I'm concerned with the topic of elliptic differential operators, especially the Laplace operator, and there is something unclear. So the definition says $L$ is called elliptic, if

$0<\lambda (x)|\xi|^2\leq a_{ij}(x)\xi_i\xi_j\leq\Lambda (x)|\xi|^2,\ \forall \xi=(\xi_1,...,\xi_n)\in\mathbb{R}^n\setminus\{0\}$

holds for $\lambda(x)$ the minimum, respectively $\Lambda(x)$ the maximum of the eigenvalues of $A(x)=(a_{ij}(x))$.

For the Laplace operator it is said that $A=I$ so $\lambda=\Lambda=1=a_{ii}$. So I have to show, that

$1|\xi|^2 = \xi_1^2+\xi_2^2+\xi_3^2\leq \xi_i\xi_i$ in $\mathbb{R}^3$. But for $(\xi_1,\xi_2,\xi_3)^t=(1,0,1)^t$ the inequality can not be shown with $\xi_i\xi_i=0$ and th left side is strict positive.

So does "$\xi=(\xi_1,...,\xi_n)\in\mathbb{R}^n\setminus\{0\}$" means "$\xi=(\xi_1,...,\xi_n)\in\mathbb{R}^n$ with $\xi_i\neq 0$ for $1\leq i\leq n$"?


1 Answer 1


Whatever source you are reading is implicitly (or perhaps explicitly explained earlier) using the Einstein summation convention, and

$$ a_{ij} \xi_i\xi_j = \sum_{i,j = 1}^n a_{ij} \xi_i\xi_j $$


$$ \xi_i\xi_i = \sum_{i = 1}^n \xi_i \xi_i $$


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