TITLE: $f_n$ uniformly converge to $f$ and $g_n$ uniformly converge to $g$ then $f_n \cdot g_n$ uniformly converge to $f\cdot g$
QUESTION [14 upvotes]: Please help me check, if 

$f_n$ uniformly converge to $f$ and $g_n$ uniformly converge to $g$ then $f_n + g_n$ uniformly converge to $f+g$
$f_n$ uniformly converge to $f$ and $g_n$ uniformly converge to $g$ then $f_n \cdot g_n$ uniformly converge to $f\cdot g$ 

What I've done:
I Guess that first statement is true, second is false. I've tried to calculate directly from definition, nothing seems correct.

REPLY [5 votes]: Let $\epsilon >0$ there's $n_1\in \mathbb{N}$, $\forall n\geq n_1$ $||f_n-f||_\infty<\epsilon/2$ and there's $n_2\in \mathbb{N}$, $\forall n\geq n_2$ $||g_n-g||_\infty<\epsilon/2$,
now let $n_0=\max(n_1,n_2)$ then $\forall n\geq n_0$ $||f_n+g_n-(f+g)||_\infty\leq ||f_n-f||_\infty+||g_n-g||_\infty <\epsilon$
For the second result if we suppose that the functions are continuous on the compact $[a,b]$ then use the same idea with the fact that $(f_n)$ is bounded since it's convergent sequence and with the inequality
$$||f_n g_n - fg||_\infty=||f_n g_n - f_n g + f_n g -fg||_\infty\leq ||f_n||_\infty||g_n-g||_\infty + ||g||_\infty||f_n-f||_\infty$$