70 Noncommutative Mathematics for Quantum Systems
Similarly, we get ImzGν(z) > 0 forz ∈C−. It follows that the
functions in front of the integrals in the definitions ofgandhare
bounded as functions of x, and therefore g and h are
square-integrable. Sincez−^1 yis bounded too, we see that Equation
(1.7.6) defines a bounded operator.
Let us now check that the operator defined in (1.7.6) is the inverse
ofz−
√
SxMy
√
Sx.
Using the notation of the previous subsection, we can writeSx
also asSx=Mx− 1 P 2 + 1 =MxP 2 +P 2 ⊥, whereP 2 ⊥is the projection
onto the orthogonal complement of the subspace of functions that
do not depend on√ y. Its square root can be written as
Sx=M√xP 2 +P 2 ⊥=M√x− 1 P 2 + 1 ; it acts as
(√
Sxψ
)
(x,y) =
(√
x− 1
)∫
R+
ψ(x,y)dν(y) +ψ(x,y)
on a functionψ∈Dom
√
Sx⊆L^2 (R+×R+,μ⊗ν).
Sincehdoes not depend ony, we have
√
Sxh=
√
xh. Forgwe
get
(√
Sx
g
z−y
)
(x) = (
√
x− 1 )
∫
R+
g(x)
z−y
dν(y) +
g(x)
z−y
=
(
(
√
x− 1 )Gν(z) +
1
z−y
)
g(x).
Setφ=ψz−+yg+h. Applying
√
Sxtoφ, we get
(√
Sxφ
)
(x,y)
=
ψ(x,y)
z−y
+
√
x−x
(z−y)
(
( 1 −x)zGν(z) +x
)
∫
R+
ψ(x,y)dν(y)
+
z(x− 1 )
(z−y)
(
( 1 −x)zGν(z) +x
)
∫
R+
ψ(x,y)
z−y
dν(y)
=
ψ(x,y) +g(x)
z−y
.
From this we get
((
z−
√
SxMy
√
Sx
)
φ
)
(x,y) = ψ(x,y)