It is a well known result (sometimes called the Weyl lemma) that the laplacian in mathbbRnmathbbRn is hypoelliptic, i.e. if ff is a distribution s.t. triangle(f)triangle(f) is smooth in an open set, than ff itself is smooth in the same open set. To establish the result one observes that:
1) finHs(mathbbRn)finHs(mathbbRn) and trianglefinHs(mathbbRn)trianglefinHs(mathbbRn) than finHs+2(mathbbRn)finHs+2(mathbbRn) (HH are ordinary L2L2 Sobolev spaces)
2) the same results holds if mathbbRnmathbbRn is replaced by an arbitrary bounded open set UU. Here one can define Hs(U):=Hs(U):={f:rhofinHs(mathbbRn)forallrhoinCinftyc(U)f:rhofinHs(mathbbRn)forallrhoinCinftyc(U)}
3) every distribution ff lies in some Hs(U)Hs(U), for some s>−inftys>−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)hattriangle(f)(xi)=−|xi|2hatf(xi) one can naturally define a square root of −triangle−triangle as hatLambda(f)(xi):=|xi|hatf(xi)hatLambda(f)(xi):=|xi|hatf(xi).
Here comes my question: Is LambdaLambda hypoelliptic?
The equivalent of 1) for LambdaLambda is trivially true, but 2) seems much harder because LambdaLambda appears to be "non local" (the values of Lambda(f)Lambda(f) on an open set seem to depend on the values of ff everywhere).
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