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Will Kwon
Will Kwon
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It is well known that the weak space $L^{p,\infty}$ has less density property contrary to standard $L^p$ space. Related to this one, I'm struggling to prove the following statement which is given in the paper of Baker-Seregin-Sverak:

Proposition. Let $u_0 \in L^{3,\infty}$ be divergence-free in the sense of distributions. Then there exists a sequence $u_0^{(k)} \in C_{0,0}^\infty(\mathbb{R}^3)$ such that $$ u_0^{(k)}\rightarrow u_0\quad \text{in } L^{3,\infty}. $$$$ u_0^{(k)}\rightarrow u_0\quad \text{weakly star in } L^{3,\infty}. $$

Here $C_{0,0}^\infty(\mathbb{R}^3)$ is the space of all smooth vector fields with compact support whose divergence is free.

I have no idea to prove the above statement. Approximation by smooth functions is easy by using mollification, but I have no idea to obtain a suitable sequence as stated in the proposition. In the case of $L^p$ with $1<p<\infty$, by using the Hahn-Banach theorem and De Rham's theorem, we can show that $$ L^p_\sigma = \left\{ u \in L^p : \int_{\mathbb{R}^3} u\cdot \nabla \phi \,dx=0\quad \text{for all } \phi \in D^{1,p'}\right\}. $$ Here $L^p_\sigma$ is the closure of $C_{0,0}^\infty$ under $L^p$-norm and $D^{1,p'}$ is the space of all functions $u$ such that $\nabla u \in L^{p'}$.

Thanks for your time.

It is well known that the weak space $L^{p,\infty}$ has less density property contrary to standard $L^p$ space. Related to this one, I'm struggling to prove the following statement which is given in the paper of Baker-Seregin-Sverak:

Proposition. Let $u_0 \in L^{3,\infty}$ be divergence-free in the sense of distributions. Then there exists a sequence $u_0^{(k)} \in C_{0,0}^\infty(\mathbb{R}^3)$ such that $$ u_0^{(k)}\rightarrow u_0\quad \text{in } L^{3,\infty}. $$

Here $C_{0,0}^\infty(\mathbb{R}^3)$ is the space of all smooth vector fields with compact support whose divergence is free.

I have no idea to prove the above statement. Approximation by smooth functions is easy by using mollification, but I have no idea to obtain a suitable sequence as stated in the proposition. In the case of $L^p$ with $1<p<\infty$, by using the Hahn-Banach theorem and De Rham's theorem, we can show that $$ L^p_\sigma = \left\{ u \in L^p : \int_{\mathbb{R}^3} u\cdot \nabla \phi \,dx=0\quad \text{for all } \phi \in D^{1,p'}\right\}. $$ Here $L^p_\sigma$ is the closure of $C_{0,0}^\infty$ under $L^p$-norm and $D^{1,p'}$ is the space of all functions $u$ such that $\nabla u \in L^{p'}$.

Thanks for your time.

It is well known that the weak space $L^{p,\infty}$ has less density property contrary to standard $L^p$ space. Related to this one, I'm struggling to prove the following statement which is given in the paper of Baker-Seregin-Sverak:

Proposition. Let $u_0 \in L^{3,\infty}$ be divergence-free in the sense of distributions. Then there exists a sequence $u_0^{(k)} \in C_{0,0}^\infty(\mathbb{R}^3)$ such that $$ u_0^{(k)}\rightarrow u_0\quad \text{weakly star in } L^{3,\infty}. $$

Here $C_{0,0}^\infty(\mathbb{R}^3)$ is the space of all smooth vector fields with compact support whose divergence is free.

I have no idea to prove the above statement. Approximation by smooth functions is easy by using mollification, but I have no idea to obtain a suitable sequence as stated in the proposition. In the case of $L^p$ with $1<p<\infty$, by using the Hahn-Banach theorem and De Rham's theorem, we can show that $$ L^p_\sigma = \left\{ u \in L^p : \int_{\mathbb{R}^3} u\cdot \nabla \phi \,dx=0\quad \text{for all } \phi \in D^{1,p'}\right\}. $$ Here $L^p_\sigma$ is the closure of $C_{0,0}^\infty$ under $L^p$-norm and $D^{1,p'}$ is the space of all functions $u$ such that $\nabla u \in L^{p'}$.

Thanks for your time.

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Will Kwon
Will Kwon
  • 323
  • 1
  • 8

Weak-star approximation of smooth functions in weak $L^p$-space

It is well known that the weak space $L^{p,\infty}$ has less density property contrary to standard $L^p$ space. Related to this one, I'm struggling to prove the following statement which is given in the paper of Baker-Seregin-Sverak:

Proposition. Let $u_0 \in L^{3,\infty}$ be divergence-free in the sense of distributions. Then there exists a sequence $u_0^{(k)} \in C_{0,0}^\infty(\mathbb{R}^3)$ such that $$ u_0^{(k)}\rightarrow u_0\quad \text{in } L^{3,\infty}. $$

Here $C_{0,0}^\infty(\mathbb{R}^3)$ is the space of all smooth vector fields with compact support whose divergence is free.

I have no idea to prove the above statement. Approximation by smooth functions is easy by using mollification, but I have no idea to obtain a suitable sequence as stated in the proposition. In the case of $L^p$ with $1<p<\infty$, by using the Hahn-Banach theorem and De Rham's theorem, we can show that $$ L^p_\sigma = \left\{ u \in L^p : \int_{\mathbb{R}^3} u\cdot \nabla \phi \,dx=0\quad \text{for all } \phi \in D^{1,p'}\right\}. $$ Here $L^p_\sigma$ is the closure of $C_{0,0}^\infty$ under $L^p$-norm and $D^{1,p'}$ is the space of all functions $u$ such that $\nabla u \in L^{p'}$.

Thanks for your time.