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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.

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1 Answer 1

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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.

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  • $\begingroup$ Thank's for the answer. I don't understand some points. First, i found that $D^k(\dfrac{1}{1+n^2x^2})= \dfrac{P_k}{(1+n^2x^2)^k}$ where $d P_k < 2k$. How you found that $D^{\alpha}(\dfrac{1}{1+n^2x^2})= \dfrac{P_k}{(1+x^2)^{k+1}}$, why $k+1$? and why this show that $D^k(\dfrac{1}{1+n^2x^2})= n^k \dfrac{P_k (nx)}{(1+n^2 x^2)^{k+1}}? $Please $\endgroup$
    – user487908
    Oct 29, 2017 at 16:15
  • $\begingroup$ If you look at the first derivative, you get $-2n^2x/(1+n^2x^2)^2$, So I think it is $k+1$, not $k$. For the inequality on the degree, find a recurrence formula for the $P_k$. If you differentiate $k$ times a function of the form $f(nx)$, you find $n^kf^{(k)}(nx)$. $\endgroup$
    – Kelenner
    Oct 29, 2017 at 16:23
  • $\begingroup$ No, we have $(\dfrac{1}{1+n^2x^2})'=\dfrac{-2n^2x}{1+n^2x^2}$ and $(\dfrac{1}{1+n^2x^2})''=\dfrac{1+n^2x^2-2n^2x}{(1+n^2x^2)^2}$, so $D^k(\dfrac{1}{1+n^2x^2})= \dfrac{P_k(x)}{(1+n^2x^2)^k}$ with degree of $P_k < 2k$. So i don't undertstand how we found your first equality. Help me please to undertand your idea. $\endgroup$
    – user487908
    Oct 29, 2017 at 16:34
  • $\begingroup$ The derivative of a function of the form $1/u$ is $-u^{\prime}/u^2$.... $\endgroup$
    – Kelenner
    Oct 29, 2017 at 16:37
  • $\begingroup$ OK for the first equality, thank you so much. Now please i don't understant why we write $D^k(\dfrac{1}{1+n^2x^2})= n^k \dfrac{P_k(nx)}{(1+n^2x^2)}^{k+1}$. I don't understant your answer in this point $\endgroup$
    – user487908
    Oct 29, 2017 at 16:47

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