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In the following question I am trying to find the pointwise limit of $f_n$ and then prove that $f_n$ does not converge uniformly

$$f_n(x)=\frac{1}{1+nx^2}$$

Then taking the limit we get that,

$\lim_{n\to\infty} f_n(x)$ is $1$ when $x=0$ and when $x\ne0$ it is $0$.

The part I am having trouble with is proving that $(f_n)$ does not converge uniformly.

My thinking would be to find the limit of this sequence then choose a $\epsilon>0$ and show that $|f_n(x)-f(x)|\geq\epsilon$ which shows a contradiction and proves that it does not converge uniformly. I am having some trouble with this part and would greatly appreciate some help, either with the way I have prescribed if it works, or another logical way.

Thanks!

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2 Answers 2

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Recall that if $\{f_n\}$ is a sequence of continuous functions such that $f_n\to f$ uniformly, then $f$ is also continuous.

In your case, each $f_n$ is continuous but the pointwise limit of the $f_n$ is not continuous, which means that the convergence cannot be uniform.

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  • $\begingroup$ how do I show that pointwise limit of $f_n$ is not continuous? $\endgroup$
    – jh123
    Apr 15, 2017 at 23:13
  • $\begingroup$ You already found the pointwise limit: $f(x)=0$ if $x\neq 0$, while $f(0)=1$. Is this a continuous function? $\endgroup$ Apr 15, 2017 at 23:13
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    $\begingroup$ It fails to be continuous at $x=0$, since for any $\delta>0$ there exists $y$ with $|y|<\delta$ such that $|f(y)-1|=1$. $\endgroup$ Apr 15, 2017 at 23:16
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For $\phi \ne D\subset \mathbb R,$ and for any functions $f,g$ from $D$ to $\mathbb R,$ let $\|f-g\|=\min (1,\sup \{|f(x)-g(x)|:x\in D\}).$

Saying that a sequence $f_n:D\to \mathbb R$ converges uniformly on $D$ to $f:D\to \mathbb R$ is equivalent to saying that $\lim_{n\to \infty} \|f_n-f\|=0.$

In your Q, we have $ D=\mathbb R.$ We have $f(x)=0$ when $x\in D$ \ $\{0\}.$ But we have $1/2=|f_n(1\sqrt n)-f(1/\sqrt n)|\leq \|f_n-f\|$.

So $(\|f_n-f\|)_n$ does not converge to $0.$

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