# Sequence which converges pointwise but not uniformly?

it might be simple but I don't find a sequence $f_n: [0,1] \rightarrow \mathbb{R}, n \in \mathbb{N}$ that converges pointwise but not uniformly.

First I thought it could be $f_n(x) = \frac{x}{n}$ but it is not right, is it?

Thanks for help!

• No, the example you suggest is not right: $|f_n(x)|\le 1/n$ for all $x$, so the sequence converges to $0$ uniformly. Oct 26, 2014 at 17:32

$f_n(x)=x^n, x\in[0,1)$ $f_n(1)=1$

It converges pointwisely to $f(x)=0, x\in[0,1)$ $f(1)=1$ but not uniformly since uniformly convergence preserves continuity.

Your example converges uniformly on $[0,1]$.

Consider a sequence of functions such as $$\int f_n(x) = 1\\ f_n(x) \to 0$$ such as $$f_n(x) = \begin{cases} n &\text{ if } nx<1 \\ 0 & \text{ otherwise} \end{cases}$$

For a different solution, pick any derivative $f$ that is not continuous, for instance, the derivative of $$g(x)=\left\{\begin{array}{cl}x^2\sin(1/x)&x\ne0,\\0&x=0.\end{array}\right.$$ We can always find a sequence of continuous functions converging to $f$ pointwise, but the convergence cannot be uniform, or $f$, being the uniform limit of continuous functions, would be continuous as well.

To see that $f$ is limit of continuous functions, recall that $f=g'$, so for any $x$, $$f(x)=\lim_{n\to\infty}\frac{g(x+1/n)-g(x)}{1/n}.$$

(This requires a small adjustment if $g$ is only defined in $[0,1]$, as adding $1/n$ to points $x$ near $1$ may bump us out of the interval. But not even such an adjustment is needed for the example I suggest above.)