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Is this right if $ k \geqslant 1$ ? Then why? $$ | \nabla f |^k \leqslant \sum_{i=1}^n | \partial_i f |^k \; $$

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up vote 1 down vote accepted

This is essentially just the triangle equality for the so called $p$-norm (in your case $p=k$) together with the monotonicity of the $k$-th root. See this wikipedia article.

Edit: I was wrong, this now doesn't seem to have to do with $p$-norms. What was I thinking? What you want to show is that for a vector $x=(x_1,\dots,x_n)\in\mathbb R^n$ with all components $\geq 0$ the following holds: $$(x_1^2+\dots+x_n^2)^{k/2}\leq x_1^k+\dots+x_n^k$$

Since everything is $\geq 0$, this inequality holds iff it holds for the squares of both sides. So you want to show $$(x_1^2+\dots+x_n^2)^k\leq(x_1^k+\dots+x_n^k)^2.$$ But this is actually not true:

Suppose $x_1=x_2=1$ and $k=3$. Then $$(x_1^2+x_2^2)^k=2^3=8\not\leq (x_1^k+x_2^k)^2=2^2=4.$$ You get a counterexample to your inequality from this by choosing a function $f:\mathbb R^2\to\mathbb R$ with $\partial f/\partial x_1=\partial f/\partial x_2=1$.

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Thank you, I have reduced that $ |\nabla f |^k = (\sum_{i=1}^{n} ( \partial_i f )^2 )^{(k/2)}$. But I don't know what to do next.. – Miau Jun 23 '12 at 10:04

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