I would like to know if I understood the following exercise (a part of a larger one) right:
Let $X \subset \mathbb{R}$ be an nonempty compact set. Prove that the function
$V: c \in \mathbb{R}^n \mapsto \{ { min_{x \in \mathbb{R}^n} \ \ c^Tx \ \ \ with \ x \in X} \}$
is concave.
I did the following approach:
The definition of concavity of a function $f: A \rightarrow B$ for $A, B \subset \mathbb{R}^n $ is:
$f(tc + (1-t)\widetilde{c}) \ge tf(c) + (1-t)f(\widetilde{c})$ for all $t \in [0,1]$ and for all $c, \widetilde{c} \in A$.
But since our given funtion $V$ maps to a set and not to an element of $\mathbb{R}^n$. I supposed that I need to show the above inequality for every arbitrary combination of solutions to our minimizing problem. So I wrote:
$V(tc + (1-t)\widetilde{c}) = min_{x \in \mathbb{R}^n} \ \ tc^Tx + (1-t)\widetilde{c}^Tx \ \ \ with \ x \in X$
and by a simple comparsion we see that
$ a \in min_{x \in \mathbb{R}^n} \ \ tc^Tx + (1-t)\widetilde{c}^Tx \ \ge \ b \in min_{x \in \mathbb{R}^n} \ \ tc^Tx $
$ a \in min_{x \in \mathbb{R}^n} \ \ tc^Tx + (1-t)\widetilde{c}^Tx \ \ge \ b \in min_{x \in \mathbb{R}^n} \ \ (1-t)\widetilde{c}^Tx $
So we may conclude that
$V(tc + (1-t)\widetilde{c}) = min_{x \in \mathbb{R}^n} \ \ tc^Tx + (1-t)\widetilde{c}^Tx \ \ \ with \ x \in X$
$\ge min_{x \in \mathbb{R}^n} \ \ tc^Tx \ \ \ with \ x \in X \ \ \ \ + \ \ \ \ min_{x \in \mathbb{R}^n} \ \ tc^Tx + (1-t)\widetilde{c}^Tx \ \ \ with \ x \in X$
$= tV(c) + (1-t)V(\widetilde{c})$
And since $c, \widetilde{c}$ were arbitrary this proves the concavity of V. Is my solution correct?
