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Based in many questions and answers like [1, 2, 3 ] and a comment a good comment here [4]. I would like to know that the space $H=H_0^1(\Omega)\cap H^2(\Omega)$ can be equiped with this norm $$\tag{1}\|\cdot\|_H=||\Delta\cdot||_{L^2}+||\nabla\cdot||_{L^2}+||\cdot||_{L^2},$$ this is the $H^2$-norm. Or, is it equipped with this one $$\tag{2}\|\cdot\|_H=||\Delta\cdot||_{L^2}+||\cdot||_{L^2}.$$ Can I say that if $H$ is equipped with (1) then, normes (1), (2) and $||\Delta\cdot||_{L^2}$ are equivalent in $H$.

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  • $\begingroup$ What are the assumptions for $\Omega$? $\endgroup$ – user9464 Nov 9 '16 at 16:21
  • $\begingroup$ @Jack, we can assume that the domain $\Omega$ is bounded in $\mathbb{R}^2$ and with Lipschitz boundary $\endgroup$ – Achaire Nov 9 '16 at 17:40
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This depends on the regularity of the domain: If $H^2$-regularity for the equation $-\Delta u=f$ is available then one has $$ \|u\|_{H^2(\Omega)} \le c \|f\|_{L^2(\Omega)} = c \|\Delta u\|_{L^2(\Omega)}, $$ and your norm is equivalent to the $H^2$-norm.

If $\Omega$ is not nice, i.e., it has a non-convex corner, then there are functions $y\in H^1_0(\Omega)$ with $\Delta y\in L^2(\Omega)$ but $y\not\in H^2(\Omega)$.

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  • $\begingroup$ Thanks for your helphol answer, the domain $\Omega$ is bounded in $\mathbb{R}^2$ and with Lipschitz boundary. if I'll use the fourth- order equation like $\Delta\Delta u= f$ where $u\in H$ is it done also??? $\endgroup$ – Achaire Nov 12 '16 at 11:45

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