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Let $\sum|g_n|$ converge uniformly in $X$. Suppose there exists $K$ such that $|f_n(x)|\leq K$, for all $n\in\mathbb{N}$ and all $x\in X$. Prove that $\sum f_ng_n$ converges absolutely and uniformly in $X$.

How can I do this problem, it's from a book that I'm using of real analysis, can someone help me with the solution D:? It use a trick or something?

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You have two uses of $k$ in the question. In $\sum f_kg_k$ if you use $|f_n(x)|\leq k$ to replace $f_k$ by $k$ –  Ross Millikan Nov 17 '11 at 5:55
@t.b.: I have satisfied my commitment with an upvote. I was happy to do so. Not that 10 points matters to you. Good answer. –  Ross Millikan Nov 17 '11 at 6:22
@Ross: Thanks!${}{}$ –  t.b. Nov 17 '11 at 6:24
I was wondering if you could tell me what's missing in my answer. Also, I noticed that you haven't accepted any answers recently. Please tell the answerers what you expect more or do accept their answers if they were helpful. –  t.b. Jan 5 '12 at 18:49

1 Answer 1

You can simply estimate $$ \left| \sum_{n=1}^\infty \phantom{|} f_n g_n\right| \leq \sum_{n=1}^\infty \phantom{|}|f_n|\,|g_n| \leq K \cdot \sum_{n=1}^\infty\phantom{|} |g_n|. $$ This gives you that $h(x) = \sum_{n=1}^\infty f_n(x)g_n(x)$ is well-defined for all $x$ and that the sum converges absolutely and uniformly.

More detail for uniform convergence: By hypothesis there is for each $\varepsilon \gt 0$ an $N$ such that for all $x$ we have $\sum_{n = N+1}^\infty |g_n(x)| \lt \varepsilon/K$. Now for all $x$ and all $M \geq N$ this gives $$ \left|h(x) - \sum_{n = 1}^{M} \phantom{|} f_n(x) g_n(x)\right| \leq \sum_{n=N+1}^\infty \phantom{|}|f_n(x)|\,|g_n(x)| \leq K \cdot \sum_{n = N+1}^\infty \phantom{|}|g_n(x)| \lt \varepsilon, $$ which shows uniform convergence of the series.

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