Wednesday, 19 September 2007

fourier analysis - Hypoellipticity of square root of laplacian

It is a well known result (sometimes called the Weyl lemma) that the laplacian in mathbbRn is hypoelliptic, i.e. if f is a distribution s.t. triangle(f) is smooth in an open set, than f itself is smooth in the same open set. To establish the result one observes that:



1) finHs(mathbbRn) and trianglefinHs(mathbbRn) than finHs+2(mathbbRn) (H are ordinary L2 Sobolev spaces)



2) the same results holds if mathbbRn is replaced by an arbitrary bounded open set U. Here one can define Hs(U):={f:rhofinHs(mathbbRn)forallrhoinCcinfty(U)}



3) every distribution f lies in some Hs(U), for some s>infty
Then a kind of bootstraping argument and Sobolev embedding gives the result.



1) is very easy to establish (just look at the Fourier transform side), while the local version 2) is harder and requires the "local nature" of the laplacian (in fact of partial differentiation operators), as is encoded for example by the Leibniz rule (Reed and Simon has a complete proof).



Since on the Fourier trasform side hattriangle(f)(xi)=|xi|2hatf(xi) one can naturally define a square root of triangle as hatLambda(f)(xi):=|xi|hatf(xi).



Here comes my question: Is Lambda hypoelliptic?



The equivalent of 1) for Lambda is trivially true, but 2) seems much harder because Lambda appears to be "non local" (the values of Lambda(f) on an open set seem to depend on the values of f everywhere).

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