Consider the function
$f: \{1 , \ldots, m\} \times \Delta^{m} \rightarrow \mathbb{R}$ where
\begin{equation}
\Delta^{m}:=\{\lambda \in \mathbb{R}^{m} \ | \ \sum_{i=1}^{m} \lambda_i = 1 , \ \lambda_j \geq 0 \text{ for } j=1, \ldots m \}
\end{equation}
\begin{equation}
f(j,\lambda)=\langle y^{(j)}, \sum _{i=1}^{m} \lambda_i x^{(i)}-z \rangle
\end{equation}
For some $y^{s} \in \mathbb{R}^{n} \text{ for } s=1, \ldots, m, \ z \in \mathbb{R}^{n}$ and $x^{(i)} \in \mathbb{R}^{n} \text{ for } i=1, \ldots m . $
Question: Is the function $F : \Delta^{m} \mapsto \mathbb{R}^{m}$
\begin{equation}
F(\lambda) = \underset{j=1, \ldots, m }{\min} f(j,\lambda)
\end{equation}
concave? Where does it attain its minimum (as a function of $\lambda$ of course)? Please consider first my attempt before answering, and tell me where I am wrong (if I am). Additional question: where does it attain its maximum?
Attempt: I know that the pointwise minimum of a set of concave function is again concave. So it would be enough to show that $f(j,\lambda)$ is concave, to then conclude that the minimum is attained at some vertex $v_i=(0, \ldots, 1, \ldots, 0)$ where $1$ is in the $i^{\text{th}}$ position. My argument for concavity is:
\begin{gather}
f(j, \alpha(\lambda^{(1)}) + (1-\alpha)(\lambda^{(2)}))=\langle y^{(j)}, \alpha \sum _{i=1}^{m} \lambda_i^{(1)} x^{(i)}+(1-\alpha)\sum _{i=1}^{m} \lambda_i ^{(2)}x^{(i)}-(1-\alpha)z-\alpha z \rangle \\
=(1-\alpha) \langle y^{(j)}, \sum _{i=1}^{m} \lambda_i^{(1)} x^{(i)}-z \rangle + \alpha \langle y^{(j)}, \sum _{i=1}^{m} \lambda_i^{(2)} x^{(i)}-z \rangle \\
= \alpha f(j, \lambda^{(1)}) + (1-\alpha)f(j,\lambda^{(2)}),
\end{gather}
where $\alpha \in [0,1]$, $\lambda^{(1)}, \lambda^{(2)} \in \Delta^{m}$.
Best Answer
$g(\lambda) = f(j,\lambda)$ is concave in $\lambda$, and the minimum of concave functions is concave.
By the maximum principle, a concave function is minimized at an extreme point of the domain, which is the argument for your choice $v_i$.