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I am trying to show that $f_n(x) = \frac{nx}{1+n^2x^2}$ where $x\in \mathbb{R}$ does not converge uniformly.

I have made an attempt and would like to make sure that I am going about it in the right way (i.e. if my thought process is correct or I am missing something).

First note that $f_n(x) \rightarrow 0$ pointwise and so if it converges uniformly it should converge to $f(x) = 0$ as well. To show this is not the case, note that $f_n(\frac{1}{n}) = 1/2$.

Hence, given $\epsilon = \frac{1}{3}$ for all $N\in \mathbb{N}$ we can find $n\geq N$ and $x\in \mathbb{R}$ where $|f_n(x)-f(x)|\geq \epsilon$, indeed we can just take $n = N$ and $x = \frac{1}{n} \in \mathbb{R}$ and $|f_n(x)-f(x)| = |f_n(\frac{1}{n})| = \frac{1}{2}>\frac{1}{3} = \epsilon$.

This shows that $(f_n)$ does not converge uniformly.

Martin Sleziak
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fosho
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  • $f_n\to 2/x$ pointwise... – bartgol Jul 29 '16 at 19:48
  • I have edited the question, thanks! – fosho Jul 29 '16 at 19:50
  • Now it's ok. And your proof is correct. – bartgol Jul 29 '16 at 19:52
  • Just a question, am I right in saying that I only really need to find an $n\geq N$ OR an $x\in \mathbb{R}$ that leads to a $|f_n(x)-f(x)|>\epsilon$? – fosho Jul 29 '16 at 19:53
  • @Dman Simply note that $\sup_{x\in \mathbb{R}}f_n(x)=\frac12 \ne 0$. – Mark Viola Jul 29 '16 at 19:55
  • @Dr.MV I did notice this, however I wanted to give it a go with the $\epsilon$-definition – fosho Jul 29 '16 at 19:57
  • You found an $\epsilon>0$, such that for all $N$ there exists an $n>N$ and an $x\in \mathbb{R}$, such that $|f_n(x)-f(x)| \ge \epsilon$ – Mark Viola Jul 29 '16 at 19:58
  • @Dr.MV but would it suffice to show there exists $n$ or an $x$ (not an $n$ AND an $x$)? For some other example say, not this one. – fosho Jul 29 '16 at 19:59
  • No, not existence of such $n$. You need to show that given any $n$ you can find some $x$ such that your function varies from its pointwise limit by at least $\epsilon$ when evaluated at $x$. – Zestylemonzi Jul 29 '16 at 20:03
  • @Zestylemonzi There exists an $\epsilon>0$, such that for all $N>0$ there exists an $n>N$ and an $x\in \mathbb{R}$ with $|f_n(x)-f(x)|\ge \epsilon$. That is, for each $N>0$ we only need one $n>N$ to negate UC. – Mark Viola Jul 29 '16 at 20:07
  • @Dr. MV Yup I agree. We're talking about different '$n$'s - I'm saying that you can not just show that there exists some $n \in \mathbb{N}$ and some $x$ such that $|f_n(x) - f(x)| > \epsilon$. – Zestylemonzi Jul 29 '16 at 20:18
  • See also https://math.stackexchange.com/questions/588976/to-prove-that-f-nx-fracnx1n2x2-does-not-uniformly-converge-to-fx – Arnaud D. Nov 25 '19 at 15:44

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The negation of uniform convergence on a set $A$ is, there exists a number $\epsilon>0$, such that for all $N$, there exists a number $n>N$ and an $x\in A$ with $|f_n(x)-f(x)|\ge \epsilon$.

If one finds a number $\epsilon>0$ such that for all $N$, there exists an $x\in A$ with $|f_n(x)-f(x)|\ge \epsilon$ for all $n>N$, then certainly this more than suffices as proof of uniform convergence on $A$.

Mark Viola
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