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Consider $\sum_{i,j=1}^n \displaystyle\int_{\mathbb{R}^n} \dfrac{\partial^2 u}{\partial^2 x_i} \overline{\dfrac{\partial^2 v}{\partial^2 x_j} } dx + \lambda \displaystyle\int_{\mathbb{R}^n} u \overline{v} dx$ and with this scalar product we can define a norm which is equivalent to the classical norm on $H^2(\mathbb{R^n})$.

How can use the Riesz theorem to prove that, for all $\lambda > 0,$ and for all $f \in H^{-2}(\mathbb{R^n})$ there exist a unique $u \in H^2(\mathbb{R^n})$ solution of the equation $(\Delta^2 + \lambda) u = f$?


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Hint (or solution...): Take the Riesz-representative of $f$.

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The Riesz-representation theorem says that: let $1 < p < \infty$ and let $\varphi \in (L^p)^*$. Then, there exists a unique function $u \in L^{p'}$ such that $$< \varphi , f > = \displaystyle\int u f, \forall f \in L^p$$ Morever $||u||_{p'}=||\varphi||_{(L^p)^*}$ and my problem is how use the Riese-Representation of $f \in H^{-2}(\mathbb{R^n})$ – jijiii Apr 26 '13 at 17:44
Okay, so let $f \in H^{-2}.$ By Riesz - Representation theorem, there exist a unique $u \in H^2$ such that $$< f , \varphi> = < f , \varphi >, \forall \varphi \in H^2$$ then, by definition of product scalar, we have $$< f , \varphi > = \displaystyle\int_{\mathbb{R}^n} \sum_{i,j=1,...,n} \dfrac{\partial^2 u}{\partial x_i^2} \dfrac{\partial^2 \varphi}{\partial x_j^2} dx + \lambda \displaystyle\int_{\mathbb{R}^n} u \varphi dx$$ – jijiii Apr 26 '13 at 21:09
Finally, how we can deduce rigouroysly that $(\Delta^2 + \lambda) u = f$? My difficult is to lie between $$< f , \varphi > = \displaystyle\int_{\mathbb{R}^n} \sum_{i,j=1,...,n} \dfrac{\partial^2 u}{\partial x_i^2} \dfrac{\partial^2 \varphi}{\partial x_j^2} dx + \lambda \displaystyle\int_{\mathbb{R}^n} u \varphi dx$$ and $$(\Delta^2 + \lambda) u = f$$ – jijiii Apr 26 '13 at 21:12
Any one can help me please – jijiii Apr 27 '13 at 9:14

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