= Solution
Use the positive variable $Y$ from part (a). Since $0\leq(S-B)^+\leq S$ and $(S-B)^+\geq S-B$,
$$
0\leq c=\mathbb E[Y(S-B)^+]\leq\mathbb E[YS]=s,
\qquad c\geq\mathbb E[Y(S-B)]=s-b.
$$
Combining these inequalities gives the <one-period call price bounds>
$$
\boxed{(s-b)^+\leq c\leq s.}
$$
For the strict assertion, both $\{S>B\}$ and $\{S<B\}$ have positive probability. Positivity of $Y$ gives
$$
c=\mathbb E[Y(S-B)^+]>0,
\qquad c-(s-b)=\mathbb E[Y(B-S)^+]>0.
$$
If $s-b\geq0$, the second inequality is the required strict bound; if $s-b<0$, the first is. Hence $\boxed{c>(s-b)^+}$. The weighted expectations are finite because $P$ and $Y$ are bounded.
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