I have a question about duality in Sobolev spaces .Let $\Omega $ be an open of $R$,Suppose that u$\in H_{0}^{2}(\Omega )\cap H^{3}(\Omega )$ and $v\in H_{0}^{2}(\Omega ).$ Is that this expression make a sens ? $\int\limits_\Omega {\partial _x^4u.vdx} $. Can i make a integration by part to get $\int\limits_\Omega {\partial _x^2u\partial _x^2vdx} $ ? Thank you ?
Dualty is Sobolev spaces
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sobolev-spaces
1 Answers
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$$u \in H_{0}^{2}(\Omega )\cap H^{3}(\Omega ),v\in H_{0}^{2}(\Omega )$$
As you said by integration by parts we get
$$\int_\Omega {\partial _x^4u ~v~\text{d}x} =\underbrace{\big[\partial_x^3 u~ v\big]_{\partial \Omega}}_{=0, ~v\in H_0^2} – \int_\Omega \partial_x^3 u~ \partial_x v ~\text{d}x=-\underbrace{\big[\partial_x^2 u~ \partial_x v\big]_{\partial \Omega}}_{=0, ~v\in H_0^2}+\int_\Omega \partial_x^2 u~ \partial_x^2 v ~\text{d}x$$
and using Cauchy-Schwarz we get that this integral exists
$$\int_\Omega \partial_x^2 u~ \partial_x^2 v ~\text{d}x \leq ||u''||_2 ||v''||_2<\infty$$
since $u \in H^2$ and $v \in H^2$ and therefore by definition $u'' \in L^2(\Omega)$ and $v'' \in L^2(\Omega)$. The integral in between also exists since we can bound it by $||u'''||_2 ||v'||_2$.
But as you mentioned does the first integral really make sense? Only implicitly by the equality to the integral on the right hand side. But just by the assumption we can't say that the fourth weak derivative of $u$ exists.