product of two uniformly continuous functions is uniformly continuous

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?

Solution 1:

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.

Solution 2:

Let $f$ and $g$ be bounded functions. Hence there are $c,d\in\mathbf{R}$ such that $c,d>0$, $\vert f(x) \vert < c$ and $\vert g(y) \vert < d$ for every $x,y\in(a,b)$. Let $\epsilon>0$.

Since $f$ is uniformly continuous on $(a,b)$, $\exists\delta_f(\epsilon)>0$ such that for all $x,y \in (a,b)$, $|x-y|<\delta_f$, $|f(x)-f(y)|<\epsilon/2d$.

Since $g$ is uniformly continuous on $(a,b)$, $\exists\delta_g(\epsilon)>0$ such that for all $x,y \in (a,b)$, $|x-y|<\delta_g$, $|g(x)-g(y)|<\epsilon/2c$.

Let $\delta = min\{\delta_f,\delta_g\}$. Hence, for all $x,y\in(a,b), \vert x-y \vert<\delta \Rightarrow |g(x)-g(y)|<\frac{\epsilon}{2c}$ and $|f(x)-f(y)|<\frac{\epsilon}{2d}$. Since $\vert f(x) \vert < c$ and $\vert g(y) \vert < d$ for every $x,y\in(a,b)$, it also implies that

$$|f(x)g(x)-f(y)g(y)|=|f(x)g(x)-f(x)g(y)+f(x)g(y)-f(y)g(y)| \leq |f(x)||g(x)-g(y)| + |g(y)||f(x)-f(y)| < c.\frac{\epsilon}{2c} + d.\frac{\epsilon}{2d} = \epsilon$$

Finally we have $|f(x)g(x)-f(y)g(y)| < \epsilon$ and $f.g$ is uniformly continuous if $f$ and $g$ are bounded functions.