Uniform quadratic-form bounds for positive definite symmetric matrices and smoothness of the inverse

(Spectral representation and the bounds (a).) By [L1] choose an orthonormal basis $e_1,\dots ,e_m$ of ${\mathbb{R}}^m$ with $B{e}_i={\lambda }_i{e}_i$ and real ${\lambda }_i$. Writing $z={\sum }_i{z}_i{e}_i$ gives $z^{\mathsf{T}}Bz={\sum }_i{\lambda }_i{z}_i^2$ and $\mathrm{\mid }z{\mathrm{\mid }}^2={\sum }_i{z}_i^2$. If some ${\lambda }_i\le 0$, then $e_i^{\mathsf{T}}B{e}_i={\lambda }_i\le 0$ with $e_i\mathrm{\ne }0$, contradicting positive definiteness; so all ${\lambda }_i>0$. Put $a:={\min }_i{\lambda }_i$ and $b:={\max }_i{\lambda }_i$; both are positive and $a\mathrm{\mid }z{\mathrm{\mid }}^2\le z^{\mathsf{T}}Bz\le b\mathrm{\mid }z{\mathrm{\mid }}^2$ follows from the representation. This is (a).

(Uniform bounds for an arbitrary compact topological parameter space (c).) If $\mathrm{\Lambda }=\mathrm{\varnothing }$, take $a=b=1$ and the conclusion is vacuous. For each $\lambda \in \mathrm{\Lambda }$, let $a_{\lambda }>0$ and $b_{\lambda }$ be the uniquely determined smallest and largest eigenvalues of $B(\lambda )$ from step 1.1. Continuity of the finitely many matrix entries gives an open neighborhood $V_{\lambda }$ on which every entry of $B(\mu )-B(\lambda )$ has absolute value less than $a_{\lambda }\mathrm{/}(2m)$. For $\mu \in V_{\lambda }$, $\mathrm{\mid }{z}^{\mathsf{T}}(B(\mu )-B(\lambda ))z\mathrm{\mid }\le \frac{a_{\lambda }}{2m}{\left({\sum }_i\mathrm{\mid }{z}_i\mathrm{\mid }\right)}^2\le \frac{a_{\lambda }}{2}\mathrm{\mid }z{\mathrm{\mid }}^2,$ where the last inequality follows by summing $2\mathrm{\mid }{z}_i{z}_j\mathrm{\mid }\le z_i^2+z_j^2$ over the finite pairs. Hence $(a_{\lambda }\mathrm{/}2)\mathrm{\mid }z{\mathrm{\mid }}^2\le z^{\mathsf{T}}B(\mu )z\le (b_{\lambda }+a_{\lambda }\mathrm{/}2)\mathrm{\mid }z{\mathrm{\mid }}^2$. The neighborhoods $V_{\lambda }$ cover $\mathrm{\Lambda }$; compactness, in its open-cover definition, gives finitely many $V_{{\lambda }_1},\dots ,V_{{\lambda }_q}$ covering it, with $q\ge 1$ in the nonempty case. Set $a={\min }_j{a}_{{\lambda }_j}\mathrm{/}2>0$ and $b={\max }_j(b_{{\lambda }_j}+a_{{\lambda }_j}\mathrm{/}2)$. The preceding inequalities give the uniform bounds at every parameter. This uses no metric on $\mathrm{\Lambda }$ and selects only a finite subcover; the eigenvalue constants are canonical, so no arbitrary-index choice of bases is needed.

(Inverse bounds (b).) In the same eigenbasis, $B^{-1}{e}_i={\lambda }_i^{-1}{e}_i$ and ${\lambda }_i^{-1}\in [b^{-1},a^{-1}]$, so $b^{-1}\mathrm{\mid }z{\mathrm{\mid }}^2\le z^{\mathsf{T}}{B}^{-1}z\le a^{-1}\mathrm{\mid }z{\mathrm{\mid }}^2$; in particular $B^{-1}$ exists.

(Smoothness of the inverse (d).) On $D$, $x\mapsto \det B(x)\ge a^m>0$: the determinant is the product of the eigenvalues, each at least $a$ by (a) applied at $x$, or equivalently $\det$ is continuous and positive. The adjugate entries are polynomial in the entries of $B$, hence smooth in $x$, and dividing by the smooth nonzero function $\det B$ is smooth by [L4]; so $B(x{)}^{-1}$ is smooth and satisfies the bound by (b).