If $a(x,\xi)$ is of $C^{\infty}$ class and positively homogeneous of degree m for $|\xi|\geq1$, i.e., $$ a(x, t\xi) = t^{m} a(x, \xi), |\xi| \geq 1, t\geq1, $$ then $a \in S^{m}_{1,0} = S^{m}$.
We need to prove that $|\partial^{\alpha}_{x}\, \partial^{\beta}_{\xi} a(x,\xi)| \leq C (1+|\xi|)^{m-|\alpha|}$, but I don't have a good idea for bounding the expression above by a polynomial.
I tried to derivate the expressions, but this doesn't help me. I tried to put the function $a$ on the left as a quotient to try to find some clue, but one more time this doesn't help.
Please, can you help me? I need this example done for my work.
By the chain rule,
$$\frac{\partial}{\partial r}a(x,r\tfrac{\xi}{|\xi|}) = \sum_{i=1}^n(\partial_{\xi_i}a)(x,r\tfrac{\xi}{|\xi|})\frac{\xi_i}{|\xi|}$$
By repeated applications of the chain rule and evaluating at $r=|\xi|$, we get
$$ \frac{\partial^N}{\partial r^N}a(x,r\tfrac{\xi}{|\xi|})\big|_{r=|\xi|} = \sum_{i_1=1}^n\dots \sum_{i_N=1}^n(\partial_{\xi_{i_1}}\dots\partial_{\xi_{i_N}}a)(x,\xi)\frac{\xi_{i_1}}{|\xi|}\cdots\frac{\xi_{i_N}}{|\xi|} \tag{*}$$
But by homogeneity,
$$\frac{\partial^N}{\partial r^N}a(x,r\tfrac{\xi}{|\xi|}) = \frac{\partial^N (r^{m})}{\partial r^N}a(x,\tfrac{\xi}{|\xi|}) = (m-1)\dots(m-N+1)r^{m-N}a(x,\tfrac{\xi}{|\xi|})$$
Now, for any multi-indices $\alpha,\beta$ with $|\beta|=N$, we may bound $ \partial^\beta_\xi a(x,\xi)$ by the right hand side of (*), since the sum includes all possible combinations of $N$th order derivatives in $\xi$. Therefore, we get
$$|\partial^\alpha_x\partial^\beta_\xi a(x,\xi)| \leq (m-1)\dots(m-N+1)|a(x,\tfrac{\xi}{|\xi|})||\xi|^{m-N}$$
But $a$ is smooth, so in particular $|\partial^\alpha_x\partial^\beta_\xi a(x,\xi)|$ is bounded near $\xi=0$ and for $x$ in any arbitrary compact set $K$. Hence we may replace $|\xi|$ with $(1+|\xi|)$ above to obtain
$$\sup_{x\in K}|\partial^\alpha_x\partial^\beta_\xi a(x,\xi)| \leq A_{m,N,K}(1+|\xi|)^{m-N}$$
Where $A_{m,N,K}$ is a constant depending on $m,N,K$ and derivatives of $a$, so that $a\in S^m$.