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

  • 3
    $\begingroup$ 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 $D^{2,2}(\mathbb{R}^N)$. $\endgroup$ – 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 self-advertizing is allowed: maybe have a look at my paper "Splines minimizing rotation-invariant semi-norms in Sobolev spaces", 1977).

| cite | improve this answer | |

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 $L^{\frac{2N}{N-4}}$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.