Where does $f_n$ converge pointwise, is it uniformly convergent?

Question

Let $f_{n}:[0,1] rightarrow mathbb{R}$ be a sequence of function defined by
$$
f_{n}(x)=left{begin{array}{cc}
n(1-n x), & text { if } 0<x<frac{1}{n} \
0, & text { if } x=0 text { or } frac{1}{n} leq x leq 1
end{array}right.
$$
The graph is a right triangle and when $nto infty$, triangle's basis vanishes. So can I say that this function converges pointwisely to $f(x)=0$?
If I can say that I can choose $$x=dfrac{n-1}{n^2}in (0,1/n)$$
For $epsilon=1/2$, $forall nin mathbb N^+$ $|f_n(x)-0|=1>epsilon$
If I cannot assume the above pointwise assumption, how can I show it is not a Cauchy sequence?

mrsamy · Answer

Hint.
You have the intuition for the pointwise convergence. But in order to give a proof, you need to show that for every $xin[0,1]$,
$$
lim_{nto 0}f_n(x)=0
$$
One way to see that the sequence is not uniformly convergent is by observing that
$$
int_0^1 f_n(x)dx=frac12
$$
for all $n$ and that $int_0^1f(x),dx=0$, where $f(x)equiv 0$.

If you want to follow the definition of uniform convergence, you need to show that there exists $epsilon>0$ such that for any $N$, there exist $xin[0,1]$ and $n>N$ such that
$$
|f_n(x)-0|ge epsilon
$$

Where does $f_n$ converge pointwise, is it uniformly convergent?

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