I've just begun my grad program and we were introduced to this problem in our Analysis I course: consider two uniformly continuous functions $f$ and $g$, say from $\mathbb R$ to $\mathbb R$, where $f$ is bounded. I've learned that the canonical way to disprove that $fg$ is uniformly continuous is countering the claim with $f(x) = \sin(x)$ and $g(x) = x$.
Upon naïve inspection in Desmos, other uniformly continuous bounded functions that are also periodic will also work for $f$ for the counterexample (like the triangle wave; I don't have sufficient math background to prove or disprove that any function like this will work, though). Here's the catch: can a counterxample be built using a non-periodic, uniformly continuous bounded function for $f$? $g$ doesn't need to be $g(x) = x$.
I thought of a very boring possiblity: from my layman understanding, you could "shrink" the graph of $\sin$ after some value $x_0$ so that it's "periodic after $x_0$". You could still make this non-periodic function be uniformly continuous and bounded and $fg$ wouldn't be uniformly continuous. I say it's boring because it still relies on periodicity as a crucial property for the counterxample. Are there any "non-boring examples" or is periodicity always required in one way or another?
You can take$$f(x)=\begin{cases}x-\lfloor x\rfloor&\text{ if }\lfloor x\rfloor\text{ is a perfect square and }x-\lfloor x\rfloor\leqslant\frac12\\1-x+\lfloor x\rfloor&\text{ if }\lfloor x\rfloor\text{ is a perfect square and }x-\lfloor x\rfloor>\frac12\\0&\text{ otherwise}\end{cases}$$(see its graph below) and $g(x)=x$.