Discontinuous limit function and uniform convergence The sequence I am working with is this, in the interval [-1,1]:
$f_n (x)= \begin{cases}-1,-1\leq x\leq-\frac{1}{n}\\[2ex]nx, -\frac{1}{n}&lt;x&lt;\frac{1}{n} \\[2ex]1, \frac{1}{n}\leq x \leq1\end{cases}$
We can show pretty easily that this converges pointwise to the function
$f (x)= \begin{cases}-1, x∈[-1,0[\\[2ex]0, x=0 \\[2ex]1, x∈]0,1]\end{cases}$
I would like to find the intervals in which this converges uniformly. It can't converge uniformly to a discontinuous function, so obviously we have no uniform convergence in the interval [-1,1]. However, when it comes to showing convergence in ]0,1], can I somehow use the uniform limit theorem here as well to show that the sequence cannot converge uniformly?
You’re right that you can use the uniform convergence theorem (in the contrapositive) to rule out uniform convergence. On all of $[-1,1]$, uniform convergence + continuity of the $f_n$ would imply continuity of $f$, which is false as you point out. To answer your question about $(0,1]$ specifically, yes the same argument works. Uniform convergence in $(0,1]$ plus ordinary convergence at $0$ would imply uniform convergence on $[0,1]$ (hopefully it’s clear why; adding a finite set does not affect the uniformity). But the limiting function is not continuous on $[0,1]$ so the convergence could not have been uniform.
hint
For $ x\in (0,1]$ and $ n $ great enough, we will have $$x&gt;\frac 1n \;and \; f_n(x)=1$$
thus
$$f(x)=\lim_{n\to+\infty}f_n(x)=1$$
then
$$|f_n(x)-f(x)|=0 \; if \; x\ge \frac 1n $$
and
$$|f_n(x)-f(x)|=1-nx \; if \; 0&lt;x&lt;\frac 1n$$