Relation of the kernels of one bounded operator and its extension

Question

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

Answer 1

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.

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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.

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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).