Thread: Real Analysis - Uniform Convergence

1. 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.

2. 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.
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. )