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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. – user31373 Jun 23 '12 at 16:52

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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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oh sorry, $\Delta$ means laplacian. – Miau Jun 23 '12 at 10:11
@Miau The laplacian is defined as $\nabla^{2}$ – Shaktal Jun 23 '12 at 10:12
Even if you define the Laplacian directly as the sum of the second partial derivatives, it is a very easy computation to show that $\Delta=\nabla^2$. – Stefan Geschke Jun 23 '12 at 10:13
sorry, I wrote that $ \nabla^2 f = ( \partial_1^2 f , \cdots , \partial_n^2 f) $. – Miau Jun 23 '12 at 10:14
This is your definition of $\nabla^2$ now? This is certainly not the usual definition. – Stefan Geschke Jun 23 '12 at 10:21

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