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Let $u \in C^2$. Somehow I tend to believe that the following identity $$ \int_{\partial B_1 (x)} \big\langle (\nabla^2 u) (x) y, y \big\rangle d\sigma_y = C \Delta u(x) $$ holds for some constant $C>0$ depending on $|\partial B_1|$. Here $(\nabla^2 u) (x)$ is the Hessian matrix of $u$ evaluated at $x$. This could depend on the fact that Laplacian is the trace of the Hessian. However, is the above identity true?

QA Ngô
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1 Answers1

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This is solved here, just substitute $M\in \mathbb R^{n\times n}$ with the Hessian of $u$ at $x$.

Jan Bohr
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