Give an example of a sequence of uniformly continuous functions on $\mathbb{R}$ that converge pointwise to a non-uniformly continuous function.
My thoughts:
I'm trying to work backwards: by choosing a non-uniformly continuous function, but I can't find anything that works. Any help would be appreciated.
Thanks
Best Answer
If you don't care about the limit function being continuous, one of the simplest examples is the sequence of functions $f_{n}(x) = x^{n}$ on $[0,1]$. This sequence converges pointwise to the function which is zero for $x \lt 1$ and $1$ for $x = 1$. This is of course an example on $[0,1]$, but you get an example on $\mathbb{R}$ by extending all the functions by zero for $x \leq 0$ and by $1$ for $x \geq 1$.
Recall that a continuous function on a compact interval is automatically uniformly continuous (prove this, in case you don't know that statement!). To get an example of a non-uniformly continuous function, we need to look for a function on an unbounded interval.
A very simple example of a continuous but not uniformly continuous function is $f(x) = x^{k}$ on $[0,\infty)$ for $k \neq 0,1$. Now simply define $f_{n}(x) = f(x)$ if $0 \leq x \leq n$ and $f_{n}(x) = f(n)$ if $x \geq n$. You should be able to check yourself that each $f_{n}$ is uniformly continuous and that the sequence $f_{n}$ converges pointwise to $f$.
To extend this example to all of $\mathbb{R}$, simply extend the functions by zero to the left.
A very similar idea works for $e^{x}$. If you want a slightly more interesting example, you can try to tackle $f(x) = \sin{(e^x)}$ with the same idea of truncating and extending constantly to the left and right.
Added: In fact, the procedure I outlined is one that always works (there are other ways but this probably is the most straightforward one). More precisely, if $f: \mathbb{R} \to \mathbb{R}$ is continuous, put $$f_{n}(x) = \begin{cases} f(x), &\text{if } |x| \leq n,\\f(n), &\text{if }x \geq n,\\f(-n), &\text{if } x \leq -n\end{cases}$$ and check that $f_{n}$ is uniformly continuous. This is because $f_{n}$ is uniformly continuous inside the compact interval $[-(n+1),n+1]$ due to the fact I mentioned above and constant outside. It is easy to see that $f_{n}(x) \to f(x)$ for all $x$, so $f_{n} \to f$ pointwise.
So to get an example of the kind you're asking about, the only thing you really need to think about is how to find a continuous but not uniformly continuous function, and I've given a few examples that should illustrate the kinds of functions you should be looking at.