Consider any vector $\mu\equiv(\mu_1,...\mu_M)\in \mathbb{R}^M$.
Take $G: \mathbb{R}^M\rightarrow \mathbb{R}$ with $G(a)\equiv \mathbb{E}_{\mathbb{P}}(\max_{k\in \{1,...,M\}}V_k+a_k)$ for any $a\equiv (a_1,...,a_M)\in \mathbb{R}^M$, where:
A1: $V\equiv(V_1,...,V_M)$ is an $M\times 1$ random vector with absolutely continuous distribution and support on an open subset of $\mathbb{R}^M$ with positive Lebesgue measure. ($\mathbb{E}_{\mathbb{P}}$ denotes expectation using probability measure $\mathbb{P}$).
Show that $$ \sup_{a\in \mathbb{R}^M}\sum_{k=1}^M \mu_k a_k- G(a)=\begin{cases} \max_{a\in \mathbb{R}^M}\sum_{k=1}^M \mu_k a_k- G(a) & \text{ if } \mu\in [0,1]^M \text{ and } \sum_{k=1}^M \mu_k =1\\ \infty & \text{ otherwise } \end{cases} $$
Could you help me with this?
NEW PART: So far I have managed to show the first part
1) Notice that, under A1, for any $y\in \{1,...,M\}$ $$ \mathbb{P}(V_y+\alpha_{xy}\geq V_k+\alpha_{xk}\text{ }\forall k=1,...,M)= \frac{\partial \mathbb{E}_{\mathbb{P} } (\max_{k\in \mathcal{Y}} V_k+\alpha_{k})}{\partial \alpha_{y}} $$
2) Fix $\mu\in \mathbb{R}^M$ and define the function $$ \alpha\in \mathbb{R}^M \mapsto H_{\mu}(\alpha)\equiv \sum_{k=1}^M a_{k} \mu_k - G(\alpha) \in \mathbb{R} $$
3) Let's start with investigating whether there exists an $\tilde{\alpha}\in \mathbb{R}^M$ solving the F.O.C., which is necessary for existence of a global maximum of $H_{\mu}$. The F.O.C. are $$ \begin{cases} \frac{\partial H_{\mu}(\alpha)}{ \partial \alpha_{1}}=0\\ ...\\ \frac{\partial H_{\mu}(\alpha)}{ \partial \alpha_{M}}=0\\ \end{cases} \Leftrightarrow \begin{cases} \mu_1=\frac{\partial G(\alpha)}{\partial \alpha_{1}}\\ ...\\ \mu_M=\frac{\partial G(\alpha)}{\partial \alpha_{M}}\\ \end{cases} \Leftrightarrow \begin{cases} \mu_1=\mathbb{P}(V_1+\alpha_{1}\geq V_k+\alpha_{k} \text{ }\forall k =1,...,M)\\ ...\\ \mu_M=\mathbb{P}(V_M+\alpha_{M}\geq V_k+\alpha_{k} \text{ }\forall k =1,...,M)\\ \end{cases} $$
By step 1), a necessary condition for existence of an $\tilde{\alpha}\in \mathbb{R}^M$ solving the system is that $\mu\in [0,1]^M$ and $\sum_{k=1}^M \mu_k=1$.
Viceversa, can we claim that if $\mu\in [0,1]^M$ and $\sum_{k=1}^M \mu_k=1$, then there exists an $\tilde{\alpha}\in \mathbb{R}^M$ solving the system? I think that this is ensured by A.1, can you confirm?
Hence, we can say that an $\tilde{\alpha}\in \mathbb{R}^M$ solving the system exists if and only if $\mu\in [0,1]^M$ and $\sum_{k=1}^M \mu_k=1$.
4) The last step is showing that $H_{\mu}(\tilde{\alpha})$ is the global maximum of $H_{\mu}$. This is the case because $H_{\mu}$ is concave since: $a\in \mathbb{R}^M\mapsto \sum_{k=1}^M a_{k} \mu_k\in \mathbb{R}$ is concave; $a\in \mathbb{R}^M\mapsto -G(\alpha)\in \mathbb{R}$ is concave; the sum of convex is convex.
What about the $\infty$ case?