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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 given $\gamma = \bar{\gamma}$, the optimized objective value for problem 1 was greater than the optimized objective value of problem 2, does the optimized objective 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 to 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?

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  • $\begingroup$ Please clarify: By optimal value, do you mean the optimal $x^*$ or the optimised value of the objective function? If its the former, you can only claim that the optimal $x^*$ from problem 1 will be WEAKLY greater than the optimal $x^*$ from Problem 2 as you increase $\gamma$. The issue is that the optimal $x^*$ for both the problems may hit the lower boundary of the interval as you increase $\gamma$. $\endgroup$
    – user28372
    Aug 15 '20 at 15:59
  • $\begingroup$ What I meqn was optimised value of the objective function, not the optimal $x^*$. I will edit $\endgroup$
    – OR_student
    Aug 15 '20 at 23:42
  • $\begingroup$ Cross-posted on Operations Research Stack Exchange: or.stackexchange.com/q/4681 $\endgroup$
    – Flux
    Aug 16 '20 at 18:04
  • $\begingroup$ For $K \leq 0$ the claim is true, but not for $K>0$ $\endgroup$
    – Bertrand
    Aug 18 '20 at 11:13
  • $\begingroup$ Are you referring to my claim in the comment or OP’s claim? $\endgroup$
    – user28372
    Aug 18 '20 at 16:54
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Take $\Omega_1=\Omega_2=[0,0.5]$.

Let $p(x)=x^4$, so that $p''(x)=12x^2>0$ on $(0,0.5)$.

Let $q(x)=0.5x$, which is linear in $x$, and $K=0.2>0$.

For $\bar\gamma=1$, both objective functions attain their respective maximum at $x=0.5$. As the following figure shows, objective function $(1)$ (blue curve) has a higher maximum than that of objective function $(2)$ (red curve).

enter image description here

But for $\gamma=5>\bar\gamma$, objective function $(1)$ attains a maximum value of approximately $0.276$, while objective function $(2)$ attains a higher maximum value of approximately $0.309$, contradicting the claim that $(1)$'s maximum would always be greater than $(2)$'s maximum for $\gamma>\bar\gamma$.

enter image description here

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