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Let $\mathcal{V}$ be the space (without topology)

$$\displaystyle \mathcal{V}=\{u\in C_0^\infty(\Omega)\mid \nabla\cdot u=0\}$$

where $\Omega$ is a nonempty open connected subset of $\mathbb{R}^n$.

It is said in the Navier-Stokes Equations by Temam that the closure of $\mathcal{V}$ in $L^2(\Omega)$ and in $H_0^1(\Omega)$ are two basic spaces in the study of the Navier-Stokes equations. While it is quite clear what the closure of $\mathcal{V}$ in $L^2(\Omega)$ means, I don’t quite understand later one.

By definition, $H_0^1(\Omega)$ is $W_0^{1,2}(\Omega)$ which is defined as

$$\displaystyle \overline{C_0^\infty(\Omega)}^{\|\|_{W^{1,2}(\Omega)}},$$

the closure of $C_0^\infty(\Omega)$ in the Sobolev space $W^{1,2}(\Omega)$.

Here are my questions:

  • What is the convention for the topology of $H_0^1(\Omega)$?

  • Is the same as the topology of $H^1(\Omega)$ so that the closure of $\mathcal{V}$ in $H_0^1(\Omega)$ is the same as the closure of $\mathcal{V}$ in $H^1(\Omega)$?

  • Could anyone come up with a cited reference regarding the topology of $H_0^1$?


[Added:] What really puzzles me is: if the closure of $\mathcal{V}$ in $H_0^1(\Omega)$ is the same as the closure of $\mathcal{V}$ in $H^1(\Omega)$, why bother mentioning $H_0^1(\Omega)$?

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    $\begingroup$ This work by Robinson may be helpful: amazon.com/…. $\endgroup$ – Albert Dec 21 '15 at 15:08
  • $\begingroup$ I'm not so sure I understand the point of the question. Since you are starting with a subspace of $C^\infty_0(\Omega)$ anyway, the closure will be a subspace of the closure of the whole space (and of course, the closure will be taken w.r.t to the $W^{1,2}$ norm). Am I missing some (more or less subtle) detail? $\endgroup$ – Thomas Dec 21 '15 at 16:42
  • $\begingroup$ Regarding your last question, the reason is that if you consider only $\mathcal{V}$ then a differential problem with the boundary condition you can formulate only classically, that considering classical differentiation. $\endgroup$ – user288972 Dec 22 '15 at 15:43
  • $\begingroup$ Here is an answer from MO by one of Temam's former students. $\endgroup$ – Jack Dec 23 '15 at 0:24
  • $\begingroup$ Sure, I have just added some detail. $\endgroup$ – user288972 Dec 23 '15 at 0:35
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Every normed space is a metric space and every metric space is obviously a topological space. You just know this, since $C_{0}^{\infty}(\Omega) \subset W^{1,2}(\Omega)$. In other words, since the closure is in general constituted by its limit points, get that the functions of $H_{0}^{1}(\Omega)$ have the property that vanish on $\partial \Omega$ since they are limit of functions that behave in this way. It's clear that consider the same norm (and then the same topology), you've also written.

If it is unclear you can consider the inclusions $\mathcal{V} \subset H_{0}^1(\Omega) \subset H^1(\Omega)$, formally the topology of $H_{0}^1(\Omega)$ is induced by the metric $d(f,g)=\left \| f-g \right \|_{H^{1}(\Omega)}$ $\forall f,g \in H_{0}^1(\Omega)$.

Now, in general, for an arbitrary topological space $(X,\mathcal{T})$, if $Y \subset X$ is closed in $X$, then all and only the closed of the topological subspace $(Y, \mathcal{T}^Y)$ (where $\mathcal{T}^Y$ is subspace topology) are the closed of $(X,\mathcal{T})$ contained in $Y$

By definition, $H_{0}^1(\Omega)$ is closed in the norm $\left \| \cdot \right \|_{H^{1}(\Omega)}$, that is, in space $H^1(\Omega)$, and then from $\mathcal{V} \subset H_0^{1}(\Omega) \subset H^1(\Omega)$ follows that the closure of $\mathcal{V}$ is closed in $H_{0}^1(\Omega)$ and in $H^1(\Omega)$, and the two closures are the same.

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Here is an answer from MO by one of Temam's students:

Yes, it is the same. I think that Roger (my advisor) wanted only to emphazise that because $\cal V$ is included in $H^1_0$, and the latter is a closed subspace of $H^1$, the closure of $\cal V$ is contained in $H^1_0$, hence its elements satisfy the boundary condition $u=0$.

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