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A symbol $p \in S^m(\Omega)$ where $\Omega \subset \mathbb{R^n}$, or its corresponding operator $p(x,D) \in \Psi^m(\Omega)$ is said to be elliptic of order m if for every compact $A\subset \Omega$ there are positive constants $c_A,C_A$ such that
$|p(x,\xi)|\ge c_A|\xi|^m$ for $x \in A$ and $|\xi|\ge C_A$.

Using this definition, I need to prove the following lemma:
If $p \in S^m(\Omega)$ is elliptic, there exists $\zeta \in C^{\infty}(\Omega \times \mathbb{R^n})$ with the following property: For any compact $A \subset \Omega$ there are positive constants $c,C$ such that for $x \in A$ we have
a. $\zeta(x,\xi)=1$ when $|\xi|\ge C$
b.$|p(x,\xi)| \ge c|\xi|^m$ when $\zeta(x,\xi) \not = 0$

My try: By definition, zeroes of $p(x,\xi)$ for $x$ in any compact set $A$ are contained within the ball $|\xi|\le C_A$ with respect to $\xi$ and we have $\frac{1}{|p(x,\xi)|} \le \frac{1}{c_AC_A^m}$ for $x \in A$ and $|\xi|\ge C_A$. I tried to use paracompactness of $\Omega$ to bound the zeroes of $p(x,\xi)$ within some ball centred at $0$ w.r.t $\xi$ (but could not attain it) so that $\zeta(x,\xi)$ can be defined to be $1$ outside such a ball and $0$ in some neighbourhood of zeroes of $\zeta(x,\xi)$.
Any help is deeply acknowledged.

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1 Answer 1

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In the first place, you can write $ \mathbf 1_\Omega=\sum_{k\ge 1}\chi_k, \ \chi_k\in C^\infty_c(\Omega). $ Let $\nu\in C^\infty(\mathbb R^n)$, vanishing on $B(0,1/2)$, equal to 1 on $B(0,1)^c$: let us define $$ \zeta(x,\xi)=\sum_{k\ge 1}\chi_k(x) \nu(\xi/R_k), $$ where the positive $R_k$ is chosen so that $p(x,\xi)\ge c_k\vert\xi\vert^m$ for $\vert \xi\vert\ge R_k$ for a positive $c_k$.

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