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Let $\Omega$ be a bounded domain with regular boundary. Assuming that functions $C_c^{\infty}(\Omega)$ are dense in $H_0^2(\Omega)$ (wwith the norm in $H^2(\Omega)$, I need to show that this is actually a norm. $$||u||_{h_0^2(\Omega)}:=\left(\int_{\Omega}|\Delta u|^2(x)dx\right)^{\frac{1}{2}}$$ Thanks in advance.

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    Did you start with verifying the three properties of a norm? You can find them here https://en.wikipedia.org/wiki/Norm_(mathematics) – 2017-02-09
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    Where are your problems with the scaling property? – 2017-02-09
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    why don't you look at the simplest case : $\Omega = [0,1]$ ? – 2017-02-09
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    I think that the real question here is not "is this a norm" but "is this norm equivalent to the $H^2$-norm on this subspace". The answer is affirmative and you can find some explanation [here](http://math.stackexchange.com/questions/2134160/boundedness-of-l2-norms-of-mixed-derivatives-of-functions-from-w2-2-math). – 2017-02-09

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See here if you don't know which properties I'm referring to

Absolute homogeneity: For $\lambda \in \mathbb{C},$ we have $$\|\lambda u\|_{h_{0}^{2}(\Omega)} = \|\Delta(\lambda u)\|_{L^{2}(\Omega)} = \|\lambda (\Delta u)\|_{L^{2}(\Omega)} = |\lambda| \|\Delta u\|_{L^{2}(\Omega)}= |\lambda|\|u\|_{h_{0}^{2}(\Omega)}$$ where I've used the abosolute homogeneity of the $L^{2}(\Omega)$-norm.

Triangle Inequality: \begin{align*}\|u+v\|_{h_{0}^{2}(\Omega)} &= \|\Delta( u+v)\|_{L^{2}(\Omega)} = \|\Delta u + \Delta v\|_{L^{2}(\Omega)} \leq \|\Delta u\|_{L^{2}(\Omega)} + \|\Delta v\|_{L^{2}(\Omega)} \\ &= \|u\|_{h_{0}^{2}(\Omega)} + \|v\|_{h_{0}^{2}(\Omega)} \end{align*} where I used the triangle inequality on $L^2(\Omega)$ (also known as Minkowski's inequality).

Finally, let $\|u\|_{h_{0}^{2}(\Omega)}= 0$. This implies $\Delta u = 0$. We have to show that this implies $u=0$. By partial integration, we obtain

$$0 = - \int_{\Omega} \Delta u \cdot u ~\mathrm{d}x = \int_{\Omega}|\nabla u|^2 \mathrm{d}x$$ which shows that $\nabla u = 0$. By the Poincaré inequality, we see that the (classical) sobolev norm of $u$ is $0$, and therefore $u$ is $0$.

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    So $H^2_0$ is the closure of $C^\infty_c$, not of $H^1_0$ (i.e. $f|_{\partial \Omega} =0$ but also $\nabla f|_{\partial \Omega} = 0$) – 2017-02-09
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    My real question : *is the closure of $C^\infty_c(\Omega)$ the same as the closure of $\{ f \in C^\infty(\Omega), f|_{\partial \Omega} = 0\}$ in $H^2(\Omega)$* ? (the closure in $H^1(\Omega)$ is $H^1_0(\Omega)$ in both case) – 2017-02-09
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    @user1952009 yes, but it is important here that $\Omega$ is bounded. You have $C_{c}^{\infty}(\Omega) \subseteq \{f \in C^{\infty}(\Omega), f|_{\partial \Omega} = 0 \} \subseteq H_{0}^{2}(\Omega) $. Now take closures of these inclusions. Note however, that if $\Omega$ is unbounded, we do not have $\{f \in C^{\infty}(\Omega), f|_{\partial \Omega} = 0\} \subseteq H_{0}^{2}(\Omega)$ – 2017-02-09
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    For the integration by parts : the definition of $H^k_0(\Omega)$ is the closure of $C^\infty_c(\Omega)$ in $H^k$. We take $f_k \in C^\infty_c(\Omega), \|f_k-f\|_{H^2} \to 0$ so that $\int_\Omega \Delta f . f = \lim_k \int_\Omega \Delta f_k . f = \lim_k\lim_n \int_\Omega \Delta f_k . f_n$ by Cauchy-Schwarz and $f,\nabla f, \Delta f \in L^2$. Also clearly $\int_\Omega \Delta f_k . f_n = -\int_\Omega \nabla f_k . \nabla f_n$ so that $\int_\Omega \Delta f . f = -\lim_k\lim_n \int_\Omega \nabla f_k . \nabla f_n =- \int_\Omega \nabla f . \nabla f $ again by Cauchy-Schwarz – 2017-02-09
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You just need to find that for every $f,g\in H^2_0(\Omega)$, for every $\lambda\in\mathbb{R},$ the following properties hold.

Absolute homogeneity. Here we use the fact that $\Delta$ and the integral are linear operators. Moreover that $\sqrt{\lambda^2}=|\lambda|$ $$ \|\lambda f\|=\left(\int_\Omega |\Delta (\lambda f)|^2\right)^{\frac{1}{2}}=|\lambda|\left(\int_\Omega |\Delta (\lambda f)|^2\right)^{\frac{1}{2}}=|\lambda\|f\| $$ Triangular inequality. It follows immediately by Minkowsky inequality

$$ \|f+g\|=\left(\int_\Omega |\Delta (f+g)|^2\right)^{\frac{1}{2}}\le \left(\int_\Omega |\Delta f|^2\right)^{\frac{1}{2}}+\left(\int_\Omega |\Delta g|^2\right)^{\frac{1}{2}}= \|f\|+\|g\| $$

Zero vector. You should check that whenever $\|f\|=0$, then $f=0$ a.e. in $\Omega$. Observe that $f$ satisfies the Dirichlet problem

$$ \begin{cases} \Delta f=0\qquad\text{in}\; \Omega\\ f=0 \qquad\text{on}\; \partial\Omega \end{cases} $$ By performing integration by parts $$ \int_\Omega |\nabla f|^2=-\int_\Omega (\Delta f ) f=0 $$ hence $f=const$ a.e. in $\Omega$. Because of the boundary conditions, we conclude $f=0$ a.e. in $\Omega$.

(I hope you can fill the gaps in the proof)

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    $f = const$ a.e. is not necessarily true if $\Omega$ is not connected. Also, "because of the boundary conditions" is not a clean way to argue, because in order to evaluate $f$ at the boundary you need to introduce the trace operator. So you have $T(f) = 0$, $\nabla u = 0$, how do you proceed? It's straightforward using the Poincaré inequality, as indicated in my answer. – 2017-02-09
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    @user159517 I don't understand something : the boundary condition $ f|_{\partial \Omega} =\nabla f|_{\partial \Omega} = 0$ is stronger than just $ f|_{\partial \Omega} =0$, right ? So how aviti91 can get $f=0$ only from $ f|_{\partial \Omega} =0$ ? I'd say integrating by parts $\int_\Omega (\Delta f ) f$ might be allowed only when we know $\nabla f|_{\partial \Omega} = 0$ – 2017-02-09
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    You get $f=const$ in each connected component of the domain (by Poincaré inequality indeed). Then because of the boundary conditions and trace theorem you can conclude – 2017-02-09
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    @user159517 you should be less pedantic. Look at your answer. Its not clear neither. We are here to help each other, not to behave like assholes – 2017-02-09
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    @user1952009 no, for the partial integration to be valid you only need $f \in H_{0}^{1}$. If $F$ is a vector field and $u$ a scalar function, we have $$\text{div}(uF) = u\text{div}(F) + \nabla u \cdot F$$. Applying Gauß' Theorem to the left hand side we see that the partial integration formula holds whenever $uF$ is orthogonal to the boundary normal vector. If either $u$ or $F$ vanishes identically, this clearly holds. In particular, setting $F = \nabla u$, we see that $u|_{\partial \Omega} = 0$ is sufficient for the argument in question to hold – 2017-02-09
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    @user1952009 note however, that the above argument doesn't really hold for arbitrary $H^1$-functions. It doesn't make much sense to write something like $u|_{\partial \Omega}$ for arbitrary functions $u \in H^{1}(\Omega)$. However, the argument above generalizes in the way one would expect. – 2017-02-09
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    @avati91 I don't much appreciate you calling me an asshole or pedantic. Feel free to point out the flaws in my answer. I downvoted yours because it was posted later than mine, was conceptually identical for the first two points and at the very least inaccurate in the final one. Hence, not useful in my opinion. – 2017-02-09
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    @aviti don't be upset, user159517 is sharing his knowledge. I upvoted your answer for peacefulness. and you might be pedantic when using Minkowsky inequality, instead of the triangle inequality for $\|\Delta f+\Delta g\|_{L^2(\Omega)}$ – 2017-02-09
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    Also you answer is inaccurate @user159517. Let the user who asked the question decide if my answer doesn't help him or not. This is not a challenge – 2017-02-09