What assumptions on $f$ guarantee that $\sup_{|u|\le r} g(u)\in L^{\infty}(\mathbb{R})$?

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a derivable real function and let $$ g(u) =\frac{[(f(u^2))^{\prime}]^2}{2}.$$

I am trying to understand under what hypotheses on $f$ we can conclude that $g$ is a continuous function such that for any $r\ge 0$ it is $$\sup_{|u|\le r} g(u)\in L^{\infty}(\mathbb{R}). $$

Clearly, if $f^{\prime}$ is continuous so it is $g$, but I don't know how to obtain the above assumption on the supremum.

Could someone please help me or give some hints?

Thank you in advance.

Answer 1

In fact $g(u) = 2u^2f'(u)^2$. You can see that \begin{equation} g \in L^{\infty}(\mathbb R) \Leftrightarrow h \in L^{\infty}(\mathbb R). \end{equation} Indeed, if $g\in L^{\infty}(\mathbb R)$, then $h(r) = \sup_{|u|\le r}g(u)\le \left\|g\right\|_{\infty}$ for almost all $r$. Reversly if $h\in L^\infty(\mathbb R)$, $g(u) \le h(|u|+1) \le \left|h\right|_{\infty}$ for almost all $u$.

Using these two the condition that you are looking for is $u\mapsto 2u^2f'(u)^2\in L^\infty\left(\mathbb R\right)$.