I came with this question while trying to get a counterexample for the statement:

if $f$ and $g$ are uniformly continuous functions, $g$ is bounded and $f$ is not necessarily bounded then $fg$ is uniformly continuous.

That is a common question in analysis books and is, indeed, already answered in this forum. But the answers, both in books and in the forum, are always the same( or slight variations of it):

$f(x) = x$ and $g(x) = sin\,x$

So, are there any other counterexamples? In others terms, can we rephrase the statement imposing that $g$ "doesn't look like" $sin\,x$ to make it true?

An example, that I have no evidence of being true, would be:

if $f$ and $g$ are uniformly continuous functions, $g$ is bounded and not periodic, $f$ is not necessarily bounded. Then $fg$ is uniformly continuous.

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    $\begingroup$ I posted the question that product of $2$ uniform continous functions is again uniform continuous in the link below. math.stackexchange.com/questions/345480/… $\endgroup$ – Idonknow Mar 10 '18 at 15:59
  • $\begingroup$ @Idonknow The functions are over $(a,b)$, not over $\mathbb R$ in that question. A uniformly continuous function over $(a,b)$ is bounded, so the issue asked about here is not relevant in that case. $\endgroup$ – Milo Brandt Mar 10 '18 at 16:09

I don't think there is really any easy way to add hypothesis to make this true. The problem is that, if $f$ is unbounded, then where $f$ is large, a small change in $g$ can cause a large change in $fg$. That is, the sensitivity of the product $fg$ to perturbations in $g$ just keeps getting bigger - but the control offered by uniform continuity cannot prevent $g$ from changing fast enough to make $fg$ not uniformly continuous.

We can make this theorem fail even with increasing functions. Let $f(x)=x$. Now, we will define a function $g$ which is constant on most intervals, but occasionally increases a small amount in an interval of length one. For convenience, define the following function: $$\Lambda(x)=\begin{cases}0 & \text{if }x\leq 0 \\ x & \text{if }0\leq x \leq 1 \\ 1 &\text{if }x\geq 1. \end{cases}$$ Then, define $$g(x)=\sum_{n=0}^{\infty}\frac{1}{2^n}\Lambda(x-4^n).$$ Observe that $g(4^n+1)-g(4^n)=\frac{1}{2^n}$. Also note that $g(x)\in [0,1)$ for every $x$. Since $f$ is increasing, we have the following inequality: $$f(4^n+1)g(4^n+1)-f(4^n)g(4^n) > f(4^n)(g(4^n+1)-g(4^n)) = 4^n\cdot \frac{1}{2^n}=2^n.$$ This implies that $fg$ is not uniformly continuous, because a uniformly continuous function would have to have the property that $\sup |f(x)-f(x+c)|$ is finite for every $c$.

  • $\begingroup$ Sorry if is somewhat trivial, but how do we know that $g$ is uniformly continuous? it is intuitive as the inclination is getting smaller as we move to infinite but that also happens for $fg$ ( or does not?) $\endgroup$ – Nícolas Pinto Mar 10 '18 at 20:08
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    $\begingroup$ @NícolasPinto There are a few ways to see that. A general one is that every increasing bounded continuous function is uniformly continuous. A more specific way is to note that $g$ is piecewise linear, with slope never exceeding $1$ in absolute value, thus is Lipschitz, thus is uniformly continuous. Lastly, one can note that, since we are summing copies of the same function that are just translated, the supremum of values $|g(x)-g(x+c)|$ for $|c|<\delta$ is no more than $2=\sum_{n=0}^{\infty}\frac{1}{2^n}$ times the same supremum for $\Lambda$, which gives a direct proof of uniform continuit $\endgroup$ – Milo Brandt Mar 10 '18 at 23:14
  • $\begingroup$ Thanks for the detailed explanation! $\endgroup$ – Nícolas Pinto Mar 11 '18 at 2:29

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