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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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up vote 6 down vote accepted

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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A little odd doubt(not regarding this post): Can we say that "$f^2$ is uniformly continuous $\Rightarrow f$ is uniformly continuous"? – Vikrant Desai 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? – Vikrant Desai 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...… – Vikrant Desai Dec 16 '15 at 15:13

$f$ 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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How do we know $f$ and $g$ are continuous at endpoint? – Idonknow Mar 29 '13 at 7:09… this might help you – jim Mar 29 '13 at 7:19

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

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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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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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