Let $f,g:\mathbb{R}^n \rightarrow \mathbb{R}$ be uniformly continuous and bounded functions. Suppose there exists $k \gt 0$ such that $|g(x)| \geq k$ for any $x\in \mathbb{R}^n$. Prove that $\frac{f}{g}$ is uniformly continuous.
I'm trying to come up with an epsilon delta proof.
Let $\epsilon \gt 0$.
By the uniform continuity of $f$ and $g$ we have that
$\exists \ \delta_1$ such that if $\| x – y \| \lt \delta_1$ then $|f(x) – f(y)| \lt \epsilon$
$\exists \ \delta_2$ such that if $\| x – y \| \lt \delta_2$ then $|g(x) – g(y)| \lt \epsilon$
Now I've tried to bound $\big|\frac{f(x)}{g(x)} – \frac{f(y)}{g(y)}\big|$ with no success:
$$\bigg|\frac{f(x)}{g(x)} – \frac{f(y)}{g(y)}\bigg|=\bigg|\frac{f(x)g(y)-f(y)g(x)}{{g(x)g(y)}}\bigg|=\frac{|f(x)g(y)-f(y)g(x)|}{|{g(x)g(y)}|} \leq \frac{|f(x)g(y)-f(y)g(x)|}{k^2}$$
but I don't know how to continue from here.
Best Answer
Using your notations and continuing what you already did, we have
$$\frac{|f(x)g(y)-f(y)g(x)|}{k^2} = \frac{|(f(x)-f(y))g(y)-f(y)(g(x)-g(y))|}{k^2}$$
We then use the triangular inequality to get $$\frac{|(f(x)-f(y))g(y)-f(y)(g(x)-g(y))|}{k^2}\leq \frac{|f(x)-f(y)|}{k^2}|g(y)|+\frac{|g(x)-g(y)|}{k^2}|f(y)|$$
And we use the hypothesis of boundedness and uniform continuity to conclude
$$\frac{|f(x)-f(y)|}{k^2}|g(y)|+\frac{|g(x)-g(y)|}{k^2}|f(y)|\leq \frac{\epsilon}{k^2}\operatorname{sup}|f|+\frac{\epsilon}{k^2}\operatorname{sup}|g|$$
This finishes the proof. If one wants to conclude it with a single $\epsilon$ at the end, then just switch $\epsilon$ with $$\epsilon':=\frac{k^2}{\operatorname{sup}|f|+\operatorname{sup}|g|}\epsilon$$ when using the hypothesis of uniform continuity.