Suppose $f:\mathbb{R} \supset E \rightarrow \mathbb{R}$ and $g: \mathbb{R} \supset E \rightarrow \mathbb{R}$ are uniformly continuous. Show that $f+g$ is uniformly continuous. What about $fg$ and $\dfrac{f}{g}$?

My Attempt

Firstly let's state the definition; a function is uniformly continuous if
$$\forall \varepsilon >0\ \ \exists \ \ \delta >0 \ \ \text{such that} \ \ |f(x)-f(y)|< \varepsilon \ \ \forall \ \ x,y \in \mathbb{R} \ \ \text{such that} \ \ |x-y|<\delta$$

Sum $f+g$

Now to to prove $f+g$ is uniformly continuous;
$\bullet$ Choose $\delta_1 >0$ such that $\forall$ $x,y \in \mathbb{R}$ $|x-y|<\delta_1$ $\implies$ $|f(x)-f(y)|< \dfrac{\epsilon}{2}$

$\bullet$ Choose $\delta_2 >0$ such that $\forall$ $x,y \in \mathbb{R}$ $|x-y|<\delta_2$ $\implies$ $|g(x)-g(y)|< \dfrac{\varepsilon}{2}$

$\bullet$ Now take $\delta := min\{ \delta_1, \delta_2\}$ Then we obtain for all $x,y \in \mathbb{R}$ $$ |x-y|<\delta \implies |f(x)+g(x)-f(y)+g(y)| < |f(x)-f(y)| + |g(x)-g(y)| < \dfrac{\varepsilon}{2}+\dfrac{\varepsilon}{2}= \varepsilon$$

Product $fg$

Now for $fg$ for this to hold both $f:E \rightarrow \mathbb{R}$ and $g:E \rightarrow \mathbb{R}$ must be bounded , if not it doesn't hold. $\bullet$ $\exists \ \ M>0 \ \ such \ that \ \ |f(x)|<M \ \ and \ \ |g(x)|<M \ \ \forall \ x \in E$

$\bullet$ Choose $\delta_1 >0$ such that $\forall$ $x,y \in \mathbb{R}$ $|x-y|<\delta_1$ $\implies$ $|f(x)-f(y)|< \dfrac{\epsilon}{2M}$

$\bullet$ Choose $\delta_2 >0$ such that $\forall$ $x,y \in \mathbb{R}$ $|x-y|<\delta_2$ $\implies$ $|g(x)-g(y)|< \dfrac{\epsilon}{2M}$

$\bullet$ Now take $\delta := min\{ \delta_1, \delta_2\}$. Then, $|x-y|<\delta$ implies for all $x,y \in \mathbb{R}$, that $$|f(x)g(x)-f(y)g(y)| \leq |g(x)||f(x)+f(y)|+|f(y)||g(x)+g(y)| \leq $$ $$ M|f(x)+f(y)| + M|g(x)+g(y)| < M \dfrac{\epsilon}{2M} + M \dfrac{\epsilon}{2M} = \epsilon$$

Are these proofs correct? I am not sure how to approach the $\dfrac{f}{g}$ case.

  • 5
    $\begingroup$ The proof for $f+g$ is correct. Neither $fg$ nor $f/g$ are uniformly continuous in general. $\endgroup$
    – user147263
    Feb 18, 2015 at 2:44
  • $\begingroup$ Your proof for the product fg is almost correct. You only missed one point.In the definition of bounded function, M can be 0 (example the constant function 0 is bounded by 0). If $M = 0, f = g = 0$ and the function $f(x)g(x) = 0$ is uniformly continuous. Then you can divide by $M$, assuming that $M \neq 0$. It is a suttle point, but important one. But anyway if 0 is the upper bound, we can find another M' = M + 1 such that is also an upper bound and your proof is correct. $\endgroup$ Jul 21, 2017 at 20:14
  • $\begingroup$ I think you mean if the functions $f$ and $g$ are continuous in a certain point $a \in \mathbb R$. This rigorous way of defining things gives you a clear idea on how to prove the statement in much finer detail. $\endgroup$ Oct 1, 2017 at 13:45
  • $\begingroup$ @user147263 Is the proof, that $fg$ is uniformly continuous here math.stackexchange.com/questions/2311565/… wrong? $\endgroup$
    – Ramanujan
    Apr 10, 2018 at 10:36
  • $\begingroup$ @ViktorGlombik: the proof in linked answer deals with closed and bounded intervals which imply that functions involved are bounded. One loses uniform continuity in case one deals with unbounded functions. $\endgroup$
    – Paramanand Singh
    Nov 19, 2019 at 13:56

4 Answers 4


A product of uniformly continuous functions is not necessarily uniformly continuous. For example, set $E=\mathbb{R}$ and choose $f(x)=g(x)=x$. Their product, $x^2$, is an example of a nonuniformlycontinuous function.


Product $fg$

Boundnness is suffient but not necessary for uniform continuity of product $fg$.

Here is an example where $f,g$ both are unbounded but their product is uniformly continuous: Let $E=[0,\infty)$ and $f(x)=g(x)=\sqrt{x}$. Here $f$ and $g$ are unbounded but their product $(fg)(x)=x$ is uniformly continuous.

$\frac fg$:
It is not true that $\frac fg$ is always uniformly continuous. Let $E=[1,\infty)$,$f(x)=x$ and $g(x)=\frac{1}{x}$ then $\left(\frac fg\right)(x)=x^2$ is not uniformly continuous on $[1,\infty)$.

  • $\begingroup$ so under what condition $f/g$ is uniformly continuous? $\endgroup$ May 17, 2017 at 12:33

Yeah, but I think you're missing a step here:

$|f(x)g(x)-f(y)g(y)| \leq |g(x)||f(x)+f(y)|+|f(y)||g(x)+g(y)| $

I think you meant

$|f(x)g(x)-f(y)g(y)| = |f(x)g(x)-f(y)g(x)+f(y)g(x)-f(y)g(y)| \leq |g(x)||f(x)-f(y)|+|f(y)||g(x)-g(y)| \leq |g(x)||f(x)+f(y)|+|f(y)||g(x)+g(y)| $

or something like that?

  • 2
    $\begingroup$ I think you want minuses where you have $f(x) + f(y)$ and so on. $\endgroup$
    – DanZimm
    Dec 17, 2014 at 10:26
  • $\begingroup$ @DanZimm Ah thanks. Is it okay now? $\endgroup$
    – BCLC
    Dec 23, 2014 at 4:53

As BCLC noted, how you got $$|f(x)g(x)-f(y)g(y)| \leq |g(x)||f(x)+f(y)|+|f(y)||g(x)+g(y)|$$ wasn't very clear. For the case $\frac{f}{g}$, you could note that $$\frac{f}{g} = f \cdot \frac{1}{g}$$ and use the previous part.

  • $\begingroup$ assuming fg is uniformly continuous? $\endgroup$
    – BCLC
    Aug 11, 2015 at 18:15

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