8
$\begingroup$

Let $u$ be in the Sobolev space $H^2$. Then the standard definition of the norm is $$ ||u||_{H^2}^2 = \sum_{|\alpha| \leq 2} ||D^{\alpha} u||_{L^2}^2. $$

However, I have also seen it defined this way $$ ||u||_{H^2}^2 = ||u||_{L^2}^2 + \sum_{|\alpha| = 2} ||D^{\alpha} u||_{L^2}^2. $$

Are these two norms equivalent? Clearly, the latter is $\leq$ the former. To get an ineqaulity the other way, is the Poincare inequality used?

$\endgroup$

3 Answers 3

10
$\begingroup$

It's possible Poincaré inequality may not be true, for example if you work on $\mathbb R^n$. However, as @timur says, we have if $u\in\mathcal D(\mathbb R^n)$ $$||D^{e_i}u||_{L^2}^2=\int_{\mathbb R^n}D^{e_i}uD^{e_i}udx=-\int_{\mathbb R^n}uD^{2e_i}udx\leq ||u||_{L^2}||D^{2e_i}u||_{L^2}\leq \frac 12(||u||_{L^2}^2+||D^{2e_i}u||^2_{L^2}),$$ hence by a density argument $$\sum_{|\alpha|=1}||D^{\alpha}u||_{L^2}^2\leq \frac n2||u||_{L^2}+\frac 12\sum_{|\alpha|=2}||D^{\alpha}u||_{L^2}^2.$$ So, if we put $N_1(u)^2:=\sum_{|\alpha|\leq 2}||D^{\alpha}u||_{L^2}^2$ and $N_2(u)^2=||u||_{L^2}^2+\sum_{|\alpha|=2}||D^{\alpha}u||_{L^2}^2$ we have $$N_2(u)\leq N_1(u)\leq N_1(u)+\frac n2||u||_{L^2}+\sum_{|\alpha|=2}||D^{\alpha}u||_{L^2}^2\leq \left(1+\max\left\{1,\frac n2\right\}\right)N_2(u)$$ for all $u\in H^2(\mathbb R^n)$.

$\endgroup$
1
  • 2
    $\begingroup$ What if we do not have vanishing on boundary?! And you do get boundary terms in applying the integration by parts? $\endgroup$ Feb 12, 2020 at 0:37
4
$\begingroup$

In the other direction, use an interpolation inequality of the form $$ \sum_{|\alpha|=1}\|D^\alpha u\|_{L^2}^2\leq A \|u\|_{L^2}^2 + B \sum_{|\alpha|=2}\|D^\alpha u\|_{L^2}^2 $$

where $A$ and $B$ are constants.

$\endgroup$
3
$\begingroup$

As both are Banach spaces with the respective norm, and the second is $\geq$ than the first, then by the Banach's theorem applied to the identity (The inverse of an invertible bounded linear operator between Banach spaces is continuous) both are equivalent.

$\endgroup$
1
  • 1
    $\begingroup$ Nice proof but I'm not convinced. You have to prove that $(L^{p,k}, || \cdot ||_{weaker})$, is a Banach space which seems to me that brings us back to day 1. $\endgroup$ Mar 19, 2021 at 19:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .