I have a problem:
\begin{align} \max_{p,q} \log_2(1+ph)\\ s.t: \log_2(1+ph)\leq \log_2(1+qf)\\ p<P\\ q<Q \end{align} where $h$ and $f$ are positive real numbers. We can rewrite the problem as:
\begin{align} \max_{p,q} \log_2(1+ph)\\ s.t: 0\leq \log_2(1+qf)-\log_2(1+ph)\\ 0<P-p\\ 0<Q-q \end{align}
is this problem non-convex due to $0\leq \log_2(1+q h)-\log_2(1+p f)$ which is the difference of two concave functions? Can we say that $\log_2(1+qf)-\log_2(1+ph)$ is a concave function in this case because the constraint requires that $\log_2(1+qf)\geq\log_2(1+ph)$?
I know that we can easily convert the constraint into an affine function by removing $\log_2$ from both sides. But, if we don't do that, is the problem still convex?
The reason I am asking this question is that if we solve the problem by using KKT conditions we mostly get the optimal solution (I am saying "mostly" because I have compared the solution with heuristic search (both the solutions were the same for the values of $f$ and $h$ I have checked), however, it is not feasible to check for all possible values, so I can not be sure if the solution will be always optimal).
$(-x^2)-(-x^4)$ is neither convex nor concave.
$(-2x^2)-(-x^2)$ is concave.
$(-x^2)-(-2x^2)$ is convex.
$(-x^2)-(-x^2)$ is both.