Exercise 6.2.8

Let’s examine the difference

We’d like to bound the rightmost term.

Theorem 6.2.6 implies $g$ is continuous, and Theorem 4.4.1 implies $g(K)$ is compact, hence $|g(K)|$ has a minimum, call it $m$. This allows us to bound $|1∕g(x)|<1∕m$.

To bound $1∕g_n$ set $𝜖$ small enough that $m-𝜖>0$ then use uniform continuity to get $N$ such that $n\ge N$ has

Since $g_n(x)\in (g(x)-𝜖,g(x)+𝜖)$ we have $|g_n(x)|>|g(x)|-𝜖$ and finally $|g(x)|>m$ implies $|g_n(x)|>m-𝜖$ thus $|1∕g_n(x)|<1∕(m-𝜖)$ and so

Given an $𝜖$, setting $N$ big enough to make $|g(x)-g_n(x)|<M∕𝜖$ gives the desired result.