Let $p(\xi) = \sum_{|\alpha|\le k} A_\alpha \xi^\alpha$ be a matrix valued polynomial in $\xi \in \Bbb R^n$. Where $\xi^{\alpha} = \xi^{\alpha_1} \cdots \xi^{\alpha_k}$. $A_\alpha$ are $\Bbb R$- matrices of fixed dimension indexed by $\alpha$, not all equal to $0$, for $|\alpha|=k$.
Suppose $\sum_{|\alpha|=k} A_\alpha \xi^\alpha =0$ if and only if $\xi=0$. Show there exists $c,R > 0$ such that $$ |p(\xi)| \ge c(1+|\xi|^2)^{k/2} $$ for all $|\xi| \ge R$. (I excluded $x$ as the coefficients are constant).
How does one obtain this?
The origin is a claim made in pg41, line +5.
Consider the function $q: \mathbb{R}^n \to \mathbb{R}$ given by $q(\xi) = |\sum_{|\alpha| = k} A_\alpha \xi^\alpha |$ and note that this function is clearly continuous and non-negative. By hypothesis we know that $q(\xi) =0$ if and only if $\xi=0$, so this implies that $q$ does not vanish when restricted to the unit sphere $\mathbb{S}^{n-1} \subset \mathbb{R}^n$. By the extreme value theorem, the minimum of $q$ over $\mathbb{S}^{n-1}$ is achieved and is positive, so there exists $B >0$ such that $q(\xi) \ge B$ for all $\xi \in \mathbb{S}^{n-1}$. Now note that for $\xi \neq 0$ we have that $\xi/|\xi| \in \mathbb{S}^{n-1}$, so $q(\xi/|\xi|) \ge B$. However, $q$ is homogeneous by construction, so $$ q(\xi) = |\xi|^k q(\xi/|\xi|) \ge B |\xi|^k. $$ This holds for $\xi \neq 0$, but it also clearly holds for $\xi =0$.
Now we use the triangle inequality to estimate $$ |p(\xi)| = |\sum_{|\alpha|=k} A_\alpha \xi^\alpha + \sum_{|\alpha| \le k-1} A_\alpha \xi^\alpha | \ge q(\xi) - | \sum_{|\alpha| \le k-1} A_\alpha \xi^\alpha| \ge B |\xi|^k - |\sum_{|\alpha| \le k-1} A_\alpha \xi^\alpha |. $$ On the other hand, numerous applications of Young's inequality (or any number of variants) show that there exists a constant $D >0$ (depending on $k$ and the matrix norms of the various matrices $A_\alpha$) such that $$ |\sum_{|\alpha| \le k-1} A_\alpha \xi^\alpha | \le D(1 + |\xi|^{k-1}). $$ Combining these then shows that $$ |p(\xi)| \ge B |\xi|^k - D(1+|\xi|^{k-1}) = \frac{B}{2} |\xi|^k + \frac{B}{2} |\xi|^k - D(1+|\xi|^{k-1}). $$ Now choose $R >2D/B$ such that $$ \left( \frac{B}{2} R - D \right) R^{k-1} \ge \frac{B}{2} +D, $$ which is possible since the left side blows up as $R \to \infty$. With this choice of $R$ we have that if $|\xi| \ge R$, then $$ \left( \frac{B}{2} |\xi| - D \right) |\xi|^{k-1} \ge \left( \frac{B}{2} R - D \right) R^{k-1} \ge \frac{B}{2} +D $$ and hence that $$ \frac{B}{2} |\xi|^k - D(1+ |\xi|^{k-1}) \ge \frac{B}{2}. $$ Thus, for this choice of $R$ we have that if $|\xi| \ge R$, then $$ |p(\xi)| \ge \frac{B}{2} |\xi|^k + \frac{B}{2} = \frac{B}{2}(1 + |\xi|^k). $$ Finally, we note that the binomial theorem and Young's inequality allow us to estimate $$ (1 + |\xi|^k) \ge E (1+|\xi|^2)^{k/2} $$ for some constant $E >0$, and so the desired estimate with $C = EB/2$ follows from the previous two inequalities.