# Show that there exists a strictly increasing function $g$ such that $\sup_{\|x\|>1}\frac{\|x\|}{g(f(x))}<\infty$

functionslimitsmultivariable-calculusreal-analysissupremum-and-infimum

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

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.