Uniform convergence of a sequence of functions on $\mathbb{R}$ Just for clarification, with the function 
$k_n(x) = \left\{
 \begin{array}{ll}
  0  & \mbox{if } x \leq n \\
  x-n & \mbox{if } x \geq n
 \end{array}
\right.$
I've shown the function is uniformly convergent on $[a,b]$, and it asks if the function is uniformly convergent on $\mathbb{R}$ or not. 
My thinking is that since $\displaystyle\lim_{n \to \infty} k_n(x) = k(x) = 0 \ \ \forall x \in \mathbb{R}$
It's sufficient to show $|k_n(x) - k (x)| = |k_n(x)| > \epsilon$ for given $\epsilon > 0 \ $ and for some $x \in \mathbb{R}$, $n \in \mathbb{N}$.
So, given some $\epsilon$ and for fixed $n$, there exists some $x \in \mathbb{R}$ such that $x \geq n + \epsilon$ giving
$|k_n(x)| = |x-n| > \epsilon$
and hence $k_n$ is not uniformly convergent at this point (or for all $p > x$) hence $k_n$ is not uniformly convergent on $\mathbb{R}$
Thanks again for all the help.
 A: Nice thinking, but I think you're getting a little confused, since you are talking about "for given $\epsilon >0$" and you say "not uniformly convergent at this point". The key idea is that uniform continuity is a global property, not a local one like regular continuity.
The mathematical statement that a series of functions $k_n$ is not uniformly convergent to a function $k$ on a set $X$ is: $\exists \epsilon > 0$ such that for all $n \in \mathbb{N} \space$, there is some $x\in X$ such that $|k_n(x)-k(x)|>\epsilon$. Notice that we are no longer talking about arbitrary $\epsilon > 0$.
Hint: As you have correctly stated, since uniform convergence implies pointwise convergence, if $k_n$ converges uniformly to some limit, it must be the pointwise one. To show that it must converge, you must exhibit some number $\epsilon > 0$ (here we may take for example, $1$) such that for any function $k_n$ we can find a real number $x$ such that $k_n(x) > \epsilon$.
A: The pointwise limit is $f(x)=0$ on $\mathbb{R}$.
If $f_n$ converged uniformly on $\mathbb{R}$, this would necessarily be to $f$ and then
$$
\lim_{n\rightarrow+\infty}\sup_{x\in\mathbb{R}}|f_n(x)|=0.
$$
But actually
$$
\sup_{x\in\mathbb{R}}|f_n(x)|=+\infty
$$
for all $n$.
Contradiction.
