we consider the sequences of function $(\psi_n)$ defined by $\psi_n(x)= \dfrac{1}{1+n^2 x^2} \varphi(x)$, where $\varphi \in \mathcal{D}(\mathbb{R})$ $\varphi$ is an test function. We suppose that $D^\alpha \varphi(0)=0, \forall \alpha \in \mathbb{N}$. The question is prouve that $D^{\alpha} \psi_n(x)$ converge uniformly to $\psi=0$.
I try to use Libniz formula. Let $\alpha \in \mathbb{N}$, we have $$ D^{\alpha} \psi_n(x)= \sup_{\beta \leq \alpha} C_{\alpha}^{\beta} D^{\alpha}(\dfrac{1}{1+n^2x^2}) D^\beta \varphi(x), $$ and we can remark that $D^\alpha (\dfrac{1}{1+n^2x^2}= \dfrac{f_n(x)}{(1+n^2 x^2)^{\alpha}}$, where $d f_n > 2\alpha$.
Then we have to prouve that $\lim_{n \to +\infty} \sup_{x \in K} |D^\alpha \psi_n(x)|=0$, where $K= Supp(\varphi)=[-a,a]$ with $a>0$. But i can't prouve it. Thank's in advance to the help.
First show that $$D^k(\frac{1}{1+x^2})=\frac{P_k(x)}{(1+x^2)^{k+1}}$$ with $P_k$ a polynomial of degree $\leq k$. This show that $$D^k(\frac{1}{1+n^2x^2})=n^k\frac{P_k(nx)}{(1+n^2x^2)^{k+1}}$$ Now, for $x\in [-a,a]$, you have $$|D^k(\frac{1}{1+n^2x^2})|\leq c_1\frac{n^{2k}}{(1+n^2x^2)^{k+1}}$$ for some constant $c_1$ independant of $x$ and $n$. Let $x\not =0$. Using the inequality $n^2x^2\leq 1+n^2x^2$, we get $|D^k(\frac{1}{1+n^2x^2})|\leq c_1\frac{1}{x^{2k}(1+n^2x^2)}$. As we have $2n|x|\leq 1+n^2x^2$, we get that $$|D^k(\frac{1}{1+n^2x^2})|\leq c_2\frac{1}{n|x|^{2k+1}}$$ Now, if $m$ is another integer, we get $$|D^k(\frac{1}{1+n^2x^2})||D^m(\varphi(x))|\leq \frac{c_2}{n}\frac{|D^m(\varphi(x))|}{|x|^{2k+1}}$$
You can now use the Lagrange-Taylor formula: $$D^{m}(\varphi)(x)=\frac{x^{2k+1}}{(2k+1)!}D^{m+2k+1}(\varphi)(d)$$ for some $d$, and with $M $ an upper bound for $|D^{m+2k}(x)|$, you finally get with $c_3=c_2M/((2k+1)!)$ that
$$|D^k(\frac{1}{1+n^2x^2})||D^m(\varphi(x)|\leq \frac{c_3}{n}$$ this is true for $x=0$ also, and of course if $x\not \in [-a,a]$, hence on $\mathbb{R}$.
It is easy to finish.