# Is strongly elliptic operator necessarily properly elliptic?

Let us consider $$A(D)u:=\sum_{|p|\leq l} a_p D^p u,$$ where $a_p$ may be complex, and define $$A_0(\xi)= \sum_{|p|=l} a_p \xi^p$$

Def 1The operator $A$ is called strongly elliptic if $$A_0(\xi)\neq 0 \quad \forall\, \xi\in \mathbb{R}^n \,\,\text{with}\,\,\xi\neq 0.$$

Def 2. The operator defined above with $l=2m$ is said to be properly elliptic if it is elliptic and if for any linearly independent $\xi$ and $\xi'\in \mathbb{R}^n$ belonging to $\mathbb{R}^n$, the polynomial $A_0(\xi+\tau \xi')$ in the complex number $\tau$ has $m$ roots with positive imaginary part.
Def 3 The operator defined above with $l=2m$ is said to be strongly elliptic if there exist a complex number $\gamma$ and a positive constant $\alpha$ such that $$\Re (\gamma A_0(\xi))\geq \alpha |\xi|^{2m} \,\,\,\,\forall\, \xi\in \mathbb{R}^n.$$

Obviously, strongyly elliptic is elliptic. If $n>2$, then we can find any elliptic operator is properly elliptic. Is strongly elliptic operator necessarily properly elliptic?

Here is my attempt to attack it. For the case $n>2$, you know that it is easy. If $n>2$, then we can pass continuously from $A(\xi+\tau \xi')$ into $A(-\xi+\tau \xi')$, because the set $\{\xi\in \mathbb{R}^n; \xi\not\in \{\mathbb{R}\xi_0\}\}$ is connected. Under such passage, we know that the numbers of roots in the upper and the lower half-plane are preserved, and these numbers are equal.

For the case $n=2$, let us add one dimension. Let us consider $$B_0:=A_0(\xi)+ \lambda^{2m} i$$ Then $B_0$ is elliptic and hence it is properly elliptic, since now the dimension is large than 0. Indeed, if $B_0(\xi_0,\lambda_0)=0$, then we have $\Re A_0(\xi_0,\lambda_0)=\Re B_0(\xi_0,\lambda_0)=0$, which implies that $\xi_0=0$. Then we have $B_0(\xi_0,\lambda_0)=i\lambda_0^p=0$.