Pointwise convergence to a uniform continuous function What can we say about a sequence of functions that is pointwise convergent (over $R$) to a uniform continuous function? Does it converge uniformly?
I have tried it using the definition but can't get any result. I hope someone can help me.
 A: No. Let $f_n(x) = \dfrac \pi2 + \arctan(x-n).$ For every value of $x$ this converges to $0$ as $n\to\infty,$ but convergence is not uniform since for each value of $n$, $f_n(x)\to\pi$ as $x\to\infty.$
A: No: the sequence $f_n(x)=1_{[n,n+1]}(x)$ converges pointwise to the uniformly continuous function $f(x)=0$, but doesn't converge uniformly.
A: A variation on the answer provided by carmichael561 which happens to be continuous (that is, each $f_n$ is continuous) is$$f_n(x)=\left\{\begin{array}{l}\sin x&\text{ if }x\in\bigl[n\pi,(n+1)\pi\bigr]\\0&\text{ otherwise.}\end{array}\right.$$
A: Take the function $f_n:[0,+ \infty) \rightarrow \mathbb{R}$ such that $f_n(x)=e^{-nx}$
Another example:
Take the sequence of functions $f_n(x)=1_{[\frac{1}{n+1},\frac{1}{n}]}(x)$ defined in $[0,1]$.
This sequence converges pointwise to the zero function (which is clearly uniformly continuous) but does not converge unformly to this function.
$$\sup_{x \in [0,1]}|f_n(x)|=1$$
Or $f_n(x)=\frac{nx}{n+1}$ defined in an unbounded subset of $\mathbb{R}$.
This sequence converges pointwise to $f(x)=x$ which is uniformly continuous but does not converge uniformly.
