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

TL;DR

This paper demonstrates the necessity of certain geometric and regularity conditions for the $L^p$-boundedness of the Cauchy-Leray integral in complex spaces, through explicit counter-examples.

Contribution

It establishes the optimality of previously known hypotheses by providing elementary counter-examples that show the necessity of strong $ ext{C}$-linear convexity and regularity conditions.

Findings

Strong $ ext{C}$-linear convexity is necessary for $L^p$-boundedness.

Regularity of order 2 is required for the $L^p$-theory.

Counter-examples demonstrate the failure of boundedness without these conditions.

Abstract

We prove the optimality of the hypotheses guaranteeing the $L^p$-boundedness for the Cauchy-Leray integral in $\mathbb C^n$, $n\geq 2$, obtained in [LS-4]. Two domains, both elementary in nature, show that the geometric requirement of strong $\mathbb C$-linear convexity, together with regularity of order 2, are both necessary.

Equations

$$|z_2{|}^2+x_1^2+y_1^4<1.$$

$$|z_2{|}^2+|x_1{|}^m+y_1^2<1,\text{where}1<m<2.$$

$$\mathbf{C}(f)(z)=\underset{bD}{\int }\frac{1}{\Delta (w,z{)}^n}f(w)d\lambda (w)z\in D.$$

$$\Delta (w,z)=⟨\partial \rho (w),w-z⟩=\sum _{j=1}^n\frac{\partial \rho (w)}{\partial w_j}(w_j-z_j),$$

$$d\lambda (w)=\frac{j^{\ast }}{(2\pi i{)}^n}\left(\partial \rho \wedge (\overline{\partial }\partial \rho {)}^{n-1}\right)$$

$$\Gamma (w)=\frac{(n-1)!}{4{\pi }^n}\frac{\mathcal{L}(w)}{|\nabla \rho (w)|},$$

$$D_0=\{z:2\mathrm{Im}{z}_2>x_1^2\},$$

$$\Delta_{0}(w,z)=\frac{1}{2}\big{(}(u_{1}-x_{1})^{2}+2i\,(u_{1}(v_{1}-y_{1})+u_{2}-x_{2}\big{)}.$$

$$\left\{\begin{array}[]{lll}\mathbf{S}&=&\left\{|u_{1}|\leq a\delta^{2}\,,\ |v_{1}|\leq\frac{1}{2}\,,\ |u_{2}|\leq a\delta^{2}\right\}\\ \\ \mathbf{S}^{\prime}&=&\left\{\delta\leq|x_{1}|\leq 2\delta\,,\ |y_{1}|\leq\frac{1}{2}\,,\ |x_{2}|\leq a\delta^{2}\right\}\end{array}\right.$$

$${\sigma }_0({\mathbf{S}}_0)\approx {\lambda }_0({\mathbf{S}}_0)\approx m(\mathbf{S})\approx {\delta }^4\mathrm{while}{\sigma }_0({\mathbf{S}}_0^{\prime })\approx {\lambda }_0({\mathbf{S}}_0^{\prime })\approx m({\mathbf{S}}^{\prime })\approx {\delta }^3$$

$$\Vert {\mathbf{C}}_0(f){\Vert }_{L^p({\mathbf{S}}_0^{\prime },d{\sigma }_0)}\le A_p\Vert f{\Vert }_{L^p(b{D}_0,d{\sigma }_0)},$$

$$\underset{{\mathbf{S}}_0}{\int }\frac{1}{|{\Delta }_0(w,z){|}^2}f(w)d{\lambda }_0(w),$$

$$\mathrm{Re}{\Delta }_0(w,z)\ge \frac{{\delta }^2}{4}\mathrm{and}|\mathrm{Im}{\Delta }_0(w,z)|\le a{\delta }^2+2a{\delta }^2=3a{\delta }^2.$$

$$\mathrm{Re}\frac{1}{{\left({\Delta }_0(w,z)\right)}^2}\ge c^{\prime }{\delta }^4.$$

$$\mathrm{Re}\big{(}\mathbf{C}_{0}(f)\big{)}(z)\geq c^{\prime}\delta^{-4}\lambda_{0}(\mathbf{S}_{0})\geq c>0\,,\quad\mathrm{for}\quad z\in\mathbf{S}_{0}^{\prime}\,.$$

$$\Vert {\mathbf{C}}_0(f){\Vert }_{L^p({\mathbf{S}}_0^{\prime })}^p\ge c{\lambda }_0({\mathbf{S}}_0^{\prime })\ge c^{\prime }{\delta }^3,$$

$$\Vert f{\Vert }_{L^p(b{D}_0)}^p\le {\lambda }_0({\mathbf{S}}_0)\le c{\delta }^4.$$

$$D=\{|z_2-i{|}^2+x_1^2+y_1^4<1\}.$$

$$\rho (z)=x_1^2+y_1^4+x_2^2+y_2^2-2y_2$$

$$\mathrm{Re}\,\Delta(w,z)=\mathrm{Re}\langle\partial\rho(w),w-z\rangle=\frac{1}{2}\big{(}\nabla\rho(w),w-z\big{)}_{\mathbb{R}}$$

$$\big{(}\nabla\rho(w),w-z\big{)}_{\mathbb{R}}\geq$$

$$\ge (x_1-u_1{)}^2+(x_2-u_2{)}^2+(y_2-v_2{)}^2+(v_1^2+y_1^2)(v_1-y_1{)}^2,$$

$$f(\beta )-f(\alpha )=(\beta -\alpha )f^{\prime }(\alpha )+\underset{\alpha }{\overset{\beta }{\int }}(\beta -\alpha )f^{\prime \prime }(t)dt$$

$$\underset{\alpha }{\overset{\beta }{\int }}(\beta -\alpha )f^{\prime \prime }(t)dt=(\beta -\alpha {)}^2\mathrm{if}{f}^{\prime \prime }(t)=2,$$

$$\underset{\alpha }{\overset{\beta }{\int }}(\beta -\alpha )f^{\prime \prime }(t)dt\ge ({\beta }^2+{\alpha }^2)(\beta -\alpha {)}^2,\mathrm{if}{f}^{\prime \prime }(t)=12t^2,$$

$$\Vert \mathbf{C}(f){\Vert }_{L^p({\mathbf{S}}^{\prime })}\le A_p\Vert f{\Vert }_{L^p}$$

$${\rho }_{ϵ}(z)=x_1^2-2y_2+{ϵ}^2(y_1^4+x_2^2+y_2^2),$$

$${\rho }_{ϵ}(z)\to {\rho }_0(z),\mathrm{and}\nabla {\rho }_{ϵ}(z)\to \nabla {\rho }_0(z)$$

$$\underset{z\in b{D}_{ϵ}}{\int }F(z)d{\lambda }_{ϵ}(z)={ϵ}^{-4}\underset{z\in b{D}_0}{\int }F({\tau }_{{ϵ}^{-1}}(z))d\lambda (z)$$

$${\tau }_{ϵ}(F)(z)=F({\tau }_{ϵ}(z))=F\circ {\tau }_{ϵ}.$$

Full text

Two domains, both elementary in nature, show that the geometric requirement of strong $\mathbb{C}$-linear convexity, together with regularity of order 2, are both necessary.

∗ Supported in part by the National Science Foundation, award DMS-1503612.

∗∗ Supported in part by the National Science Foundation, award DMS-1265524.

2000 Mathematics Subject Classification: 30E20, 31A10, 32A26, 32A25, 32A50, 32A55, 42B20, 46E22, 47B34, 31B10

Keywords: Cauchy-Leray integral; Hardy space; Cauchy Integral; Cauchy-Szegő projection; Lebesgue space; pseudoconvex domain; minimal smoothness; Leray-Levi measure

1. Introduction

The purpose of this paper is to present counter-examples that show that the $L^{p}$ results for the Cauchy-Leray integral obtained in [LS-4] are essentially optimal. A second paper in this series will deal with counter-examples for the Cauchy-Szegő projection, relevant to [LS-5].

Recall that in the case of the unit ball $\mathbb{B}$ in $\mathbb{C}^{n}$, the Cauchy-Leray integral and the Cauchy-Szegő projection agree, and the same holds for the unbounded realization $U_{0}$ of $\mathbb{B}$. However when $n\geq 2$ for more general domains, these two operators are quite different. For the former, the result obtained in [LS-4] states:

Suppose $D$ is a bounded domain in $\mathbb{C}^{n}$ whose boundary is of class $C^{1,1}$ and which is strongly $\mathbb{C}$-linearly convex. Then the induced Cauchy-Leray transform is bounded on $L^{p}(bD,d\sigma)$ for $1<p<\infty$.

Our counter-examples show that these two restrictions on the boundary of $D$, the geometric condition of strong $\mathbb{C}$-linear convexity, and the regularity of degree 2, are in fact necessary.

We consider the following simple bounded domains in $\mathbb{C}^{2}$. Here $z_{j}=x_{j}+i\,y_{j}$, $j=1,2$. The first domain is defined by

[TABLE]

This is the domain given up to a translation by (9) in Section 3. It has a $C^{\infty}$ (in fact real-analytic) boundary, is strongly pseudo-convex, but not strongly $\mathbb{C}$-linearly convex.

The second example is

[TABLE]

This domain is given (up to a translation) by (20) in Section 5. The domain has a boundary of class $C^{2-\epsilon}$, ($\epsilon=2-m$), is strongly $\mathbb{C}$-linearly convex (and hence strongly pseudoconvex).

We show that for each domain the induced Cauchy-Leray transform is not bounded on $L^{p}$ for any $p$, $1\leq p\leq\infty$.

The idea of the proof is as follows: we assume that $L^{p}$-boundedness holds for one of these domains. Then an appropriate scaling and limiting argument shows that this positive result implies a corresponding conclusion in a limiting model domain, where it is much easier to supply an explicit counter-example.

For the first example the limiting domain is the unbounded half-space $2\,\mathrm{Im}\,z_{2}\,>\,x_{1}^{2}$ (called $D_{0}$ in (5)), which is holomorphically equivalent to the more familiar half-space $U_{0}$, $2\,\mathrm{Im}z_{2}\,>\,|z_{1}|^{2}$, which itself is holomorphically equivalent to the unit ball. Note that the last two are strongly $\mathbb{C}$-linearly convex, but $D_{0}$ is not. The hint that one might be led to a counter-example for $D_{0}$, and then for the domain (1) is that 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 $2\,\mathrm{Im}\,z_{2}\,>|x_{1}|^{m}$, see (21).

A remark is in order about the family of boundary measures $\{d\mu_{a}\}_{a}$ given in (29) that define the $L^{p}$ spaces, $L^{p}=L^{p}(bD,d\mu_{a})$ in the above results. Three examples of $d\mu_{a}$ are significant in many circumstances. First, the induced Lebesgue measure $d\sigma$; second, the Leray-Levi measure $d\lambda$, see (4); and third, the Fefferman measure [B2], [F-1], [G]. When the domain is smooth and strongly pseudo-convex the three measures give the same $L^{p}$ 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] where counter-examples are given for Cauchy-Leray integrals. However the less explicit and more complex nature of their construction and proof limit the results to the case $p=2$. It should be stressed that when $n\geq 2$ in general the Cauchy-Leray transform is far from “self-adjoint” and so failure of $L^{2}$-boundedness does not imply failure for any $p$, $p\neq 2$.

2. The Cauchy-Leray integral; the model domain $D_{0}$

For a bounded domain (say $C^{2}$) domain $D$ in $\mathbb{C}^{n}$, with a defining function $\rho$, the corresponding Cauchy-Leray integral is

[TABLE]

Here $f$ is (say) a bounded function on $bD$,

[TABLE]

and

[TABLE]

is the Leray-Levi measure (with $j^{*}$ the pull-back under the inclusion: $bD\hookrightarrow\mathbb{C}^{n}$). We have $d\lambda(w)=\Gamma(w)\,d\sigma(w)\,,$ with $d\sigma$ the induced Lebesgue measure on $bD$ and

[TABLE]

where $\mathcal{L}(w)$ is the determinant of the Levi-form acting on the tangent space at $w\in bD$. (See e.g., [Ra, Ch. 7].)

It is worth recalling two intrinsic properties of (3):

• •

Its invariance under changes of coordinates that are given either by translations or unitary mappings of the space $\mathbb{C}^{n}$, see [Bo].

• •

The independence of the Cauchy-Leray integral (3) of the particular defining function $\rho$ of $D$.

We consider first the unbounded “model domain” $D_{0}$ in $\mathbb{C}^{2}$.

With $z=(z_{1},z_{2})$, $z_{j}=x_{j}+i\,y_{j}$, it is defined by

[TABLE]

which is to be compared with the more familiar form of $D_{0}$ given by

$\{z\,:\ 2\,\mathrm{Im}z_{2}>|z_{1}|^{2}\}$. These two domains are biholomorphically equivalent via the mappings $z_{1}\leftrightarrow z_{1}$, $z_{2}\leftrightarrow z_{2}\pm i\,z_{1}^{2}$. Now the complex tangent space of these domains at the origin is the subspace $\{(z_{1},0)\}$. So since $|z_{1}|^{2}=x_{1}^{2}+y_{1}^{2}$ , and on $\mathbb{R}^{2}$ this is positive definite, this implies that the second domain is strongly $\mathbb{C}$-linearly convex. However because the form $x_{1}^{2}$ is degenerate along the direction $y_{1}$, it follows that strong $\mathbb{C}$-linear convexity fails for the domain (4). (For more about these convexities see [APS]; [Hö]; [LS-3, Sect. 3.3].)

With $\rho_{0}(z)=2y_{2}-x_{1}^{2}$, and $\Delta_{0}(w,z)=\langle\partial\rho_{0}(w),w-z\rangle$, and if we write $w=(w_{1},w_{2})$, $w_{j}=u_{j}+i\,v_{j}$, $j=1,2$, a simple calculation gives that with $w$ and $z\in bD_{0}$,

[TABLE]

Now for fixed $z\in bD_{0}$, observe that $\Delta_{0}(w,z)$ vanishes on the one-dimensional variety given by $u_{1}=x_{1}$, $u_{1}(v_{1}-y_{1})+u_{2}=x_{2}$. Also the Leray-Levi measure $d\lambda_{0}$ corresponding to $\rho_{0}$ is $du_{1}\,dv_{1}\,du_{2}/(4\pi^{2})\approx d\sigma_{0}$, the induced Lebesgue measure on $bD_{0}$, if we are near the origin.

To construct a counterexample for the model domain $D_{0}$ we first choose a small constant $a$, which we will keep fixed throughout ($a=1/12$ will do). Then for any $\delta$, $0<\delta<1$, we define:

[TABLE]

with $w\in bD_{0}$ written as $(u_{1}+i\,v_{1},u_{2}+i\,u^{2}_{1}/2)$ and $z\in bD_{0}$ written as $(x_{1}+i\,y_{1},x_{2}+i\,x^{2}_{1}/2)$ and $\mathbf{S}_{0}$ and $\mathbf{S}_{0}^{\prime}$ being the corresponding sets for $w\in bD_{0}$, $z\in bD_{0}$ via (7), then it follows that $\Delta_{0}(w,z)\neq 0$ whenever $w\in\mathbf{S}_{0}$ and $z\in\mathbf{S}_{0}^{\prime}$. Also

[TABLE]

where $\sigma_{0}$ denotes the induced Lebesgue measure on $bD_{0}$ and $m$ is the standard Euclidean measure on the three-dimensional Euclidean space.

We shall now test the presumed inequality

[TABLE]

when $f$ is assumed to have support in $\mathbf{S}_{0}$. Under these circumstances $\mathbf{C}_{0}(f)(z)$ is well-defined as an absolutely convergent integral

[TABLE]

when $z\in\mathbf{S}_{0}^{\prime}$, in view of the non-vanishing of $\Delta_{0}(w,z)$ for these $w$ and $z$.

Proposition 1**.**

The inequality (8) fails for every $p$, $1\leq p\leq\infty$.

The constant $A_{p}$ is of course assumed to be independent of $f$.

In fact it is clear by (6) and (7) that we have:

[TABLE]

Then $\left(\mathrm{Re}\,\Delta_{0}(w,z)\right)^{2}\geq 2\,|\mathrm{Im}\,\Delta_{0}(w,z)|^{2}$, (which holds if $a\leq 1/12$). But $|\Delta_{0}(w,z)|\leq c\delta^{2}$ for some $c>0$, thus

[TABLE]

Now take $f=\chi_{\mathbf{S}_{0}}$, the characteristic function of $\mathbf{S}_{0}$. Therefore

[TABLE]

So

[TABLE]

while

[TABLE]

Since $\delta^{3}$ is not $O(\delta^{4})$ for small $\delta$, (8) cannot hold, when $p<\infty$. The case $p=\infty$ requires a separate but simpler argument which we omit.

3. The first counter-example

We turn to the domain (1) in $\mathbb{C}^{2}$ and it is useful to consider a translate of it, given by

[TABLE]

We will show after rescaling and a passage to the limit, that we can reduce consideration of $D$ to $D_{0}$. From (9) it is clear that

[TABLE]

is a defining function for $D$, and that $\rho$ is strongly pluri-subharmonic, so $D$ is strongly pseudo-convex. Moreover, since each of the four one-variable functions, $x_{1}^{2}$, $y_{1}^{4}$, $x_{2}^{2}$, $y_{2}^{2}-2y_{2}$ are strictly convex, the domain $D$ is itself strictly convex. We note that

[TABLE]

where $(\cdot,\cdot)_{\mathbb{R}}$ is the real inner product induced on $\mathbb{R}^{4}=\mathbb{C}^{2}$ from $\langle\cdot,\cdot\rangle$. With $w=(w_{1},w_{2})$, $w_{j}-u_{j}+i\,v_{j}\,,$ $j=1,2$, we claim that

[TABLE]

[TABLE]

when $w,z\in bD$.

To prove (10) we use the identity

[TABLE]

for the functions $f_{1}=u_{1}^{2}$, $f_{2}=u_{2}^{2}$, $f_{3}=v_{2}^{2}-2v_{2}$, and $f_{4}=v_{1}^{4}$. Similarly, we replace $w=(u_{1}+i\,v_{1},u_{2}+i\,v_{2})$ by $z=(x_{1}+i\,y_{1},x_{2}+i\,y_{2})$ and add the corresponding identities. Taking into account that $\rho(w)=f_{1}+f_{2}+f_{3}+f_{4}$ wih $\rho(w)=0$ and the similar fact for $\rho(z)$, together with the observations that

[TABLE]

and

[TABLE]

then yields (10).

Now (10) shows that if $w$ and $z\in bD$, $w\neq z$, then $z$ lies on one side of the (real) tangent plane to $bD$ at $w$. By convexity of $D$, the same holds for $z\in\overline{D}\setminus\{w\}$.

Turning to the Cauchy-Leray integral of $D$ we recall two preliminary facts. First $\mathbf{C}(f)(z)$ is holomorphic in $z\in D$, if $f$ is say an integrable function on $bD$. Second, whenever $F$ is a holomorphic function in $D$ which is continuous in $\overline{D}$, and f=F\big{|}_{bD}, then $\mathbf{C}(f)(z)=F(z),\ z\in D$. The latter fact follows from the Cauchy-Fantappiè formalism (see [Ra] and [LS-3]).

Note that when $f$ is a bounded function supported in a closed set $\mathbf{S}$ in $bD$, then $\mathbf{C}(f)(z)$ is well-defined as a convergent integral whenever $z$ is outside the support $\mathbf{S}$. So certainly the extendability of $\mathbf{C}$ to a bounded operator on $L^{p}$ would imply

[TABLE]

whenever $\mathbf{S}^{\prime}$ is disjoint from $\mathbf{S}$, with $A_{p}$ independent of $f$, $\mathbf{S}$, and $\mathbf{S}^{\prime}$.

Theorem 2**.**

For any $p$, $1\leq p\leq\infty$, the presumed inequality (11) fails when tested for a bound $A_{p}$ independent of $f$ and its support $\mathbf{S}$, and with $\mathbf{S}^{\prime}$ disjoint from $\mathbf{S}$.

A further statement along these lines is made possible by the following fact (whose proof will appear elsewhere): whenever $f$ is of class $C^{1}$ on the boundary $bD$, $\mathbf{C}(f)$ is extendable to a continuous function on $\overline{D}$. With this we can define the Cauchy-Leray transform $\mathcal{C}(f)$ for such $f$, by \mathcal{C}(f)=\mathbf{C}(f)\big{|}_{bD}.

Corollary 3**.**

The mapping $f\mapsto\mathcal{C}(f)$, initially defined for $C^{1}$ functions $f$ is not extendable to a bounded operator on $L^{p}(bD)$.

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 $\tau_{\epsilon}$, $\epsilon>0$, on $\mathbb{C}^{2}$ by $\tau_{\epsilon}(z_{1},z_{2})=(\epsilon z_{1},\epsilon^{2}z_{2})$ and set $D_{\epsilon}=\tau_{\epsilon^{-1}}(D)$, with $D$ as in (9). Then $\rho(z)=x_{1}^{2}+y_{1}^{4}+x_{2}^{2}+y_{2}^{2}-2y_{2}$ is a defining function for $D$ and $\rho_{\epsilon}(z)=\epsilon^{-2}\rho(\tau_{\epsilon}(z))$ is a defining function for $D_{\epsilon}$. Note that

[TABLE]

from which it is clear that the domains $D_{\epsilon}$ increase as $\epsilon$ decreases, with limit our model domain $D_{0}=\{\rho_{0}(z)<0\}$ and $\rho_{0}(z)=x_{1}^{2}-2y_{2}$.

Observe also that

[TABLE]

uniformly on compact subsets of $\mathbb{C}^{2}$.

We next define the “transported” measures $d\lambda_{\epsilon}$ on $bD_{\epsilon}$ by

[TABLE]

for any continuous functions $F$ defined on $bD_{\epsilon}$.

Here $d\lambda$ is the Leray-Levi measure on $bD$, but note that $d\lambda_{\epsilon}$ is not the Leray-Levi measure on $bD_{\epsilon}$: as a result the operator $\mathbf{C}_{\epsilon}$ defined below is not the Cauchy-Leray integral of $bD_{\epsilon}$. We can also define the corresponding action of $\tau_{\epsilon}$ on functions by

[TABLE]

Then by what has been said above, $\tau_{\epsilon}$ maps $L^{p}(bD,d\lambda)$ to $L^{p}(bD_{\epsilon},d\lambda_{\epsilon})$ and we have the “isometry”:

[TABLE]

Now let $\Delta_{\epsilon}(w,z)=\langle\partial\rho_{\epsilon}(w),w-z\rangle$. Then

[TABLE]

Hence

[TABLE]

We now define the operator $\mathbf{C}_{\epsilon}$, by setting

[TABLE]

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

[TABLE]

for bounded $f$ on $bD_{\epsilon}$, if $z\in bD_{\epsilon}$ lies outside the support of $f$. In fact going back to the definition (3), and using (12) and (14), we see that

[TABLE]

[TABLE]

showing (16).

Next, if (11) held, then by (13) we would also have

[TABLE]

whenever $F$ is a bounded function on $bD_{\epsilon}$ and $\mathbf{S}_{\epsilon}^{\prime}$ is a closed subset of $bD_{\epsilon}$, disjoint from the support of $F$.

At this point we restrict the attention to the unit ball $\mathbb{B}$ in $\mathbb{C}^{2}$, and we exploit the common coordinate system for $bD_{0}\cap\mathbb{B}$, and $bD_{\epsilon}\cap\mathbb{B}$, when $\epsilon$ is small. That is in this ball, $bD_{0}$ is the graph over $(x_{1}+i\,y_{1},x_{2})$ given by $(x_{1}+i\,y_{1},x_{2}+i\,x_{1}^{2}/2)$, while $bD_{\epsilon}$ is the graph given by

$(x_{1}+i\,y_{1},x_{2}+i\,\left(x_{1}^{2}/2+\Phi_{\epsilon}(x_{1},y_{1},x_{2})\right))$, with $\Phi_{\epsilon}(x_{1},y_{1},x_{2})=O(\epsilon^{2})$.

Now if in these coordinates we write $d\lambda=\Lambda(x_{1},y_{1},x_{2})\,dx_{1}\,dy_{1}\,dx_{2}$ and $d\lambda_{\epsilon}=\Lambda_{\epsilon}(x_{1},y_{1},x_{2})\,dx_{1}\,dy_{1}\,dx_{2}$, then by (12) we have

[TABLE]

Finally we take $f$ to be the function $\chi_{\mathbf{S}}$, the characteristic function of the set $\mathbf{S}$ in (7), and lift $f$ to functions on $bD_{\epsilon}$ and $bD_{0}$ respectively, by setting $f_{\epsilon}(z_{\epsilon})=f(x_{1},y_{1},x_{2})$ when $z_{\epsilon}=(x_{1}+i\,y_{1},x_{2}+i\,\left(x_{1}^{2}/2+\Phi_{\epsilon}(x_{1},y_{1},x_{2})\right))$, and $f_{0}(z_{0})=f(x_{1}+i\,y_{1},x_{2}+i\,x_{1}^{2}/2)$.

We also lift the sets $\mathbf{S}$ and $\mathbf{S}^{\prime}$ to $\mathbf{S}_{\epsilon}$ and $\mathbf{S}_{\epsilon}^{\prime}$ (subsets of $bD_{\epsilon}$) in the same way. Our claim is with that notation

[TABLE]

In fact,

[TABLE]

However by (18) $\Lambda_{\epsilon}(u_{1},v_{1},u_{2})\to\Lambda(0,0,0)$, and moreover $\Delta_{\epsilon}(w,z_{\epsilon})\to\Delta_{0}(w,z_{0})$ because $\nabla\rho_{\epsilon}\to\nabla\rho_{0}$, while $\Delta_{0}(w,z_{0})\neq 0$ if $w\in\mathbf{S}_{0}$ and $z_{0}\in\mathbf{S}_{0}^{\prime}$. This gives (19).

As a result (17) with $F=f_{\epsilon}$ leads to

[TABLE]

which contradicts Proposition 1, proving the theorem.

5. The second counter-example

Here the domain will be taken to be

[TABLE]

Its model domain is

[TABLE]

For any $f$ that is bounded on $bD$, the Cauchy-Leray integral $\mathbf{C}(f)(z)$ 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 $f\mapsto\mathbf{C}(f)$ fails to be bounded in $L^{p}$ in the sense that the proposed inequality

[TABLE]

cannot hold. Here $\mathbf{S}^{\prime}$ is any set disjoint from the support of $f$, and the bound $A_{p}$ is assumed independent of $f$ and $\mathbf{S}^{\prime}$.

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 $\rho(z)=|x_{1}|^{m}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}-2y_{2}$, and that of $D_{0}$ is $\rho_{0}(z)=|x_{1}|^{m}-2y_{2}$. Note that both domains are of class $C^{2-\alpha}$, with $\alpha=2-m$. Also since $|x_{1}|^{m}\,,\ y_{1}^{2}\,,\ x_{2}^{2}\,,\ y_{2}^{2}-2y_{2}$ are strongly convex functions of one variable, the domain $D$ is strongly convex, hence strongly $\mathbb{C}$-linearly convex, and in fact strongly pseudoconvex in the following sense: the domain $D$ is exhausted by an increasing family of smooth domains $\{D_{\gamma}\}_{\gamma}$ which are uniformly strongly pseudo-convex, with defining function: $\rho_{\gamma}(z)=(x_{1}^{2}+\gamma)^{m/2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}-2y_{2}$.

This convexity implies that $\mathrm{Re}\,\Delta(w,z)>0$ for $w\in bD$ and $z\in\overline{D}$, except when $z=w$.

Returning to the model domain, if $\Delta_{0}(w,z)=\langle\partial\rho_{0}(w),w-z\rangle$ a calculation gives

[TABLE]

[TABLE]

Here we have used the notation

[TABLE]

Now we set

[TABLE]

and let $\mathbf{S}_{0}$ and $\mathbf{S}_{0}^{\prime}$ be the corresponding induced sets on $bD_{0}$.

We have that near the origin

[TABLE]

where $d\lambda_{0}$ and $d\sigma_{0}$ are the Leray-Levi measure and the induced Lebesgue measure on $bD_{0}$. Thus

[TABLE]

[TABLE]

By (23) it follows that

[TABLE]

and if we choose $a$ sufficiently small, then

[TABLE]

Taking $f_{0}$ to be the characteristic function of $\mathbf{S}_{0}$, we therefore get $\mathrm{Re}\,\mathbf{C}_{0}(f_{0})(z)\geq c>0$, for $z\in\mathbf{S}_{0}^{\prime}$. Hence this gives a contradiction to (22) in the case when $\mathbf{C}_{0}$ is the Cauchy-Leray integral of the model domain $D_{0}$.

To pass to the domain $D$ we carry out the scaling via

[TABLE]

The domain $D_{\epsilon}=\tau_{\epsilon^{-1}}(D)$ has a defining function

[TABLE]

which converges to $\rho_{0}(z)=|x_{1}|^{m}-2y_{2}$.

We also set $\Delta_{\epsilon}(w,z)=\langle\partial\rho_{\epsilon}(w),w-z\rangle$. The transported measure $d\lambda_{\epsilon}$ on $D_{\epsilon}$ is now defined by the identity

[TABLE]

(compare with (12)).

Finally, if we assumed that (22) held for the domain $D$ we can then see by the reasoning in Section 4 that the corresponding result would hold for the model domain $D_{0}$ achieving our desired contradiction.

We point out in closing that (22) fails not only for the $L^{p}$ norms taken with the induced Lebesgue measure but others as well. Consider the measures $d\mu_{a}$ on $bD$ given by

[TABLE]

with $\mathcal{L}$ the determinant of the Levi-form, and $d\sigma$ the induced Lebesgue measure.

Then the argument above shows that (22) fails when

[TABLE]

(because $\mathcal{L}^{a}\approx|x_{1}|^{(m-2)a}$). Here the factor $\epsilon^{-2m}$ in (28) is replaced with $\epsilon^{a(2-m)-2-m}$ to reflect the new transported measures $d\mu_{a,\epsilon}$.

The case $a=0$ corresponds to induced Lebesgue measure; the case $a=1$, to the Leray-Levi measure; and the case $a=1/3$ to the Fefferman measure [B2], [F-1], [G].