# Completion of $C_0^{\infty}(\mathbb{R}^N)$ with norm $\|u\|= \Bigg(\int_{{\mathbb{R}}^N} |\Delta u |^2 \, \mathrm{d}x \Bigg)^{\frac{1}{2}}.$

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 $$D^{2,2}(\mathbb{R}^N) \subset H$$.

• Integrate by parts a couple of times. – Deane Yang Aug 19 '15 at 12:55
• 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 $D^{2,2}(\mathbb{R}^N)$. – Hheepp Aug 19 '15 at 13:04

"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 $$u\in C_0^\infty(\mathbb{R}^N)$$, 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 $$L^{\frac{2N}{N-4}}$$.