− 00χ u0 + β0 1− β = θχ f(e)− χ + β0(1) 1− β(1) approaches +∞ as e→ f−1(χ) and decreases strictly in e. Thus either ∂S(1, e)/∂e is strictly positive for all e ∈ (f−1(χ), e), in which case S(1, e) crosses zero only once, or ∂S(1, e)/∂e changes sign once at a value e < e, from positive for e < e to negative for e > e. In the latter case, S(1, e) ≥ S(1, e) > 0 for e ≥ e, so S(1, e) cannot cross ze...