Real Analysis - Uniform Convergence

**ajj86** · January 21st 2009, 01:01 AM

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.

**ThePerfectHacker** · January 21st 2009, 09:39 AM

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

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

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