Wednesday, January 1, 2014

A relationship on real functions

Given two functions f, g, 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 . ≫ is obviously a strict order on ℝ. The following are additional elementary facts about this relationship.
Proposition. Let f, g, f', g', f+, g-, h, r : such that f g, f' g', f+(x) f(x) and g(x) g-(x) for all xh 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}.
provided the expressions actually exist for all x in . 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 .
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),
hence fmin and gmax are well defined and  fmin(x) > gmax(x) for all x. Conversely if fmin(x) > gmax(x), let us take
h(x) := (fmin(x) + gmax(x))/2;
fmin and gmax are obviously nondecreasing, thus h is also nondecreasing, and for all x in
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.

1. Proposition 2 might not hold true in both directions.

Consider
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

(f >> g) => (fmin(x) > gmax(x))

1. Hi, actually gmax(x) =max{g(y) : y ≤ x}=max{y : y ≤ x}=x, not infinity. Note the ≤ in the definition of gmax.

2. Proposition 2 might not hold true in both directions.

Consider
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