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*f*_{min},*f*_{max}:*f*

_{min}(

*x*) := min{

*f*(

*y*) :

*y*≥

*x*},

*f*

_{max}(

*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

*f*

_{min}and

*g*

_{max}exist and

*f*

_{min}(

*x*) >

*g*

_{max}(

*x*) for all

*x*in ℝ.

**Proof.**If

*f*≫

*g*then there is a nondecreasing function

*h*such that

min{

max{

*f*(*y*) :*y*≥*x*} > min{*h*(*y*) :*y*≥*x*} =*h*(*x*),max{

*g*(*y*) :*y*≤*x*} < max{*h*(*y*) :*y*≤*x*} =*h*(*x*),*f*

_{min}and

*g*

_{max}are well defined and

*f*

_{min}(

*x*) >

*g*

_{max}(

*x*) for all

*x*. Conversely if

*f*

_{min}(

*x*) >

*g*

_{max}(

*x*), let us take

*h*(

*x*) := (

*f*

_{min}(

*x*) +

*g*

_{max}(

*x*))/2;

*f*

_{min}and

*g*

_{max}are obviously nondecreasing, thus

*h*is also nondecreasing, and for all

*x*in ℝ

*f*(

*x*) ≥

*f*

_{min}(

*x*)

*> h*(

*x*) >

*g*

_{max}(

*x*) ≥

*g*(

*x*).

As a corollary, if

*f*≫*g*then lim_{x→+∞}*f*(*x*) is lower bounded and lim_{x→-∞}*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))