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Let us consider $\mathscr D(\mathbb R^3;\mathbb R^3)$ the space of smooth compactly supported vector fields in $\mathbb R^3$ and let us define $\widetilde{\mathscr D}=\mathscr D(\mathbb R^3;\mathbb R^3)\cap \ker\text{div}$ the space of smooth compactly supported vector fields in $\mathbb R^3$ with null divergence.

Question 1: is it true that $ \widetilde{\mathscr D}=\text{curl } \mathscr D(\mathbb R^3;\mathbb R^3) $?

Question 2: is it true that the topological dual ${(\widetilde{\mathscr D})}{'}$ of $\widetilde{\mathscr D}$ satisfies $$ {(\widetilde{\mathscr D})}{'}=\mathscr D'(\mathbb R^3;\mathbb R^3)/\text{grad } \mathscr D'(\mathbb R^3;\mathbb R)? $$ These questions are implicitly important to define weak solutions of Euler or Navier-Stokes equations. As suggested by the answers below (thanks), the first question should have a positive answer, thanks to the Poincaré lemma and the ellipticity of the exterior differentiation.

For the second question, the dual space of this subspace of $\mathscr D(\mathbb R^{3};\mathbb R^{3})$ should be the quotient $$ \mathscr D'(\mathbb R^{3};\mathbb R^{3})/\bigl(\text{curl} \mathscr D(\mathbb R^{3};\mathbb R^{3})\bigr)^{\perp}. $$ The orthogonal of $\text{curl} \mathscr D(\mathbb R^{3};\mathbb R^{3})$ is made with distributions $T$ such that, with duality brackets $$\forall \phi\in \mathscr D(\mathbb R^{3};\mathbb R^{3}),\quad 0=\langle{T},{\text{curl } \phi}\rangle=\langle{\text{curl }T},{\phi}\rangle, $$ so it is the kernel of the curl , that is $\text{grad } \mathscr D'(\mathbb R^{3};\mathbb R)$. Thus we should be able to infer that $$ \bigl(\mathscr D(\mathbb R^{3};\mathbb R^{3})\cap \ker\text{div}\bigr)^{*}=\mathscr D'(\mathbb R^{3};\mathbb R^{3}){\large/}\bigl(\text{grad }\mathscr D'(\mathbb R^{3};\mathbb R)\bigr). $$ Interestingly, if this is true, it means that when you formulate weakly the Navier-Stokes equations and you restrict the test functions to be with null divergence, as it is done in the proof of the classical Leray theorem, you obtain a solution modulo a gradient, which may be bothering (?) A few words about the latter topic: If you test only against smooth compactly supported vector fields with null divergence, you obtain that the weak limit obtained by Leray's theorem satisfies $$ \partial_tu+\mathbb P (u\cdot \nabla) u+ \nu \text{ curl}^2 u=\nabla q, $$ where $\mathbb P$ is the Leray projector. Of course, you are saved by the fact that you know that the weak limit has also null divergence, so that applying $\mathbb P$, you get $\partial_tu+\mathbb P (u\cdot \nabla) u+ \nu \text{ curl}^2 u=0$. My point is that this additional piece of information (div $u=0$) is due to a (linear) limit of approximate solutions, which is somehow independent of this idea of testing only against vf with null divergence. More generally, I want to point out that some information is lost when you test only against vf with null divergence.

Let us consider $\mathscr D(\mathbb R^3;\mathbb R^3)$ the space of smooth compactly supported vector fields in $\mathbb R^3$ and let us define $\widetilde{\mathscr D}=\mathscr D(\mathbb R^3;\mathbb R^3)\cap \ker\text{div}$ the space of smooth compactly supported vector fields in $\mathbb R^3$ with null divergence.

Question 1: is it true that $ \widetilde{\mathscr D}=\text{curl } \mathscr D(\mathbb R^3;\mathbb R^3) $?

Question 2: is it true that the topological dual ${(\widetilde{\mathscr D})}{'}$ of $\widetilde{\mathscr D}$ satisfies $$ {(\widetilde{\mathscr D})}{'}=\mathscr D'(\mathbb R^3;\mathbb R^3)/\text{grad } \mathscr D'(\mathbb R^3;\mathbb R)? $$ These questions are implicitly important to define weak solutions of Euler or Navier-Stokes equations. As suggested by the answers below (thanks), the first question should have a positive answer, thanks to the Poincaré lemma and the ellipticity of the exterior differentiation.

For the second question, the dual space of this subspace of $\mathscr D(\mathbb R^{3};\mathbb R^{3})$ should be the quotient $$ \mathscr D'(\mathbb R^{3};\mathbb R^{3})/\bigl(\text{curl} \mathscr D(\mathbb R^{3};\mathbb R^{3})\bigr)^{\perp}. $$ The orthogonal of $\text{curl} \mathscr D(\mathbb R^{3};\mathbb R^{3})$ is made with distributions $T$ such that, with duality brackets $$\forall \phi\in \mathscr D(\mathbb R^{3};\mathbb R^{3}),\quad 0=\langle{T},{\text{curl } \phi}\rangle=\langle{\text{curl }T},{\phi}\rangle, $$ so it is the kernel of the curl , that is $\text{grad } \mathscr D'(\mathbb R^{3};\mathbb R)$. Thus we should be able to infer that $$ \bigl(\mathscr D(\mathbb R^{3};\mathbb R^{3})\cap \ker\text{div}\bigr)^{*}=\mathscr D'(\mathbb R^{3};\mathbb R^{3}){\large/}\bigl(\text{grad }\mathscr D'(\mathbb R^{3};\mathbb R)\bigr). $$ Interestingly, if this is true, it means that when you formulate weakly the Navier-Stokes equations and you restrict the test functions to be with null divergence, as it is done in the proof of the classical Leray theorem, you obtain a solution modulo a gradient, which may be bothering (?) A few words about the latter topic: If you test only against smooth compactly supported vector fields with null divergence, you obtain that the weak limit obtained by Leray's theorem satisfies $$ \partial_tu+\mathbb P (u\cdot \nabla) u+ \nu \text{ curl}^2 u=\nabla q, $$ where $\mathbb P$ is the Leray projector. Of course, you are rescued by the fact that you know that the weak limit has also null divergence, so that applying $\mathbb P$, you get $\partial_tu+\mathbb P (u\cdot \nabla) u+ \nu \text{ curl}^2 u=0$. My point is that this additional piece of information (div $u=0$) is due to a (linear) limit of approximate solutions, which is somehow independent of this idea of testing only against vf with null divergence. More generally, I want to point out that some information is lost when you test only against vf with null divergence.

Let us consider $\mathscr D(\mathbb R^3;\mathbb R^3)$ the space of smooth compactly supported vector fields in $\mathbb R^3$ and let us define $\widetilde{\mathscr D}=\mathscr D(\mathbb R^3;\mathbb R^3)\cap \ker\text{div}$ the space of smooth compactly supported vector fields in $\mathbb R^3$ with null divergence.

Question 1: is it true that $ \widetilde{\mathscr D}=\text{curl } \mathscr D(\mathbb R^3;\mathbb R^3) $?

Question 2: is it true that the topological dual ${(\widetilde{\mathscr D})}{'}$ of $\widetilde{\mathscr D}$ satisfies $$ {(\widetilde{\mathscr D})}{'}=\mathscr D'(\mathbb R^3;\mathbb R^3)/\text{grad } \mathscr D'(\mathbb R^3;\mathbb R)? $$ These questions are implicitly important to define weak solutions of Euler or Navier-Stokes equations. As suggested by the answers below (thanks), the first question should have a positive answer, thanks to the Poincaré lemma and the ellipticity of the exterior differentiation.

For the second question, the dual space of this subspace of $\mathscr D(\mathbb R^{3};\mathbb R^{3})$ should be the quotient $$ \mathscr D'(\mathbb R^{3};\mathbb R^{3})/\bigl(\text{curl} \mathscr D(\mathbb R^{3};\mathbb R^{3})\bigr)^{\perp}. $$ The orthogonal of $\text{curl} \mathscr D(\mathbb R^{3};\mathbb R^{3})$ is made with distributions $T$ such that, with duality brackets $$\forall \phi\in \mathscr D(\mathbb R^{3};\mathbb R^{3}),\quad 0=\langle{T},{\text{curl } \phi}\rangle=\langle{\text{curl }T},{\phi}\rangle, $$ so it is the kernel of the curl , that is $\text{grad } \mathscr D'(\mathbb R^{3};\mathbb R)$. Thus we should be able to infer that $$ \bigl(\mathscr D(\mathbb R^{3};\mathbb R^{3})\cap \ker\text{div}\bigr)^{*}=\mathscr D'(\mathbb R^{3};\mathbb R^{3}){\large/}\bigl(\text{grad }\mathscr D'(\mathbb R^{3};\mathbb R)\bigr). $$ Interestingly, if this is true, it means that when you formulate weakly the Navier-Stokes equations and you restrict the test functions to be with null divergence, as it is done in the proof of the classical Leray theorem, you obtain a solution modulo a gradient, which may be bothering (?)

Let us consider $\mathscr D(\mathbb R^3;\mathbb R^3)$ the space of smooth compactly supported vector fields in $\mathbb R^3$ and let us define $\widetilde{\mathscr D}=\mathscr D(\mathbb R^3;\mathbb R^3)\cap \ker\text{div}$ the space of smooth compactly supported vector fields in $\mathbb R^3$ with null divergence.

Question 1: is it true that $ \widetilde{\mathscr D}=\text{curl } \mathscr D(\mathbb R^3;\mathbb R^3) $?

Question 2: is it true that the topological dual ${(\widetilde{\mathscr D})}{'}$ of $\widetilde{\mathscr D}$ satisfies $$ {(\widetilde{\mathscr D})}{'}=\mathscr D'(\mathbb R^3;\mathbb R^3)/\text{grad } \mathscr D'(\mathbb R^3;\mathbb R)? $$ These questions are implicitly important to define weak solutions of Euler or Navier-Stokes equations. As suggested by the answers below (thanks), the first question should have a positive answer, thanks to the Poincaré lemma and the ellipticity of the exterior differentiation.

For the second question, the dual space of this subspace of $\mathscr D(\mathbb R^{3};\mathbb R^{3})$ should be the quotient $$ \mathscr D'(\mathbb R^{3};\mathbb R^{3})/\bigl(\text{curl} \mathscr D(\mathbb R^{3};\mathbb R^{3})\bigr)^{\perp}. $$ The orthogonal of $\text{curl} \mathscr D(\mathbb R^{3};\mathbb R^{3})$ is made with distributions $T$ such that, with duality brackets $$\forall \phi\in \mathscr D(\mathbb R^{3};\mathbb R^{3}),\quad 0=\langle{T},{\text{curl } \phi}\rangle=\langle{\text{curl }T},{\phi}\rangle, $$ so it is the kernel of the curl , that is $\text{grad } \mathscr D'(\mathbb R^{3};\mathbb R)$. Thus we should be able to infer that $$ \bigl(\mathscr D(\mathbb R^{3};\mathbb R^{3})\cap \ker\text{div}\bigr)^{*}=\mathscr D'(\mathbb R^{3};\mathbb R^{3}){\large/}\bigl(\text{grad }\mathscr D'(\mathbb R^{3};\mathbb R)\bigr). $$ Interestingly, if this is true, it means that when you formulate weakly the Navier-Stokes equations and you restrict the test functions to be with null divergence, as it is done in the proof of the classical Leray theorem, you obtain a solution modulo a gradient, which may be bothering (?) A few words about the latter topic: If you test only against smooth compactly supported vector fields with null divergence, you obtain that the weak limit obtained by Leray's theorem satisfies $$ \partial_tu+\mathbb P (u\cdot \nabla) u+ \nu \text{ curl}^2 u=\nabla q, $$ where $\mathbb P$ is the Leray projector. Of course, you are saved by the fact that you know that the weak limit has also null divergence, so that applying $\mathbb P$, you get $\partial_tu+\mathbb P (u\cdot \nabla) u+ \nu \text{ curl}^2 u=0$. My point is that this additional piece of information (div $u=0$) is due to a (linear) limit of approximate solutions, which is somehow independent of this idea of testing only against vf with null divergence. More generally, I want to point out that some information is lost when you test only against vf with null divergence.