# product of two uniformly continuous functions is uniformly continuous

Suppose that $f$ and $g$ are uniformly continuous functions defined on $(a,b)$. Prove that $fg$ is also uniformly continuous on $(a,b)$.

My attempt: Since $f$ is uniformly continuous on $(a,b)$, for all $\epsilon>0$, we have $\delta_f(\epsilon)>0$ such that for all $x,y \in (a,b)$, $|x-y|<\delta_f$, $|f(x)-f(y)|<\epsilon$

Since $g$ is uniformly continuous on $(a,b)$, for all $\epsilon>0$, we have $\delta_g(\epsilon)>0$ such that for all $x,y \in (a,b)$, $|x-y|<\delta_g$, $|g(x)-g(y)|<\epsilon$

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

Here I don't know how to bound $|f(x)|$ and $|g(y)|$. I have proven that uniformly continuous functions preserve boundedness of an interval , i.e. $f$ is bounded on $(a,b)$. Can anyone help me?

• Is "product of two uniformly continuous functions is uniformly continuous" correct? Where do you find this conclusion? I can give an counterexample. $f(x)=x$ is uniformly continuous but $f(x)=x^2$ is not! – winston Nov 18 '19 at 8:42

There is a nice way:

Hint: Try to show that if f, g are Uniformly continuous, so are $f \pm g$ and $f^2$. Then observe that $fg = 0.5((f+g)^2 - f^2 -g^2)$. Hope this helps.

• A little odd doubt(not regarding this post): Can we say that "$f^2$ is uniformly continuous $\Rightarrow f$ is uniformly continuous"? – Error 404 Dec 16 '15 at 7:46
• On $[0,1]$, let $f(x)=1$ if $x$ is rational and $-1$ otherwise. Then $f$ is not continuous (let alone uniformly), but $f^2$ is constant!! – Gautam Shenoy Dec 16 '15 at 9:30
• Oh nice!! But If "$f$ is continuous" is also an extra hypothesis along with $f^2$ being uniformly continuous, then can we deduce that $f$ is uniformly continuous? – Error 404 Dec 16 '15 at 11:21
• Have you checked this site for a solution? If you don't find the solution, try posting this question giving your workouts etc. – Gautam Shenoy Dec 16 '15 at 14:28
• Here is the link about my question and my workout... math.stackexchange.com/questions/1576885/… – Error 404 Dec 16 '15 at 15:13

Let $$f$$ and $$g$$ be bounded functions. Hence there are $$c,d\in\mathbf{R}$$ such that $$c,d>0$$, $$\vert f(x) \vert < c$$ and $$\vert g(y) \vert < d$$ for every $$x,y\in(a,b)$$. Let $$\epsilon>0$$.

Since $$f$$ is uniformly continuous on $$(a,b)$$, $$\exists\delta_f(\epsilon)>0$$ such that for all $$x,y \in (a,b)$$, $$|x-y|<\delta_f$$, $$|f(x)-f(y)|<\epsilon/2d$$.

Since $$g$$ is uniformly continuous on $$(a,b)$$, $$\exists\delta_g(\epsilon)>0$$ such that for all $$x,y \in (a,b)$$, $$|x-y|<\delta_g$$, $$|g(x)-g(y)|<\epsilon/2c$$.

Let $$\delta = min\{\delta_f,\delta_g\}$$. Hence, for all $$x,y\in(a,b), \vert x-y \vert<\delta \Rightarrow |g(x)-g(y)|<\frac{\epsilon}{2c}$$ and $$|f(x)-f(y)|<\frac{\epsilon}{2d}$$. Since $$\vert f(x) \vert < c$$ and $$\vert g(y) \vert < d$$ for every $$x,y\in(a,b)$$, it also implies that

$$|f(x)g(x)-f(y)g(y)|=|f(x)g(x)-f(x)g(y)+f(x)g(y)-f(y)g(y)| \leq |f(x)||g(x)-g(y)| + |g(y)||f(x)-f(y)| < c.\frac{\epsilon}{2c} + d.\frac{\epsilon}{2d} = \epsilon$$

Finally we have $$|f(x)g(x)-f(y)g(y)| < \epsilon$$ and $$f.g$$ is uniformly continuous if $$f$$ and $$g$$ are bounded functions.

$f$ and $g$ are continuous on $[a,b]$ hence bounded

try to show :$\lim_{x\rightarrow a+} f(x)$ and $\lim_{x\rightarrow b-} f(x)$ exist as finite limits.

• How do we know $f$ and $g$ are continuous at endpoint? – Idonknow Mar 29 '13 at 7:09
• math.stackexchange.com/questions/241825/… this might help you – jim Mar 29 '13 at 7:19
• We get this from the extreme value theorem right? – Max von Hippel Oct 10 '17 at 1:35

$f$ uniform continuous on $(a,b)$ implies $\exists \varepsilon > 0$ such that $|f(x)-f(y)|< 1$ whenever $|x - y| < \varepsilon$. Pick a $N \in \mathbb{N}$ such that $\frac{b-a}{N} < \varepsilon$, we then have:

$$\min_{i=1 \ldots N-1} f(a + \frac{i}{N})- 1 < f(x) < \max_{i=1 \ldots N-1} f(a + \frac{i}{N}) +1$$ because every $x \in (a,b)$ is at a distance $< \varepsilon$ from one of the $a + \frac{i}{N}, i=1 \ldots N-1$.

For the product of two uniformly continuous functions to be uniformly continuous, the two functions need to be bounded.

• The OP forgot the add the assumption of compactness to the problem statement. – Max von Hippel Oct 10 '17 at 1:35

the product of two uniformly continuous functions is not necessarily uniformly continuous for example $f(x)=x$ and $g(x)= \sin x$ are uniformly continuous on $(0,1)$ but $f\cdot g is not. • That is not quite true. Your example has a$C^1$extension to$\mathbb{R}$and is therefore lipschitz on every bounded set. Thus, it is uniformely continuous as well. – Severin Schraven Jul 24 '17 at 7:52 • can you give a counter example for$f, g$both uniformly continuous on some interval (open. close. half-close), so that$fg$isn't uniformly continuous? – Jneven Jan 7 '19 at 8:09 Multiplication of two uniformly continuous function is not uniformly continuous function. Let f (x)=x ,g (x)=sinx They are both uniformly continuous function on (0,x). But f(x).g(x) is not uniformly continuous function on (0,x) ,for all x\in \Bbb R^+ • The OP is considering functions on a bounded interval$(a,b)\$, as stated. – Mikhail Katz May 10 '17 at 8:20
• Oh. Sorry. Multiplication mistake. .. – gobinda chandra May 10 '17 at 8:22