# How can this convex optimization problem be proved?

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?

• 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? – Richard Aug 15 at 19:00
• 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. – 이명훈 Aug 15 at 23:39
• For small $\gamma$, optimal value of problem 2 cqn be greater than problem 1. – 이명훈 Aug 15 at 23:40
• Cross-posted on Economics Stack Exchange: economics.stackexchange.com/q/39224 – Flux Aug 16 at 18:03