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  • $(f_n)$ converges pointwise to $f$ on $A$
  • $(g_n)$ converges pointwise to $g$ on $A$

Prove $(f_n g_n)$ converges pointwise to $fg$.

I am using the identity: $f_n g_n - fg = (f_n - f)(g_n - g) + f(g_n - g) + g(f_n - f)$.

\begin{align*} |f_ng_n - fg| &= |(f_n - f)(g_n - g) + f(g_n - g) + g(f_n - f)| \\ &\leq |(f_n - f)(g_n - g)| + |f(g_n - g)| + |g(f_n - f)| \end{align*}

but I don't know where to go from here. Any help will be appreciated!

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    This really shouldn't require work... You should already be familiar with this statement: let $u_n$ and $v_n$ be two sequences of real numbers that converge respectively to $u$ and $v$, then $u_n v_n$ converges to $uv$. – 2017-01-08
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    $(f_n)$ and $(g_n)$ are sequences of functions – 2017-01-09
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    @DDRM but if you see pointwise convergence, (f_ng_n) (x) is a numerical sequence – 2017-01-09
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    @wnitram I see. That makes sense. Then the conclusion would follow immediately by the Algebraic Limit Theorem for Sequences. I guess my work above could be used for proving uniform convergence. I believe $f_n$ and $g_n$ would have to be bounded though. Thanks for the clarification. – 2017-01-10

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