Only pointwise convergence or also uniform convergence?

Question

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?

Answer 1

For each $n\in\Bbb N$, $f_n\left(\sqrt n\right)=e^{-1}$. So, the convergence is not uniform: if $\varepsilon=1-e^{-1}$, there is some $x\in\Bbb R$ such that $\bigl|f_n(x)-1\bigr|\geqslant\varepsilon$.

Concerning your approach, the function $\bigl|f_n(x)-1\bigr|$ attains its minimum (not its maximum) at $0$.