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I hope to prove this, $ | \Delta f | \leqslant n | \nabla^2 f| $ where $f : \mathbb R^n \to \mathbb R $.

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    -1 for unexplained notation in the question.2012-06-23

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Use Cauchy-Schwarz inequality $ |\Delta f|=\left|\sum\limits_{i=1}^n \partial_i^2 f\right|= \left|\sum\limits_{i=1}^n 1\cdot\partial_i^2 f\right|\leq \left(\sum\limits_{i=1}^n 1^2\right)^{1/2}\left(\sum\limits_{i=1}^n (\partial_i^2 f)^2\right)^{1/2}= \sqrt{n}|\nabla^2 f|\leq n|\nabla^2 f| $

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I don't understand. Isn't $\Delta f=\nabla^2f$? This is assuming that the first $\nabla$ means divergence, the second gradient.

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    Moreover, I'd probably write $\partial_{ii}^2$ for the second derivative with respect to the $i$-th variable.2012-06-23