Let $f\colon \mathbb{R}^n\to [0,\infty)$ be continuous and radially unbounded, that is, for all $c>0$, there exists $r>0$ such that if $\|x\|>r$, then $f(x)>c.$ Assume that for all $x\in\mathbb{R}^n\backslash\{0\}$, $f(x)>0$. Prove or disprove that there exists a strictly increasing function $g\colon [0,\infty) \to [0,\infty)$ such that

\begin{equation}

\sup_{\|x\|>1}\frac{\|x\|}{g(f(x))}<\infty.

\end{equation}

Here is my attempt: As $\|x\|\to\infty$, if $f(x)$ goes to $\infty$ faster than $\|x\|$ then we can prove the claim easily. So I think the difficulty is the case where $f(x)$ goes to $\infty$ slower than $\|x\|$. For example if $f(x)=\ln(1+\|x\|)$, then $g(y)=e^y$ satisfy the claim. So I think the claim is true because it seems that we can always increase the growth rate of the denominator using an increasing $g$ but not sure how to prove the claim using these ideas. Any help is appreciated.

## Best Answer

Let $h(y) := \sup\{\|z\| : f(z) \le y\}$. The fact that $f$ is radially unbounded ensures that $h(y) < \infty$ for all $y \in [0,\infty)$. Then $h$ is non-decreasing because we are always taking the sup of a larger set, and $h(f(x)) = \sup\{\|z\| : f(z) \le f(x)\} \ge \|x\|$ because we trivially have $f(x) \le f(x)$.

To make the function strictly increasing, let $g(y) := h(y) + |y|$. Then for all $x \in \mathbb{R}^n$, we have \begin{align*} \frac{\|x\|}{g(f(x))} &= \frac{\|x\|}{h(f(x)) + |f(x)|} \le \frac{\|x\|}{\|x\|+ |f(x)|} \le 1 < \infty, \end{align*} so $g$ satisfies the condition in the problem.