By using the Fourier transform prove that if $u \in C_0^2(\mathbb{R}^2)$ [twice differentiable and vanishes at $\infty$] then
\begin{equation} \int_{\mathbb{R}^2} \det H(u) \,dx\,dy = 0 \end{equation} where $H$ is the Hessian.
Attempt: We have $H(u) = u_{xx}u_{yy} - u_{xy}^2$ then
\begin{equation} \int_{\mathbb{R}^2} \det H(u) \,dx\,dy = \int_{\mathbb{R}^2} u_{xx}u_{yy} - u_{xy}^2 \end{equation} Then by Plancherel's theorem we have
\begin{equation} \int_{\mathbb{R}^2} u_{xy}^2 = \int_{\mathbb{R}^2} |\widehat{u_{xy}}|^2 \,d\xi = \int_{\mathbb{R}^2} \xi_1^2\xi_2^2|\hat{u}|^2 = \int_{\mathbb{R}^2} \xi_1^2\hat{u}\xi_2^2\overline{\hat{u}} = \int_{\mathbb{R}^2} \widehat{u_{xx}}\overline{\widehat{u_{yy}}} \end{equation} However I am unsure of how to use this identity to show that \begin{equation} \int_{\mathbb{R}^2} \widehat{u_{xx}}\widehat{u_{yy}} \,d\xi = \int_{\mathbb{R}^2} u_{xx}u_{yy} \,dx\,dy \end{equation}
Any help would be greatly appreciated. The book also claims it holds for arbitrary dimensions in the sense that if $u \in C_0^2(\mathbb{R}^d)$ then \begin{equation} \int_{\mathbb{R}^d} \det\text{H}(u) = 0 \end{equation}