Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was going through the derivation of Wigner distribution properties, and encountered a certain step in the proof I could not justify.

Namely, the step requires the following equality to be true:

$\int d^2 \lambda \frac{\partial}{\partial \lambda} \left( \exp(-\lambda \alpha^* + \lambda^* \alpha) f(\lambda) \right) \equiv 0$,

where $\alpha$ and $\lambda = x + iy$ are complex numbers, $\int d^2 \lambda \equiv \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} dx dy$, and the differential is a Wirtinger one:

$\frac{\partial}{\partial \lambda} \equiv \frac{1}{2} \left( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right)$.

The function $f$ is bounded, continuous and infinitely differentiable (w.r.t. $x$ and $y$), but not necessarily going to zero on the infinity.

As far as I understand, this is a Fourier transform, and the equality would reduce to a known property (Fourier transform of derivative), if $\lim_{x \rightarrow \infty} f = 0$ and $\lim_{y \rightarrow \infty} f = 0$, which is, unfortunately, not the case. Could anyone point me to the idea behind the proof in such circumstances?

share|improve this question

1 Answer 1

I have never seen the derivation of this but, if $f$ is entire and bounded (is not it?) then by Liouville's theorem it is constant. This seems to be enough to get the result you are looking for.


share|improve this answer
Oh, I completely forgot about this question. No, the problem was that $f$ was not entire (that's why I only used Wirtinger differentials). I have managed to strengthen "$f$ is bounded" to "$f$ goes to zero on the infinity" since then, for which the proof is trivial. I should probably close the question now, as I'm not even sure this equality can be proved with my initial conditions. –  fjarri Oct 25 '13 at 10:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.