Convex optimization - comparing optimal values (any counter example?)

Question

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 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 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?

Answer 1

Suppose $p(x)=x^2$ and $q(x)=x$, $\Omega_1=\Omega_2=[0,\infty)$. Then, for $\gamma\in[0,1]$:

1. Maximizer is $x_1=\dfrac{1}{2\gamma}$, maximum is $y_1=\dfrac{1}{4\gamma}$
2. Maximizer is $x_2=\dfrac{1-\gamma}{2\gamma}$, maximum is $y_2=\dfrac{1}{4\gamma}+\dfrac{\gamma+4K-2}{4}$

We have

$$y_1-y_2=\frac{2-4K-\gamma}{4}$$

so that $y_1\geq y_2$ if and only if

$$\gamma\leq 4K-2.$$

We have $4K-2\in[0,1]$ if and only if $K\in[1/2,3/4]$.

If, for example, $K=5/8$, then $y_1>y_2$ if and only if $\gamma\in[0,1/2)$ which is the opposite of what you have conjectured.