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If we let $f_n(x)=\frac{x^2+nx}{n}$ for $x\in \mathbb{R}$. I have proved that $f_n$ converges pointwise to $f(x)=x$

Now I am trying to prove that $f_n$ does not converge uniformly on $\mathbb{R}$

$$|f_n(x)-f(x) | = \sup\left|\frac{x^2+nx}{n}-x\right| = \sup \left|\frac {x^2}{n}\right|\rightarrow0$$

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  • $\begingroup$ The problem in your last step is that $$\sup_{x\in\Bbb R}\frac{x^2}n$$ is not finite so its limit when $\;n\to\infty\;$ isn't zero, as needed to have uniform convergence. $\endgroup$ – DonAntonio Feb 16 '17 at 23:16
  • $\begingroup$ so this solution and logic is incorrect? $\endgroup$ – user284 Feb 16 '17 at 23:17
  • $\begingroup$ @DonAntonio how is limit not 0? $\endgroup$ – user284 Feb 16 '17 at 23:24
  • $\begingroup$ Before taking the limit you must have the supremum, which is.... There you go!! $\endgroup$ – DonAntonio Feb 16 '17 at 23:28
  • $\begingroup$ @DonAntonio looking at the answer that has been provided, does my comment to this answer make sense? $\endgroup$ – user284 Feb 16 '17 at 23:30
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Let $\epsilon=1$. Then, for any $n$, we select $x=\sqrt{n}$ and find that

$$\left|\frac{x^2+nx}{n}-x\right|=\left|\frac{x^2}{n}\right|\ge 1$$

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  • $\begingroup$ does this prove along with my answer provided that it is not uniformly convergent? in my answer I have said that the limit must go to 0. but this answer you have provided shows a contradiction so its not uniformly convergent? $\endgroup$ – user284 Feb 16 '17 at 23:26
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    $\begingroup$ @user284 In your question you put some strange things, like "...using the fact that $\;x^2\rightarrow\sqrt n\;$ " (??), and then some $\;x_n\;$ ...(?). I think the above answer is what you need in the spirit of what you wanted. +1 $\endgroup$ – DonAntonio Feb 16 '17 at 23:33
  • $\begingroup$ @user284 This is precisely the negation of uniform convergence for $x\in\mathbb{R}$. $\endgroup$ – Mark Viola Feb 16 '17 at 23:41

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