Set $S \subset \mathbb R^2$ is compact and convex. A typical element of $S$ is $s=(s_1,s_2) \in S$. Also, $d \in \mathbb R^2$ is a fixed element such that there exists $s \in S$ such that $s \gt d$. Here $s \gt d$ means $s_1 \gt d_1$ and $s_2 \gt d_2$
$$\underset{(d_1,d_2)\le(s_1,s_2) \in S}{\overset{}{\operatorname{arg \;\; max}}} \; (s_1-d_1)(s_2-d_2)$$
Let $H(s_1,s_2)=(s_1-d_1)(s_2-d_2)$. Since $H$ is strictly quasi concave on $\{s \in S: s \gt d \}$, there exists $s \in S$ such that $s \gt d$ and $S$ is convex, so the maximizer of $H$ is unique.
My question: I did not get the last part. Why is maximizer of $H$ unique?
Best Answer
Strict quasiconcavity means (at least to me) that for any two points $p,q$ in the domain of $H$ and for every $t\in (0,1)$ $$H(tp+(1-t)q)>\min(H(p),H(q)) \tag1$$ (With nonstrict inequality this is simply quasiconcavity).
If $H$ attains its maximum at two distinct points $p,q$, then (1) immediately yields a contradiction.