The Cauchy-Leray integral: counter-examples to the $L^p$-theory

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by  Loredana Lanzani

THE CAUCHY-LERAY INTEGRAL:
COUNTER-EXAMPLES TO THE Lp-THEORY
7
1 ∗ ∗∗
LOREDANA LANZANI AND ELIAS M. STEIN
0
2
n Abstract. We prove the optimality of the hypotheses guaran-
a teeing the Lp-boundedness for the Cauchy-Leray integral in Cn,
J n≥2, obtained in [LS-4].

3 Twodomains,bothelementaryinnature,showthatthegeomet-
1 ric requirement of strong C-linear convexity, together with regu-

larity of order 2, are both necessary.
]
V
C
.
h
t
a 1. Introduction
m
The purpose of this paper is to present counter-examples that show
[
that the Lp results for the Cauchy-Leray integral obtained in [LS-4]
1
v are essentially optimal. A second paper in this series will deal with
2 counter-examples for the Cauchy-Szeg˝o projection, relevant to [LS-5].
1 Recall that in the case of the unit ball B in Cn, the Cauchy-Leray
8
3 integral and the Cauchy-Szeg˝o projection agree, and the same holds for
0 the unbounded realization U of B. However when n ≥ 2 for more gen-
. 0
1 eral domains, these two operators are quite different. For the former,
0
the result obtained in [LS-4] states:
7
1
: Suppose D is a bounded domain in Cn whose boundary is of class
v
i C1,1 and which is strongly C-linearly convex. Then the induced Cauchy-
X
Leray transform is bounded on Lp(bD,dσ) for 1 < p < ∞.
r
a
Ourcounter-examples showthatthesetwo restrictionsonthebound-
ary of D, the geometric condition of strong C-linear convexity, and the
regularity of degree 2, are in fact necessary.
Date: January 17, 2017.
∗
Supported in part by the National Science Foundation, award DMS-1503612.
∗∗
Supported in partby the National Science Foundation, awardDMS-1265524.
2000Mathematics Subject Classification: 30E20, 31A10, 32A26, 32A25, 32A50,
32A55, 42B20, 46E22, 47B34, 31B10.
Keywords: Cauchy-Leray integral; Hardy space; Cauchy Integral; Cauchy-Szeg˝o
projection; Lebesgue space; pseudoconvex domain; minimal smoothness; Leray-Levi
measure.
1
2 LANZANIANDSTEIN
We consider the following simple bounded domains in C2. Here z =
j
x +iy , j = 1,2. The first domain is defined by
j j
(1) |z |2 +x2 +y4 < 1.
2 1 1
This is the domain given up to a translation by (9) in Section 3. It
has a C∞ (in fact real-analytic) boundary, is strongly pseudo-convex,
but not strongly C-linearly convex.
The second example is
(2) |z |2 +|x |m +y2 < 1, where 1 < m < 2.
2 1 1
This domain is given (up to a translation) by (20) in Section 5. The
domainhasa boundaryofclass C2−ǫ, (ǫ = 2−m), isstronglyC-linearly
convex (and hence strongly pseudoconvex).
We show that for each domain the induced Cauchy-Leray transform
is not bounded on Lp for any p, 1 ≤ p ≤ ∞.
The idea of the proof is as follows: we assume that Lp-boundedness
holds for one of these domains. Then an appropriate scaling and lim-
iting argument shows that this positive result implies a corresponding
conclusioninalimitingmodeldomain, whereitismucheasiertosupply
an explicit counter-example.
Forthefirstexamplethelimitingdomainistheunboundedhalf-space
2Imz > x2 (called D in (5)), which is holomorphically equivalent
2 1 0
to the more familiar half-space U , 2Imz > |z |2, which itself is holo-
0 2 1
morphically equivalent to the unit ball. Note that the last two are
strongly C-linearly convex, but D is not. The hint that one might be
0
led to a counter-example for D , and then for the domain (1) is that
0
its Cauchy-Leray operator is not “pseudo-local”; (see (6) which shows
that the kernel is singular away from the diagonal).
The analysis of the second domain, represented by (2), is parallel
to that of the first domain. For example, the corresponding limiting
domain is 2Imz > |x |m, see (21).
2 1
A remark is in order about the family of boundary measures {dµ }
a a
given in (29) that define the Lp spaces, Lp = Lp(bD,dµ ) in the above
a
results. Three examples of dµ are significant in many circumstances.
a
First, the induced Lebesgue measure dσ; second, the Leray-Levi mea-
sure dλ, see (4); and third, the Fefferman measure [B2], [F-1], [G].
Whenthedomainissmoothandstronglypseudo-convex thethreemea-
sures give the same Lp spaces, so that the counter-example holds for
the domain (1) for all such measures. In example (2) these measures
are essentially different, yet the counter-example still holds in all cases.
We should also call the reader’s attention to the earlier relevant work
in[BaLa]wherecounter-examples aregivenforCauchy-Leray integrals.
EXAMPLES CAUCHY-LERAY 3
However thelessexplicit andmorecomplexnatureoftheirconstruction
and proof limit the results to the case p = 2. It should be stressed that
when n ≥ 2 in general the Cauchy-Leray transform is far from “self-
adjoint” and so failure of L2-boundedness does not imply failure for
any p, p 6= 2.
2. The Cauchy-Leray integral; the model domain D
0
For a bounded domain (say C2) domain D in Cn, with a defining
function ρ, the corresponding Cauchy-Leray integral is
1
(3) C(f)(z) = f(w)dλ(w) z ∈ D.
Z ∆(w,z)n
bD
Here f is (say) a bounded function on bD,
n
∂ρ(w)
∆(w,z) = h∂ρ(w),w −zi = (w −z ),
j j
∂w
Xj=1 j
and
j∗
(4) dλ(w) = ∂ρ∧(∂∂ρ)n−1
(2πi)n
(cid:0) (cid:1)
is the Leray-Levi measure (with j∗ the pull-back under the inclusion:
bD ֒→ Cn). We have dλ(w) = Γ(w)dσ(w), with dσ the induced
Lebesgue measure on bD and
(n−1)! L(w)
Γ(w) = ,
4πn |∇ρ(w)|
where L(w) is the determinant of the Levi-form acting on the tangent
space at w ∈ bD. (See e.g., [Ra, Ch. 7].)
It is worth recalling two intrinsic properties of (3):
• Its invarianceunder changes of coordinatesthat aregiven either
by translations or unitary mappings of the space Cn, see [Bo].
• The independence of the Cauchy-Leray integral (3) of the par-
ticular defining function ρ of D.
We consider first the unbounded “model domain” D in C2.
0
With z = (z ,z ), z = x +iy , it is defined by
1 2 j j j
(5) D = {z : 2Imz > x2},
0 2 1
which is to be compared with the more familiar form of D given by
0
{z : 2Imz > |z |2}. These two domains are biholomorphically equiv-
2 1
alent via the mappings z ↔ z , z ↔ z ± iz2. Now the complex
1 1 2 2 1
tangent space of these domains at the origin is the subspace {(z ,0)}.
1
4 LANZANIANDSTEIN
So since |z |2 = x2 + y2 , and on R2 this is positive definite, this im-
1 1 1
plies that the second domain is strongly C-linearly convex. However
because the form x2 is degenerate along the direction y , it follows that
1 1
strong C-linear convexity fails for the domain (4). (For more about
these convexities see [APS]; [H¨o]; [LS-3, Sect. 3.3].)
With ρ (z) = 2y − x2, and ∆ (w,z) = h∂ρ (w),w − zi, and if we
0 2 1 0 0
write w = (w ,w ), w = u +iv , j = 1,2, a simple calculation gives
1 2 j j j
that with w and z ∈ bD ,
0
1
(6) ∆ (w,z) = (u −x )2 +2i(u (v −y )+u −x .
0 1 1 1 1 1 2 2
2
(cid:0) (cid:1)
Now for fixed z ∈ bD , observe that ∆ (w,z) vanishes on the one-
0 0
dimensional variety given by u = x , u (v −y )+u = x . Also the
1 1 1 1 1 2 2
Leray-Levimeasuredλ correspondingtoρ isdu dv du /(4π2) ≈ dσ ,
0 0 1 1 2 0
the induced Lebesgue measure on bD , if we are near the origin.
0
To construct a counterexample for the model domain D we first
0
choose a small constant a, which we will keep fixed throughout (a =
1/12 will do). Then for any δ, 0 < δ < 1, we define:
S = |u | ≤ aδ2, |v | ≤ 1 , |u | ≤ aδ2
1 1 2 2
(7)  (cid:8) (cid:9)
 S′ = δ ≤ |x | ≤ 2δ, |y | ≤ 1 , |x | ≤ aδ2
1 1 2 2
(cid:8) (cid:9)
with w ∈ bD written as (u +iv ,u +iu2/2) and z ∈ bD written as
0 1 1 2 1 0
(x +iy ,x +ix2/2) and S and S′ being the corresponding sets for
1 1 2 1 0 0
w ∈ bD , z ∈ bD via (7), then it follows that ∆ (w,z) 6= 0 whenever
0 0 0
w ∈ S and z ∈ S′. Also
0 0
σ (S ) ≈ λ (S ) ≈ m(S) ≈ δ4 while σ (S′) ≈ λ (S′) ≈ m(S′) ≈ δ3
0 0 0 0 0 0 0 0
where σ denotes the induced Lebesgue measure on bD and m is the
0 0
standard Euclidean measure on the three-dimensional Euclidean space.
We shall now test the presumed inequality
(8) kC0(f)kLp(S′0,dσ0) ≤ ApkfkLp(bD0,dσ0),
when f is assumed to have support in S . Under these circumstances
0
C (f)(z) is well-defined as an absolutely convergent integral
0
1
f(w)dλ (w),
Z |∆ (w,z)|2 0
0
S
0
when z ∈ S′, in view of the non-vanishing of ∆ (w,z) for these w and
0 0
z.
Proposition 1. The inequality (8) fails for every p, 1 ≤ p ≤ ∞.
EXAMPLES CAUCHY-LERAY 5
The constant A is of course assumed to be independent of f.
p
In fact it is clear by (6) and (7) that we have:
δ2
Re∆ (w,z) ≥ and |Im∆ (w,z)| ≤ aδ2 +2aδ2 = 3aδ2.
0 0
4
Then (Re∆ (w,z))2 ≥ 2|Im∆ (w,z)|2, (which holds if a ≤ 1/12).
0 0
But |∆ (w,z)| ≤ cδ2 for some c > 0, thus
0
1
Re ≥ c′δ4.
(∆ (w,z))2
0
Now take f = χS0, the characteristic function of S0. Therefore
Re C (f) (z) ≥ c′δ−4λ (S ) ≥ c > 0, for z ∈ S′ .
0 0 0 0
(cid:0) (cid:1)
So
kC (f)kp ≥ cλ (S′) ≥ c′δ3,
0 Lp(S′0) 0 0
while
kfkp ≤ λ (S ) ≤ cδ4.
Lp(bD0) 0 0
Since δ3 is not O(δ4) for small δ, (8) cannot hold, when p < ∞. The
case p = ∞ requires a separate but simpler argument which we omit.
3. The first counter-example
We turn to the domain (1) in C2 and it is useful to consider a trans-
late of it, given by
(9) D = {|z −i|2 +x2 +y4 < 1}.
2 1 1
We will show after rescaling and a passage to the limit, that we can
reduce consideration of D to D . From (9) it is clear that
0
ρ(z) = x2 +y4 +x2 +y2 −2y
1 1 2 2 2
is a defining function for D, and that ρ is strongly pluri-subharmonic,
so D is strongly pseudo-convex. Moreover, since each of the four one-
variable functions, x2, y4, x2, y2 −2y are strictly convex, the domain
1 1 2 2 2
D is itself strictly convex. We note that
1
Re∆(w,z) = Reh∂ρ(w),w−zi = ∇ρ(w),w−z
2 R
(cid:0) (cid:1)
where (·,·)R is the real inner product induced on R4 = C2 from h·,·i.
With w = (w ,w ), w −u +iv , j = 1,2, we claim that
1 2 j j j
(10) ∇ρ(w),w−z ≥
R
(cid:0) (cid:1)
≥ (x −u )2 +(x −u )2 +(y −v )2 +(v2 +y2)(v −y )2,
1 1 2 2 2 2 1 1 1 1
when w,z ∈ bD.
6 LANZANIANDSTEIN
To prove (10) we use the identity
β
′ ′′
f(β)−f(α) = (β −α)f (α)+ (β −α)f (t)dt
Z
α
forthefunctionsf = u2,f = u2, f = v2−2v , andf = v4. Similarly,
1 1 2 2 3 2 2 4 1
we replace w = (u + iv ,u + iv ) by z = (x + iy ,x + iy ) and
1 1 2 2 1 1 2 2
add the corresponding identities. Taking into account that ρ(w) =
f +f +f +f wih ρ(w) = 0 and the similar fact for ρ(z), together
1 2 3 4
with the observations that
β
(β −α)f′′(t)dt = (β −α)2 if f′′(t) = 2,
Z
α
and
β
(β −α)f′′(t)dt ≥ (β2 +α2)(β −α)2, if f′′(t) = 12t2,
Z
α
then yields (10).
Now (10) shows that if w and z ∈ bD, w 6= z, then z lies on one side
of the (real) tangent plane to bD at w. By convexity of D, the same
holds for z ∈ D \{w}.
Turning to the Cauchy-Leray integral of D we recall two preliminary
facts. First C(f)(z) is holomorphic in z ∈ D, if f is say an integrable
function on bD. Second, whenever F is a holomorphic function in D
which is continuous in D, and f = F , then C(f)(z) = F(z), z ∈ D.
bD
The latter fact follows from the Cauc(cid:12)hy-Fantappi`e formalism (see [Ra]
(cid:12)
and [LS-3]).
Note that when f is a bounded function supported in a closed set S
in bD, then C(f)(z) is well-defined as a convergent integral whenever
z is outside the support S. So certainly the extendability of C to a
bounded operator on Lp would imply
(11) kC(f)kLp(S′) ≤ ApkfkLp
whenever S′ is disjoint from S, with A independent of f, S, and S′.
p
Theorem 2. For any p, 1 ≤ p ≤ ∞, the presumed inequality (11) fails
when tested for a bound A independent of f and its support S, and
p
with S′ disjoint from S.
A further statement along these lines is made possible by the follow-
ing fact (whose proof will appear elsewhere): whenever f is of class C1
on the boundary bD, C(f) is extendable to a continuous function on
EXAMPLES CAUCHY-LERAY 7
D. With this we can define the Cauchy-Leray transform C(f) for such
f, by C(f) = C(f) .
bD
(cid:12)
Corollary 3. The(cid:12)mapping f 7→ C(f), initially defined for C1 func-
tions f is not extendable to a bounded operator on Lp(bD).
4. Proof of Theorem 2
We shall obtain a contradiction to (11) by a scaling argument that
leads us back to Proposition 1.
We define the maps τ , ǫ > 0, on C2 by τ (z ,z ) = (ǫz ,ǫ2z ) and set
ǫ ǫ 1 2 1 2
Dǫ = τǫ−1(D), with D as in (9). Then ρ(z) = x21+y14+x22+y22−2y2 is
a defining function for D and ρ (z) = ǫ−2ρ(τ (z)) is a defining function
ǫ ǫ
for D . Note that
ǫ
ρ (z) = x2 −2y +ǫ2(y4+x2 +y2),
ǫ 1 2 1 2 2
from which it is clear that the domains D increase as ǫ decreases, with
ǫ
limit our model domain D = {ρ (z) < 0} and ρ (z) = x2 −2y .
0 0 0 1 2
Observe also that
ρ (z) → ρ (z), and ∇ρ (z) → ∇ρ (z)
ǫ 0 ǫ 0
uniformly on compact subsets of C2.
We next define the “transported” measures dλ on bD by
ǫ ǫ
(12) F(z)dλǫ(z) = ǫ−4 F(τǫ−1(z))dλ(z)
Z Z
z∈bDǫ z∈bD0
for any continuous functions F defined on bD .
ǫ
Here dλ is the Leray-Levi measure on bD, but note that dλ is not
ǫ
the Leray-Levi measure on bD : as a result the operator C defined
ǫ ǫ
below is not the Cauchy-Leray integral of bD . We can also define the
ǫ
corresponding action of τ on functions by
ǫ
τ (F)(z) = F(τ (z)) = F ◦τ .
ǫ ǫ ǫ
Then by what has been said above, τ maps Lp(bD,dλ) to Lp(bD ,dλ )
ǫ ǫ ǫ
and we have the “isometry”:
(13) kFk = ǫ4/pkτ (F)k .
LP(bD,dλ) ǫ Lp(bDǫ,dλǫ)
Now let ∆ (w,z) = h∂ρ (w),w−zi. Then
ǫ ǫ
∆(τ (w),τ (z)) = h∂ρ(τ w),τ (w−z)i = ǫ2h∂ρ (w),w−zi = ǫ2∆ (w,z).
ǫ ǫ ǫ ǫ ǫ ǫ
Hence
(14) ∆ (w,z) = ǫ−2∆(τ (w),τ (z)).
ǫ ǫ ǫ
8 LANZANIANDSTEIN
We now define the operator C , by setting
ǫ
1
(15) C (f)(z) = f(w)dλ (w)
ǫ Z ∆ (w,z)2 ǫ
ǫ
bDǫ
which is well-defined as the integral above for any bounded function f,
as long as z is outside the support of f. Our next claim is that
(16) Cǫ(f)(z) = C(f ◦τǫ−1)(τǫ(z)),
for bounded f on bD , if z ∈ bD lies outside the support of f. In fact
ǫ ǫ
going back to the definition (3), and using (12) and (14), we see that
1
C(f ◦τǫ−1)(τǫ(z)) = Z ∆(w,τ (z))2 f(τǫ−1(w))dλ(w) =
ǫ
bD
1 1
ǫ4 f(w)dλ (w) = f(w)dλ (w)
Z ∆(τ (w),τ (z))2 ǫ Z ∆ (w,z)2 ǫ
ǫ ǫ ǫ
bDǫ bDǫ
showing (16).
Next, if (11) held, then by (13) we would also have
(17) kCǫ(F)kLp(S′ǫ,dλǫ) ≤ ApkFkLp(bDǫ,dλǫ)
whenever F is a bounded function on bD and S′ is a closed subset of
ǫ ǫ
bD , disjoint from the support of F.
ǫ
At this point we restrict the attention to the unit ball B in C2, and
we exploit the common coordinate system for bD ∩ B, and bD ∩ B,
0 ǫ
when ǫ is small. That is in this ball, bD is the graph over (x +iy ,x )
0 1 1 2
given by (x +iy ,x +ix2/2), while bD is the graph given by
1 1 2 1 ǫ
(x +iy ,x +i (x2/2+Φ (x ,y ,x ))), with Φ (x ,y ,x ) = O(ǫ2).
1 1 2 1 ǫ 1 1 2 ǫ 1 1 2
Now if in these coordinates we write dλ = Λ(x ,y ,x )dx dy dx
1 1 2 1 1 2
and dλ = Λ (x ,y ,x )dx dy dx , then by (12) we have
ǫ ǫ 1 1 2 1 1 2
(18) Λ (x ,y ,x ) = Λ(ǫx ,ǫy ,ǫ2x ).
ǫ 1 1 2 1 1 2
Finally we take f to be the function χS, the characteristic function
of the set S in (7), and lift f to functions on bD and bD respec-
ǫ 0
tively, by setting f (z ) = f(x ,y ,x ) when z = (x + iy ,x +
ǫ ǫ 1 1 2 ǫ 1 1 2
i (x2/2+Φ (x ,y ,x ))), and f (z ) = f(x +iy ,x +ix2/2).
1 ǫ 1 1 2 0 0 1 1 2 1
We also lift the sets S and S′ to S and S′ (subsets of bD ) in the
ǫ ǫ ǫ
same way. Our claim is with that notation
(19) C (f )(z ) → C (f )(z ), if z ∈ S′ .
ǫ ǫ ǫ 0 0 0 0 0
In fact,
1
C (f )(z ) = f(u ,v ,u )Λ (u ,v ,u )du dv du .
ǫ ǫ ǫ Z ∆ (w,z )2 1 1 2 ǫ 1 1 2 1 1 2
ǫ ǫ
S
EXAMPLES CAUCHY-LERAY 9
However by (18) Λ (u ,v ,u ) → Λ(0,0,0), and moreover ∆ (w,z ) →
ǫ 1 1 2 ǫ ǫ
∆ (w,z ) because ∇ρ → ∇ρ , while ∆ (w,z ) 6= 0 if w ∈ S and
0 0 ǫ 0 0 0 0
z ∈ S′. This gives (19).
0 0
As a result (17) with F = f leads to
ǫ
kC0(f0)kLp(S′0,dλ0) ≤ Apkf0kLp(bD0,dλ0)
which contradicts Proposition 1, proving the theorem.
5. The second counter-example
Here the domain will be taken to be
(20) D = {|z −i|2 +|x |m +y2 < 1}, with 1 < m < 2.
2 1 2
Its model domain is
(21) D = {2Imz > |x |m}.
0 2 1
For any f that is bounded on bD, the Cauchy-Leray integral C(f)(z)
is well-defined for z that lies in the complement of the support of f.
As in the previous sections we will show that the mapping f 7→ C(f)
fails to be bounded in Lp in the sense that the proposed inequality
(22) kC(f)kLp(S′,dσ) ≤ ApkfkLp(bD,dσ)
cannot hold. Here S′ is any set disjoint from the support of f, and the
bound A is assumed independent of f and S′.
p
The proof of this assertion follows the same lines as in Sections 2-4
for the domain (9), and so we will only discuss the minor differences
that occur.
The defining function of D is ρ(z) = |x |m+y2+x2+y2−2y , and
1 1 2 2 2
that of D is ρ (z) = |x |m−2y . Note that both domains are of class
0 0 1 2
C2−α, with α = 2−m. Also since |x |m, y2, x2, y2−2y are strongly
1 1 2 2 2
convex functions of one variable, the domain D is strongly convex,
hence strongly C-linearly convex, and in fact strongly pseudoconvex in
the following sense: the domain D is exhausted by an increasing family
of smoothdomains {D } which areuniformly strongly pseudo-convex,
γ γ
with defining function: ρ (z) = (x2 +γ)m/2 +y2 +x2 +y2 −2y .
γ 1 1 2 2 2
This convexity implies that Re∆(w,z) > 0 for w ∈ bD and z ∈ D,
except when z = w.
Returning to the model domain, if ∆ (w,z) = h∂ρ (w),w − zi a
0 0
calculation gives
(23) ∆ (w,z) =
0
= |x |m−|u |m+m[u ]m−1(u −x ) + i m[u ]m−1(v −y )+2(x −y ) .
1 1 1 1 1 1 1 1 2 2
(cid:0) (cid:1)
10 LANZANIANDSTEIN
Here we have used the notation
1 d
[u ]m−1 = |u |m = |u |m−1signu .
1 1 1 1
m du
1
Now we set
S = {|u | ≤ aδ2, |v | ≤ δ2−m, |u | ≤ δm}
1 1 2
(24) 
 S′ = {δ ≤ |x | ≤ 2δ, |y | ≤ δ2−m, |x | ≤ δm}
1 1 2
and let Sand S′ be the corresponding induced sets on bD .
0 0 0
We have that near the origin
dλ ≈ |u |m−2du dv du , and dσ ≈ du dv du ,
0 1 1 1 2 0 1 1 2
wheredλ anddσ aretheLeray-LevimeasureandtheinducedLebesgue
0 0
measure on bD . Thus
0
(25) λ (S ) ≈ δ2m
0 0
(26) σ (S ) ≈ δ4, and σ (S′) ≈ δ3.
0 0 0 0
By (23) it follows that
Re∆ (w,z) & δm, while |Im∆ (w,z)| . δm,
0 0
and if we choose a sufficiently small, then
1
Re ≥ cδ−2m if w ∈ S , while z ∈ S′ .
2 0 0
∆ (w,z)
0
(cid:0) (cid:1)
Taking f to be the characteristic function of S , we therefore get
0 0
ReC (f )(z) ≥ c > 0, for z ∈ S′. Hence this gives a contradiction
0 0 0
to (22) in the case when C is the Cauchy-Leray integral of the model
0
domain D .
0
To pass to the domain D we carry out the scaling via
(27) τ (z ,z ) = (ǫz ,ǫmz )
ǫ 1 2 1 2
The domain Dǫ = τǫ−1(D) has a defining function
ρ (z) = ǫ−mρ(τ (z)) = ρ (z)+ǫm(x2 +y2)+ǫ2−my2,
ǫ ǫ 0 2 2 1
which converges to ρ (z) = |x |m −2y .
0 1 2
We also set ∆ (w,z) = h∂ρ (w),w − zi. The transported measure
ǫ ǫ
dλ on D is now defined by the identity
ǫ ǫ
(28) F(z)dλǫ(z) = ǫ−2m F(τǫ−1(z))dλ(z),
Z Z
bDǫ bD
(compare with (12)).