2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 '13 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. $\endgroup$ – Delio Mugnolo Mar 16 '13 at 4:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.