I'd like to check if the following statements are true about optimizing a linear objective function $f$ over a convex space. For the purpose of the problem, the space over which we are optimizing is the convex hull of the points $P_1$, $P_2, \cdots, P_n$.
If $Q = \lambda_1 P_{q_1} + \lambda_2 P_{q_2} + \cdots +\lambda_k P_{q_k}$ (where $\sum_{i=1}^k \lambda_i = 1$ and $\lambda_i > 0$) is an optimal solution, then it must be the case that all the $P_{q_i}$ are also all optimal solutions. My justification for this comes from the fact that optimal solutions to LP must occur at extreme points, and if there were to be an optimal solution that were not an extreme point, it must be the case that the gradient of the objective function is perpendicular to the convex hull of the extreme points that are optimal solutions. For sake of demonstration, if $P_{q_1}$ were optimal and $P_{q_2}$ were not, then $\nabla f \cdot (P_{q_1} - P_{q_2}) > 0$, and hence $Q$ cannot be optimal either as $Q' = (\lambda_1 + \lambda_2) P_{q_1} + 0 \cdot P_{q_2} + \cdots + \lambda_k P_{q_k}$ would be a better solution.
The converse of the previous statement: if $P_{q_1}, P_{q_2}, \cdots, P_{q_k}$ are all optimal solutions (which necessarily means that $f$ attains the same value at these extreme points), then any convex combination of the $P_{q_i}$ must be an optimal solution. The justification for this statement would be that since $f$ is linear, it is "monotonic" in some sense (looking for a better way to formalize this), and hence must be constant along the convex hull of the $P_{q_i}$.
Could someone verify if these statements and justifications are correct?
For your first part, to construct the contradiction, in fact, we do not require $P_{q_1}$ to be optimal, to construct a better solution, we just require $c^TP_{q_1}< c^TP_{q_2}$, where $c$ is the linear function and we assume that we are working with minimization problem.
In that case, let's compare the objective value obtained from $Q'$ and $Q$.
$$c^TQ=\sum_{i=1}^kc^T\lambda_iP_{q_i}$$
$$c^TQ'=c^T(\lambda_1+\lambda_2)P_{q_1}+\sum_{i=3}^kc^TP_{q_1}$$
Hence $$c^TQ-c^TQ'=\lambda_2(c^TP_{q_2}-c^TP_{q_1})>0$$ which contradicts optimality of $Q$.
Hence for $Q$ to be optimal, all $P_{q_i}$ are optimal solutions.
For the second part, let's use the linearity property directly, if $c^TP_{q_i}$ are all optimal and it is equal to $f^*$, then
$$c^TQ = c^T\sum_{i=1}^k \lambda_iP_{q_i}=\sum_{i=1}^k\lambda_i (c^TP_{q_i})=f^*\sum_{i=1}^k \lambda_i =f^*$$
Hence $Q$ is an optimal solution as well.