Mean values of polynomial and holomorphic matrices

Question

Lemma. Assume $H: \mathbb{R} \to \mathbb{R}^{d \times d}$ is a polynomial of degree $m$, such that for all $x \in \mathbb{R}$, $H(x)$ is a symmetric semidefinite matrix. For all $n \geq 0$ and real polynomials $p(x)$ of degree $n$, $$ \sup_{x \in [-1,1]} \sqrt{p^THp} \leq {1 \over 2}\left({1 \over 2}m+n+1\right)^2 \int_{-1}^1 \sqrt{p^THp}. $$

Proof. Define $q(x) = p^T(x)H(x)p(x)$, a polynomial of degree $m+2n$. Since $q(x) \geq 0$, all the roots must be in conjugate pairs. Collecting one root from each pair, we see that $q(x) = |r(x)|^2$, where $r(x)$ is a polynomial of degree $m/2+n$. From the book of Rahman and Schmeisser, (13.3.2), or by hand, we find that $\sup_{x \in [-1,1]} |r(x)| \leq {1 \over 2}\left({1 \over 2}m+n+1\right)^2 \int_{-1}^1 |r(x)|$.

Question. I would like instead to assume that $H : [-1,1] \to \mathbb{R}^{d \times d}$ is a symmetric semidefinite-valued function that is furthermore analytic on a complex neighborhood of $[-1,1]$. This case obviously doesn't have an $m$, so the estimate would instead read something like: $$ \sup_{x \in [-1,1]} \sqrt{p^THp} \leq C\left(n+1\right)^2 \int_{-1}^1 \sqrt{p^THp}. $$ The constant $C$ presumably depends on $H$ but is otherwise independent of $n$ and $p$. I'm trying to convince the Weierstrass preparation theorem that it's true but it hasn't worked yet.

Answer 1

I hate to post answers so soon after asking, but I think I solved the above problem, so I don't want to waste anyone's time.

Polynomial approximation of $H$. Let $x_0$ be a root of $x \to \det H(x)$ of multiplicity $1 \leq \nu$. Since $H(x)$ is bounded, we thus see that the smallest eigenvalue of $H(x)$ must be at least $\epsilon|x-x_0|^\nu$ for some $\epsilon>0$. Consider the power series of $H(x)$, and the Taylor polynomial $P(x)$ of order $m = \max\{\nu,d\}$: $$ H(x) = \sum_{k=0}^{\infty} H_k (x-x_0)^k \text{ and } P(x) = \sum_{k=0}^{m} H_k (x-x_0)^k. $$ Hopefully, $x \to \det P(x)$ also has a root of order $\nu$ at $x_0$ (this hopefully follows from the fact that $\det X$ is a polynomial of degree $d$ in the entries of the matrix $X$, so the first $m\geq \nu$ terms of $\det P(x)$ coincide with the first $m$ terms of the Taylor series of $\det H(x)$ near $x=x_0$).

$P$ is an "optimal preconditioner" for $H$. We also have the estimate $$ \|H-P\| \leq C|x-x_0|^{m+1}. $$ in a neighborhood of $x_0$. Thus, in a punctured neighborhood of $x_0$ and nonzero vector $v$, $$ \frac{v^TPv}{v^THv} = 1-\frac{v^T(H-P)v}{v^THv} = 1-R, $$ $$ |R| \leq {C \over \epsilon}|x-x_0|^{1+m-\nu} \leq 0.5. $$ This leads to our key estimate $$ 0.5v^TPv \leq v^THv \leq 1.5v^TPv, $$ in some neighborhood of $x_0$, say $x \in (a,b)$ and for arbitrary vector $v$.

The main result. Let $q(x)$ be an arbitrary polynomial of degree $n$. Note that $q^TPq$ is a polynomial of degree $m+2n$. From Rahman and Schmeisser, (13.3.2), which apparently is usable for the "$p$-norm" with $p=1/2$, we find that \begin{align} \sup_{(a,b)} q^THq & \leq 1.5\sup_{(a,b)} q^TPq \\\\ & \leq 1.5(2n+m+1)^4 \left({1 \over b-a}\int_a^b \sqrt{q^TPq}\right)^2 \\\\ & \leq 3(2n+m+1)^4 \left({1 \over b-a}\int_a^b \sqrt{q^THq}\right)^2. \end{align} The result follows by partitioning $[0,1]$ into finitely many intervals, and each interval $(a_k,b_k)$ is either a suitably small neighborhood of the finitely many roots of $\det H(x)$, or $H(x)$ is uniformly bounded below and above on $(a_k,b_k)$.

Commentary. The above was a simplified version of my actual problem so I'm not sure my solution will be usable. It turns out that my function $H(x)$ is also parametrized by some other variable, i.e. $H(x,y)$, but the supremum and integral are w.r.t. $x$. For that, it feels like some matrix version of the Weierstrass preparation theorem would be needed, but I'm currently stuck. I guess I've made the additional assumption that $\det H(x)$ is not uniformly $0$.

Edit. Hm. I just discovered this paper on matrix-valued preparation theorems. Maybe that is what I need.