$\mathbf{Question}$: Suppose we have $T\in \mathcal{R}^{p\times n},W\in \mathcal{R}^{p\times m}$ and $h\in \mathcal{R}^p$.
For $Q:\mathcal{R}^n \times \mathcal{R}^m \rightarrow \mathcal{R}\cup \{\pm\infty\}$ we have $Q(\xi,\eta)=\min\limits_y \{\eta^T y|Wy=h-T\xi,y\geq 0\}$.
Now, determine whether the function is convex, concave, or neither of the two in the following cases (and justify why):
1). For a fixed $\xi=\hat{\xi}$.
2). For a fixed $\eta=\hat{\eta}$ such that the polyhedral set $P=\{x|W^Tx\leq\hat{\eta}\}$ is nonempty.
3).$Q(\xi,\eta)$.
$\mathbf{Answer}$: For part (2), my claim is that $Q$ is convex, from the theorem:
$\textbf{Theorem}$: Given $A\in \mathcal{R}^{m\times n}$ and $c\in \mathcal{R}^n$, suppose that the polyhedral set $P:=\{u:A^Tu\leq c\}$ is non-empty. Then the function $\hat{\theta}=\min\limits_x \{c^Tx:Ax=\hat{b},x\geq 0\}$ is convex.
For (2) can such a reasoning be drawn directly from the theorem? Also for part (1), my gut feeling is telling me that its concave, and I tried proving my claim along the definition of an epigraph but am unable to make any progress..
$\textbf{Update on (1):}$ with @LinAlg's and @prubin's advice, since $\eta$ is not fixed, for $\lambda\in[0,1]$, $y\geq0$,
$$\min\limits_y\{f(y),f(y')\}\leq \lambda f(y)+(1-\lambda)f(y')\leq f(\lambda y+(1-\lambda)y')$$
$\forall y,y'$ in the domain of $f$. So it follows that $f$ is concave and quasi-concave. Thus, $Q(\hat{\xi},\eta)$ is concave.
$\textbf{End of update to (1)}$
Is this the right way about doing this? (clarification needed)
For (3), I have no clue as to how I can approach this, but a couple of my pals advised me that it may be neither concave nor convex.
Hence, some clarifications and hints with regards to the above questions will be deeply appreciated.
If there is a more intuitive way of approaching problems such as the above, I'll love to hear it too!
Thanks.
In the theorem, $c$ is fixed, so $\hat{\theta}$ is a function of just $\hat{b}$. After mapping the notation of the theorem to the notation of your problem, you can get part 2 directly from the theorem by exploiting the fact that the composition of a convex function with a linear function is convex. (If you haven't seen that derived, the proof is straightforward from the definitions of linearity and convexity.)