on heat extensions commuting with multiplication

Discussion in 'Math Research' started by Prabhu, Dec 2, 2011.

  1. Prabhu

    Prabhu Guest

    On June 8, I had started a thread with the following question:

    Quote

    Let f and g be two complex-valued L^2 functions on the plane. Let F, G
    and H be the heat extensions of f, g and fg respectively.

    Are there ever non-trivial f and g such that H = FG ?

    Unquote

    There was some discussion but the problem remained unsettled. I think
    the following might be an easy proof, except it needs to be completed
    by a real-analysis step. If someone could help, thanks.

    Let P_t denote the heat kernel and

    let K denote the Fourier transform operator.

    F = P_t f, G = P_t g and H = P_t (fg).

    K(P_t f)(\xi) = e^{-t |\xi|^2} Kf(\xi)
    K(P_t (fg))(\xi) = e^{-t |\xi|^2} K(fg)(\xi)
    K(P_tf P_tg)) = K(P_t f) \star K(P_t g) where \star denotes
    convolution.

    Since the hypothesis is that H = FG, we have

    K(fg)(\xi) = e^{t|\xi|^2} [K(P_t f) \star K(P_t g)](\xi)

    is independent of t for all \xi. We would like to claim that this
    implies K(fg) is identically 0, which would then imply our claim in a
    few steps more.

    Calculating the right side gives the following integral:

    \int_{\bC} e^{-2t (|\eta|^2 - \eta\cdot \xi)} Kf(\xi - \eta) Kg(\eta)
    dm(\eta)

    This is supposed to be independent of t, for all \xi. Is it easy to
    see from here that the value must be 0 therefore?

    Thanks.

    Prabhu

    PS. \int fg = 0 is a known fact.
     
    Prabhu, Dec 2, 2011
    #1
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