I was asked to show that given two functions $f:\mathbb{R}\rightarrow \mathbb{R}$ and $g:\mathbb{R}\rightarrow \mathbb{R}$ which are both uniformly continuous, to show that the product $fg:\mathbb{R}\rightarrow \mathbb{R}$ was not always necessarily uniformly continuous. Rather than just give a counter example, I wanted to try showing it directly assuming that it was always uniformly continuous and see where the proof gets hairy or where it seems to fail. Only problem is that I seemed to have accidentally shown myself that it is always true, so I wanted to show everyone so you could show me where I went wrong!
Proof
Let $\{u_n\}$ and $\{v_n\}$ be sequences in $\mathbb{R}$ such that
$\lim_{n \rightarrow \infty } [u_{n} – v_{n}]=0$
If we apply the function to our sequences, then we have
$\lim_{n \rightarrow \infty } [(fg)(u_{n}) – (fg)(v_{n})]$
$\lim_{n \rightarrow \infty } [f(u_{n})g(u_{n}) – f(v_{n})g(v_{n})]$
But since f and g were uniformly continuous, then limit of f(u) = f(v) and limit g(u) =g(v)
So if we let the f's converge to a, and the g's converge to b, then it would seem that using the product rule for limits, we would wind up with
ab – ab which indeed equals zero, and would meet our criterion for uniform continuity.
My guess is that I went wrong because I assumed that $fg(u_n) = f(u_n)g(u_n)$
Best Answer
Try $f(x)=g(x)=x$. For $f$, $g$, we get that $\delta=\varepsilon$, in the $\varepsilon-\delta$ definition of uniform continuity, but for their product $h$, if $x_n=n$ and $y_n=n+1/n$ then $$ y_n-x_n=\frac{1}{n}, \quad\text{while}\quad h(y_n)-h(x_n)=2+\frac{1}{n^2}. $$ Hence $h$ is not uniformly continuous.