Proof. (Lemma 8.1) Take any point (x, x) on the diagonal, belonging to the
domain of F .Forα> 0 small enough
U1 (x, x) ' U(x + α, x) - U(x, x) and L1(x, x) ' (x, x) - L(x - α, x). (17)
Hence, from A2:
U(x + α, x) - U(x, x) > L(x, x) - L(x - α, x). (18)
From A1 we know that
U(x + α, x - α) - U(x, x - α) ≥ U(x + α, x) - U(x, x). (19)
Take bε> 0 such that
U(x + α, x) - U (x, x) ≥ L(x, x) - L(x - α, x)+bε. (20)
From continuity of U(x, y) in x (for fixed y = x - α), it follows that
bε
∀0 < ε ≤ -, ∃δ > 0 such that
bε
|x — (x — α)| = α ≤ δ and |U(x, x — α) — U(x — α,x — α)| ≤ ε ≤ —.
Moreover,
bε
∀0 < ε ≤ -, ∃δ > 0 such that
bε
|x + α — x| = α ≤ δ and |U(x + α,x — α) — U(x, x — α)| ≤ ε ≤ —.
This allows us to establish that:
(U(x + α, x — α) — U(x, x — α)) — (U (x, x — α) — U(x — α, x — α)) ≤ 2ε ≤ bε
(21)
since, for |A| <εand |B| <ε,thenA — B<2ε.
Rewriting (21),
U(x + α, x — α) — U(x, x — α) ≤ U(x, x — α) — U(x — α, x — α)+bε. (22)
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