Winter 1993

Problem Statement: Suppose that \{f_n\} converges to f uniformly and \{g_n\} converges to g uniformly on [a,b] where each f_n,g_n is continuous on [a,b]. Prove that f_ng_n converges uniformly to fg on [a,b].

Proof: Since [a,b] is a compact set and f_n,g_n are each continuous on [a,b] it follows that each f_n,g_n is bounded on [a,b]. So, there exist M,P\in\mathbb{R} such that |f_n(x)|\leq M and |g_n(x)|\leq P for every x\in [a,b] and for every n\in\mathbb{N}.

Let \varepsilon>0. Then there exists N_1\in\mathbb{N} such that |f_n(x)-f(x)|<\dfrac{\varepsilon}{2P} for every n\geq N_1 and for every x\in [a,b] since f_n\rightarrow f uniformly. Similarly, there exists N_2\in\mathbb{N} such that |g_n(x)-g(x)|<\dfrac{\varepsilon}{2M} for every n\geq N_2 and for every x\in [a,b].

Let N=max\{N_1,N_2\}. Then for n\geq N and x\in [a,b] it follows that

|f_ng_n(x)-fg(x)|=|(f_n(x)-f(x))g_n(x)+f(x)(g_n(x)-g(x))|

\leq |f_n(x)-f(x)|P+M|g_n(x)-g(x)| for every x\in [a,b]

<\dfrac{\varepsilon}{2P}P+\dfrac{\varepsilon}{2M}M

=\dfrac{\varepsilon}{2}+\dfrac{\varepsilon}{2}=\varepsilon

Thus, f_ng_n converges to fg uniformly on [a,b].

\Box

Reflection: This one really followed straight from the definitions. The key being in that we knew each of the functions was bounded. Also, we used a little bit of “math magic” to make the inside of the absolute values look like things we knew about. The thing to keep in mind when doing all of these problem is what you know information about. You can never make a conclusion about some inequality if you don’t have any information about it!

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This entry was posted in Analysis, Continuity, Math, Sequence of Functions, Uniform Convergence. Bookmark the permalink.

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