# If f and g are uniformly continuous then fg is uniformly continuous [duplicate]

Given that $f,g:(a,b) \rightarrow \mathbb{R}$ and both uniformly continuous on $(a,b)$ .... Can we say that $fg$ is uniformly continuous? .... If not please provide a counter-example and if true please provide me a hint

Thank you

• Hint: a uniformly continuous function on $(a,b)$ extends to a continuous function on $[a,b]$. (This takes a little work to prove, so it might be "a sledgehammer to kill a fly", but it will work.) – Ian Nov 7 '16 at 17:28
• you only need to show that $f$ and $g$ are bounded... – user251257 Nov 7 '16 at 17:39
• Omar, this question is $(1)$ a duplicate, and $(2)$ off topic for the reason that you failed to provide context in your question (where did you encounter this question? What can you add in your question that ensures us you've put some effort into answering your own question. I.e., what do you have to say? – Namaste Nov 17 '16 at 0:16
• @amWhy (1) I think your attitude is a bit off topic as I only asked for a hint. I asked no one to do the problem for me. (2) I encountered this problem in my first exam of analysis I at my university. – Omar Qasem Nov 22 '16 at 0:20
• Sorry because it was a duplicate I did in fact look for it but I didn't happen to find it. – Omar Qasem Nov 22 '16 at 0:21

## 1 Answer

If $f$ is uniformly continuous on $(a,b)$ then $f$ is bounded. To see this find a $\delta$ such that $|f(x)-f(y)| < 1$ whenever $|x-y| < \delta$. Then pick some $n$ such that ${b-a\over n} < \delta$ and take the points $x_k = a+ k {b-a\over n}$ for $k=1,....,n-1$. Note that every point in $(a,b)$ is less than $\delta$ away from one of the $x_k$. Then let $B = \max_k |f(x_k)|+1$ and check that $|f(x)| \le B$ for all $x \in (a,b)$.

Hence $f,g$ are bounded by some $B$.

Given $\epsilon>0$. Choose $\delta>0$ such that if $|x-y| < \delta$, then $|f(x)-f(y)| < {1 \over 2B} \epsilon$ and $|f(x)-f(y)| < {1 \over 2B} \epsilon$.

Then \begin{eqnarray} |f(x)g(x)-f(y)g(y)| &\le& |f(x)(g(x)-g(y))| + |g(y)(f(x)-f(y))| \\ &\le& B|g(x)-g(y)| + B|f(x)-f(y)| \\ &<& \epsilon \end{eqnarray}