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  • December 6, 2023
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I stumbled upon a theorem today that I feel like I’ve needed in the past, though I can’t remember any particular applications. I’m writing it up here as a note to my future self should the need reappear.

The theorem gives sufficient conditions to conclude

f(g(x)) ≤ g(f(x))

and uses this to prove, for example, that

arcsin( sinh(x/2) ) ≤ sinh( arcsin(x)/2 )

on the interval [0, 1].

If you think of any applications, please leave a comment.

Here’s the theorem, found in [1].

Let f be continuous with domain 0 ≤ x < 1 or 0 ≤ x ≤ 1, f(0) = 0, f(1) > 1 (including the possibility that f(1) = +∞); let g be continuous with domain the range of f, and g(1) ≤ 1. Let f(x)/x and g(x)/x be strictly increasing on their domains. Finally let f(x) ≠ x for 0 < x < 1. Then f(g(x)) ≤ g(f(x)) for 0 < x < 1.

[1] Ralph P. Boas. Inequalities for a Collection. Mathematics Magazine, January 1979, Vol. 52, No. 1, pp. 28–31

The post f(g(x)) versus g(f(x)) first appeared on John D. Cook.

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