Pseudo Differential Operator example

Question

I was reading a introduction about Pseudo differential operators, and the definition of that type of operator was:

$$|\partial_x^\alpha \partial_\xi^\beta f(x,\xi) | \le C_{\alpha,\beta}(1 + |\xi|)^{n-|\alpha|}$$ where $x,\xi$ $\in$ $\!R^m\times\!R^p$ ,$C_{\alpha,\beta}$ is a constant that depends only on $\alpha$ and $\beta$ , n is the order of the Pseudo differential operator $f(x,\xi)$.I was trying to prove that $\lt \xi\gt = (1 + |\xi|^2)^{m/2}$ is a Pseudo differential operator, the hint was to use the fact that $f(a,x)=(a^2 + |\xi|^2)^{m/2}$ have the homogeneity property, that is $f(ra,r\xi)=r^mf(a,\xi)$ but i was not able to prove using that.$$ $$ $\alpha=(\alpha_1,..,\alpha_m) \quad \partial_x^\alpha f(x)=\partial_{x_1}^{\alpha_1}...\partial_{x_m}^{\alpha_m}f(x) \quad |\alpha|= |\alpha_1| +....+ |\alpha_m| $

Answer 1

Note that $f(x,\xi)=\langle\xi\rangle=(1+|\xi|^2)^{m/2}$ does not depend on $x$, so you only need to prove it for $\alpha=0$. Note that each time you do a $\dfrac{\partial}{\partial\xi_j}$ on $\xi^\gamma(1+|\xi|^2)^k$ one of two things happen

• you killed one of the $\xi_j$ occuring in $\xi^\gamma$ and so $|\gamma|$ goes down by $1$.
• you operate on the $(1+|\xi|^2)^k$ factor, gaining a factor $2k\xi_j(1+|\xi|^2)^{-1}$, so $|\gamma|$ goes up by $1$ but the power of $(1+|\xi|^2)$ goes down by $1$

So the upshot is, after you apply $\partial_\xi^\beta$ to $f$, the result is a sum of terms $c_\gamma\xi^\gamma(1+|\xi|^2)^{\leq \frac{m}{2}-|\gamma|}$ with $|\gamma|\leq|\beta|$. Furthermore, you can bound the coefficient $|c_\gamma|\leq C(\beta,m)$ and of course the bound $|\xi|<(1+|\xi|^2)^{1/2}\leq 1+|\xi|$, so that gives the bound $$ |\partial_\xi^\beta f|\leq C_\beta (1+|\xi|^2)^{\frac{m}{2}} \leq C_\beta (1+|\xi|)^m $$