Let $S\subseteq \mathbb{R}^2$, $d:=(d_1,d_2) \in S$, and $s:=(s_1,s_2)$ a generic point of $S$. Assume that there exists $s \in S$ such that $s_1>d_1$ and $s_2 >d_2$. Consider the following maximization problem
$$
\begin{align*}
\max_{(s_1,s_2)\in S} & (s_1-d_1)(s_2-d_2)\\
\text{s.t.}\quad & s_1\geq d_1 \text{ and }s_2\geq d_2
\end{align*}$$
Does the problem have a unique solution under certain conditions? In positive case, one of the necessary conditions is $S$ compact and convex?
Best Answer
I think it's enough to focus on $d$ being the origin as the original problem can be restated at origin using translation. So, we are given a subset $S$ of the positive quadrant, and we would like to maximize $s_1\cdot s_2$. Think of the level sets of $f(s) = s_1\cdot s_2$: if we have two points lying on the same level set, then by taking the straight line between them we can find the point lying on higher-level sets. Thus, this gives us the intuition that convexity implies uniqueness. Can you proceed from here? At the same time, it's not yet a formal statement, so please be careful - perhaps while trying to show that you'll come up with a counterexample.
I think compactness is irrelevant for uniqueness, and is more about the existence.
Regarding uniqueness, suppose that $S$ is just convex and that there exist $s'\neq s''\in S$ such that $f(s') = f(s'') = f^* = \sup_{s\in S}f(s)$. Consider $\hat s = \frac12(s'+s'')\in S$; it satisfies \begin{align} f(\hat s) &= \frac14(s'_1+s'_2)(s''_1+s''_2) = \frac12f^* +\frac14(s'_1s''_2+s''_1s'_2) \\ &= \frac12f^* + \frac14f^*\left(\frac{s'_1}{s''_1} + \frac{s''_1}{s'_1}\right) > f^* \end{align} since $s'_1 \neq s''_1$: recall that for $x>0$ we have $x+\frac1x>2$ unless $x = 1$. Hence, there exists $\hat s\in S$ at which $f$ is greater than $f^*$, contradiction. Thus, if $S$ is convex and there exists a point of maximum, such point is unique.