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Is it the case that $$|v|_{H^2(\Omega)}^2 = |v|_{H^1(\Omega)}^2 + \int_\Omega\sum_{i=1}^n |\nabla v_{x_i}|^2?$$ I think yes but I have never seen anybody write it like this. I guess generalisations are possible for $H^k$ functions. Is there any nicer way to write the RHS, so I want to make no reference to partial differentiation but gradients are OK..

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The two norms are definitely equivalent, and might be equal, depending on your exact definition of $|v|_{H^2(\Omega)}$. To show the equivalence, you might need to use the Cauchy-Schwartz inequality and the fact that $\frac{\partial}{\partial x_i} \left( \frac{\partial v}{\partial x_j} \right) = \frac{\partial^2 v}{\partial x_i \partial x_j}$ which is immediate from the definition (as typically a priori the left hand and the right hand have different definitions).

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Thanks. Do you know any way to make the RHS better looking? I want only grad terms involved and no partials. –  mr. x Dec 9 '12 at 19:33
    
Nothing that is not a cosmetic makeup. That is, the right hand side should contain in some way the second partial derivatives. You can hide them as $||\nabla v||_{H^1(\Omega)}$ if you interpret the norm of a vector-valued Sobolev function appropriately, but this just hides away the second partial derivatives inside $||\nabla v||_{H^1(\Omega)}$. –  levap Dec 9 '12 at 19:44

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