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I have a question that it seems simple but I can not solve it.

Let $\Omega$ be the unit ball centered at zero in $\mathbb{R}^N$, $N>4$. Assume that $C_{0,rad}^{\infty}(\Omega)$ is the space of all radial symmetric smooth functions with compact support.

What is the completion of $C_{0,rad}^{\infty}(\Omega)$ with respect to the following norm

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

Is it $W_0^{2,2}(\Omega)$ or $W_{0,rad}^{2,2}(\Omega)$.

I think that the answer is $W_0^{2,2}(\Omega)$ but I have no idea to prove it.

I will be thanked to any useful hint or answer.

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    $\begingroup$ I don't see how it can possibly be $W^{2,2}_0(\Omega)$. For instance, the completion certainly can't contain any function whose Laplacian isn't radially symmetric. $\endgroup$
    – Nate Eldredge
    Commented Jan 11, 2016 at 3:56
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    $\begingroup$ Maybe you already know that, but by integration by parts your norm is equal to $\left(\int_\Omega|D^2u|^2dx\right)^{1/2}$ since $\langle \nabla u,\nabla\Delta u\rangle=-|D^2u|^2+\tfrac{1}{2}\Delta|\nabla u|^2$ where $|D^2u|^2$ is the Frobenius norm of the Hessian. $\endgroup$
    – Thomas Richard
    Commented Jan 11, 2016 at 10:30
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    $\begingroup$ Thomas's remark, along with the definitions of the two spaces you're trying to decide between, immediately gives the answer. This is more appropriate for math.stackexchange.com $\endgroup$
    – Deane Yang
    Commented Jan 11, 2016 at 13:51

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