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!
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 $$|f_n (x)|\le \frac {1}{n}. $$