I am a physicist and my basic maths skills are a bit rusty, I thank in advance anyone who will be kind enough to give me at least some hints :)
My problem is checking the monotonicity of a function $C(x) = h(f(x)) - h(g(x))$, where the functions $h,f,g$ have the following properties:
- $\forall x \quad f(x) > g(x), \quad f(x) >1, \quad g(x) >1$
- $\forall x \quad f'(x) >0, \quad g'(x) >0 $ (monotonically increasing)
- $\forall y \quad h'(y) > 0, \quad h''(y) <0 $ (monotonically increasing and concave)
- $ \lim_{y \to \infty} h(y) = + \infty$ (no horizontal asymptotes)
Basically I'd like to know the conditions under which I can reduce the study of $C(x)$ to that of $f(x)-g(x)$. Ideally I would like to know under which conditions I can say that $ f'(x) > g'(x) \implies C'(x) >0$, i.e. if $f$ grows faster then $g$ then my function is monotonically increasing.
I don't know if this is useful, but I think I have proven the converse as follows:
$ C'(x) = h'(f(x)) f'(x) - h'(g(x)) g'(x) < h'(g(x)) \left[ f'(x) - g'(x) \right] $, since $h'(f(x)) < h'(g(x))$ for the concavity property of $h$.
P.S. I actually have a specific form for $h(y)=\frac{y+1}{2} \log \left(\frac{y+1}{2} \right) -\frac{y-1}{2} \log \left(\frac{y-1}{2} \right) $, but I am curious about the general case and I think there must be some general rule.
EDIT: I noticed that my proof for converse is a proof in the sense I need if $h''(x)>0$, which is quite easy to understand intuitively.
I rewrote part of the questione because I became convinced that there must be some kind of condition involving $f'(x) - g'(x)$ and $h''(y)$.
Intuitively I would guess something like $\forall x \quad \forall y \quad f'(x) - g'(x) > -h''(y) $, but I was not able to prove anything similar. I hope some real mathematician can help me!