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If $f$, $\nabla_x \cdot \nabla_y f \in L^2(\mathbb{R}^d_x\times \mathbb{R}^d_y)$, what can be said about vanishing at infinity of $\nabla_x f$, $\nabla_y f$ or if they even belong to $L^2_{x,y}$?

It is clear to me that $(\nabla_x^2 + \nabla_y^2) f \in L^2_{x,y}$ implies $\nabla_x, \nabla_y f \in L^2_{x,y}$ (see e.g. Is a function in $L^2$ which second derivative is in $L^2$ in $H^2$? or Does existence of the second weak derivative of $f\in L^2$ imply existence of the first?).

Note that in this setting I only claim control over the mixed second derivative $\nabla_x \cdot\nabla_y f $.

In particular I am intrigued about partial integration in the sense whether

$$\int (\overline{\nabla_x f}) \nabla_x \cdot \nabla_y (\nabla_x f) = \int (\nabla_x f) \nabla_x \cdot \nabla_y (\overline{\nabla_x f}) $$

or

$$\int (\overline{\nabla_y f}) \nabla_x \cdot \nabla_y (\nabla_y f) = \int (\nabla_y f) \nabla_x \cdot \nabla_y (\overline{\nabla_y f})$$

holds.

EDIT: By considering Fourier transforms, it is clear that we have no control over $\int (|\xi|^2+|\eta|^2) |\widehat{f}(\xi,\eta)|^2d\xi d\eta$ if we only have control over $\int (1+|\xi|^2|\eta|^2) |\widehat{f}(\xi,\eta)|^2d\xi d\eta$. Therefore $\nabla_x f,\nabla_y f$ will not lie in $L^2$ in general.

EDIT 2: It was pointed out that in the second part of the question, the quantities with three derivatives acting on $f$ might not be well-defined. Therefore a slight change of the question:

Does the control over $f$ and $\nabla_x\cdot \nabla_y f$ in $L^2_{x,y}$ allow for enough decay for $\nabla_x f, \nabla_y f$ at infinity such that we have

$$\int \overline{f} \nabla_x \cdot \nabla_y f = \int f \nabla_x \cdot \nabla_y \overline{f}$$

Otherwise could we just make additional assumptions on $f$ such that the two integrals above are well-defined and the question remains?

Jakob Elias
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    In general, for sure they do not always belong to $L^2$. Consider $d=1$: by Plancherel’s theorem, you have $\widehat f\in L^2(\mathbb R^2)$ and $\xi\eta\widehat f(\xi,\eta)\in L^2(\mathbb R^2)$, where $\xi,\eta$ are the Fourier variables corresponding to $x,y$. With the above conditions, there is no way of controlling $\xi\widehat f(\xi,\eta)$ or $\eta\widehat f(\xi,\eta)$ to make them lie in $L^2$. Despite that, you have good control on the low frequencies of $f$, so you should still have some decay in some sense – Lorenzo Pompili Nov 09 '23 at 23:40
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    In the identities below the question, you have three derivations acting on $f$. Could you be a little more precise on how the above question relates to those identities? In particular, how is any of those quantities well defined? – Lorenzo Pompili Nov 09 '23 at 23:42
  • Concerning your first comment, I had this same thought yesterday, but forgot to change the question. For the second, yes, maybe for a start instead of $\int \overline{\nabla_x f} \nabla_x \cdot \nabla_y(\nabla_x f)$ let's focus on something like $\int \overline{f} \nabla_x \cdot \nabla_y f$, which should be well-defined by the assumptions on $f$ and $\nabla_y\cdot \nabla_y f$. – Jakob Elias Nov 10 '23 at 08:00

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