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I have the following sequences of functions $f_n = n^{-1} \chi_{[0,n)}$. I want to show that this function converges uniformly and pointwise to $0$. For the uniform convergence, my proof is the following:

Fix some $\epsilon > 0$ and let $N =\frac{1}{\epsilon}$ such that for all $n \geq N$ we have that $|n^{-1}\chi_{[0,n)} -0 | = |n^{-1}\chi_{[0,n)}| \leq |\frac{1}{n}| \leq |\frac{1}{N}| < \epsilon$ and this is true for all $x \in \mathbb{R}$.

I understand that for showing pointwise convergence, we have to come up with the value of $N$ that depends on $\epsilon$ and $x$. My confusion comes up when thinking about this $N$. I have thinking about this for a long time but I haven't been able to come up with one. Any suggestions or hints to show this will be highly appreciated! Thanks!

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    $\begingroup$ Fun fact: somewhere in your lecture notes or textbook must be the lemma, observation, or theorem that uniform convergence implies pointwise convergence. $\endgroup$ – Clement C. Feb 28 '18 at 20:16
  • $\begingroup$ @ClementC you really made me laugh! :D $\endgroup$ – Netchaiev Feb 28 '18 at 20:21
  • $\begingroup$ @ClementC. I rather hope it is an observation and not a theorem $\endgroup$ – Tashi Walde Feb 28 '18 at 20:25
  • $\begingroup$ We don't "have to" : in general the N depends on $x$ for it is sufficient to prove pointwise convergence, but it does not have to (or, in other way to consider it, you can make it depends on $x$, since you can define $N(x):=N$ for every $x$ ; but that would be point...less) $\endgroup$ – Netchaiev Feb 28 '18 at 20:26
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For $x <0,$ $f_n(x)=0/n=0$

For $x=0,$ $f_n (0)=1/n $

For $x>0$ and large enough $n,$

$f_n (x)=\frac {1}{n} $ since $x\in [0,n) $.

in all cases, $$\lim_{n\to+\infty}f_n (x)=0$$

There is pointwise convergence to zero function at $\Bbb R$

for the uniforme convergence, observe that $$|f_n (x)|\le \frac {1}{n}. $$

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