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Math Help - Fourier transform of operators

  1. #1
    Member Mauritzvdworm's Avatar
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    Fourier transform of operators

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

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

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

    Any help will be greatly appreciated
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  2. #2
    Member Mauritzvdworm's Avatar
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    Since \hat{F} acts on \hat{f}(k) as follows
    \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
    <br />
Ff(x)=\sum_k \hat{F}\hat{f}(k)e^{ikx}<br />
=\sum_{k\geq0}\hat{f}(k)e^{ikx}-\sum_{k<0}\hat{f}(k)e^{ikx}<br />
    =\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
    =\int^{1}_{-1}f(y)\left(1+\sum_{k>0}\left(e^{-ik(y-x)}-e^{ik(y-x)}\right)\right)dy
    =\int^{1}_{-1}f(y)\left(1-2i\sum_{k>0}\sin k(y-x)\right)dy

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

    where \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

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