4
$\begingroup$

Sorry for this long and formal post. The application in PDEs is mentioned just at the end.

Let $$V \hookrightarrow H \text{ and } Q_H' \hookrightarrow Q',$$ where $V$ and $Q$ are Banach and $H$ and $Q_H$ are Hilbert spaces. The hooked arrow $\hookrightarrow$ denotes the continuous embedding, which is basically $V \subset H$ and $\|v\|_V \geq \|v\|_H$ for all $v\in V$. The same with $Q_H'\hookrightarrow Q'$.

Also, assume that the embeddings are dense, i.e. $\overline V = H$ and $\overline {Q_H'} = Q'$, where the overline denotes the closure of the space with respect to the norm of its superspace.

Consider the linear bounded operator $$J\colon V \to Q_H'.$$ Then $V_0:=\ker(J)$ is a closed subspace of $V$. From an inf-sup condition, I have that $$\|Jv\|_{Q_H'} \geq \gamma \|v\|_V\quad (1)$$ for all $v \in V_1$, where $V_1$ is complementary to $V_0$, i.e. $V=V_0 \oplus V_1$.

Furthermore I have that $J\colon V\subset H\to Q'$ is bounded, so that one can define the natural extension $\bar J\colon H \to Q'$, using that $V$ is dense in $H$, that is bounded as well. Also for $\bar J$ I assume this boundedness from below, see $(1)$, for functions that are not in the kernel of $\bar J$.

Now my question is: Is the kernel of $J$ dense in the kernel of $\bar J$?

Or, equivalently, is $\overline V_0 = H_0$, where $H_0$ is the kernel of $\bar J$?

What I have tried so far:

  • I have shown that $\overline{V_0} \subset H_0$. To show the converse direction, I thought of taking $h \in H_0$ and show that there is a sequence $\{v_{0,n}\} \subset V_0$ that goes to $h_0$ (in the norm of $H$).

  • Since $\overline V = H$, there is $\{v_n\} \subset V$ that goes to $h_0$ (in the norm of $H$).

  • Because of $(1)$ there is a bounded projector $P_V\colon V \to V$, with $P(V)=V_0$. Then one can split up every $v_n$ into $v_{0,n}:=Pv_n$ and the remainder $v_{1,n}$ that is in $V_1$.

  • Now I want to show, that $\{v_{1,n}\}$ goes to $0$ (in $H$) what would make $\{v_{0,n}\} \subset V_0$ approaching $h_0$.

......

In terms of PDEs, this would answer the questions, whether the (sub)space of divergence free elements of $H_0^1(\Omega)^3$ is dense in the (sub)space of these functions in $L^2(\Omega)^3$. In this case:

  • $J:=div$
  • $V:= H_0^1(\Omega)^3$ and $H:=L^2(\Omega)^3$
  • $Q_H := L^2(\Omega)/\mathbb R$ and $Q' = (H^1(\Omega)/\mathbb R)'$

And the question becomes: Is $$ \{v\in H_0^1(\Omega)^3:\text{div } v = 0 \in L^2(\Omega)/\mathbb R \} \text{ dense in } \{v \in L^2(\Omega)^3:\text{div } v = 0 \in (H^1(\Omega)/\mathbb R)' \} $$

$\endgroup$
  • $\begingroup$ Divergence free vector fields are not dense in $L^2(\Omega)^3$. $\endgroup$ – Shuhao Cao May 15 '13 at 13:56
  • $\begingroup$ Right. But maybe dense in some subspaces... I have edited the end of my post to make this clearer. $\endgroup$ – Jan May 15 '13 at 14:04
2
$\begingroup$

Some updates: Somehow I think the argument doesn't need to be so long. Actually the divergence free vector field in $H(\mathrm{div})$ can be defined as

$L^2$-divergence free vector fields are the closure of $C^{\infty}$-divergence free vector fields in $L^2$-norm.

This is in Luc Tartar's book page 35 here. The the density argument of $C^{\infty}\subset H^1 \subset L^2$ would imply the density of the $H^1$-divergence free vector fields in $L^2$-divergence free vector fields. The proof I presented was just try to replicate what Tartar did for the density of $C^{\infty}$-divergence free vector fields in $H^1$-divergence free vector fields.


Tool we use to prove density: Suppose subspace $\mathscr{X} \subset X$, they are both Banach. Define $$X^{\perp} = \{l\in X': \langle l,v\rangle =0 \;\forall v\in X\},$$ and $$\mathscr{X}^{\perp} = \{l\in X': \langle l,v\rangle =0 \;\forall v\in \mathscr{X}\},$$ where $X'$ is the set of all bounded linear functional on $X$. Then we have

Claim: $\mathscr{X}$ is dense in $X$. $\Longleftrightarrow$ $\mathscr{X}^{\perp} =X^{\perp} $.

Sketch of the proof: First $\mathscr{X}^{\perp} \supset X^{\perp} $ holds always. "$\Rightarrow$" is like standard exercise. For "$\Leftarrow$", we want to prove $\mathscr{X}^{\perp} \subset X^{\perp}$ implying the left: suppose the density does not hold, then we could find an open subset $Z\subset X$ so that $\overline{\mathscr{X}} \cap Z =\varnothing $. Choose $z\in Z$, we can find a non-zero bounded linear functional $g\in X'$ such that $\langle g,z\rangle \neq 0$. Consider a functional $L$ on $\overline{\mathscr{X}} + \{z\} $: $$ \langle L,x+tz\rangle = \langle l,x\rangle + t\langle g,z\rangle, \quad \text{ for } x\in \mathscr{X}, t\in \mathbb{R}, l\in \mathscr{X}^{\perp}, $$ then we can extend $L$ to whole $X$. It can be checked that $L\in \mathscr{X}^{\perp}$, but $\langle L,z \rangle = \langle g,z \rangle\neq 0$ implies $L\notin X^{\perp}$. Thus $\mathscr{X}^{\perp} \not\subset X^{\perp}$ and the claim.


Now we move on to prove

Divergence free vector fields in $H^1$ is dense in divergence free vector fields in $L^2$.

Denote $$ V :=H_0^1(\Omega)^3,\quad V_0 := \{v\in H_0^1(\Omega)^3:\mathrm{div}\, v = 0\}, $$ and $$ H := L^2(\Omega)^3,\quad H_0 := \{v \in L^2(\Omega)^3:\mathrm{div}\, v = 0 \}, $$ then what you wanted to show is:

(A) $V_0$ is dense in $H_0$.

We can prove this using above claim. Define $$ H(\mathrm{div}) = \{v \in L^2(\Omega)^3,\mathrm{div}\, v \in L^2(\Omega) \}, $$ and we can check this is a Hilbert space under the norm: $$ \|\cdot\|_{H(\mathrm{div})}^2 = \|\cdot \|_{L^2(\Omega)^3}^2 + \|\mathrm{div}(\cdot)\|_{ L^2(\Omega)}^2. $$ Now all the relevant spaces are Hilbert now and we can associate the bounded linear functional with a specific inner product.

First Let $l\in H(\mathrm{div})'$, representation theorem in Hilbert space says there is some $u_l \in H(\mathrm{div}) \subset L^2(\Omega)^3$ : $$ \langle l,v\rangle = \int_{\Omega} u_l \,v + \int_{\Omega}(\mathrm{div} \,u_l )\,(\mathrm{div}\, v). $$

Consider some $l$ vanishes on $V_0$: $$V_0^{\perp} = \{l\in H(\mathrm{div})': \langle l,v\rangle =0 \;\forall v\in V_0\}\subset \{l\in (H_0^1(\Omega)^3)': \langle l,v\rangle =0 \;\forall v\in V_0\}. $$ We also know that $$ \mathrm{div}: H_0^1(\Omega)^3 \to L^2(\Omega),\quad \text{ and }\quad \mathrm{div}^* = -\nabla : (L^2(\Omega))'\simeq L^2(\Omega) \to ( H_0^1(\Omega)^3)'. $$ Closed range theorem reads: $$ R(-\nabla ) = (\mathrm{ker}(\mathrm{div}))^{\perp} = \{l\in ( H_0^1(\Omega)^3)': \langle l,v\rangle =0 \;\forall v\in \mathrm{ker}(\mathrm{div}) = V_0 \} \supset V_0^{\perp}, $$ and this means $\langle l,v\rangle =0 $ for any $v\in V_0$, then $u_l = \nabla \phi$ for some $\phi\in L^2(\Omega)/\mathbb{R}$ in the sense of isomorphism: $$ \langle l,v\rangle = \int_{\Omega} u_l \,v = \int_{\Omega} \nabla \phi \,v, $$ for $v\in H^1_0(\Omega)^3$ and divergence free.

Now we want to show $$ V_0^{\perp}\subset \{l\in H(\mathrm{div})': \langle l,v\rangle =0 \;\forall v\in H_0\} = H_0^{\perp}. $$ For the above $l$ that vanishes on $V_0$, $u_l = \nabla \phi$, for $u_l \in H(\mathrm{div})\subset L^2(\Omega)^3$, we can pin down this $\phi\in H^1_0(\Omega)$ by solving: $$ \int_{\Omega} \nabla \phi \cdot \nabla v = \int_{\Omega} u_l \cdot \nabla v,\quad \text{ for } \forall v\in H^1_0(\Omega). $$ We can use Green's identity which is valid for $u\in H(\mathrm{div})$ and $\phi \in H^1$, this result can be found in Tartar's book: for $u\in H_0\subset H(\mathrm{div})$ $$ \langle l,u\rangle = \int_{\Omega} \nabla \phi \cdot u = -\int_{\Omega} \phi\,\mathrm{div}\,u + \int_{\partial \Omega} (u\cdot n)\phi \,dS, $$ and the boundary term vanish for $\phi \in H^1_0(\Omega)$. Therefore $\langle l,u\rangle = 0$ for $u\in H_0$, and we have:

(B) $V_0^{\perp}\subset H_0^{\perp}$.

By the claim, we have (A).

$\endgroup$
  • $\begingroup$ Thanks so much for this thorough answer. I still have to come to terms with the boundary integral of $(u\cdot n)\phi$. Whether it is defined, if $(u\cdot n)$ is not the zero in $L^2(\partial \Omega)$. Maybe the 'right' $H_0$ is to be defined with the additional condition $u\cdot n=0$. $\endgroup$ – Jan May 30 '13 at 14:24
  • $\begingroup$ I will now use your blueprint to tackle the more formal case, with more general operators. $\endgroup$ – Jan May 30 '13 at 14:27
  • $\begingroup$ @Jan Yeah, you are right, $\displaystyle \int_{\partial \Omega}(u\cdot n)\phi$ is in general not zero. Consider $\phi + c$ where $c\in \mathbb{R}$: $\displaystyle \int_{\partial \Omega}(u\cdot n)(\phi+c) = \displaystyle \int_{\partial \Omega}(u\cdot n)\phi$, because $\displaystyle \int_{\partial \Omega}(u\cdot n)$ vanishes in the distribution sense when $u$ is divergence free, $H^1(\Omega)/\mathbb{R}$ does not help much, then modifying the procedure to pin down a specific $v$ using boundary condition will do, please see my edit. $\endgroup$ – Shuhao Cao May 30 '13 at 19:15
  • $\begingroup$ @Jan Today I realized that we can always borrow the result of the density of the divergence free smooth vector fields in $L^2$-divergence free vector fields. Please see my update about Tartar's book page 29-34. $\endgroup$ – Shuhao Cao Jul 18 '13 at 16:09

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.