I am considering the quadratic form $u^T H(x) u$ where $H$ is the Hessian as a function of the spatial coordinate $x$. I am interested in finding a bound of the form: $|u^T H(x) u| \le Ku^Tu$ where $K$ is a constant to be determined.
Suppose that $x$ is fixed, i.e. $H(x) = H$. We know that since $H = Q^T \Lambda Q$ (where $Q$ is orthogonal and $\Lambda$ is diagonal), $u^T H u = u^TQ^T \Lambda Qu = y^T\Lambda y = \sum_k \lambda_iy_i^2$ where $\lambda_i$ is the $i$-th diagonal entry of $\Lambda$.
Now, $u^THu \le (\text{max}|\lambda_i| )\,||y||^2 = K ||u||^2$ where $K$ is the maximum absolute eigenvalue.
How does this extend if $H$ is a function of $x$? Can I obtain a $K$ independent of $x$? The assumption here is that the second order partial derivatives are bounded over the domain that I am considering. Since the eigenvalues are computed from entries of $H(x)$, and since entries of $H(x)$ are bounded over the domain, is what I'm trying to show trivial?
Let $x\in D\rightarrow f(x)$ be a $C^2$ function, $H=D^2f$ be its Hessian and $(\lambda_i(x))_i=spectrum(H(x))$. Then $x\rightarrow H(x)$ is continuous and, consequently, $\phi:x\in D \rightarrow sup_i |\lambda_i(x)|$ is continuous. Assume that $D$ is compact; then $\phi$ is bounded by $K$ and, for every $u\in\mathbb{R}^n,x\in D$, one has $|u^TH(x)u|\leq K||u||^2$.