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While studying elliptic operators, I encountered the following problem, which I'm having problems to prove or give a counter-example:

Let $\Omega$ be an open subset of $\mathbb{R}^m$, and suppose that for every element $x\in\Omega$ there is associated a positive symmetric matrix $A(x)=[a^{ij}(x)]_{i,j=1}^n\in M_n(\mathbb{R})$. Let $\lambda(x)$ and $\Lambda(x)$ denote, respectively, the minimum and maximum eigenvalues of $A(x)$. Suppose that

(i) $\sup_{x\in\Omega}|a^{ii}(x)|<\infty$; and

(ii) $\inf_{x\in\Omega}\lambda(x)>0$, that is, there exists a positive number $\lambda_0\in(0,+\infty)$ such that, for every $x\in\Omega$, $\lambda_0<\lambda(x)$.

Can one prove that $\sup_{x\in\Omega}|\Lambda(x)|<\infty$?

I couldn't relate the maximum and minimum eigenvalues in a good way. I've tried using the fact that all norms on finite-dimensional spaces are equivalent, so I could use something like the spectral norm, but I didn't succeed.

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Let $\sup_{x,i} |a_{ii}(x)| = M$ and let $\lambda(x) = \lambda_1(x) \le \dots \le \lambda_n(x) = \Lambda(x)$ be the eigenvalues of $A(x)$. Then $$ 0 \le \Lambda(x) \le \sum_i \lambda_i(x) = \sum_i a_{ii}(x) \le nM \, $$ The assumption that $\lambda(\cdot)$ is bounded away from $0$ is not needed.

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