Under what conditions a rational function has bounded derivative?
This question arise to me when c
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 where I is an interval then: f is globally lipschitz
So taking rational function we have . My view I think I should assume that so that (however this doesn't seem to be necesary). And then perhaps a condition on the degree guarantees boundedness...
Answer & Explanation
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, has no non-removable discontinuities (that is, that any zeros of occur as zeros of with at least the same multiplicity). Otherwise, is unbounded at any zeros of and has unbounded derivative. I'll leave it to you to prove that these two conditions are sufficient. Comment: is also a rational function, so this is basically just asking when a rational function is bounded on . Looking in the complex plane, it means that any poles of have to occur off the real axis, so the denominator never vanishes on . Furthermore, 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