Let say we have a square symmetric matrix $A(\mathbf{x})$ of size $n\times n$ where each element $a_{ij} = a_{ij}(\mathbf{x})$ is a real-value function with $n$ real variables $\mathbf{x} = (x_1,\dots, x_n)$. Suppose that there exists two positive constants $C_1$ and $C_2$ such that for every vector $v\in \mathbb{R}^n$, the following inequality holds. $$ C_1\Vert v\Vert_2^2 \leq v^TA(\mathbf{x})v \leq C_2\Vert v\Vert_2^2. $$
I have three following questions:
(1) Do the “eigenvalues” of $A$ are functions of $\mathbf{x}$?
(2) Can we conclude that all eigenvalues of $A$ are bounded within the range $(C_1, C_2)$?
(3) How about the eigenvalues (or perhaps eigenfunctions) of the following matrix $$ D = \left[\dfrac{\partial a_{ij}}{\partial x_i}\right]_{i,j=1}^n $$ can we find the range of those eigenvalues?
The concept of a matrix that takes indices as functions is quite unfamiliar to me, so it would be great if someone gives me a brief summary so I can understand.
Edit: I forgot to mention that $A$ is symmetric, and I changed question (3) by a more specific way.
The way to think of it is, for every particular value of $\bf x$, $A({\bf x})$ is an ordinary real symmetric matrix, and has all the usual properties of such matrices. In particular, (2) is true because it is true for real symmetric matrices. It turns out that the eigenvalues are continuous functions of $\bf x$, if the matrix elements are continuous, and at least as much differentiable as the matrix elements are (this would not be the case if $A({\bf x})$ was not symmetric).