Pseudodifferential operators on spaces with boundary Consider the upper half space $\mathbb{R}^n_{+} = \{x = (x_1,..,x_n) \in \mathbb{R}^n : x_n \geq 0\}$. Consider the Laplacian on this space with either the Dirichlet boundary condition or the Neumann boundary condition. My question is: can the Laplacian $\Delta$ under such boundary conditions be treated as a pseudodifferential operator of order 2? I can understand that the usual trick of Fourier inversion does not work, but if we still forcibly write
$$-\Delta f(x) = \int p(x, \xi)\hat{f}(\xi)e^{ix.\xi}d\xi$$ and assume that $p(x, \xi)$ belongs to some symbol class $S^m_{\rho, \delta}$, what can go wrong?
 A: What are pseudo-differential operators (pseudors) on a closed half-space $H$ or on a manifold with boundary? It is not adequate to only define them as restrictions from some open neighbourhood of $H$ to $H$. A pseudor $P$ does not map $C_c^\infty(H)$ into $C^\infty(H)$ (smoothness up to the boundary!) unless $P$ satisfies the transmission condition; see Boutet de Monvel' paper in Acta. Math. of 1971 or section 18.2 in Hörmander's volume III. Differential operators and the parametrices of elliptic differential operators satisfy the transmission condition, but a (scalar) square root of the Laplacian does not. (Dirac operators are non-scalar square roots of Laplacians, and they are differential operators.)
Much work (by Boutet de Monvel, Melrose, Schulze, and many others) has been done in developing pseudo-differential calculi for manifolds with boundary (possibly with corners). Already finding the right function (distribution) spaces, on with the pseudors act as an algebra of operators, is an important issue. For example, if functions $u$ are extended somehow from $H$ to all space, then one might wish to define $Pu$ in such a way that it does not depend on the particular extension. This is no problem if $P$ is a differential operator because differential operators are local. But for genuine pseudors it is; well, it was before any theory of pseudo-differential boundary calculi existed.
A: The formula will work just fine (with no pseudo required), provided you interpret $\widehat{f}$ suitably. Since $f$ is only a function on a half space, we have to make an agreement on what exactly we mean by $\widehat{f}$.
To obtain the Dirichlet Laplacian, define $\widehat{f}$ as the Fourier transform of the odd extension $f(x,-y)=-f(x,y)$ of $f$ (I use the notation $x=(x_1,\ldots ,x_{n-1})$, $y=x_n$). Then
$$
\widehat{-\Delta f} = |\xi|^2 \widehat{f} ,
$$
as expected. Moreover, the domains also get handled correctly by this formula: $f\in D(-\Delta)=W^2_0$ precisely if the odd extension of $f$ satisfies $|\xi|^2\widehat{f}\in L^2(\mathbb R^n)$, or, equivalently, if (in $L^2$ sense)
$$
f(x,y) = \int_{\mathbb R^{n-1}} d\xi' \int_0^{\infty} dk\, g(\xi',k) e^{i\xi' x}\sin ky
$$
for some $g\in L^2$ with $(|\xi'|^2+y^2)g\in L^2$.
For the Neumann Laplacian, work with even extensions instead.
