In the proof of Theorem 3.2.1(i), consider the set of c satisfying (c)

In the proof of Theorem 3.2.1(i), consider the set of c satisfying

α

(c) ≤ α ≤ α(c − 0). If there is only one such

c, c is unique; otherwise, there is an interval of such values [c1, c2]. Argue that, in this case, if α

(c) is continuous at c2, then Pi(C) = 0 for i = 0, 1, where

= 
x : p0(x) > 0 and c1 <
p1(x)
p0(x) ≤ c2 
.
If α

(c) is not continuous at c2, then the result is false.