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.

  • 2
    $\begingroup$ You can write $f(x)g(y)-f(y)g(x)$ as $(f(x)-f(y))g(y)-(g(x)-g(y))f(y)$. This trick is used for instance when proving that the product of two continuous functions is again continuous. $\endgroup$
    – Suzet
    Jul 29, 2018 at 23:08
  • $\begingroup$ @Suzet great I think I got it. Write it as an answer if you want so I can accept it. $\endgroup$
    – Yagger
    Jul 29, 2018 at 23:33
  • $\begingroup$ I'm writing it right now :) $\endgroup$
    – Suzet
    Jul 30, 2018 at 0:09

1 Answer 1


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.


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