Example:$f_n\rightarrow f$, $f$ is continuous but $f_n\nrightarrow f$ uniformly. I am trying to find a counter example to the converse of the statement "uniform limit of continuous functions is continuous."
 A: On noncompact domains there are all sorts of crazy things that you can do, but that's not really much of a surprise when you think about it the right way.
On a compact domain we can put together an example this way: start with $x^n$. The pointwise limit here isn't continuous, but the problem is just at $x=1$. So multiply by $1-x$. The maximum of this new function occurs at $x=\frac{n}{n+1}$. Note that $\lim_{n \to \infty} \left ( \frac{n}{n+1} \right )^n=1/e$ and $1-\frac{n}{n+1}=\frac{1}{n+1}$. Thus the whole thing is on the order of $1/n$. Thus $\| n(1-x)x^n \|_\infty$ behaves like $1/e$ as $n \to \infty$, yet the whole thing converges pointwise to the continuous function zero.
A: $x^n$ on open $(0,1)$ will work, but maybe you'd like one on a compact set.
Actually, the compact case is one that is any interesting.
A: Consider $f_n$, on the reals, defined by $f_n(x)= x/n$. 
A: How about a triangle moving off to infinity? Then $f\equiv 0 $ and $\|f-f_n\|_{\infty}=1 $ for all $n$.
A: One interesting example is $f_n(x)=nxe^{-nx}$ on the interval $[0,\infty)$ or $[0,1]$. It is clear that $f_n(x)\to0$ for all $x\in[0,\infty)$ since the exponential dependence on $n$ will eventually dominate, but notice that $sup(f_n(x))=f(\frac{1}{n})=\frac{1}{e}$ for any $n$.
A: On a compact domain: consider the function on $[0,1]$ giving  the isosceles triangle of area $1$ being "squeezed" to the left. In particular, the vertices of the triangle for $f_n$ are at $(0, 0)$, $(1/2^n, 0)$, and $(1/2^{n+1}, 2^{n+1})$ and the function is zero to the right of $1/2^n$. This does not converge uniformly but the limit function is zero.
