Is it true that if $\sum_{n=0}^{\infty} f_n(x)=f(x)$, then $\sum_{n=0}^{\infty} f_n(x)\cdot g(x)=f(x)\cdot g(x)$

Question

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.

Answer 1

• 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.