# Multiplication of uniformly continuous functions

• Apr 3rd 2011, 04:45 AM
moses
Multiplication of uniformly continuous functions
I'm trying to prove:

If $\displaystyle f$:$\displaystyle A --> R$ and $\displaystyle g$:$\displaystyle A --> R$ are uniformly continuous functions, then $\displaystyle f * g$ is uniformly continuous.

I've gotten to the fact that
|f(x)g(x) - f(y)g(y)| <= |f(x)| * |g(x)-g(y)| + |g(y)| * |f(x) - f(y)|

But I'm not sure where to go from here...
• Apr 3rd 2011, 07:02 AM
TheEmptySet
Quote:

Originally Posted by moses
I'm trying to prove:

If $\displaystyle f$:$\displaystyle A --> R$ and $\displaystyle g$:$\displaystyle A --> R$ are uniformly continuous functions, then $\displaystyle f * g$ is uniformly continuous.

I've gotten to the fact that
|f(x)g(x) - f(y)g(y)| <= |f(x)| * |g(x)-g(y)| + |g(y)| * |f(x) - f(y)|

But I'm not sure where to go from here...

We really need some more information to solve this problem. We need some information about the set $\displaystyle A$.

For example let $\displaystyle f(x)=g(x)=x$ both of these are uniformly continuous on the set $\displaystyle [0,\infty)$ let $\displaystyle \delta =\epsilon$

Now consider their prouduct $\displaystyle h:[0,\infty) \to \mathbb{R},h(x)=f(x)g(x)=x^2$

on this unbounded set this is not uniformly continuous.

So if A is compact. The proof is easy! because continuous functions on compact sets are bounded.

If A is an open interval $\displaystyle (a,b)$ the function is uniformly continuous if and only if it can be continuously extended to the closed compact set $\displaystyle [a,b]$