# Real Analysis - Uniform Convergence

• Jan 21st 2009, 01:01 AM
ajj86
Real Analysis - Uniform Convergence
c. Prove that if there exists an M > 0 such that |fn| <= M and |gn| <= M for all n Є N, then (fn*gn) does converge uniformly.

Any help with this would be greatly appreciated. I've been working on it all night and haven't really come up with much of anything.(Worried)
• Jan 21st 2009, 09:39 AM
ThePerfectHacker
Quote:

Originally Posted by ajj86
c. Prove that if there exists an M > 0 such that |fn| <= M and |gn| <= M for all n Є N, then (fn*gn) does converge uniformly.

Any help with this would be greatly appreciated. I've been working on it all night and haven't really come up with much of anything.(Worried)

I guess you are told that $\displaystyle f_n \to f$ and $\displaystyle g_n\to g$ converge uniformly on a set $\displaystyle S$. And you want to show $\displaystyle f_ng_n$ converge uniformly on $\displaystyle S$ too. We will argue that $\displaystyle f_ng_n\to fg$. Thus, we need to show $\displaystyle |f_n(x)g_n(x) - f(x)g(x) | < \epsilon$ for all $\displaystyle x\in S$ if $\displaystyle n\geq N$. To prove this notice that $\displaystyle |f_n(x)g_n(x) - f(x)g(x)| = |f_n(x)g_n(x) - f_n(x)g(x) + f_n(x)g(x) - f(x)g(x)|$.
And apply the triangle inequality. Here you will use the fact that $\displaystyle f_n$ is bounded.

(It appears that we also need to know that $\displaystyle g$ is bounded on $\displaystyle S$. But we do not need to worry since $\displaystyle g_n$ are all bounded it means $\displaystyle g$ is bounded too. )