Consider a finite set $\mathcal{Y}\equiv \{1,...,L\}$ and a function $u:\mathcal{Y}\rightarrow \mathbb{R}$. Let $\Delta(\mathcal{Y})$ be the set of probability distributions over $\mathcal{Y}$, that is $$ \Delta(\mathcal{Y})\equiv \{P\in \mathbb{R}^L: P(y)\geq 0 \forall y\in \mathcal{Y}, \sum_{y\in \mathcal{Y}}P(y)=1\} $$ Consider the set $$ \mathcal{Q}\equiv \{P\in \Delta(\mathcal{Y}): \forall y\in \mathcal{Y}\text{, IF }P(y)>0\text{, THEN } u(y)-u(\tilde{y})\geq 0 \text{ }\forall \tilde{y}\neq y\} $$ Consider the set $\mathcal{Y}^*\equiv argmax_{y\in \mathcal{Y}}u(y)$. For each $y\in \mathcal{Y}^*$, take $P\in \Delta(\mathcal{Y})$ such that $P(y)=1$. Collect all such $|\mathcal{Y}^*|$ vectors into the set $\mathcal{B}$. For example. let $\mathcal{Y}^*=\{1,2\}$ and $L=3$. Then, $$ \mathcal{B}\equiv \{(1,0,0),(0,1,0)\} $$ Let $co\{\mathcal{B}\}$ denote the convex hull of $\mathcal{B}$.
Question: are $\mathcal{Q}$ and $co\{\mathcal{B}\}$ different or equal? How can we show it?
My thoughts and sources of confusion:
1) The definition of $\mathcal{Q}$ seems to suggest that $\mathcal{Q}$ is not convex, because it is defined by an "IF ... THEN ..." constraint which is not linear. Instead, $co\{\mathcal{B}\}$ is convex by definition. Hence, it should be that $\mathcal{Q}$ and $co\{\mathcal{B}\}$ are different.
2) $co\{\mathcal{B}\}\subseteq \mathcal{Q}$ because every element of $co\{\mathcal{B}\}$ satisfies the "IF ... THEN ..." constraint defining $\mathcal{Q}$.
3) if 1) and 2) hold, then it should be $\mathcal{Q}\supset co\{\mathcal{B}\}$. However, I cannot think of an element in $\mathcal{Q}$ that is not part of $co\{\mathcal{B}\}$.
Could you help to clarify?
We have that $\cal Q = co (\cal B)$. To see this, it remains to be shown that $\cal Q \subseteq co (\cal B)$.
Let $P \in \cal Q$, and write $P = (p_1,\ldots,p_L)$. Moreover, denote by $e_i$ the vector in $\mathbb R^L$ that is equal to $1$ in the $i^{\text{th}}$ entry and $0$ otherwise. Thus, we can write $$ P = \sum_{i=1}^L p_ie_i. $$
Note, however, that $p_i > 0$ only if $e_i \in \cal B$ (or, equivalently, $i \in \cal Y^*$). This follows from the definition of $\cal Q$. Thus, $P \in co(\cal B)$.