Show $(1+|\xi|^2)^{m/2}\in S^m$ (pseudo-differential operators) I am trying to show that $p(\xi)=(1+|\xi|^2)^{m/2}\in S^m$, i.e., that for all $m\in\mathbb{R}$ and all multi indices $\alpha,\beta$
$$|D^{\beta}_{x} D^{\alpha}_{\xi}p(\xi)|\leq C_{\alpha,\beta}(1+|\xi|^2)^{m/2-|\alpha|}$$
for all $x,\xi\in \mathbb{R}^n$. By multi indices we mean $\alpha=(\alpha_1,...,\alpha_n)$ where $\alpha_j\in\mathbb{N}$ and by $D^{\alpha}_{\xi}$ we mean 
$$D^{\alpha}_{\xi}=D^{\alpha_1}_{\xi_1}\dots D^{\alpha_n}_{\xi_n},\ \ D^{\alpha_j}_{\xi_j}=\frac{1}{i}\frac{\partial^{\alpha_j}}{\partial\xi_{j}^{\alpha_j}}$$
$\textbf{My attempt:}$ Since $p$ does not depend on $x$ it's trivial that the above holds for any multi index $\beta >0$ thus we focus on the case $\beta=0$, i.e.,
$$|D^{\alpha}_{\xi}p(\xi)|\leq C_{\alpha}(1+|\xi|^2)^{m/2-|\alpha|}$$
My attempt is to prove this by induction on the multi index $\alpha$. It's true that for $\alpha=0$ that the above holds. We now assume that the above holds for all multi indices $\alpha$ with a length of at most $l$, i.e., $|\alpha|\leq l$. Now, consider a multi index $\gamma$ with length $|\gamma|=l+1$. The goal is now to show that 
$$|D^{\alpha}_{\xi}p(\xi)|\leq C_{\gamma}(1+|\xi|^2)^{m/2-|\gamma|}$$
One can notice that $D^{\gamma}_{\xi}=D^{\alpha}_{\xi}D_{\xi_j}$ for some $j=1,...,n$. This yields that
$$|D^{\gamma}_{\xi}p(\xi)|=|D^{\alpha}_{\xi}(D_{\xi_j}p(\xi))|=|D^{\alpha}_{\xi}m\xi_j(1+|\xi|^2)^{m/2-1}|$$
Now one can use Leibniz' formula on the above to obtain
$$D^{\alpha}_{\xi}m\xi_j(1+|\xi|^2)^{m/2-1}=m\sum_{\delta\leq\alpha}{\alpha\choose \delta}(\partial^{\delta}\xi_j)(\partial^{\alpha-\delta}(1+|\xi|^2)^{m/2-1})$$
This is where I am having difficulties proceeding. I've thought about using that $\partial^{\delta}\xi_j$ is bounded above by $(1+|\xi|^2)$ with some exponent. Also, I believe that I can use my induction hypothesis in the sense that 
$$\partial^{\alpha-\delta}(1+|\xi|^2)^{m/2-1}\leq C_{\alpha,\delta}(1+|\xi|^2)^{m/2-1-|\alpha-\delta|}$$
but this is about as far as I can come. Any help/hints are highly appreciated!
Thanks in advance.
 A: I'll be using different notation; I hope it will be clear.
Claim: If $|\alpha| = N,$ then $D^\alpha p (x)$ is a finite linear combination of terms of the form
$$\tag 1 (1+|x|^2)^{m/2 -(N-k)}x^{\beta_k},$$
where $\beta_k$ is a multi-index of order $N-2k.$ Here $k$ is an integer running through the values $0,1,\dots ,\lfloor {N/2}\rfloor.$
Note that the absolute value of $(1)$ is bounded above by
$$(1+|x|^2)^{m/2 -(N-k)}\cdot [(1+|x|^2)^{1/2}]^{|\beta_k|}=(1+|x|^2)^{m/2 -N}.$$
Ths implies the desired result.
Proof of the claim: It certainly holds for $N=0.$ Suppose it holds for $N.$ Apply $D_j$ to the expression in $(1).$ We get
$$\tag 2 (1+|x|^2)^{m/2 -(N-k)-1}2x^{\beta_k+e_j} + (1+|x|^2)^{m/2 -(N-k)}D_j(x^{\beta_k}).$$
Now if $\beta_k$ is nonzero in the $j$th component, then $D_j(x^{\beta_k})$ is a constant times $x^{\beta_k-e_j}.$ If the $j$th component of $\beta_k$ is $0,$ then $D_j(x^{\beta_k})$ is just $0.$ In both cases we have $(2)$ as the sum of two terms of the desired form in $(1)$ for $N+1.$
