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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

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1 Answer 1

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.

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I know it's totally useless (and it obfuscates the obvious), but if $g$ is the characteristic function of $[0,1]$, then you can write $f_n(x) = \left(\int_0^x g \right)^n$ in your first example. For those who just love formulas... –  Joel Cohen May 12 '11 at 17:27
    
@Joel: I had to chuckle at that. I don't know exactly how useless it is, though (think of distributions :)). But probably for the present exercise it's really really really point- and useless... Thanks! –  t.b. May 12 '11 at 17:33

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