I have a differential operator defined by its Fourier transform:


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


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.



Fourier transform of a differential operator
Let me change slightly your notations and consider the quadratic form in $\mathbb R\_{\xi,\eta}^2$
$$
Q(\xi,\eta)=\alpha \xi ^2+2\gamma \xi \eta+\beta \eta^2,
$$
where $\alpha, \beta$ are real parameters.
This is a Fourier multiplier :
$
Fourier\bigl(Q(D\_x,D\_y)u\bigr)(\xi,\eta)=Q(\xi,\eta)\hat u(\xi,\eta).
$
Let us for instance assume that $\gamma^2<\alpha\beta, \alpha >0$,
i.e. $Q$ is positive-definite. $Q$ has positive characteristic roots and can be transformed by an orthogonal transformation into
$\lambda \xi ^2+\mu \eta^2,\lambda, \mu >0.$
Changing scales leads to $\xi ^2+\eta^2$. This multiplier is the $-$Laplace operator in two dimensions, with the fundamental solution
$$
\frac{-1}{2\pi}\ln(\sqrt{x^2+y^2})=E(x,y).
$$
So the inverse of the Fourier multiplier $D\_x^2+D\_y^2$ is the convolution by $E$. Some analogous things can be done when $\gamma^2\ge\alpha\beta$.


Bazin.


