# Thread: Fourier transform of operators

1. ## Fourier transform of operators

Say we let the operator $\displaystyle D:=\frac{1}{i}\patial$
It is then possible to wire this operator in the form $\displaystyle D=F|D|$ like the polar decomposition of the bounded operators on some Hilbert space.
Let $\displaystyle A$ be the algebra of continuous functions on the circle and consider a function $\displaystyle f\in A$ with Fourier transform
$\displaystyle \hat{f}(k)=\int^{1}_{-1}f(x)e^{-ikx}dx$

Then we have
$\displaystyle \hat{Df}(k)=\hat{D}\hat{f}=\frac{1}{i}\int^{1}_{-1}\frac{\partial}{\partial x}f(x)e^{-ikx}dx$

I can show that
$\displaystyle \hat{D}\hat{f}(k)=k\hat{f}(k)$
Then defining $\displaystyle |\hat{D}|\hat{f}(k):=|k|\hat{f}(k)$
we can find $\displaystyle \hat{F}$ which we can just think of as a matrix with $\displaystyle \pm 1$ along the diagonal

I would like to find the inverse Fourier transform of $\displaystyle \hat{F}$, it should resemble something that looks like a derivative, but alas I cannot find it.

Any help will be greatly appreciated

2. Since $\displaystyle \hat{F}$ acts on $\displaystyle \hat{f}(k)$ as follows
$\displaystyle \hat{F}\hat{f}(k)=\left\{\begin{array}{cc}\hat{f}( k) &,k\geq 0 \\ -\hat{f}(k) &,k<0 \end{array}\right.$

we can find the inverse Fourier transform
$\displaystyle Ff(x)=\sum_k \hat{F}\hat{f}(k)e^{ikx} =\sum_{k\geq0}\hat{f}(k)e^{ikx}-\sum_{k<0}\hat{f}(k)e^{ikx}$
$\displaystyle =\sum_{k\geq0}\int^{1}_{-1}f(y)e^{-ik(y-x)}dy-\sum_{k<0}\int^{1}_{-1}f(y)e^{-ik(y-x)}dy$
$\displaystyle =\int^{1}_{-1}f(y)\left(1+\sum_{k>0}\left(e^{-ik(y-x)}-e^{ik(y-x)}\right)\right)dy$
$\displaystyle =\int^{1}_{-1}f(y)\left(1-2i\sum_{k>0}\sin k(y-x)\right)dy$

The derivative should come frome the commutator $\displaystyle [F,\pi(f)]\psi$ where $\displaystyle \pi: C(\mathbb{T})\rightarrow B(L^2(\mathbb{T}))$

where $\displaystyle \pi(f)\psi(x)=f(x)\psi(x)$

However, the details still elude me... my guess is that we should have a result that looks more or less like

$\displaystyle [F,\pi(f)]\psi \sim \frac{\psi(x)-\psi(y)}{x-y}$
together with some multiplicative factor.