Assume that $f_n\to f$ uniformly and $g_n \to g$ uniformly on $S$. Also, assume that each $f_n$ and $g_n$ is bounded on $S$. Prove that $f_n(x)g_n(x)\to f(x)g(x)$ uniformly.

I am a little bit unsure about how to deal with the absolute values. Here's what I got:


The first part is bounded, and I can see that. how do I deal with $|f(x)g_n(x)-g(x)f_n(x)|$? Is it true that $|f(x)g_n(x)-g(x)f_n(x)|\le|f(x)g_n(x)|+|g(x)f_n(x)|<M_fG+M_gF$, where $M_f, M_g$ are bounds for $f$ and $g$, and $G,F$ are bounds for $g_n$ and $f_n$? I feel like I'm missing something...



  1. If $f_n\to f$ uniformly, and $f_n$ is bounded, then $f$ is bounded and $f_n$ is uniformly bounded, that is, $$\exists M>0, |f_n(x)|\leq M \quad \forall n\in \mathbb{N}$$

  2. $$\begin{align} |f_n(x)g_n(x)-f(x)g(x)|&=|f_n(x)g_n(x)-f(x)g_n(x)+f(x)g_n(x)-f(x)g(x)|\\ &=|\left(f_n(x)-f(x)\right)g_n(x)+f(x)\left(g_n(x)-g(x)\right)|\\&\leq |f_n(x)-f(x)||g_n(x)|+|f(x)||g_n(x)-g(x)| \end{align}$$

  • $\begingroup$ How did you get the second equality? Simply by observation? $\endgroup$ – 3x89g2 Nov 4 '14 at 3:23
  • $\begingroup$ @Misakov No. I got the second equality because I intended to do so. That's why we add $-f(x)g_n(x)+f(x)g_n(x)$. We want to group $f_n(x)-f(x)$ and $g_n(x)-g(x)$ and make use of their convergence result. $\endgroup$ – John Nov 4 '14 at 3:27
  • $\begingroup$ I see, seems like my rearrangement is over complicated. $\endgroup$ – 3x89g2 Nov 4 '14 at 3:33

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