# If $f$ and $g$ are bounded and uniformly continuous prove that $fg$ is uniformly continuous [duplicate]

Suppose $f(x)$ and $g(x)$ are bounded and uniformly continuous, then we obtain

$|x-y|<\delta \rightarrow |f(x)-f(y)|<\sqrt{M}\sqrt{|x-y|}$

and

$|x-y|<\delta \rightarrow |g(x)-g(y)|<\sqrt{M}\sqrt{|x-y|}$

$|x-y|<\delta \rightarrow |f(x)-f(y)||g(x)-g(y)|<\sqrt{m}\sqrt{|x-y|}\sqrt{m}\sqrt{|x-y|}=M|x-y|$
that satisfy Lipschitz condition, therefore $f(x)g(x)$ is uniformly continuous.
If $h=fg$ , you haven't shown that $h$ is Lipschitz, i.e. that there exists $C$ such that $\vert h(x)-h(y) \vert \leq C\vert x-y\vert$ for any $x,y$. Further, you never define $\delta$ or $M$ in your proof.
We can prove the result using the definition of uniform continuity. Suppose $f,g$ are uniformly continuous on $[a,b]$ and bounded. Let $M$ be the maximum value obtained by either $\vert f \vert$ or $\vert g \vert$. To show that $fg$ is uniformly continuous, let $\varepsilon > 0$. Then there exists $\delta > 0$ such that whenever $\vert x-y\vert < \delta$, we have $$\vert f(x)-f(y)\vert < \frac{\varepsilon/2}{M+1}$$
$$\vert g(x)-g(y)\vert < \frac{\varepsilon/2}{M+1}$$ Then, if $\vert x-y \vert < \delta$, $$\vert f(x)g(x)-f(y)g(y)\vert = \vert f(x)g(x)-f(x)g(y)+f(x)g(y)-f(y)g(y)\vert$$ $$\leq \vert f(x)\vert\vert g(x)-g(y)\vert + \vert g(y)\vert \vert f(x)-f(y)\vert$$ $$< \frac{\varepsilon}{2}\left(\frac{\vert f(x)\vert}{M+1}+\frac{\vert g(x)\vert}{M+1} \right) < \frac{\varepsilon}{2}(1+1)= \varepsilon$$ Thus completing the proof.