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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?

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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 $$\Vert u\Vert=\left(\Vert \hat u\Vert_{L^2}^2+\Vert |\xi|^2 \hat u\Vert_{L^2}^2\right)^{\frac{1}{2}}.$$

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  • $\begingroup$ 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 $L^2$. $\endgroup$
    – Sue
    Mar 14, 2013 at 11:04
  • $\begingroup$ Peter Michor's suggestion is correct, but you might want to check Gårding's inequality for a possibly even easier argument. $\endgroup$ Mar 16, 2013 at 4:54
  • $\begingroup$ Does this holds on bounded domains? $\endgroup$ Dec 6, 2019 at 15:31
  • $\begingroup$ @Afonso yes, provided either the boundary is $C^{1,1}$ or the domain is convex. $\endgroup$
    – username
    Apr 5, 2021 at 13:44

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