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.
Best Answer
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.