fourier analysis - Hypoellipticity of square root of laplacian

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).