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Let $f_n(x) = x$ and $f(x) = x$ for $x \in \mathbb{R}$ and let $g_n(x) = \frac{1}{n}$ and $g(x) = 0$ for $x \in \mathbb{R}$. Prove that $f_n \to f$ uniformly on $\mathbb{R}$ and $g_n \to g$ uniformly on $\mathbb{R}$ but $f_{n}g_n$ foes not converge uniformly to$fg$ on $\mathbb{R}$

My Work:

So {$f_n$} converges uniformly to $f(x) = x$, {$g_n$} converges to $0$ and $f_{n}g_n(x) = \frac{x}{n} \to 0$ for every $x \in \mathbb{R}$, but for $\epsilon = 1$ there is an $x_n = n$ such that $f_ng_n(x_n) = \frac{n}{n} = 1$ and $|1-0| = 1 \geq \epsilon $ so $f_ng_n $ does not converge uniformly

is this proof correct? Any help on this would be appreciated. Thank you!

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  • $\begingroup$ You should say a few more words on why $f_n \to f$ and $g_n \to g$ uniformly. The part that $f_ng_n$ does not converge uniformly to $fg$ is fine. $\endgroup$ – Daniel Fischer Apr 7 '14 at 20:37
  • $\begingroup$ yeah do you have any suggestions on how I would prove that. It just seems sort of obvious @DanielFischer $\endgroup$ – user3182418 Apr 7 '14 at 20:39
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    $\begingroup$ Just write down $\sup \{ \lvert f_n(x) - f(x)\rvert : x \in \mathbb{R}\} = \dotsc$. Well, for $f_n$, you could also say that it is a constant sequence, and only for $g_n$ use the $\sup \{ \lvert g_n(x) g(x)\rvert\}$. It is really obvious, but it needs a few words. $\endgroup$ – Daniel Fischer Apr 7 '14 at 20:42

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