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I have a question that I could not find it any where.

Is the completion of $C_0^{\infty}(\mathbb{R}^N)$ with the respect to norm

$$\|u\|= \Bigg(\int_{{\mathbb{R}}^N} |\Delta u |^2 \, \mathrm{d}x \Bigg)^{\frac{1}{2}}, $$

well-known space?

I know that completion of $C_0^{\infty}(\mathbb{R}^N)$ with the respect to norm below $$\|u\|= \Bigg(\int_{{\mathbb{R}}^N} | \nabla u |^2 \, \mathrm{d}x \Bigg)^{\frac{1}{2}}, $$ is well-known space $D^{1,2}(\mathbb{R}^N)=\Big \{ u \in L^{\frac{2N}{N-2}}(\mathbb{R}^N); \,D^{\alpha}u \in L^2(\mathbb{R}^N) \,\,\,\,for \, all \, |\alpha|=1\} $

and also I know that completion of $C_0^{\infty}(\mathbb{R}^N)$ with the respect to norm below $$\|u\|= \Bigg(\int_{{\mathbb{R}}^N} | D^2 u |^2 \, \mathrm{d}x \Bigg)^{\frac{1}{2}}, $$ is $D^{2,2}(\mathbb{R}^N)=\Big \{ u \in L^{\frac{2N}{N-4}}(\mathbb{R}^N); \,D^{\alpha}u \in L^2(\mathbb{R}^N) \,\,\,\,for \, all \, |\alpha|=2\}. $

It is obvious if I name the completion space with $H$ then $D^{2,2}(\mathbb{R}^N) \subset H$.

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    $\begingroup$ Integrate by parts a couple of times. $\endgroup$ – Deane Yang Aug 19 '15 at 12:55
  • $\begingroup$ Did you mean this norm is equivalence with $\|u\|= \Bigg(\int_{{\mathbb{R}}^N} | D^2 u |^2 \, \mathrm{d}x \Bigg)^{\frac{1}{2}}$, so Completion is $D^{2,2}(\mathbb{R}^N)$. $\endgroup$ – Finish Aug 19 '15 at 13:04
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"The" completion is not always a space of functions, for $N=1$ or $2$ for example it is a quotient $D^{-2}L^2/P_1$ (equivalence classes of functions $u\in H^2_{loc}$ with $\partial_i \partial_j u\in L^2$ with $u\sim v$ iff $u(x)-v(x)=a\cdot x+b$.

For $N=3$ and $4$ it is $D^{-1}\dot{H}^1/P_0$ where the "homogeneous Sobolev space" $\dot{H}^1$ is "the" completion of $C_0^\infty$ with respect to $\int|\nabla v|^2$ (and $P_0$ is the space of constants).

Only for $N\ge 5$ is "the" completion with respect to $\int (\Delta v)^2$ a function space: $\dot{H}^2$, the space of tempered distributions $v$ (in fact, functions in $L^{2N/(N-4)}$) whose Fourier transform $\hat{v}\in L^2_{loc}$ with $\int [|\xi|^2\hat{v}(\xi)]^2<\infty$. Same space as your $D^{2,2}$, because $|\xi|^4=\sum(\xi_i \xi_j)^2$.

(If self-advertizing is allowed: maybe have a look at my paper "Splines minimizing rotation-invariant semi-norms in Sobolev spaces", 1977).

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If $u\in C_0^\infty(\mathbb{R}^N)$, then the integration by parts (twice) yields \begin{split} &\int_{\mathbb{R}^N}|\Delta u|^2 = \int_{\mathbb{R}^N}\left(\sum_i\frac{\partial^2 u}{\partial x_i^2}\right)\left(\sum_j\frac{\partial^2 u}{\partial x_j^2}\right) =\sum_{i,j}\int_{\mathbb{R}^N}\frac{\partial^2 u}{\partial x_i^2}\frac{\partial^2 u}{\partial x_j^2}\\ &= -\sum_{i,j}\int_{\mathbb{R}^N}\frac{\partial^3 u}{\partial x_j\partial x_i^2}\frac{\partial u}{\partial x_j} = \sum_{i,j}\int_{\mathbb{R}^N}\frac{\partial^2 u}{\partial x_j\partial x_i}\frac{\partial^2 u}{\partial x_i\partial x_j} =\sum_{i,j}\int_{\mathbb{R}^N}\left|\frac{\partial^2 u}{\partial x_j\partial x_i}\right|^2\\ &= \int_{\mathbb{R}^N}|D^2u|^2. \end{split} Therefore if $N\geq 5$, you get that the space coincides with $D^{2,2}$. You need $N\geq 5$ to have the embedding into $L^{\frac{2N}{N-4}}$.

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