# on heat extensions commuting with multiplication

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

1. ### PrabhuGuest

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