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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.

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    $\begingroup$ Do you assume the matrix $A$ to be symmetric? If $\sigma$ denotes symmetrization, one typically has $\sigma(A^0)\neq(\sigma A)^0$. $\endgroup$
    – Joonas Ilmavirta
    Jul 14, 2015 at 11:50
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    $\begingroup$ Yes, sure. I have forgotten that. Thanks a lot. So, I have forgotten to add the assumption that $A$ is symmetric matrix. $\endgroup$
    – Darko
    Jul 14, 2015 at 13:10

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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 $-1$.

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