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!


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 $fg(u_n) = f(u_n)g(u_n)$

  • 1
    $\begingroup$ What you show is more continuity than uniform continuity, as you do not estimate $|f(x)g(x)-f(y)g(y)|$ in terms of $|x-y|$. As for counterexample, take $f(x)=x$, $g(x)=x$. $\endgroup$ – Milly Oct 22 '14 at 5:21
  • $\begingroup$ "If I attempt to prove $X$, the proof gets hairy" is not a valid method to prove $\neg X$. $\endgroup$ – Hagen von Eitzen Oct 22 '14 at 6:06

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.


By definition $(fg)(u_n)=f(u_n)g(u_n)$; that’s not the problem. The problem is that you can have $\lim_{n\to\infty}(u_n-v_n)=0$ even when the sequences $\langle u_n:n\in\Bbb N\rangle$ and $\langle v_n:n\in\Bbb N\rangle$ don’t converge. (For instance, you might have $u_n=n$ and $v_n=n+2^{-n}$.) When you wrote this:

But since f and g were uniformly continuous, then limit of f(u) = f(v) and limit g(u) =g(v)

you were implicitly assuming that these sequences were convergent to real numbers $u$ and $v$, respectively.


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