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I have a differential operator defined by its Fourier transform:

$\left(\alpha k_x k_x^2 + \beta k_y k_y^2 + \gamma k_x k_y \right)^\delta \hspace{20pt} \alpha,\beta,\gamma,\delta \in \mathbb{R}$

I don't know how to do the inverse transform, but I know that it is impossible to compute in the most general case. However I don't need it, as I would only like to demonstrate that this operator, in real space (i.e. after antitrasforming) when applied to a constant term, yields zero. This is just my guess, but seeing how the operator is written, I suppose that after anti-transforming (if that would be possible) then it would be a function of $\frac{\partial}{\partial x}$, $\frac{\partial}{\partial y}$ and $\frac{\partial ^2}{\partial x \partial y}$, so being null when operating on constants, I suppose.

Is my guess correct? If so, how can I prove my guess?

Thank you very much!

Disclaimer: I am not a mathematician, I am a (wannabe) physicist, so some of my definitions could be inaccurate, I hope the question is clear, though.

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# Fourier transform of a differential operator

I have a differential operator defined by its Fourier transform:

$\left(\alpha k_x + \beta k_y + \gamma k_x k_y \right)^\delta \hspace{20pt} \alpha,\beta,\gamma,\delta \in \mathbb{R}$

I don't know how to do the inverse transform, but I know that it is impossible to compute in the most general case. However I don't need it, as I would only like to demonstrate that this operator, in real space (i.e. after antitrasforming) when applied to a constant term, yields zero. This is just my guess, but seeing how the operator is written, I suppose that after anti-transforming (if that would be possible) then it would be a function of $\frac{\partial}{\partial x}$, $\frac{\partial}{\partial y}$ and $\frac{\partial ^2}{\partial x \partial y}$, so being null when operating on constants, I suppose.

Is my guess correct? If so, how can I prove my guess?

Thank you very much!

Disclaimer: I am not a mathematician, I am a (wannabe) physicist, so some of my definitions could be inaccurate, I hope the question is clear, though.