# Logarithm of the Fourier transform?

I've found this paper on the logarithm of the discrete fourier transform which proves that

$$log F = 1/4 i \pi (I - (1 +i)F + F^2 - (1 - i)F^3)$$

where $$F$$ is the unitary discrete Fourier transform operator.

Is there a similar analogous version known for the infinite dimensional Fourier transform?

I am asking this from the point of view of trying to better understand cumulants, whose generating function is the logarithm of the characteristic function. The characteristic function is the fourier transform of the PDF of a random variable, thus motivating this question.

EDIT: Please take as many "niceness" assumptions as needed to answer the question regarding convergence.

• First one has to worry about convergence of the logarithm in the infinite-dimensional setting …. – LSpice Mar 12 '20 at 20:08
• The spectrum of the Fourier transform on $L^2(\mathbb{R})$ is $\{\pm 1, \pm i\}$, so it should suffice to choose any polynomial which agrees with (your favorite branch of) the complex logarithm on those four points. – Nate Eldredge Mar 12 '20 at 20:34
• @NateEldredge I'm really unfamiliar with the "proper" fourier transform. I've only used it from a "physics" standpoint. Could you please expand your comment into an answer, as I do not really understand what you've said? That would be very helpful for me :) – Siddharth Bhat Mar 17 '20 at 23:58
• Reading your question again, though, I think you may be mixing up two different things. The cumulant is the log of the Fourier transform of a specific function. But the paper you link, and my comment, are about taking the log of the Fourier transform operator in the sense of spectral theory. It's the difference between $\log(F[f])$ and $(\log F)[f]$. It's like the fact that if $A$ is a matrix and $v$ is a vector, $A^2 v$ is not the same as taking $Av$ and squaring each entry. – Nate Eldredge Mar 18 '20 at 14:57

Let us start with the identity $$F^4=I$$. As a result, we have formally \begin{align} \ln F&=\ln(I+F-I)=\sum_{k\ge 1}\frac{(-1)^{k-1}(F-I)^k}{k} \\&= \sum_{1\le k\le 3}\frac{(-1)^{k-1}(F-I)^k}{k}+ \sum_{k\ge 4}\frac{(-1)^{k-1}}{k}\sum_{0\le l\le 3}\binom{k}{l} F^l(-1)^{k-l} \\&\hskip94pt+ \sum_{k\ge 4}\frac{(-1)^{k-1}}{k}\sum_{k\ge l\ge 4}\binom{k}{l} F^l(-1)^{k-l}, \end{align} and since in the last sum $$F^l=F^j$$ with $$j\equiv l\mod 4$$, we get a polynomial in $$F$$ with degree 3, that is $$\ln F= a_0+a_1 F+a_2 F^2 +a_3 F^3.$$ I guess that it is straightforward to get the expressions of the $$a_j$$ from the above identity. Also you can use the fonction $$g(x)=e^{-π x^2}$$ which is such that $$(F-I) g=0$$ implying that $$0=(\ln F)(g)=(a_0+a_1 +a_2 +a_3)(g)\Longrightarrow \sum_{0\le j\le 3}a_j=0,$$ and in fine you will get the same coefficients as yours.
$$F=e^{\frac{1}{4} \pi i \left(D^2-x^2+1\right)}$$
For $$F$$ any linear map such that $$F^4=Id$$ then $$F= \sum_{k=0}^3 i^k T_k$$ where $$T_k =\frac14 \sum_{m=0}^3 i^{-mk} F^m$$, $$T_k^2=T_k,T_k T_l = 0$$ which gives that $$(\sum_{k=0}^3 c_k T_k)^n = \sum_{k=0}^3 c_k^n T_k$$.
With $$L=\sum_{k=0}^3 \frac{ki\pi}{2} T_k$$ we get that $$\exp(L)=\sum_{n=0}^\infty \sum_{k=0}^3 \frac{(\frac{ki\pi}{2})^n}{n!} T_k=\sum_{k=0}^3 \exp(\frac{ki\pi}{2}) T_k= F$$.