I am writing a model which has the following constraints.
$$x \leq f(x) + f(y) \\ y \leq f(x) + f(y) \\ x \geq 0 \\ y \geq 0$$
My question is: Does concavity of $f(\cdot)$ guarantee that there is a maximum value of $x$ that satisfies this set of constraints?
Another way of wording that question is: Is there a finite solution to the problem
$$\max x$$
subject to the four constraints above?
I have the intuition that concavity might be enough based on working out examples with a class of concave functions. For example, if $$f(x) = \sqrt{x}$$ then the maximum value of $x$ or $y$ is $(4,4)$ and you cannot have $x > 4$ regardless of what you choose for $y$ (and vice-versa).
I tried writing out the KKT conditions for the Lagrangian of the maximization problem when $$f(x) = x^q, q< 1$$ and I believe (based on plugging in values of q into Wolfram Alpha) that the solution when the first two constraints bind (and the non-negative $x, y$ constraints don't bind) is given by $$ y = x = 2^{(1/(1-q))} $$ which is finite for $q < 1$.
Is there a way to show that x is finite if $f(\cdot)$ is concave? If not, is there a stronger property than concavity that would guarantee an upper bound for $x$?
I know that, fixing any value of $y$, there is an upper bound for $x$ (and vice-versa), but I'd like to know the conditions where there is an upper bound if we let $y$ vary freely (except that the constraint on $y$ must be satisfied).
If you have a suggestion on what next steps I should take to solve this problem, that would be greatly appreciated. Thanks for your time.
Would $f(x) = x$ not work given non-strict concavity?
For non-strict, you can add an eventually positive strictly concave function to get something like $x + \log x > x$.