$ satisfies $t = \int_0^\infty sp(t,s)ds, \forall t\in [0,\infty)$;
I conjecture that $f(p;t,k)\le 1, \forall p\in P, (t,k)\in R^{+2}$. What is a proof or counterexample?
I suppose this can be accomplished with calculus of variation, or just linear programming, with Lagrangian multipliers to deal with the constraints. But currently the partial derivative of $t$ gives me pause.