Preimage under a continuous positive definite function

Question

Consider a continuous positive definite function: $$ f:D \rightarrow \mathbb{R}_+\\ f(0) = 0\\ \forall x \in D \setminus \{0\}: f(x) > 0 $$ where $0 \in D \subseteq \mathbb{R}^n$ and $D$ is unbounded.

I want to show that $$ \exists \epsilon > 0: \exists r> 0: \forall x \in D:(f(x)=\epsilon \Rightarrow \Vert x \Vert \leq r) $$ which says that there exists $\epsilon > 0$ such that the preimage $f^{-1}[\epsilon]$ is bounded. This is a seemingly true statement but I don't know how to show that. The continuity gives that $f^{-1}[\epsilon]$ is closed in $D$ for all $\epsilon > 0$ but can we say that it is also bounded under the above conditions?

Answer 1

You can do this in 1D. Start with $y=x^2$, then cut it off by multiplying by $e^{-x^2}$. Then add $x^2\sin^2 x$. Voila. The function $f(x)=x^2e^{-x^2}+x^2\sin^2(x)$ is clearly non-negative, and is only zero at $x=0$. But for any $\varepsilon>0$ there there will be $N$ large enough that for any $x>N$ we have $x^2e^{-x^2}<\varepsilon$, while $x^2>\varepsilon$, so that the function will cross below and above $\varepsilon$ with every oscillation (i.e., at $x= \pi k$ the value is $x^2e^{-x^2}< \varepsilon$, at $x=\pi/2+\pi k$ the value is at least $x^2>\varepsilon$).

Answer 2

The statement is not true. Let us give a counterexample in ${\mathbb R}^2$ by defining a function in polar coordinates \begin{equation} f(x, y) = f(r \cos\theta, r\sin\theta) = \lambda(r) a(\theta) + \mu(r) \end{equation} Let us choose $a(\theta) = 2 + \sin \theta$, so that $a(\theta)$ takes any value in $[1, 3]$ when $\theta$ varies in $(-\pi, +\pi]$. Let us choose $\lambda(r)$ and $\mu(r)$ such that \begin{equation} \begin{array}\cr \lambda(r) + \mu(r) = \displaystyle\frac{r}{1 +r^2}\cr 3 \lambda(r) +\mu(r) = r(1 + r^2) \end{array} \end{equation} On the circle $C(0, r)$, $f$ takes all the values in $[\frac{r}{1+r^2}, r(1+r^2)]$, hence if $r$ is large enough, any given value $\epsilon > 0$ is reached on that circle.