I recently came across an interesting (mis-)use of formal equivalencies. First, the uncontroversial bits.

By the Fourier derivative theorem, it is straightforward that the 2D Laplace operator can be written as $$\Delta_{\text{2D}}f = -\mathcal{F}_{\text{2D}}^{-1}\left[(k_x^2 + k_y^2)\mathcal{F}_{\text{2D}}(f)\right],$$ if $f$ is integrable and twice differentiable and if the given inverse Fourier transform converges—which is the case if this operator is defined on a Schwartz space $\mathcal{S}(\mathbb{R}^2)$.

Furthermore, it is straightforward to generalize this to arbitrary polynomials (symbols) of differential operators on a Schwartz space, which, as I understand, is the starting point for the construction of pseudo-differential operators: $$L f := \sum_{n} \alpha_n \left(\frac{\partial}{\partial x_{i_n}}\right)^{m_n} f = \mathcal{F}\left[\sum_{n} \alpha_n\left(i k_{i_n}\right)^{m_n} \mathcal{F}(f)\right].$$

But in a document I recently read (written by non-mathematicians), this equivalency was used to define the following operator: $$D f := -\mathcal{F}_{\text{2D}}^{-1}\left[\frac{1}{k_x^2 + k_y^2}\mathcal{F}_{\text{2D}}(f)\right], $$ and then claim that $D$ is a well-defined operator which is the inverse of $\Delta_\text{2D}$ (where we directly see that this is flawed since 0 is in the spectrum of $\Delta_{\text{2D}}$).

The inverse equivalency can easily be shown to be formally true from the composition $D \circ \Delta_{\text{2D}}$ by assuming everything is well-behaved and by canceling the Fourier transforms with their inverses and polynomial with its inverse, but the assumption that everything is well-behaved is manifestly over-optimistic. $D$ itself is not convergent for a very large number of Schwartz space functions due to the singularity; as far as I can tell, this operator is basically only defined on the image $\Delta_{\text{2D}}(\mathcal{S}(\mathbb{R}^2))$.

What can be said about this "inverse" of the Laplace operator? Am I missing something and it is more well-defined than I think it is? I understand that pseudo-differential operators in general have a useful and rigorous theory, it just seems to me that it is incorrect to apply the concept in this ill-defined way (where we are integrating over the singularity with no regard to convergence).

  • 1
    $\begingroup$ "But in a document I recently read" - surely, a reference to the document would help to answer this question. Unless this document has restricted access. In any case, some context could be helpful $\endgroup$
    – Yuriy S
    Dec 20, 2020 at 2:04
  • $\begingroup$ Well, I didn't want to cast shade on the work of other people; it's from a dissertation on (experimental) x-ray phase contrast spectroscopy and they use this operator in the derivation of certain phase contrast equations. Paganin's Coherent X-Ray Optics is cited as a reference, but I haven't yet had access to that. At any rate, I was interested in the mathematical points of their procedure, which clearly was not on the top of their list, as they are primarily interested in the application. $\endgroup$ Dec 20, 2020 at 2:50
  • 1
    $\begingroup$ Thank for clarifying. I don't think there's anything wrong with asking questions about or even disagreeing with others' research methods. In any case, an interesting question and I hope someone manages to answer it, as it's beyond my own scope $\endgroup$
    – Yuriy S
    Dec 20, 2020 at 2:58

1 Answer 1


Division by $k_x^2 + k_y^2$ is only allowed if this expression is not zero. So in some sense, this operator is not well defined. As you correctly note, there are also some subtle convergence issues.

That said, there are plenty of ways to make sense of all the integrals involved if e.g. $\mathcal F_{2D} f(0) = \int f(x) dx = 0$ and $f$ is sufficiently smooth (e.g. H^2, but you could probably get away with even less regularity if you generalize what the expression should mean some more). Whether an operator is "well-behaved" depends not only on the space that it should act on, but also whether the author is from a more or less applied field.

If you're interested in integrals like this, have a look at the field of "singular integrals". This kind of expression is indeed "more well-defined" than you might think - as is the case with many constructs used in physics.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .