Uniform convergence for the sequence $\psi_n(x)=n^{-1}e^{-n^2/(n^2-x^2)}$ How can I prove that the sequence of functions:
\begin{equation}
\psi_n(x)=\begin{cases} n^{-1}e^{-n^2/(n^2-x^2)}, & |x|\leq n \\ 0, & |x|\geq n \end{cases}
\end{equation}
convergences uniformly to zero and that the same happens for all the sequences which are defined as the $\psi_n^{(k)}(x), k\in \mathbb{Z^+}$, derivatives of $\psi_n(x)$? And one last thing, why the sequence $\psi_n(x)$ does not converge to the zero element of the space $\mathcal{D}$? 
I am having difficulties with uniform convergence, therefore your assistance would be greatly appreciated!
Thank you!
 A: Graphically, what this function looks like is essentially a bump. On the set $|x|\leq n$, it looks like a plateau that quickly drops off to 0 as $|x|\to n$. So naturally, as $n\to \infty$, the function is essentially going to approach a constant function, and since the height of the plateau is proportional to $1/n$, this has to approach the $0$ function (see here for a picture).
Fix $\epsilon>0$. We want to find an $N$ such that $||\psi_n||_u<\epsilon$ for all $n\geq N$, where $||\cdot ||_u$ is the supremum norm on continuous functions. It's not too hard to see that $||\psi_n||_u=\frac{1}{n e}$, so obviously the result holds. The case for the derivatives is similar.
To see that $\psi_n\not\to 0$ in $\mathcal{D}(\mathbb{R})$, note that there is no compact set $K\subset\mathbb{R}$ such that $\text{supp} \psi_n\subset K$ for all $n$ ( $\text{supp}\psi_n=[-n,n]$, so $\bigcup_{n=1}^{\infty}[-n,n]=\mathbb{R}\subset K$, which is impossible, so $K$ cannot exist) so by definition there is no convergence.
