# Under what conditions a rational function has bounded derivative?

Under what conditions a rational function has bounded derivative?

This question arise to me when considering the following theorem:

If $f \in C^1(I,\mathbb{R})$ where $I$ is an interval then:

$f$ is globally lipschitz $\iff \exists L \ge 0.\forall t \in I.|f'(t)| \le L$

So taking rational function $f(x) = \frac{p(x)}{q(x)}$ we have $f'(x) = \frac{p'(x)q(x)-p(x)q'(x)}{q(x)^2}$.

My view

I think I should assume that $f:\mathbb{R} \to \mathbb{R}$ so that $\forall x \in \mathbb{R}.q'(x) \neq 0$ (however this doesn't seem to be necesary). And then perhaps a condition on the degree guarantees boundedness...

• The quotient rule has $q(x)^2$ in the denominator, not $q'$. – user296602 Mar 6 '18 at 20:04
• @user296602 thanks it is corrected now – Javier Mar 6 '18 at 20:05

• 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.
Comment: $f'$ is also a rational function, so this is basically just asking when a rational function is bounded on $\mathbb{R}$. 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 $\mathbb{R}$. 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 $\infty$.