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Let $\{f_{n}:\mathbb{R}\to\mathbb{R}\}$ be a sequence of functions with uniform limit $f:\mathbb{R}\to\mathbb{R}$, and let $\{c_{n}\}$ be a sequence defined by $$c_{n}:=\lim_{x\to\infty}f_{n}(x),$$ where the limit $\displaystyle\lim_{x\to\infty}f_{n}(x)$ exists in $\mathbb{R}$ for each $n\in\mathbb{N}$. Prove that $\{c_{n}\}$ is a convergent sequence.

The following are my two attempts:

(1) Since $\{f_{n}\}$ converges uniformly to $f$ on $\mathbb{R}$, for any $\varepsilon>0$, there exists $K\in\mathbb{N}$ such that $$|f_{m}(t)-f_{n}(t)|<\frac{\varepsilon}{3}$$ whenever $m,n\ge K$ and $t\in\mathbb{R}$.

Now, for each $n\in\mathbb{N}$, there exists $t_{n}\in\mathbb{R}$ large enough so that $$|f_{n}(x)-c_{n}|<\frac{\varepsilon}{3}$$ whenever $x\ge t_{n}$.

So, for any $m,n\ge K$, choose $a=\max\{t_{m},t_{n}\}$, then $$|c_{m}-c_{n}|\le|c_{m}-f_{m}(a)|+|f_{m}(a)-f_{n}(a)|+|f_{n}(a)-c_{n}|<\varepsilon,$$ i.e., $\{c_{n}\}$ is a Cauchy sequence in $\mathbb{R}$. Hence, $\{c_{n}\}$ is a convergent sequence. $\square$

(2) Since $\{f_{n}\}$ converges uniformly to $f$ on $\mathbb{R}$, for any $\varepsilon>0$, there exists $N\in\mathbb{N}$ such that $$|f_{m}(x)-f_{n}(x)|<\varepsilon$$ whenever $m,n\ge N$ and $x\in\mathbb{R}$.

So, if $m,n\ge N$, we have $$|c_{m}-c_{n}|=\left|\lim_{x\to\infty}f_{m}(x)-\lim_{x\to\infty}f_{n}(x)\right|=\lim_{x\to\infty}\left|f_{m}(x)-f_{n}(x)\right|\le\varepsilon.$$

The first proof seems to be no problem. However, the second proof doesn't seem rigorous.

Can anyone criticize my proofs? Give some advice. Thank you!

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    $\begingroup$ Are the functions all continuous? $\endgroup$ – cmk Oct 26 at 19:07
  • $\begingroup$ @cmk No, without assuming continuity $\endgroup$ – Primavera Oct 26 at 19:12
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I do not think that the statement is true at all,even if continuity is assumed.

Take the sequence $f_n=x+\frac{1}{n}$

It converges uniformly to $x$

But $c_n=+\infty,\forall n \in \Bbb{N}$ as $x \to +\infty$.

If each $f_n$ was bounded for every $n \in \Bbb{N}$ or more strongly,the sequence was uniformly bounded then your attempts are correct.

If the limits are assumed to exist and be real numbers,then the first attempt is is correct..Just instead of taking the maximum of $t_n,t_m$ you can put $t_m$ in $f_m$ and $t_n$ in $f_n.$

Also the second is correct since $\lim_{x \to +\infty}|f_n(x)-f_m(x)|=\limsup_{x \to +\infty}|f_n(x)-f_m(x)| \leq \sup_{x \in \Bbb{R}}|f_n(x)-f_m(x)| \leq \epsilon$

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  • $\begingroup$ In this problem, the existence of $\lim_{x\to\infty}f_{n}(x)$ is assumed. $\endgroup$ – Primavera Oct 26 at 19:15
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    $\begingroup$ @Primavera..... You have to mention it then in your post,because it is not clear $\endgroup$ – Marios Gretsas Oct 26 at 19:17
  • $\begingroup$ Sorry, I thought it would be sufficient $\{c_{n}\}$ is a real sequence. $\endgroup$ – Primavera Oct 26 at 19:24
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    $\begingroup$ @Primavera i edited though my answer.. $\endgroup$ – Marios Gretsas Oct 26 at 19:25
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    $\begingroup$ Thank you for concerning. I've also edited the posting. $\endgroup$ – Primavera Oct 26 at 19:28

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