Question

Prove that for all multi-indices a and all functions φ in S, x^(a)φ∈S and ∂ ^(a)φ∈S.

Answer 1

1) $\lim\limits_{|x|\to+\infty}x^u\frac{\partial^{|\beta|}}{\partial x^\beta}\bigl(\frac{\partial^{|\alpha|}\varphi}{\partial x^\alpha}\bigr)=\lim\limits_{|x|\to+\infty}x^u\frac{\partial^{|\alpha+\beta|}\varphi}{\partial x^{\alpha+\beta}}$. Since $\varphi\in S$, we have $\lim\limits_{|x|\to+\infty}x^u\frac{\partial^{|\alpha+\beta|}\varphi}{\partial x^{\alpha+\beta}}=0$, so $\frac{\partial^{|\alpha|}\varphi}{\partial x^\alpha}\in S$

2)Prove by induction on $|\beta|$ that $\frac{\partial^{|\beta|}}{\partial x^\beta}\bigl(x^\alpha\varphi\bigr)=\sum\limits_{\gamma\le\inf\{\alpha,\beta\}}\sum\limits_{\delta\le\beta}C_{\gamma\delta}x^{\alpha-\gamma}\frac{\partial^{|\delta|}\varphi}{\partial x^\delta}$ for some $C_{\gamma\delta}$ (if $\varepsilon$ and $\rho$ are multiindices, then $\varepsilon\le\rho\Leftrightarrow\forall i=1,\ldots,n\ \varepsilon(i)\le\rho(i)$ ).

a)If $|\beta|=0$, then $\gamma=0$ and $\delta=0$, so $C_{00}=1$

b)Let it is true for all $|\beta|$, where $|\beta|\le k$.

c)Consider $\frac{\partial^{|\beta|}}{\partial x^\beta}\bigl(x^\alpha\varphi\bigr)$, where $|\beta|=k+1$. Then $\beta(s)=\begin{cases} \beta'(s),&\text{if $s\neq t$}\\ \beta'(s)+1,&\text{if $s=t$} \end{cases}$ for some $t$ and $\beta'$, where $|\beta'|=k$.

So $\frac{\partial^{|\beta|}}{\partial x^\beta}\bigl(x^\alpha\varphi\bigr)=\frac{\partial}{\partial x_t}\left(\frac{\partial^{|\beta'|}}{\partial x^{\beta'}}\bigl(x^\alpha\varphi\bigr)\right)$ . By induction hypothesis we have $\frac{\partial^{|\beta|}}{\partial x^\beta}\bigl(x^\alpha\varphi\bigr)=\frac{\partial}{\partial x_t}\left(\sum\limits_{\gamma'\le\inf\{\alpha,\beta'\}}\sum\limits_{\delta'\le\beta'}C_{\gamma'\delta'}x^{\alpha-\gamma'}\frac{\partial^{|\delta'|}\varphi}{\partial x^{\delta'}}\right)=$

$\sum\limits_{\gamma'\le\inf\{\alpha,\beta'\}}\sum\limits_{\delta'\le\beta'}C_{\gamma'\delta'}\left(\frac{\partial x^{\alpha-\gamma'}}{\partial x_t}\frac{\partial^{|\delta'|}\varphi}{\partial x^{\delta'}}+x^{\alpha-\gamma'}\frac{\partial^{|\delta'+1|}\varphi}{\partial x^{\delta'}\partial x_t}\right)$

Let $\gamma(s)=\begin{cases} \gamma'(s),&\text{if $s\neq t$}\\ \gamma'(s)+1,&\text{if $s=t$} \end{cases}$ and $\delta(s)=\begin{cases} \delta'(s),&\text{if $s\neq t$}\\ \delta'(s)+1,&\text{if $s=t$} \end{cases}$ , then $\gamma\le\beta$ and $\delta\le\beta$

We have $\frac{\partial x^{\alpha-\gamma'}}{\partial x_t}=\begin{cases} (\alpha(t)-\gamma'(t))x^{\alpha-\gamma},&\text{if $\gamma\le\alpha$}\\ 0,&\text{if $\gamma\le\alpha$ is false} \end{cases}$ , so $\frac{\partial x^{\alpha-\gamma'}}{\partial x_t}\frac{\partial^{|\delta'|}\varphi}{\partial x^{\delta'}}=0$ if $\gamma\le\alpha$ is false, and $\frac{\partial x^{\alpha-\gamma'}}{\partial x_t}\frac{\partial^{|\delta'|}\varphi}{\partial x^{\delta'}}=(\alpha(t)-\gamma(t))\frac{\partial x^{\alpha-\gamma}}{\partial x_t}\frac{\partial^{|\delta'|}\varphi}{\partial x^{\delta'}}$ if $\gamma\le\alpha$ . Also we have $\gamma\le\beta$ and $\delta'\le\delta\le\beta$.

Next, $x^{\alpha-\gamma'}\frac{\partial^{|\delta'+1|}\varphi}{\partial x^{\delta'}\partial x_t}=x^{\alpha-\gamma'}\frac{\partial^{|\delta|}\varphi}{\partial x^{\delta}}$ , where $\gamma'\le\gamma\le\beta$, $\gamma'\le\alpha$ and $\delta\le\beta$ .

So we obtain that $\frac{\partial^{|\beta|}}{\partial x^\beta}\bigl(x^\alpha\varphi\bigr)$ is linear combination of $x^{\alpha-\gamma}\frac{\partial^{|\delta|}\varphi}{\partial x^\delta}$ , where $\gamma\le\inf\{\alpha,\beta\}$ and $\delta\le\beta$.

By principle of mathematical induction we have that $\frac{\partial^{|\beta|}}{\partial x^\beta}\bigl(x^\alpha\varphi\bigr)=\sum\limits_{\gamma\le\inf\{\alpha,\beta\}}\sum\limits_{\delta\le\beta}C_{\gamma\delta}x^{\alpha-\gamma}\frac{\partial^{|\delta|}\varphi}{\partial x^\delta}$ for some $C_{\gamma\delta}$ .

Then $\lim\limits_{|x|\to+\infty}x^u\frac{\partial^{|\beta|}}{\partial x^\beta}\bigl(x^\alpha\varphi\bigr)=\lim\limits_{|x|\to+\infty}x^u\sum\limits_{\gamma\le\inf\{\alpha,\beta\}}\sum\limits_{\delta\le\beta}C_{\gamma\delta}x^{\alpha-\gamma}\frac{\partial^{|\delta|}\varphi}{\partial x^\delta}=$

$=\sum\limits_{\gamma\le\inf\{\alpha,\beta\}}\sum\limits_{\delta\le\beta}C_{\gamma\delta}\lim\limits_{|x|\to+\infty}x^{\alpha-\gamma+u}\frac{\partial^{|\delta|}\varphi}{\partial x^\delta}$

Since $\varphi\in S$, for every $\gamma$ and $\delta$ from sum we have $\lim\limits_{|x|\to+\infty}x^{\alpha-\gamma+u}\frac{\partial^{|\delta|}\varphi}{\partial x^\delta}=0$, so $\lim\limits_{|x|\to+\infty}x^u\frac{\partial^{|\beta|}}{\partial x^\beta}\bigl(x^\alpha\varphi\bigr)=0$. That is $x^\alpha\varphi\in S$.