Consider the following maximization problems:

  1. $\max_{x} x -\gamma p(x)$ subject to $x \in \Omega_1$

  2. $\max_{x} x-\gamma (p(x) + q(x) )+K$ subject to $x \in \Omega_2$

where $\Omega_1 $ and $ \Omega_2$ are convex sets, $p(x) \geq 0 $ and $q(x) \geq 0$ for all $x\in \Omega_2$. Also, $p''(x)>0$ and $q(x)$ is linear in $x$ and $K>0$ is a constant.

If for a given $\gamma = \bar{\gamma}$, the optimal value for problem 1 was greater than the optimal value of problem 2, does the optimal value for the problem 1 always greater than that of problem 2 for all $\gamma > \bar{\gamma}$?

Prove or provide counter example for (1) $\Omega_1= \Omega_2$ and (2) $\Omega_1 \subset \Omega_2$.

Since the higher penalty proportional to $\gamma$ is imposed on the objective function of problem 2, this claim seems right. I tried using contradiction, in which assuming there exists $\gamma'>\bar{\gamma}$ such that optimized value for problem 2 is greater than that of problem 1, but struggling. How can this be proved?

  • $\begingroup$ Actually, I am not sure this is easy to prove because of the constant $K$. Suppose $K=1$ and $\gamma = 0$. Then the objective function value of problem 2 is larger than of problem 1. Or am I missing something? $\endgroup$ – Richard Aug 15 at 19:00
  • $\begingroup$ I'd like to prove that if the objective value of problem 1 exceeds the problem 2, then it also yields greater value for all $\gamma$ greater than that. $\endgroup$ – 이명훈 Aug 15 at 23:39
  • $\begingroup$ For small $\gamma$, optimal value of problem 2 cqn be greater than problem 1. $\endgroup$ – 이명훈 Aug 15 at 23:40
  • $\begingroup$ Cross-posted on Economics Stack Exchange: economics.stackexchange.com/q/39224 $\endgroup$ – Flux Aug 16 at 18:03

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