How can this be proved? (Convex optimization) Consider the following maximization problems:

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

$\max_{x} x-\gamma (p(x) + q(x) )+K$
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, $p''(x)&gt;0$ and $q(x)$ is linear in $x$ and $K&gt;0$ 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 $\gamma &gt; \bar{\gamma}$?
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 $\gamma'&gt;\bar{\gamma}$ such that optimized value for problem 2 is greater than that of problem 1, but struggling. How can this be proved?
Take $\Omega_1=\Omega_2=[0,0.5]$.
Let $p(x)=x^4$, so that $p''(x)=12x^2&gt;0$ on $(0,0.5)$.
Let $q(x)=0.5x$, which is linear in $x$, and $K=0.2&gt;0$.
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).

But for $\gamma=5&gt;\bar\gamma$, 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 $\gamma&gt;\bar\gamma$.