Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

It is a well known result (sometimes called the Weyl lemma) that the laplacian in $\mathbb{R}^n$ 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) $f \in H^s(\mathbb{R}^n)$ and $\triangle{f} \in H^s(\mathbb{R}^n)$ than $f \in H^{s+2}( \mathbb{R}^n)$ ($H$ are ordinary $L^2$ Sobolev spaces)

2) the same results holds if $\mathbb{R}^n$ is replaced by an arbitrary bounded open set $U$. Here one can define $H^s(U):=${$f:\rho f \in H^s(\mathbb{R}^n) \forall \rho \in C^{\infty}_c(U)$}

3) every distribution $f$ lies in some $H^s(U)$, 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 $\hat \triangle(f)(\xi) = -|\xi|^2 \hat{f}(\xi)$ one can naturally define a square root of $-\triangle$ as $\hat{\Lambda(f)}(\xi) :=|\xi|\hat{f}(\xi)$.

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).

share|improve this question
    
Have you looked at Bismut's work on hypoelliptic Dirac operators? –  Steve Huntsman Feb 16 '10 at 20:41
    
Too bad the absolute value function isn't smooth, or we'd have an elliptic pseudo-differential operator... –  Akhil Mathew Feb 16 '10 at 21:29
add comment

2 Answers

Yes, the square root of the Laplacian is an elliptic pseudodifferential operator of order $1$. Actually, strictly speaking, its symbol $|\xi|$ is not smooth at the origin, but if you insert a cutoff function, i.e. write $|\xi| = \chi(\xi)|\xi| + (1-\chi(\xi))|\xi|$, where $\chi$ is smooth, compactly supported and equals $1$ near $\xi = 0$, then taking the inverse Fourier transform gives the Schwartz kernel of the operator you are interested in; this decomposition using $\chi$ shows that it is equal to the sum of a smoothing operator (i.e. one with $\mathcal C^\infty$ Schwartz kernel) and a classical elliptic pseudodifferential operator of order $1$. This is the procedure on $\mathbb R^n$; on a manifold the construction is much the same but slightly more elaborate. In any case, any elliptic pseudodifferential operator has the type of regularity property that you are asking, i.e. $\Lambda u = f \in H^s$ implies $u \in H^{s+m}$ where $m$ is the order of $\Lambda$. For details, one good reference is Shubin, Pseudodifferential Operators and Spectral Theory (Springer). Any elliptic operator is hypoelliptic (but there are many hypoelliptic operators which are not elliptic -- e.g. the heat operator).

share|improve this answer
add comment

There is other (formal)square root of the laplacien (Dirac operator) see http://en.wikipedia.org/wiki/Dirac_operator.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.