If $f,g$ are uniformly continuous prove $f+g$ is uniformly continuous but $fg$ and $\dfrac{f}{g}$ are not

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)|

$$\bullet$$ 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.

• The proof for $f+g$ is correct. Neither $fg$ nor $f/g$ are uniformly continuous in general.
– user147263
Commented Feb 18, 2015 at 2:44
• 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. Commented Jul 21, 2017 at 20:14
• 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. Commented Oct 1, 2017 at 13:45
• @user147263 Is the proof, that $fg$ is uniformly continuous here math.stackexchange.com/questions/2311565/… wrong? Commented Apr 10, 2018 at 10:36
• @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. Commented Nov 19, 2019 at 13:56

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.

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?

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

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 $[1,\infty)$.

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

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.

• assuming fg is uniformly continuous?
– BCLC
Commented Aug 11, 2015 at 18:15