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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?

Martin Sleziak
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Idonknow
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    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 08:42
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    @winston Domain is different – Messi Lio Dec 14 '22 at 05:38

7 Answers7

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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.

Gautam Shenoy
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  • 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 07:46
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    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 09:30
  • 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
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    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
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    Here is the link about my question and my workout... http://math.stackexchange.com/questions/1576885/let-f-bbb-r-to-0-infty-be-a-continuous-function-such-that-gx-fx2?noredirect=1#comment3212845_1576885 – Error 404 Dec 16 '15 at 15:13
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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.

Mathronaut
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Lucas Camargo
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3

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

Sohom
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$f$ 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.

jim
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$f$ 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 $a + \frac{i}{N}, i=1 \ldots N-1$.

achille hui
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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.

user88595
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sudhanshu aggarwal
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    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 07:52
  • 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? – Noa Even Jan 07 '19 at 08:09
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    @Jneven I think we can take $f = g = x$ on $[1, \infty)$, and we know $fg = x^2$ is not uniformly continuous on $[1, \infty)$. – Newton Apr 04 '23 at 20:25
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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^+

gobinda chandra
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