Let $\Omega $ be an open set in $\mathbb{C}^n$, $u$ be a smooth function on $\Omega$ and let $\Phi$ be a non negative test function on $\Omega$. Prove that for all $b=(b_1,..., b_n)\in \mathbb{C^n}$ $\int_\Omega u(z)\sum_{j,k}^n\frac{\partial^2\Phi}{\partial z_j\partial \bar{z_k}}b_j\bar{b_k} d\lambda(z)=\int_\Omega \Phi(z)\sum_{j,k}^n\frac{\partial^2 u}{\partial z_j\partial \bar{z_k}}b_j\bar{b_k} d\lambda(z)$.
$\sum_{j,k}^n\frac{\partial^2\Phi}{\partial z_j\partial \bar{z_k}}b_j\bar{b_k}$ is called the Levi form of $\Phi$. As $\Phi\in C_0^\infty$, it is zero on $\partial\Omega$. What I wanted to prove is $\int_\Omega u(z)\frac{\partial^2\Phi}{\partial z_j\partial \bar{z_k}} d\lambda(z)=\int_\Omega \Phi(z)\frac{\partial^2u}{\partial z_j\partial \bar{z_k}} d\lambda(z)$. But how? Any help is appreciated.