Find the pointwise limit of the sequence of functions and decide whether it converges uniformly or not. 
For each $n\in\mathbb{N}$, define $f_{n}$ on $\mathbb{R}$ by
  \begin{equation*}
f_{n}(x) = \begin{cases}
1 &\text{if } |x|\geq 1/n\\
n|x| &\text{if } |x|<1/n
\end{cases}
\end{equation*}
  $(a)$ Find the pointwise limit of $(f_{n})$ on $\mathbb{R}$ and decide whether or not the convergence is uniform.

Hi all.
Can't find the way to start the solution, and don't know how to deal with the intervals containing $n$.
In fact I'm not strong at real analysis especially sequences of functions.
So can anybody help me solving this?
Thanks in advance.
 A: Let $|x| > 0$. Then, there is $n$ such that $|x| > \frac 1n$. For $m > n$, $|x| > \frac 1m$, so it follows that $f_m(x) = 1$ for all sufficiently large $m$. Hence, pointwise convegence to $1$ occurs. On the other hand, if $x = 0$,  then $f_n(x) = 0$ for all $n$, so the pointwise limit is zero.
Hence the pointwise limit is:
$$
f(x) =
\begin{cases} 
0 & x=0 \\
1 & x \neq 0
\end{cases}
$$
Actually, the $f_n$ are continuous,  so by uniform limit theorem, if the convergence were uniform, $f$ should be continuous, but it is not, so the convergence isn't uniform.
If  you want a constructive proof, then we have to contradict uniform convergence. To contradict it, we need the definition of uniform convergence:

For all $\epsilon > 0$, there is $N \in \mathbb N$ such that $m > N \implies |f_m(x) -f(x)| < \epsilon$ for all $x$.

To contradict this, we need to do the following:

There exists $\epsilon >0$, such that for all $N \in \mathbb N$, there is $m > N$ and a point $x$ such that $|f_m(x) - f(x)| > \epsilon$.

Pick $\epsilon = \frac 12$. Let $x \neq 0$. Now, note that $|f_m(x) - f(x)| = 0$ if $|x| > \frac 1m$, and $1-m|x|$ otherwise.
Let $N \in \mathbb N$. Let $m = N+1$, and let $x = \frac 1{4(N+1)}$. Then, note that $1-m|x| = \frac 34 > \frac 12$. Since this happens for all $N \in \mathbb N$, we get that the convergence is not uniform.
