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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?

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  • $\begingroup$ 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! $\endgroup$
    – winston
    Nov 18, 2019 at 8:42

7 Answers 7

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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.

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  • $\begingroup$ A little odd doubt(not regarding this post): Can we say that "$f^2$ is uniformly continuous $\Rightarrow f$ is uniformly continuous"? $\endgroup$
    – Error 404
    Dec 16, 2015 at 7:46
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    $\begingroup$ 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!! $\endgroup$ Dec 16, 2015 at 9:30
  • $\begingroup$ 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? $\endgroup$
    – Error 404
    Dec 16, 2015 at 11:21
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    $\begingroup$ Have you checked this site for a solution? If you don't find the solution, try posting this question giving your workouts etc. $\endgroup$ Dec 16, 2015 at 14:28
  • $\begingroup$ Here is the link about my question and my workout... math.stackexchange.com/questions/1576885/… $\endgroup$
    – Error 404
    Dec 16, 2015 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.

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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.

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  • $\begingroup$ How do we know $f$ and $g$ are continuous at endpoint? $\endgroup$
    – Idonknow
    Mar 29, 2013 at 7:09
  • $\begingroup$ math.stackexchange.com/questions/241825/… this might help you $\endgroup$
    – jim
    Mar 29, 2013 at 7:19
  • $\begingroup$ We get this from the extreme value theorem right? $\endgroup$ Oct 10, 2017 at 1:35
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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 $a + \frac{i}{N}, i=1 \ldots N-1$.

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

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  • $\begingroup$ The OP forgot the add the assumption of compactness to the problem statement. $\endgroup$ Oct 10, 2017 at 1:35
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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.

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    $\begingroup$ 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. $\endgroup$ Jul 24, 2017 at 7:52
  • $\begingroup$ 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? $\endgroup$
    – Jneven
    Jan 7, 2019 at 8:09
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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^+

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  • $\begingroup$ The OP is considering functions on a bounded interval $(a,b)$, as stated. $\endgroup$ May 10, 2017 at 8:20
  • $\begingroup$ Oh. Sorry. Multiplication mistake. .. $\endgroup$ May 10, 2017 at 8:22

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