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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.

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Well, then just integrate by parts, there's nothing else to it... Of course, you need to assume that the surface terms go to zero and this gives you a condition on (derivatives of) $f$. –  Marek Jul 25 '11 at 11:24
    
In the paper $f$ is supposed to be a test function (it's the only thing that @olumide has to add here). –  Davide Giraudo Jul 25 '11 at 11:31

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up vote 4 down vote accepted

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.

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Such a clear explanation! Thanks @gridav. –  Olumide Jul 27 '11 at 9:40

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