Given $I\subset \mathbb R^N$ open bounded. Then we define two quantities $$ Q_1=\sup\left\{ \int_\Omega u\,\text{div}\phi\,dx, \,\phi\in C_c^\infty(\Omega;\,\mathbb R^2), \,\|\phi\|_{L^\infty}\leq 1\right\} \tag 1 $$ and $$ Q_2=\sup\left\{ \int_\Omega u\,\text{div}^2\varphi\,dx, \,\varphi\in C_c^\infty(\Omega;\mathbb S^{2\times 2}), \,\|\text{div}\varphi\|_{L^\infty}\leq 1\right\} \tag 2 $$ where by $\mathbb S^{2\times 2}$ I mean $2\times 2$ symmetric metric. And if we have $$ \varphi= \begin{bmatrix} \varphi_{11} & \varphi_{12}\\ \varphi_{21} & \varphi_{22} \end{bmatrix}, $$ we write $\text{div}^2\varphi:=\text{div}[\text{div}(\varphi_{11},\varphi_{12}),\text{div}(\varphi_{21},\varphi_{22})]$

I am trying to prove that $Q_1=Q_2$. My argument is that whenever I have $\phi\in C_c^\infty(\Omega;\,\mathbb R^2)$, I can always have a $\varphi\in C_c^\infty(\Omega;\,\mathbb S^{2\times 2})$ by setting $\varphi_{12}=\varphi_{21}=0$ and $\partial_1\varphi_{11}=\phi_1$ and $\partial_2\varphi_{22}=\phi_2$. That is, the question $$ u=\text{div }v $$ is always solvable for $u$, $v\in C_c^\infty$.

Hence $Q_1\leq Q_2$. Similarly, I have $Q_2\leq Q_1$ and I done.

  • $\begingroup$ That $Q_2\le Q_1$ is clear, but I don't think your argument for $Q_1\le Q_2$ works. Having defined $\varphi_{11}$ so that $\partial_1\varphi_{11}=\phi_1$, you are not guaranteed that $\varphi_{11}$ will have compact support. Indeed, if $\varphi_1\ge 0$ (which could happen) then it won't. I don't see a way to fix this. You can't write an arbitrary compactly supported function as the divergence of a compactly supported field. $\endgroup$ – user147263 Jul 19 '15 at 6:51
  • $\begingroup$ @NormalHuman I see. You are right. I need to find another way. Thank you! $\endgroup$ – spatially Jul 23 '15 at 15:41

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