Showing that $\int_{\mathbb{R}^2} \frac{df^2 }{dx^2} \frac{df^2 }{dy^2} dx dy = \int_{\mathbb{R}^2} \left( \frac{df^2 }{dx dy} \right)^2 dx dy$ I'd appreciate help showing that
$$
\int_{\mathbb{R}^2} \frac{\partial f^2 }{\partial x^2} \frac{\partial f^2 }{\partial y^2} dx dy  = \int_{\mathbb{R}^2} \left( \frac{\partial f^2 }{\partial x \partial y} \right)^2 dx dy
$$
by integration by parts.
This relation is from page 5, equation 18, of the paper Generalized Sampling: A Variational Approach. Part I: Theory.
 A: We have $$\int_{\mathbb R^2} \dfrac{\partial^2 f}{\partial x^2}\dfrac{\partial^2 f}{\partial y^2}dx dy = \int_{\mathbb R}\left(\int_{\mathbb{R}}\dfrac{\partial^2 f}{\partial x^2}\dfrac{\partial^2 f}{\partial y^2}dy\right)dx$$ 
and 
$$\int_{\mathbb{R}}\dfrac{\partial^2 f}{\partial x^2}\dfrac{\partial^2 f}{\partial y^2}dy = \left[\dfrac{\partial^2 f}{\partial x^2}\dfrac{\partial f}{\partial y}\right]_{y=-\infty}^{y=+\infty}-\int_{\mathbb R}\dfrac{\partial^3 f}{\partial x^2\partial y}\dfrac{\partial f}{\partial y}dy $$
hence 
$$\int_{\mathbb R^2} \dfrac{\partial^2 f}{\partial x^2}\dfrac{\partial^2 f}{\partial y^2}dx dy =-\int_{\mathbb R}\left(\int_{\mathbb R}\dfrac{\partial^3 f}{\partial x^2\partial y}\dfrac{\partial f}{\partial y}dx\right)dy.$$
To conclude, we notice that 
$$\int_{\mathbb R}\dfrac{\partial^3 f}{\partial x^2\partial y}\dfrac{\partial f}{\partial y}dx=\left[\dfrac{\partial^2 f }{\partial x\partial y}\dfrac{\partial f}{\partial y}\right]_{x=-\infty}^{x=+\infty} -\int_{\mathbb R}\dfrac{\partial^2 f }{\partial x\partial y}\dfrac{\partial^2 f }{\partial y\partial x}.$$
Added later: of course, the brackets vanish since $f$ and its partial derivatives have a compact support.
