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$$ f_n(x) = \begin{cases} -n^3x+2n^2, & \text{if }0<x<1/(2n) \\ n/(n+1), & \text{if }1/(2n)≤x<1 \end{cases} $$ $ n\in \mathbb{N}$

How can I prove that $f_n$ is pointwise convergent/uniform convergent? What I have tried so far: I think the limiting function $f(x)$ is: $$ f(x) = \begin{cases} +\infty, & \text{if }0<x<0 \text{ (???)}\\ 1, & \text{if }1/(2n)≤x<1 \end{cases} $$ But I'm not sure about that and as $n/(n+1)$ converges to 1 for n to $\infty$ I thought that $f_n$ is pointwise convergent, because $f(1)=1$ as well.

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    $\begingroup$ $0<x<0$ makes no sense, even if you attach to it any number of question marks. $\endgroup$ – uniquesolution Jan 11 '18 at 18:17
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    $\begingroup$ I find three down-votes here. Why? One thing I suspect is that once one person down-votes a question, there are certain persons present who feel they must also down-vote it. One mistaken down-vote or one good-faith down-vote whose motive is mysterious to all but the voter can happen, but how do we explain multiple down-votes without any readily apparent reason? $\endgroup$ – Michael Hardy Jan 11 '18 at 18:52
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    $\begingroup$ Might it be that some people think one should down-vote a question because it contains mathematical errors? $\endgroup$ – Michael Hardy Jan 11 '18 at 18:56
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It's not that simple. You can't write $0<x<0$ or even $=+\infty$, it's not a number !

In fact, the thing that disturbs you ( and that you noticed ) is that $$ \frac{1}{2n} \underset{n \rightarrow +\infty}{\rightarrow }0 $$ Hence for all $x>0$ it exists $N$ so that if $\ n > N$ then $\displaystyle \frac{1}{2n}<x$. And then, still for $n>N$ $$ f_{n}\left(x\right)=\frac{n}{n+1} $$ Then if you consider the subsequence $\displaystyle \left(f_n\right)_{n \geq N}$, it converges to the function $\displaystyle x \mapsto 1$.

What can you conclude ?

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  • $\begingroup$ One can have a sequence approaching $+\infty.$ But that is normally not considered convergent. One might speak of "diverging to $+\infty$" or even of "converging pointwise to $+\infty$", but such "convergence" cannot be uniform in the simple sense. (However, that does not apply to this particular sequence of functions: there is no number $x$ at which it approaches $+\infty. \qquad$ $\endgroup$ – Michael Hardy Jan 11 '18 at 18:54
  • $\begingroup$ I did not talk about uniform convergence $\endgroup$ – Atmos Jan 11 '18 at 18:59
  • $\begingroup$ But the original poster did. $\endgroup$ – Michael Hardy Jan 11 '18 at 19:12
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\begin{equation} f(x) = \begin{cases} +\infty, & \text{if }0<x<0 (???)\\ 1, & \text{if }1/(2n)≤x<1 \end{cases} \end{equation}

Note that your limiting function $f$ depends on $n$, which does not make sense.

Limit of $f_n$ is $f(x)$ = 1, a constant function.

Take any point $x\in(0,1)$, and show that $|f_n(x) - f(x)| = |f_n(x) - 1| \to 0$.

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Clearly $f_n:(0,1)\to\mathbb{R}$. You have written $0<x<0 $ ??? Indeed: what should be on the right side? Nothing. Since $1/(2n)$ can be aribtrary close to $0$, it holds that for every $x\in (0,1)$ there is some $N$ such that for all $n>N$ we have: $f_n(x)=\frac{n}{n+1}$. And that implies $f(x)=\lim_{n\to\infty}f_n(x)=1$ for all $x\in (0,1)$.

That was pointwise convergence. What about uniform? It does not converge uniformly, take for example $x_n=\frac{1}{4n}<\frac{1}{2n}$. Now we have: \begin{align} \sup_{x\in (0,1)}|f_n(x)-1|\geq |f_n(x_n)-1|=|-n^2/4+2n^2-1|\to \infty \end{align} as $n\to\infty$ hence the convergence cannot be uniform.

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  • $\begingroup$ Feel free to tell me where I can correct the language errors. I think there are some, but I'm not sure where due to my bad English. $\endgroup$ – Shashi Jan 11 '18 at 19:30
  • $\begingroup$ Your English is great I'd say. But I think at the end of your answer it should be $+2n^2$ instead of $+n^2$ $\endgroup$ – alex403 Jan 11 '18 at 19:49
  • $\begingroup$ @alex403 glad to hear that! Thank you! I have corrected it. $\endgroup$ – Shashi Jan 11 '18 at 19:50

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