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Let $E$ be a subset of $\mathbb{R}$.

Let $\{f_n\}$ be a sequence of functions on $E$ such that $\sum f_n$ converges uniformly on $E$.

Let $\{g_n\}$ be a sequence of functions on $E$ such that $||g||_\infty < M$ and $g_{n+1}(x)\leq g_n(x)$ for all $x\in E$.

How do I prove that $\sum f_ng_n$ converges uniformly on $E$?

If $E$ is compact, then I can prove it by using that $\{g_n\}$ is uniformly convergent.

However, since $E$ is arbitrary, I tried to use summation by parts and prove it directly, but I failed.

How can I prove this?

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  • 2
    $\begingroup$ Summation by parts works if you choose the right way. How did you try it? $\endgroup$ – Daniel Fischer Nov 17 '14 at 12:32
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Define

$$R_k(x) = \sum_{n=k+1}^\infty f_n(x).$$

The uniform convergence of $\sum f_n$ yields the well-definedness of $R_k$, and the uniform convergence $R_k \to 0$ (on $E$). Then we check that the partial sums of $\sum f_n g_n$ are a Cauchy sequence in the supremum norm: For an arbitrary $x\in E$, we have

\begin{align} \left\lvert \sum_{n=k+1}^m f_n(x)g_n(x)\right\rvert &=\left\lvert\sum_{n=k+1}^m (R_{n-1}(x) - R_{n}(x))g_n(x)\right\rvert\\ &= \left\lvert\sum_{n=k}^{m-1} R_{n}(x) g_{n+1}(x) - \sum_{n=k+1}^m R_n(x) g_n(x)\right\rvert\\ &= \left\lvert R_k(x) g_{k+1}(x) - R_m(x) g_m(x) + \sum_{n=k+1}^{m-1} R_n(x) (g_{n+1}(x)-g_n(x))\right\rvert\\ &\leqslant \lVert R_k g_{k+1}\rVert_\infty + \lVert R_m g_m\rVert_\infty + \sum_{n=k+1}^{m-1} \lvert R_n(x)(g_{n+1}(x)-g_n(x))\rvert\\ &\leqslant M\lVert R_k\rVert_\infty + M\lVert R_m\rVert_\infty + \sum_{n=k+1}^{m-1} \lVert R_n\rVert_\infty\cdot \lvert g_{n+1}(x) - g_n(x)\rvert\\ &\leqslant M\lVert R_k\rVert_\infty + M\lVert R_m\rVert_\infty + \left(\sup_{n \geqslant k+1} \lVert R_n\rVert_\infty\right)\sum_{n=k+1}^{m-1} \lvert g_{n+1}(x)-g_n(x)\rvert\\ &= M\lVert R_k\rVert_\infty + M\lVert R_m\rVert_\infty + \left(\sup_{n \geqslant k+1} \lVert R_n\rVert_\infty\right)\lvert g_m(x) - g_{k+1}(x)\rvert\tag{m}\\ &\leqslant M\lVert R_k\rVert_\infty + M\lVert R_m\rVert_\infty + \left(\sup_{n \geqslant k+1} \lVert R_n\rVert_\infty\right)\lVert g_m - g_{k+1}\rVert_\infty\\ &\leqslant 4M \sup_{n \geqslant k} \lVert R_n\rVert_\infty, \end{align}

where in $(\mathrm{m})$, the monotonicity of the sequence $(g_n)$ was used. Thus

$$\left\lVert \sum_{n=k+1}^m f_n g_n\right\rVert_\infty \leqslant 4M \sup_{n\geqslant k} \lVert R_n\rVert_\infty,$$

and the right hand side converges to $0$ for $k\to \infty$, showing that $\sum f_n g_n$ converges uniformly.

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