As t(x) = (∑m+1 Ti(x)} \x we have
t'(x)
d ∙ t(x) + h
m+1
^ d ∙ Ti(x) + h ∙ x
i=0
x
Ti(x)
m+1 z------------------^------------------∖
5 d ∙ max {0, min{xi — xi-1, x — xi-1}} ∙ti
i=0
/ ” ∖
m+1
h ∙ У? max{0, min{xi — xi-1, x — xi-1}}
i=0
x
m + 1 / >i ∖
max max {0, min{xi — χi-1,χ — xi-1}} ∙ (dti + h)
i=0
Lemma 2 (Pfahler) Let T(x) a unidimensional tax schedule and Ti(x),i = 1,2,3 the Pfahler-
based linear cuts with the same aggregate total tax liability. Then α(x) = α1(x) < α3(x) <
α2(x) and ψ(x) = ψ2 (x) < ψ3 (x) < ψ1(x).
Proposition 4 (Jakobsson) Let T1(x) and T2(x)be two dual tax schedules. For any pre-tax
income distribution x = (x1, ...,xn) of a (finite) set of tax-payers,
1. T1(x) = (T1(x1),..., T1 (xn)) is LD by T2(x) = (T2(x1),..., T2(xn)) if and only if α1(x) >
α2(x) for all x ≥ 0.
2. Vι(x) = (xι — Tι(xι), ...,xn — Tι(xn,))is LD by ½(x) = (xι — T2(x1), ...,xn — T2(xn)) if
and only if ψ1(x) > ψ2(x) for all x ≥ 0.
It is important to point out that Jakobsson’s proof (1976) assumes implicitly that a
more restrictive constraint than α2 (x0) < α2(x0), for all x0, holds. In fact, he assumes that
α2(x0, x1) < α2(x0, x1) for all x0 < x1, given an income distribution x = {x0 ≤ x1 ≤ x2 ≤ ...}.
T (xι)-T (xo)
----------------------------------------- ∙
x1-x0
However, when x0 ≈ x1 both conditions almost coincide. Indeed, α(x0, x1) =
31