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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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  • $\begingroup$ Did you start with verifying the three properties of a norm? You can find them here en.wikipedia.org/wiki/Norm_(mathematics) $\endgroup$ – avati91 Feb 9 '17 at 11:08
  • $\begingroup$ Where are your problems with the scaling property? $\endgroup$ – user159517 Feb 9 '17 at 11:11
  • $\begingroup$ why don't you look at the simplest case : $\Omega = [0,1]$ ? $\endgroup$ – reuns Feb 9 '17 at 11:12
  • $\begingroup$ 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. $\endgroup$ – Michał Miśkiewicz Feb 9 '17 at 22:09
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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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  • $\begingroup$ So $H^2_0$ is the closure of $C^\infty_c$, not of $H^1_0$ (i.e. $f|_{\partial \Omega} =0$ but also $\nabla f|_{\partial \Omega} = 0$) $\endgroup$ – reuns Feb 9 '17 at 12:07
  • $\begingroup$ 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) $\endgroup$ – reuns Feb 9 '17 at 13:48
  • $\begingroup$ @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 $\{f \in C^{\infty}(\Omega), f|_{\partial \Omega} = 0\} \subseteq H_{0}^{2}(\Omega)$ $\endgroup$ – user159517 Feb 9 '17 at 14:02
  • $\begingroup$ 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 $\endgroup$ – reuns Feb 9 '17 at 14:22
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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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  • $\begingroup$ $f = const$ 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. $\endgroup$ – user159517 Feb 9 '17 at 11:58
  • $\begingroup$ @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 $\nabla f|_{\partial \Omega} = 0$ $\endgroup$ – reuns Feb 9 '17 at 12:12
  • $\begingroup$ 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 $\endgroup$ – avati91 Feb 9 '17 at 12:28
  • $\begingroup$ @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 $\endgroup$ – avati91 Feb 9 '17 at 12:30
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    $\begingroup$ @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 $\|\Delta f+\Delta g\|_{L^2(\Omega)}$ $\endgroup$ – reuns Feb 9 '17 at 13:57

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