Does $f_{n}(x)=e^\frac{-x^2}{n}$, $x\in \mathbb{R}$ converge uniformly?
It converges pointwise to $1$. For uniform convergence I need $|f_{n}(x)-f(x)|<\epsilon$.
$\sup|f_{n}(x)-1|$ has to converge to $0$. If you put in $x=0$, which should be the sup (I believe), you get $0$. So it converges uniformly.
My classmate says I'm wrong. He told me it only converges pointwise (not uniformly). Where is my mistake?