the original height of the square and its length is not bigger than α, as defined in
Lemma 8.1. We will now show that for points in the vertices of such rectangles
the thesis holds and a fortiori it is possible to obtain the conclusion for the
whole square. Figure 6 illustrates the proof.
Let x - z = kα, where k ∈ R,andα>0 is small enough. Now consider the
rectangle defined by the following points (x, x), (x, x - α), (z, x) and (z, x - α) .
From Lemma 8.1 we know that
F(x, x - α) - F(x - α, x - α) ≥ F(x, x) - F(x - α, x)
Also, from A1 we know that
F(x - α, x - α) - F(z, x - α) ≥ F(x - α, x) - F (z, x)
Adding these two inequalities we obtain that
F(x, x - α) - F(z, x - α) ≥ F(x, x) - F (z, x) (24)
Repeating the procedure, consider the rectangle defined by: (x, x - α), (x, x - 2α),
(z, x - α), (z, x - 2α). From A1 we know that:
F (x, x - 2α) - F (x - α, x - 2α) ≥ F(x, x - α) - F(x - α, x - α)
and also
F (x - 2α, x - 2α) - F (z, x - 2α) ≥ F(x - 2α, x - α) - F(z, x - α).
Using Lemma 8.1 we know that
F(x - α, x - 2α) - F(x - 2α, x - 2α) ≥ F(x - α, x - α) - F (x - 2α, x - α)
Adding the three inequalities we obtain:
F (x, x - 2α) - F (z, x - 2α) ≥ F(x, x - α) - F(z, x - α)
From (24) we obtain
F (x, x - 2α) - F (z,x - 2α) ≥ F(x, x) - F (z, x)
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