Under what conditions a rational function has bounded derivative? This question arise to me when c

Cara Duke

Cara Duke

Answered question

2022-05-19

Under what conditions a rational function has bounded derivative?
This question arise to me when considering the following theorem:
If f C 1 ( I , R ) where I is an interval then:
f is globally lipschitz
L 0. t I . | f ( t ) | L
So taking rational function f ( x ) = p ( x ) q ( x ) we have f ( x ) = p ( x ) q ( x ) p ( x ) q ( x ) q ( x ) 2 .
My view
I think I should assume that f : R R so that x R . q ( x ) 0 (however this doesn't seem to be necesary). And then perhaps a condition on the degree guarantees boundedness...

Answer & Explanation

coquinarq1

coquinarq1

Beginner2022-05-20Added 14 answers

There are two immediate necessary conditions.
1. The degree of the numerator must be at most one more than the degree of the denominator; otherwise, the function has unbounded derivative at infinity.
2. There cannot be any vertical asymptotes; thus, f has no non-removable discontinuities (that is, that any zeros of q occur as zeros of p with at least the same multiplicity). Otherwise, f is unbounded at any zeros of q and has unbounded derivative.
I'll leave it to you to prove that these two conditions are sufficient.
Comment: f is also a rational function, so this is basically just asking when a rational function is bounded on R . Looking in the complex plane, it means that any poles of f have to occur off the real axis, so the denominator never vanishes on R . Furthermore, f has to be bounded at infinity which means that we cannot have a pole there either.
Also, have a look at the Laurent series at any pole; it cannot have any non-principal part while being bounded at

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