While reading a proof I stumbled over the following equation:
\begin{align} \int_{B_1} \langle A,D\psi(x)\rangle dx=0 \quad \forall \psi\in W^{1,\infty}_0(B_1,\mathbb{R}^2)\forall A\in \mathbb{R}^{2\times 2} \end{align} I denote by $\langle\ \ ,\ \ \rangle$ the Frobenius inner product. This looks like something derived by partial integration, which I have never seen in that Frobenius setting. Also in the proof it is implied that $\int_{B_1} \langle D\psi(x),D\psi(x)\rangle dx=0$ does not hold. I would appreciate any hint why the equation holds and if one can adapt partial integration in that setting.
You can literally just write out the inner product and use the well-know integration by parts formula: $$ \langle A, D\phi \rangle=A_{11}\partial_1\phi_1+A_{12}\partial_1\phi_2+A_{21}\partial_2\phi_1+A_{22}\partial_2\phi_2 $$ Since your matrix is constant you can just swap the derivates using integration by parts; ill do it for 1 term as an example $$ \int_{B_1} A_{11}\partial_1\phi_1 dx=-\int_{B_1} \phi_1\partial_1 A_{11}=0 $$ Summing over 4 zero terms is still zero. Boundary terms are getting dropped since your function vanishes on the boundary by assumption.
The second eqzuality only holds when $D\phi=0$ by properties of the scalar product.