Suppose that $f$ and $g$ are uniformly continuous functions defined on $(a,b)$. Prove that $fg$ is also uniformly continuous on $(a,b)$.
My attempt: Since $f$ is uniformly continuous on $(a,b)$, for all $\epsilon>0$, we have $\delta_f(\epsilon)>0$ such that for all $x,y \in (a,b)$, $|x-y|<\delta_f$, $|f(x)-f(y)|<\epsilon$
Since $g$ is uniformly continuous on $(a,b)$, for all $\epsilon>0$, we have $\delta_g(\epsilon)>0$ such that for all $x,y \in (a,b)$, $|x-y|<\delta_g$, $|g(x)-g(y)|<\epsilon$
Notice that $$|f(x)g(x)-f(y)g(y)|=|f(x)g(x)-f(x)g(y)+f(x)g(y)-f(y)g(y)| \leq |f(x)||g(x)-g(y)| + |g(y)||f(x)-f(y)|$$
Here I don't know how to bound $|f(x)|$ and $|g(y)|$. I have proven that uniformly continuous functions preserve boundedness of an interval , i.e. $f$ is bounded on $(a,b)$. Can anyone help me?
Best Answer
There is a nice way:
Hint: Try to show that if f, g are Uniformly continuous, so are $f \pm g$ and $f^2$. Then observe that $fg = 0.5((f+g)^2 - f^2 -g^2)$. Hope this helps.