derivation is bounded

Question

Let $f\in C^1([a,b])$, and let $p>1$. Assume that there exist constants $A, B\ge0$ such that $A|f'(x)|^p\le B|f'(x)|$ for all $x\in (a,b)$.

How to prove that there exists a constant $C\ge 0$ such that $|f'(x)|\le C$ for all $x\in (a,b)$?

Answer 1

You have $|f(x)'|^{p-1}\leq B/A$ this is equivalent to saying that $exp((p-1)ln(|f'(x)|))\leq B/A$ we deduce that $(p-1)ln(|f'(x)|)\leq ln(B/A)$ and $|f'(x)|\leq exp({{ln(B/A)}\over{p-1}}).$

Answer 2

Let $x\in (a,b) $. Wlog, we can suopose that $$A>0 ; B>0; | f'(x)|>0.$$

then $$|f'(x)|^{p-1}\le \frac {B}{A} $$ and $$|f'(x)|\le (\frac {B}{A})^\frac{1}{p-1}$$

hence, $C=.. $