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I tried: Here $\sum_{n=0}^{\infty} f_n(x)=f(x)$ that means $S_n(x)=f_1(x)+f_2(x)+\cdots+f_n(x)$ converges to f(x) uniformly. So we only have to prove that \begin{align*} G_n(x)&=f_1(x)\cdot g(x)+f_2(x)\cdot g(x)+\cdots+f_n(x)\cdot g(x)\\ &=(f_1(x)+f_2(x)+\cdots+f_n(x))\cdot g(x)\\ \end{align*} but don't know when this will converge to $f(x)\cdot g(x)$ uniformly. If g is uniformly bounded then it is easy. But if it is continuous then how to proceed don't know. Any help is much appreciated.

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    $\sum_{n=0}^{\infty} f_n(x)=f(x)$ does not generally mean that the partial sums converge *uniformly*. Are you sure that this is what it's supposed to mean in your context? – 2017-02-08
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    It seems to me that it is practically impossible to formulate the exact condition for $g$ under which the series $\sum_{n=0}^{\infty} f_n g$ converges uniformly to $fg$. Assuming that $(f_n)$ are defined on $\Bbb{R}$, if there is $R>0$ such that $f_n$ vanishes outside $[-R, R]$ for all $n$ then the uniform convergence is guaranteed. If this is not the case, I guess I can always find a $g$ that fails the uniform convergence of $\sum_n f_n g$. – 2017-02-08
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    @SangchulLee It is written that if $\sum_{i\in I}f_i=1$, $f_i$ are $C^{\infty}$ functions with compact support and $f$ is another $C^{\infty}$ function with compact support then $\sum_{i\in I}f_i\cdot f=f$. I thought this might be a particular case of what I asked. – 2017-02-08
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    In your case, $f$ is bounded and this does not harm the uniform convergence. – 2017-02-08

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  • If your hypothesis is $\sum f_n$ converges pointwise to $f$, then $\sum f_n g$ converges pointwise to $fg$. That's just because the product is continuous:

$$\sum_{n=0}^N f_n(x) g(x) = \left(\sum_{n=0}^N f_n(x)\right) g(x) \underset{N \to \infty}{\longrightarrow} f(x)g(x)$$

  • If $\sum f_n$ converges uniformly to $f$, then in general $\sum f_n g$ does not converge uniformly to $fg$. It is true however if $g$ is bounded.