Continuous map on $\mathbb{R}^d$ We consider a continuous map $g$ on $\mathbb{R}^d$  if and only if $x$ belongs to $K$  where $K$ is a compact set.
I want to prove that for all $\epsilon_1>0$ there is  $\epsilon_2>0$ such that \begin{align*}
|x|\le2,\;\;g(x)\ge \epsilon_1\Rightarrow |g(x)|\ge \epsilon_2\;.
\end{align*}
Please help me to do so. Thanks in advance.
 A: First an example to illustrate why the additional restrictions on $x$ are needed:
Let $K=\{0\}$ and $g(x) = {|x| \over 1+x^2}$. Note that $K= g^{-1}(\{0\})$. However,
for any $\epsilon>0$, it is clear that $\inf_{|x|\ge \epsilon} |g(x)| = 0$. (Note
that $d(x,K) = |x|$.)
Note that the function $\phi(x) = d(x,K)$ is continuous, hence for any $\epsilon_1 >0$, the
set $C_{\epsilon_1}=\{ x | d(x,K) \ge \epsilon_1, {3 \over 4} \le |x| \le {8 \over 3} \}$
is closed and bounded and so is compact. Furthermore, $K \cap C_{\epsilon_1}= \emptyset$.
Hence either $C_{\epsilon_1}$ is empty, in which case the statement is vacuously true, or $\min_{x \in C_{\epsilon_1}} |g(x)| >0$, in which case we can find a
suitable $\epsilon_2>0$.
A: Assume to the contrary that there is an $\epsilon_1>0$ such that for all $\epsilon_2>0$ there is an $x\in\mathbb{R}^d$ with $3/4\le|x|\le8/3, d(x,K)\ge\epsilon_1$ and $|g(x)|<\epsilon_2$. By picking $1/n$ as $\epsilon_2$ for each $n\in\mathbb{N}$, we obtain a sequence $(x_n)_n$ such that
$$3/4\le|x_n|\le8/3,\ d(x_n,K)\ge\epsilon_1,\ |g(x_n)|<1/n.$$
In particular, $g(x_n)\rightarrow0$. By the first condition, $(x_n)_n$ is bounded, so by the Bolzano-Weierstraß Theorem, it contains a convergent subsequence $(x_{n_k})_k\rightarrow x$. Clearly, we still have $g(x_{n_k})\rightarrow0$, but $x_{n_k}\rightarrow x$, so, by continuity of $g$, $g(x)=0$, whence $x\in K$. However,
$$d(x,K)=\lim_{k\rightarrow\infty}d(x_{n_k},K)\ge\epsilon_1,$$
so we achieve a contradiction and conclude. Note that the lower bound $3/4\le|x|$ was superfluous and the argument still goes through as long as $|x|<M$ for some upper bound $M$.
A: Suppose not. Then, there is an $\epsilon>0$ and a sequence $(x_n)$ such that $3/4\le|x_n|\le8/3,\ d(x_n,K)\ge \epsilon$ and $g(x_n)<1/n.$ Since $K_1=\{x: 3/4\le|x|\le8/3\}$ is compact, $(x_n)$ has a convergent subsequence, which we still call $(x_n)$ for convenience. Then, $x_n\to x \in K_1$. But $g$ is continuous so $g(x)=0$. That is, $x\in K$, which is a contradiction. 
