example of a sequence of uniformly continuous functions on a compact domain converging, not uniformly, to a uniformly continuous function

Question

I am looking for an example of a sequence of uniformly continuous functions, defined on a compact domain that converge pointwise, but NOT uniformly, to a uniformly continuous function.

At first, I considered the function $f(x)=sin^{n}(x)$ where $x\in[0,\pi]$.

I thought this was a good example, however when $x=\pi/2$ then $\underset{n\rightarrow\infty}{lim}(f(x))=1$ and therefore it does not converge to a continuous function since when $x\neq\pi/2$, $\underset{n\rightarrow\infty}{lim}(f(x))=0$.

I then considered the function $f(x)=1_{[n,n+1]}(x)$ for $x\in [a,b]$. Where, $1_{[n,n+1]}(x)$ is the characteristic function, ie $f(x)=1$ if $x\in[n,n+1]$ and $f(x)=0$ otherwise.

This also seemed like a good idea, until I realized that when $n>b$ then $f(x)=0$ regardless of $x$.

Any suggestions ?

Answer 1

It should be sufficient to take a sequence of continuous (and therefore uniformly continuous) "triangle functions" on $[0, 1]$ where the $n$th function is supported on $[0, \frac{1}{n}]$, but the heights of the triangles do not approach 0 as $n \to \infty$. Then the pointwise limit is 0 everywhere (for $x=0$ because each triangle function has value 0 there, and for $x>0$ because eventually the functions become 0 for such an $x$); and therefore the pointwise limit is continuous. However, because of the condition on the heights of the triangles, the convergence to 0 will not be uniform.

Answer 2

$f_n(x) : [0,1] \to \mathbb R$

$$f_n(x) = \left\{ \begin{align} nx & \mbox{ if } x \in [0, \frac{1}{n}) \\ 2-nx & \mbox{ if } x \in [\frac{1}{n}, \frac{2}{n}) \\ 0 & \mbox{ otherwise } \end{align} \right. $$