TITLE: Estimation on elliptic operator
QUESTION [5 upvotes]: Assume the strongly elliptic property, i.e. 
$$\sum_{|\alpha|= m}a_\alpha \xi^\alpha\neq0,\ \forall \xi \in \mathbb{R}^d\backslash\{0\},$$
and
$$\sum_{|\alpha|\leq m}a_\alpha i^{|\alpha|}\xi^\alpha\neq0,\ \forall \xi \in \mathbb{R}^d,$$
where $a_\alpha$ are complext constants and $\alpha$ are multiindices.
Is the following estimation valid?
$$|\sum_{|\alpha|\leq m}a_\alpha i^{|\alpha|}\xi^\alpha|\geq \mathcal{K}(1+|\xi|^2)^{m/2},\ \forall \xi\in\mathbb{R}^d.$$
If so, how to prove?
I am thinking about 
$$f(\tilde{\xi}) = f(t,\xi):=\sum_{|\alpha|\leq m}a_\alpha i^{|\alpha|}t^{m-|\alpha|}\xi^\alpha,$$
which is positive-homogeneous of order $m$ and $t^2+|\xi|^2 = |\tilde{\xi}|^2$.
Now the question would be how to argue that the value of $f(t,\xi)$ on the unit sphere in $\mathbb{R}^{d+1}$ is strictly possitive? Will it do the job?

REPLY [2 votes]: As noted in the comments it is enough to prove that $f>0$ in the unit sphere of $\mathbb{R}^{d+1}$: By continuity of $f$ and compctness of said sphere, $f>\mathcal{K}>0$ there so that, if $r^2=t^2+|\xi|^2$ then
$$
f(t,\xi)=r^mf \left( \frac{t}{r}, \frac{ \xi}{r} \right) > \mathcal{K}r^m,
$$
and taking $t=1$ gives the desired inequality.
To prove that $f>0$ on the unit sphere notice that, by your first condition of ellipticity, 
$$
f(0,\xi)=\left| \sum_{|\alpha|=m} a_\alpha i^m \xi^{\alpha} \right| = \left| \sum_{|\alpha|=m} a_\alpha \xi^{\alpha} \right|>0.
$$
when $\xi\neq0$. If $t\neq 0$ then
$$
f(t, \xi)= |t|^mf \left( 1, \frac{\xi}{t} \right) >0 
$$
by your second condition, since $f(1, \xi)>0$ for all $\xi\in \mathbb{R}^d$.
This implies that $f(t, \xi)>0$ in $\mathbb{R}^{d+1} \setminus  \{0\}$. Since the sphere is contained in this set we're done.