Take a convex set $A$ that is a subset of the $(n-1)$-dimensional simplex. In particular, $\forall a\equiv (a_1,...,a_n) \in A$, $a_i\in [0,1]$ and $\sum_{i=1}^n a_i=1$.
Quoting from a book: it follows that $$ (\diamond) \hspace{1cm} a^*\in A\Leftrightarrow b'a^*-\underbrace{\sup_{a\in A} b'a}_{\text{support function}} \leq 0 \text{ }\forall b \in \mathcal{B} $$ where $\mathcal{B}\equiv \{b\in \mathbb{R}^n: b'b\leq 1\}$.
Question 1: Why are we checking $b'a^*-\sup_{a\in A} b'a \leq 0$ only for the $b$s in the unit ball? Is this related with the fact that $A$ is a subset of the $(n-1)$-dimensional simplex?
Question 2: Is $(\diamond)$ equivalent to $$ (\diamond \diamond) \hspace{1cm} a^*\in A\Leftrightarrow b'a^*-\sup_{a\in A} b'a \leq 0 \text{ }\forall b \in \mathcal{S} $$ where $\mathcal{S}\equiv \{b\in \mathbb{R}^n: b'b= 1\}$? That is, is it sufficient to check the $b$s in the unit sphere?
Both questions in effect ask the same, so one answer suffices.
Answers $1\& 2$.
(should be : "only for $b'$s in the unit ball, not $b$s). This because the support function is homogeneous, so that in the inequality $$b'a*\leq\sup_{a\in A}b'a$$ we can multiply both sides by any positive number without changing the inequality. In particular, we can divide by the norm of $b'$, thus obtaining a unit vector. This is not related to the specific structure of the set $A$.