Let $A(x)=[a\_{ij}(x)]\_{i,j=1,\dots,n}$, $x\in {\bf R}^n$, be a symmetric non-negative definite matrix:
$$
\langle A(x) \xi,\xi \rangle \geq 0 \ \ \forall x,\xi \in {\bf R}^n.
$$ Assume that
$$
a\_{ij}\in C(K), \ \ K\subset {\bf R}^d,
$$ where $K$ is a compact set.


Let 
$$
a\_{ij}^0=
\begin{cases}
\min\limits\_{x\in K} a\_{ij}(x), \ \ i\neq j\\
\max\limits\_{x\in K} a\_{ij}(x), \ \ i= j
\end{cases}
$$ Let 
$$
A^0=[a\_{ij}^0]\_{i,j=1,\dots,n}.
$$ I believe that it holds
$$
\langle A(x) \xi,\xi \rangle \leq n \langle A^0 \xi,\xi \rangle, \ \ \xi\in {\bf R}^n,
$$ but I cannot prove it. I am very grateful for any hint. 



Majorate semidefinite continuous matrix by a constant matrix
This is false. In particular, $A^0$ need not be positive semidefinite. For example, take $n=3$, $K = \{1,2,3\}$, let $v(x)$ be the column vector with a $1$ at position $x$ and $-1$ elsewhere, and let $A(x)=v(x)v(x)^T$. Then
\[
A^0 = \begin{bmatrix} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1\end{bmatrix},
\]
and the all ones vector is an eigenvector with eigenvalue $-1$.


