Given two functions f, g : ℝ → ℝ, we say that f is strongly greater than g, denoted f ≫ g, iff there exists a nondecreasing function h such that f(x) > h(x) > g(x) for all x in ℝ. ≫ is obviously a strict order on ℝ → ℝ. The following are additional elementary facts about this relationship.
Proposition. Let f, g, f', g', f+, g-, h, r : ℝ → ℝ such that f ≫ g, f' ≫ g', f+(x) ≥ f(x) and g(x) ≥ g-(x) for all x, h is strictly increasing and r is nondecreasing, and let k be a positive real number. We have
- (f + f') ≫ (g + g'),
- f+ ≫ g-,
- h ∘ f ≫ h ∘ g,
- k(f + r) ≫ k(g + r).
For any function f we introduce the following associated functions fmin, fmax:
fmin(x) := min{f(y) : y ≥ x},
fmax(x) := max{f(y) : y ≤ x}.
fmax(x) := max{f(y) : y ≤ x}.
provided the expressions actually exist for all x in ℝ. These auxiliary functions give us an alternative way of defining ≫.
Proposition. f ≫ g iff fmin and gmax exist and fmin(x) > gmax(x) for all x in ℝ.
Proof. If f ≫ g then there is a nondecreasing function h such that
min{f(y) : y ≥ x} > min{h(y) : y ≥ x} = h(x),
max{g(y) : y ≤ x} < max{h(y) : y ≤ x} = h(x),
max{g(y) : y ≤ x} < max{h(y) : y ≤ x} = h(x),
h(x) := (fmin(x) + gmax(x))/2;
fmin and gmax are obviously nondecreasing, thus h is also nondecreasing, and for all x in ℝ
f(x) ≥ fmin(x) > h(x) > gmax(x) ≥ g(x).
As a corollary, if f ≫ g then limx→+∞f(x) is lower bounded and limx→-∞g(x) is upper bounded.
Proposition 2 might not hold true in both directions.
ReplyDeleteConsider
f := x + 2
g := x
Clearly, f >> g with h := x + 1
But from
fmin = f
gmax = infinity
follows that
fmin(x) < gmax(x) for some x
which contradicts
(f >> g) => (fmin(x) > gmax(x))
Hi, actually gmax(x) =max{g(y) : y ≤ x}=max{y : y ≤ x}=x, not infinity. Note the ≤ in the definition of gmax.
DeleteProposition 2 might not hold true in both directions.
ReplyDeleteConsider
f := x + 2
g := x
Clearly, f >> g with h := x + 1
But from
fmin = f
gmax = infinity
follows that
fmin(x) < gmax(x) for some x
which contradicts
(f >> g) => (fmin(x) > gmax(x))