Following are two definitions of constant coefficient elliptic operator:
An operator of order m is called elliptic
Definition 1: if the top-order symbol $p_m(\xi)$ has no real zeroes except $\xi = 0$.
Definition 2: if the full symbol satisfies $|p(\xi)| \geq c|\xi|^m $ for $|\xi|\geq A$, for some positive constants $c$ and $A$.
I need to see the equivalence of the above two definitions. Any help will be deeply acknowledged.
2026-02-22 19:29:02.1771788542
Definition of constant coefficient elliptic operator
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Take your first definition. If $P(D)$ is elliptic, since $P_m(\xi)$ is a continuous function, by Weierstrass theorem we have that \begin{align*} \exists \eta \in S^{n-1}:=\lbrace \xi \in \mathbb{R}^n : |\xi|=1 \rbrace : |P_m(\xi)| \geq |P_m(\eta)| , \forall \xi \in S^{n-1} \end{align*} and $\eta \neq 0$, then $\mu:=|P_m(\eta)| > 0$, now let $\xi \in \mathbb{R}^n \setminus \lbrace 0 \rbrace$ and put $\xi /|\xi| \in S^{n-1}$, therefore \begin{align*} \mu \leq \left| P_m \left( \frac{1}{|\xi|}\xi \right) \right| = \frac{|P_m(\xi)|}{|\xi|^{m}} \end{align*} and $Q(\xi):=P(\xi) - P_m(\xi)$ is a polynomial of degree $|\alpha| \leq m-1$, that is \begin{align*} \exists d > 0 : |Q(\xi)| \leq d |\xi|^{m-1} , \forall |\xi| \geq 1 \end{align*} by these two estimates follows that \begin{align*} |P(\xi)| \geq |P_m(\xi)| - |Q(\xi)| \geq \mu |\xi|^m -d|\xi|^{m-1} = \left( \mu - \frac{d}{|\xi|} \right) |\xi|^m \end{align*} Choosing $C=\mu -d/R$, that is $R > d/\mu$ and $R \geq 1$, follows the first implication or your second definition. Let's assume the hypothesis that $\forall |\xi| \geq R >0$ there is $C >0$ such that $|P(\xi)| \geq C |\xi|^m$. If by absurdity $P(D)$ is not elliptic then $\exists \xi \in \mathbb{R}^n \setminus \lbrace 0 \rbrace$ : $P_m(\xi)=0$, therefore $\forall t \in \mathbb{R}^n$ we have that $P_m(t\xi)=t^m P_m(\xi)=0$,then $P(t\xi)=P_m(t\xi)+Q(t\xi)=Q(t\xi)$ is a polynomial in the variable $t \in \mathbb{R}^n$ of degree $|\alpha| < m$, contrary to the fact that $|P(\xi)| \geq C|\xi|^m$ $\forall \xi \in \mathbb{R}^n$ such that $|\xi| \geq R$.