# Doubt on norm of the Sobolev space $H^2(\mathbb{R}^3)$

If I consider the Sobolev space $H^2(\mathbb{R}^3)$ I have the norm $$\Vert u\Vert_{H^2(\mathbb{R}^3)}=\bigg(\sum_{|\alpha|\leq 2}\Vert D^\alpha u\Vert^2_{L^2(\mathbb{R}^3)}\bigg)^\frac{1}{2}.$$ Is this norm is equivalent to $$\Vert u\Vert=\left(\Vert u\Vert_{L^2}^2+\Vert\Delta u\Vert_{L^2}^2\right)^{\frac{1}{2}}?$$ In this way if I have to show that a function is in $H^2(\mathbb{R}^3)$ it suffices to study the function and its Laplacian,doesn't it?

You should write: $$\Vert u\Vert=\left(\Vert u\Vert_{L^2}^2+\Vert\Delta u\Vert_{L^2}^2\right)^{\frac{1}{2}}.$$ Then the two norms are equivalent. You can see this by either using interpolation inequalities or by going to the Fourier transform where the estimates are easy: $$\Vert u\Vert_{H^2(\mathbb{R}^3)}=\bigg(\sum_{|\alpha|\leq 2}\Vert |\xi^{\alpha}|\hat u\Vert^2_{L^2(\mathbb{R}^3)}\bigg)^\frac{1}{2}$$ and $$\Vert u\Vert=\left(\Vert \hat u\Vert_{L^2}^2+\Vert |\xi|^2 \hat u\Vert_{L^2}^2\right)^{\frac{1}{2}}.$$
• Yes I forgot the square in my question. So in this way when I have to verify that a function is in $H^2$ I can ignore the derivative of order $1$ and just verify that $f$ anf $\Delta f$ are in $L^2$. – Sue Mar 14 '13 at 11:04
• @Afonso yes, provided either the boundary is $C^{1,1}$ or the domain is convex. – username Apr 5 at 13:44