# B Monotony of composite functions

1. Mar 12, 2016

### anachin6000

So, it is known and easy to prove that if you have f : D -> G and g : G -> B then
-if both f and g have the same monotony => fοg is increasing
-if f and g have different monotony => fοg is decreasing
But the reciprocal of this is not always true (easy to prove with a contradicting example).
Though, it came to my mind that, if we have a function h : D -> D, a kind of of reciprocal might be valid for hοh.
I think that if hοh is monotonic it results that h is either decreasing or increasing, but I am not sure if it is true or not, neither how to prove or disprove it. This is actually my question, is it true and how you prove that?

2. Mar 12, 2016

### Samy_A

Consider h, a real, not identically zero, smooth function with support in [3,4] and |h(x)|≤1 for all x ∈ ℝ.

3. Mar 12, 2016

### Samy_A

This being said, if you add the condition that h is continuously differentiable, I think you can prove that if hοh is a monotone function while h is not, then hοh must be constant.

EDIT: continuity of h may well be sufficient.

EDIT2: no, not true, sorry.

Counter example:
Define h as follows:
for x≤0, f(x)=-x²
for x≥0, f(x)=g(x) where g is a non constant smooth function with support in [3,4] and range in [0,1].

Then, for x≤0, hoh(x)=h(-x²)=-x4.
For x>0, hoh(x)=h(h(x))=0.

Last edited: Mar 12, 2016
4. Mar 12, 2016

### FactChecker

Start with r1<r2 and track it through. I think that you will find that h monotonic => h(h) monotonic increasing.

But h(h) being monotonic implies nothing about monotonicity of h. Consider the function h(r) = r if r is rational; h(r)=-r otherwise.

5. Mar 13, 2016

### Samy_A

Correct. If h is not continous, then h(h) being monotone implies nothing about monotonicity of h.

Although that was not what the OP asked for, I wondered about what happens if h is continuous.
As shown in posts #2 and #3, then also h(h) can be monotone without h being monotone. But the following seems true for continuous h:
On intervals where h(h) is strictly monotone, h will also be strictly monotone.

Proof (tentative):
Let's assume, wlog, that f=h(h) is a monotone increasing function.

1) Let's note that an injective continuous function on an interval is strictly monotone.

2) Let [a,b] be an interval in ℝ where f is strictly increasing.
For x,y ∈ [a,b], x≠y, h(x)≠h(y), for else f(x)=f(y), and f would be constant on [x,y]. It follows from 1) that h is strictly monotone on [a,b].

3) Now let's assume that we have an interval [a,b] where f is strictly increasing, and let's fix (again wlog) on the case where h is also strictly increasing on [a,b].
Take any c>b. If h(c)<h(b), the intermediate value theorem implies that there exist an x<b and an y>b such that h(x)=h(y). But then f would be constant on [x,y], contradicting the hypothesis that f is strictly increasing in [a,b].
Hence, ∀c>b, h(c)≥h(b).

4) Let [c,d] be another interval where f is strictly increasing, with c>b.
From 2) we know that h will be strictly monotone on [c,d]. From 3) we know that h(c)≥h(b), h(d)≥h(b).
If h were strictly decreasing on [c,d], h(c)>h(d)≥h(b). The intermediate value theorem then implies that ∃x ∈ [b,c[ satisfying h(x)=h(d).
That makes f constant on [x,d], a contradiction with f being strictly increasing on [c,d].
Hence h is also strictly increasing on [c,d].

Probably this proof can be made shorter.

Last edited: Mar 13, 2016