Kernels of a bounded operator’s extension

This is a post recording my answer to a question on Math StackExchange1 with some references in L. Tartar’s book2.

Question

Whether the (sub)space of divergence free elements of $H_0^1(\Omega)^3$ is dense in the (sub)space of divergence free functions in $L^2(\Omega)^3$?

In this case:

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

And the question becomes:

$\text{Show } \;\{v\in H_0^1(\Omega)^3:\text{div } v = 0 \in L^2(\Omega)/\mathbb R \} \text{ dense in } \{v \in L^2(\Omega)^3:\text{div } v = 0 \in (H^1(\Omega)/\mathbb R)' \}$

Proof

A simple extension theorem for functionals

Main tool we use to prove density is the following lemma.

Suppose subspace $\mathcal{X} \subset X$, they are both Banach. Define $$X^{\perp} = \{l\in X': \langle l,v\rangle =0 \;\forall v\in X\},$$ and $$\mathcal{X}^{\perp} = \{l\in X': \langle l,v\rangle =0 \;\forall v\in \mathcal{X}\},$$ where $X'$ is the set of all bounded linear functional on $X$. Then we have $$\mathcal{X} \text{ is dense in } X. \Longleftrightarrow \mathcal{X}^{\perp} = X^{\perp}.$$
Sketch of the proof: First $\mathcal{X}^{\perp} \supset X^{\perp}$ holds always. "$\Rightarrow$" is like standard exercise. For "$\Leftarrow$", we want to prove $\mathcal{X}^{\perp} \subset X^{\perp}$ implying the left: suppose the density does not hold, then we could find an open subset $Z\subset X$ so that $\overline{\mathcal{X}} \cap Z =\varnothing$. 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 $\overline{\mathcal{X}} + \{z\}$: $$\langle L,x+tz\rangle = \langle l,x\rangle + t\langle g,z\rangle, \quad \text{ for } x\in \mathcal{X}, t\in \mathbb{R}, l\in \mathcal{X}^{\perp},$$ then we can extend $L$ to whole $X$. It can be checked that $L\in \mathcal{X}^{\perp}$, but $\langle L,z \rangle = \langle g,z \rangle\neq 0$ implies $L\notin X^{\perp}$. Thus $\mathcal{X}^{\perp} \not\subset X^{\perp}$ and the claim.

Main result

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: $$V_0 \text{ is dense in } H_0. \tag{A}$$ 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: $$V_0^{\perp}\subset H_0^{\perp}.$$ By the claim, we have (A).

1. https://math.stackexchange.com/q/392322

2. Tartar, Luc. An introduction to Sobolev spaces and interpolation spaces. Vol. 3. Springer Science & Business Media, 2007.

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