# 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 Feb 18 '15 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. – Richard Clare Jul 21 '17 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. – Anonymous196 Oct 1 '17 at 13:45
• @user147263 Is the proof, that $fg$ is uniformly continuous here math.stackexchange.com/questions/2311565/… wrong? – Viktor Glombik Apr 10 '18 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. – Paramanand Singh Nov 19 '19 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. – DanZimm Dec 17 '14 at 10:26
• @DanZimm Ah thanks. Is it okay now? – BCLC Dec 23 '14 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? – Akash Patalwanshi May 17 '17 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 Aug 11 '15 at 18:15