Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Monotony of composite functions

  1. Mar 12, 2016 #1
    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. jcsd
  3. Mar 12, 2016 #2


    User Avatar
    Science Advisor
    Homework Helper

    Consider h, a real, not identically zero, smooth function with support in [3,4] and |h(x)|≤1 for all x ∈ ℝ.
  4. Mar 12, 2016 #3


    User Avatar
    Science Advisor
    Homework Helper

    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
  5. Mar 12, 2016 #4


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    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.
  6. Mar 13, 2016 #5


    User Avatar
    Science Advisor
    Homework Helper

    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. :oldsmile:
    Last edited: Mar 13, 2016
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted