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What is the pointwise limit of the function $f_n$ which increases linearly on $[0,1/n]$ with $$ f_n(0)=0,\\ f_n\left(1/n\right)=n, $$ and is not defined elsewhere, that is, $f_n:[0,1/n] \to \Bbb R$.

I think the limit is the discontinuous function $f:\{0\} \to \{0,\infty\}$, with $$ f(0+)=0,\\ f(0-)=\infty, $$ where $$ f(0+)=\lim_{x \uparrow 0} f(x). $$ But this seems strange.

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Strange indeed. What is $f(0)$ anyway? If you want $f$ to be a function in the first place, you have to choose a value. Your sequence converges pointwise to $f$ on $\{0\}$ where $f(0)=0$. –  1015 Feb 2 '13 at 18:39
    
@julien As $f_n(0)=0$, then $f_n(0) \to f(0)=0$, as you said. But, I'm reluctant to throw away everything before $x=0$ to define $f:\{0\} \to \{0\}$. Can we throw away? –  Nicolas Essis-Breton Feb 2 '13 at 18:44
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If you want to talk about pointwise convergence, you have to restrict yourself, to begin with, to the intersection of the domains of your $f_n$'s. That is $\{0\}$ in this case. –  1015 Feb 2 '13 at 18:46
    
@julien It's very true. Thanks. Can you turn your two comments, as is, into an answer, I will accept it. –  Nicolas Essis-Breton Feb 2 '13 at 18:51

2 Answers 2

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Comment transfered to answer by request.

If you want to talk about pointwise convergence, you have to restrict yourself, to begin with, to the intersection of the domains of your $f_n$'s. In this case, this is $\{0\}$.

Now $f_n(0)=0$ for all $n$, so $f_n$ converges pointwise to $0$ on $\{0\}$.

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It's clear that $$\lim_{n\to\infty}f_n(0)=\lim_{n\to\infty}0=0,$$ so the pointwise limit at $x=0$ is $0$. None of the $f_n$ are defined at negative $x$-values, so the domain of the pointwise limit function will consist only of nonnegative numbers, one of which will obviously be $0$. However, for any $x>0$, there is some $N$ such that $0<\frac1n<x$ for all $n\geq N$--meaning in particular that $f_n(x)$ is undefined for all $n\geq N$, so there isn't a pointwise limit at any positive $x$. Thus (as you've concluded), the domain of the pointwise limit function is $\{0\}$, as is the range (by the work above).

It makes no sense to talk about either $f(0+)$ or $f(0-)$. Neither one of them is defined, since $f(x)$ is undefined for $x\neq 0$.

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I've been too hasty in asking julien to turn his comments into an answer. I guess you were typing your answer at that time. I will give him the point, but your answer confirm our thoughts. Thanks. –  Nicolas Essis-Breton Feb 2 '13 at 19:08
    
I did upvote your answer, right after reading it. –  Nicolas Essis-Breton Feb 2 '13 at 19:18

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