two functions are uniformly continuous on some interval I and each is bounded on I then their product is also uniformly continuous on I . [closed]

prove that if two real valued functions are uniformly continuous on some interval $I$ and each is bounded on $I$ then their product $f\cdot g=f(x)\cdot g(x)$ is also uniformly continuous on $I$ . Is boundedness of each function on $I$ is necessary for the product

Hint: Let $x_0\in I$, using the definition of uniform continuity and the boundedness of each function we prove that $fg$ is uniformly continuous at $x_0$ (and then on $I$) by the inequality

$$|f(x)g(x)-f(x_0)g(x_0)|=|f(x)g(x)-f(x)g(x_0)+f(x)g(x_0)-f(x_0)g(x_0)|\\ \leq|f(x)||g(x)-g(x_0)|+|g(x_0)||f(x)-f(x_0)|\\\leq M|g(x)-g(x_0)|+M'|f(x)-f(x_0)|$$

Yes. For example. $f(x)=x$ is uniformly continuous on $\mathbb{R}$, but not bounded. $f^{2}(x)=x^{2}$ is not uniformly continuous on $\mathbb{R}$

Let $f$ be uniformly continuous on a real interval $I$ with endpoints $0$ and $1$.

For any $\epsilon>0$ there is a $\delta>0$ such that for all $x,y\in I$, $|x-y|<\delta\implies |f(x)-f(y)|<\epsilon$.

In particular, by the Archimedean property, there is an $n\in\Bbb Z_+$ such that for all $x,y\in I$, $|x-y|<\frac2n\implies|f(x)-f(y)|<1$. Then for any $x\in I$, $|f(x)-f(0)|<n$. Or something like that.