How to prove this, $| \Delta f | \leqslant n | \nabla^2 f|$

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

Use Cauchy-Schwarz inequality $$|\Delta f|=\left|\sum\limits_{i=1}^n \partial_i^2 f\right|= \left|\sum\limits_{i=1}^n 1\cdot\partial_i^2 f\right|\leq \left(\sum\limits_{i=1}^n 1^2\right)^{1/2}\left(\sum\limits_{i=1}^n (\partial_i^2 f)^2\right)^{1/2}= \sqrt{n}|\nabla^2 f|\leq n|\nabla^2 f|$$
I don't understand. Isn't $\Delta f=\nabla^2f$? This is assuming that the first $\nabla$ means divergence, the second gradient.
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