L^2 boundedness of the Cauchy Integral on Lipschitz curves, proof I

I’ve been accepted to a Summer School in Bonn organized by prof. Christoph Thiele and prof. Diogo Oliveira e Silva. The subjects are ${T(1)}$ and ${T(b)}$ theorems, which give a characterization of ${L^2}$ boundedness for Calderon-Zygmund operators. I’m currently learning the basics so I can’t say much more, but for the summer school each student has been given an article to read which will have to be lectured to the other participants. My article is about ${L^2}$ boundedness of the Cauchy integral operator for a Lipschitz curve (a lipschitz graph, really), here’s the reference:

Two elementary proofs of the L2 boundedness of Cauchy integrals on Lipschitz curves.

by Coifman, R. R.; Jones, Peter W.; Semmes, Stephen

J. Amer. Math. Soc. 2 (1989), no. 3, 553-564.

The article features two nice proofs of the same result which use different instruments (the first one relies on complex analysis, the second one on the construction of a suitable Haar-like basis), but are similar in spirit. Interesting parallels can be made between the two.

As I’m working out all the details in the paper, I thought that it might be good to share it. This first blog post is about the first proof of the theorem. A second blog post will include the second proof, as well as a comparison between the two.

1. Main theorem

Here ${\Gamma}$ is a Lipschitz graph in ${\mathbb{C}}$, that is

$\displaystyle \Gamma(x)=x+iA(x),$

with ${\|A'\|_{L^\infty}<+\infty}$. The epigraph is called ${\Omega_{+}}$ and the hypograph ${\Omega_{-}}$.

One has the ${L^2}$-norm on the curve

$\displaystyle \|g\|_{L^2(\Gamma)}^2 = \int_{\Gamma}{|g|^2}\,ds = \int_{\mathbb{R}}{|g(\Gamma(x))|^2 |\Gamma'(x)|dx}=\int_{\mathbb{R}}{|g(\Gamma(x))|^2 \sqrt{1+A'(x)^2}dx},$

and by lipschitz property ${\sqrt{1+A'(x)^2}\sim 1}$, so that

$\displaystyle \|g\|_{L^2(\Gamma)}\sim_{A} \int_{\mathbb{R}}{|g(x+iA(x))|^2}\,dx.$

The Cauchy-integral operator ${C}$ is defined by

$\displaystyle Cg(z):=\int_{\Gamma}{\frac{g(\zeta)}{z-\zeta}}\,d\zeta$

when ${z\in\Omega_{+}}$, and when ${z\in\Gamma}$

$\displaystyle Cg(z)=\lim_{y\rightarrow 0^{+}}{Cg(z+iy)}.$

Notice that the Cauchy integral is missing a factor of ${(2\pi i)^{-1}}$ from Cauchy’s integral formula for holomorphic functions, but ${g}$ isn’t holomorphic and the factor is only a burden we don’t want to carry on here.

Remark 1 There’s an issue in the definition of ${Cg}$, namely that the (pointwise) limit along the imaginary axis might not be defined. I’ll deal with this later on, for now suppose it does indeed exist.

The main theorem is

Theorem 1 (${L^2}$-boundedness of ${C}$)

$\displaystyle \|Cg\|_{L^2(\Gamma)}\lesssim_{A} \|g\|_{L^2(\Gamma)}$

and the constant depends on ${\|A'\|_{\infty}}$ only, rather than ${A}$.

We introduce a norm on the measurable ${\mathbb{C}}$-valued functions on ${\Omega_{+}}$:

$\displaystyle \|f\|_{\Omega_{+}}:= -\frac{i}{2}\int_{\Omega_{+}}{|f(z)|^2 d(z)}\,dz\wedge d\overline{z},$

where ${d(z) = dist(z,\Gamma)}$. Notice that ${-\frac{i}{2}dz\wedge d\overline{z} = dx \wedge dy}$, the Lebesgue volume element. Norm ${\|\cdot\|_{\Omega_{+}}}$ comes from an hermitian product, obviously, and the associated Hilbert space is denoted by ${\Omega_{+}^2}$.

2. First proof

We’ll suppose ${g}$ and ${A}$ to be in ${C^{\infty}_c}$, since the conclusion for general ${g}$ (and ${A}$) can then be derived by taking the limit over a ${C^{\infty}_c}$ approximation.

First proof requires two lemmas, whose nature is that of Littlewood-Paley theory:

Lemma 2 Let ${F}$ be a holomorphic function in ${\Omega_{+}}$ such that ${F \rightarrow 0}$ at infinity. Then

$\displaystyle \|F\|_{L^2(\Gamma)} \lesssim \|A'\|_{L^\infty} \|F'\|_{\Omega_{+}} \ \ \ \ \ (1)$

Lemma 3 Let ${T}$ be the operator

$\displaystyle Tf(z):= \int_{\Omega_{+}}{\frac{f(\zeta)d(\zeta)}{(z-\zeta)^2}}dx\wedge dy.$

Then ${T}$ is ${\Omega_{+}^2 \rightarrow L^2(\Gamma)}$ bounded:

$\displaystyle \|Tf\|_{L^2(\Gamma)}\lesssim \|A'\|_{L^\infty} \|f\|_{\Omega_{+}}. \ \ \ \ \ (2)$

With these two lemmas the proof is formally easy:

Proof: (of the Main Theorem)

Applying Lemma 2,

$\displaystyle \|Cg\|_{L^2(\Gamma)} \lesssim \|A'\|_{L^\infty} \|(Cg)'\|_{\Omega_{+}} = \|A'\|_{L^\infty} \sup_{\|f\|_{\Omega_{+}}\leq 1}{\left|\left\langle(Cg)',f\right\rangle_{\Omega_{+}}\right|},$

and applying Lemma 3

$\displaystyle \left|\left\langle(Cg)',f\right\rangle_{\Omega_{+}}\right| = \left|\int_{\Omega_{+}}{(Cg)'(z)\overline{f(z)}d(z)}\,dx\wedge dy\right|$

$\displaystyle = \left|\int_{\Omega_{+}}{\int_{\Gamma}{-\frac{g(\zeta)\overline{f(z)}}{(z-\zeta)^2}d(z)}\,d\zeta}\,dx\wedge dy\right|$

$\displaystyle = \left|\int_{\Gamma}{g(\zeta)\int_{\Omega_{+}}{\frac{\overline{f(z)}d(z)}{(z-\zeta)^2}}\,dx\wedge dy}\,d\zeta\right|$

$\displaystyle = \left|\int_{\Gamma}{g(\zeta)T\overline{f}(\zeta)}\,d\zeta\right|$

$\displaystyle \leq \|g\|_{L^2(\Gamma)} \|T\overline{f}\|_{L^2(\Gamma)} \lesssim \|A'\|_{L^\infty} \|g\|_{L^2(\Gamma)}\|f\|_{\Omega_{+}}\lesssim_{A'} \|g\|_{L^2(\Gamma)},$

where we used Fubini to pass from 2nd to 3rd line and Cauchy-Schwarz in the end. $\Box$

Remark 2 ${(Cg)'}$ is well defined, we can take the (complex) derivative inside the integral thanks to the hypothesis that ${g}$ is of compact support (and thus all integrability issues disappear).

Proof: (of Lemma 2)

The following will be needed in the proof: let ${f,g}$ be holomorphic functions in ${\mathbb{H}^{+}}$, such that ${f(z) \xrightarrow{z\rightarrow \infty} 0}$ and ${|g|\leq C}$. Then

$\displaystyle \int_{\mathbb{R}\times\{0\}}{|f|^2}dx = 4 \int_{\mathbb{H}^{+}}{|f'|^2 y }\;dx\wedge dy \ \ \ \ \ (3)$

$\displaystyle \int_{\mathbb{H}^{+}}{|f g'|^2 y }\;dx\wedge dy \leq C^2 \int_{\mathbb{R}\times\{0\}}{|f|^2}dx. \ \ \ \ \ (4)$

(${\partial\mathbb{H}^{+} = \mathbb{R}\times\{0\}}$). Notice the factor ${y}$ in the ${\Omega_{+}}$-integral that makes formulas `dimensionally consistent’.

To prove the first one it’s enough to use Green’s identity. But first, notice that writing ${f=u+iv}$,

$\displaystyle |f|^2 = u^2 + v^2$

$\displaystyle |f'|^2 =\left|\frac{\partial f}{\partial x}\right|^2 = \left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial v}{\partial x}\right)^2 = \left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial u}{\partial y}\right)^2 = |\nabla u|^2 = |\nabla v|^2$

where we exploited the Cauchy-Riemann relations.

Now, by Green’s identity applied to functions ${y}$ and ${u^2}$ on the domain ${\mathbb{H}^{+}}$ one has

$\displaystyle \int_{\mathbb{H}^{+}}{y \Delta(u^2)- u^2\Delta y}\;dx\wedge dy = \int_{\partial\mathbb{H}^{+}}{-y\frac{\partial(u^2)}{\partial y}+u^2 \frac{\partial y}{\partial y}}\;dx,$

and ${\Delta(u^2) = 2 u \Delta u + 2 |\nabla u|^2= 2 |\nabla u|^2}$ since ${u}$ is harmonic, ${\Delta y= 0}$, ${y\equiv 0}$ on ${\partial\mathbb{H}^{+}}$, so that

$\displaystyle 2\int_{\mathbb{H}^{+}}{|\nabla u|^2 y}\;dx\wedge dy = \int_{\partial\mathbb{H}^{+}}{u^2}\;dx.$

Apply the same identity with ${v^2}$ in place of ${u^2}$, and by the previous relations summing the two of them we get (3). Notice we applied Green’s identity to an unbounded domain, which is justified by the fact that ${u,v \rightarrow 0}$ at infinity.

As for (4), apply (3) to ${fg}$ (which satisfies the suitable conditions),

$\displaystyle \int_{\mathbb{H}^{+}}{|f'g + fg'|^2 y }\;dx\wedge dy = \frac{1}{4}\int_{\partial\mathbb{H}^{+}}{|fg|^2}\;dx \leq \frac{C^2}{4}\int_{\partial\mathbb{H}^{+}}{|f|^2}\;dx,$

since ${|g|\leq C}$, and then (we use triangle inequality together with ${(a+b)^2 \leq 2a^2+2b^2}$)

$\displaystyle \int_{\mathbb{H}^{+}}{|f g'|^2 y}\;dx\wedge dy \leq 2\int_{\mathbb{H}^{+}}{|fg'+f'g|^2 y}\;dx\wedge dy + 2\int_{\mathbb{H}^{+}}{|f'g|^2 y}\;dx\wedge dy$

$\displaystyle \leq \frac{C^2}{2}\int_{\partial\mathbb{H}^{+}}{|f|^2}\;dx + \frac{C^2}{2}\int_{\partial\mathbb{H}^{+}}{|f|^2}\;dx$

$\displaystyle = C^2 \int_{\partial\mathbb{H}^{+}}{|f|^2}\;dx,$

where we used (3) again.

Let’s turn to the proof of Lemma 2. Observe how, by the Riemann mapping theorem, there exists a conformal (see footnote [1]) map ${\Phi}$ that maps ${\mathbb{H}^{+}}$ into ${\Omega_{+}}$ holomorphically. We’ll use it to change variables. Writing the norms explicitly, one has to prove

$\displaystyle \int_{\Gamma}{|F(\Gamma(x))|^2 |\Gamma'(x)|}\,dx \leq -\frac{i}{2} C \int_{\Omega_{+}}{|F'(z)|^2 d(z)}\, dz\wedge d\overline{z}$

for some constant ${C}$. Change variables using ${\Phi}$, writing ${G=F\circ \Phi}$ to get

$\displaystyle \int_{\mathbb{R}\times\{0\}}{|G(x+i0)|^2 |\Phi'(x)|}\;dx \leq C \int_{\mathbb{H}^{+}}{|G^\prime(z)|^2 d(\Phi(z))}\;dx\wedge dy,$

since ${\det D\Phi = |\Phi'|^2}$ (here ${D\Phi}$ is the Jacobian of ${\Phi}$ interpreted as a function from the upper half-plane of ${\mathbb{R}^2}$ to the real domain ${\Omega_{+}}$). The factor ${d(\Phi(z))}$ is a bit annoying, but we can find a substitute for it using Köbe’s theorem, which we briefly state:

Theorem 4 (Köbe’s one-quarter theorem) Let ${f}$ be a holomorphic injective function (see footnote [2]) on the disk ${\mathbb{D}}$. Then the image of the function contains a ball of radius proportional to ${|f'(0)|}$; more precisely

$\displaystyle f(\mathbb{D}) \supset B\left(f(0), \frac{|f'(0)|}{4}\right).$

Notice that if ${f}$ is defined on ${\delta\mathbb{D}}$, that is a disk (see footnote [3]) of radius ${\delta}$, then the image ${f(\delta\mathbb{D})}$ contains the ball of radius ${\delta\frac{|f'(0)|}{4}}$ centered in ${f(0)}$. In our case, ${d(\Phi(x,y))=dist(\Phi(x,y),\Gamma)}$; if we consider the restriction of ${\Phi}$ to the ball ${B((x,y),y)}$, which is injective since ${\Phi}$ is, then we have that the image contains ${B\left(\Phi(x,y), y\frac{|\Phi'(x,y)|}{4}\right)}$, and thus ${d(\Phi(x,y))\gtrsim |\Phi'(x,y)|\,y}$. On the other hand, it follows from Schwarz’s lemma (see footnote [4]) that ${d(\Phi(x,y))\leq |\Phi'(x,y)|y}$, thus ${d(\Phi(x,y))\sim |\Phi'(x,y)|\,y}$.

So, what we’re interested in proving is

$\displaystyle \int_{\partial\mathbb{H}^{+}}{|G(x)|^2|\Phi'(x)|}\;dx \leq C \int_{\mathbb{H}^{+}}{|G^\prime(z)|^2 |\Phi'(z)|\,y}\;dx\wedge dy.$

Before going further, observe that an inequality of the form

$\displaystyle A\lesssim B$

is equivalent to one of the form

$\displaystyle A\lesssim B+A^{1/2}B^{1/2}.$

This is because ${A\lesssim B+A^{1/2}B^{1/2}}$ implies that either ${A/2\lesssim B}$ or ${A/2 \lesssim A^{1/2}B^{1/2}}$, both equivalent to ${A\lesssim B}$ (if we don’t care about constants).

Observe that since ${\Phi}$ is angle-preserving and ${\Gamma}$ is a Lipschitz graph, one has on the boundary ${\|\arg \Phi^\prime\|_{L^\infty(\Gamma)} < \frac{\pi}{2}-\delta}$ for some ${\frac{\pi}{2}>\delta>0}$. This means that there isn’t much cancellation in phase and so

$\displaystyle \int_{\partial\mathbb{H}^{+}}{|G(x)|^2|\Phi'(x)|}\;dx \sim \left|\int_{\partial\mathbb{H}^{+}}{|G(x)|^2\Phi'(x)}\;dx \right|$

Now we apply Green’s identity as we did before, to the functions ${y}$ and ${|G|^2\Phi'}$ (unboundedness of domain is again justified by the decay at infinity)

$\displaystyle \int_{\mathbb{R}\times\{0\}}{|G|^2\Phi'}\;dx \stackrel{y\equiv 0}{=} \int_{\partial\mathbb{H}^{+}}{|G|^2\Phi' \frac{\partial y}{\partial y} - y \frac{\partial}{\partial y}(|G|^2)\Phi'}\;dx$

$\displaystyle = \int_{\mathbb{H}^{+}}{\left[y \Delta(|G|^2\Phi')-|G|^2\Phi'\Delta y \right]}\;dx\wedge dy = \int_{\mathbb{H}^{+}}{y\Delta(|G|^2\Phi')}\;dx\wedge dy.$

Since ${\Delta = 4 \partial\overline{\partial}}$ and both ${G}$ and ${\Phi'}$ are holomorphic, a lot of terms get canceled (see footnote [5]):

$\displaystyle \Delta(|G|^2\Phi')= 4\partial\overline{\partial}(G\overline{G}\Phi') = 4\partial \left(\overline{G}\Phi'\overline{\partial} G + G \Phi'\overline{\partial}\overline{G} + G \overline{G} \overline{\partial} \Phi'\right)$

$\displaystyle = 4\partial (G \Phi'\overline{\partial}\overline{G}) = 4 \left(\Phi'\overline{\partial}\overline{G}\partial G + \Phi' G \partial\overline{\partial}\overline{G} + \overline{\partial}\overline{G} G \partial\Phi'\right)$

$\displaystyle = 4\left(\Phi' |G'|^2 + G \overline{G'}\Phi''\right).$

Thus

$\displaystyle \left|\int_{\mathbb{R}\times\{0\}}{|G|^2\Phi'}\;dx\right| \lesssim 4\left| \int_{\mathbb{H}^{+}}{\left(|G'|^2\Phi' + G \overline{G'}\Phi''\right)y}\;dx\wedge dy\right|$

$\displaystyle \lesssim \int_{\mathbb{H}^{+}}{|G'|^2|\Phi'|y}\;dx\wedge dy + \int_{\mathbb{H}^{+}}{|G \overline{G'}\Phi''|y}\;dx\wedge dy,$

and we recognize that the first term in the right hand side is the one we’re looking for: it’s the ${B}$ in ${A \lesssim B + A^{1/2}B^{1/2}}$. As for the second one, since ${\Phi'\neq 0}$ it has a well defined logarithm, and we can write ${\Phi'(z) = e^{V(z)}}$. Then ${\Phi'' = V' \Phi'}$. Moreover ${|\Im V(z)|\leq \frac{\pi}{2}}$, which implies ${\left|e^{iV}\right|\leq e^{\frac{\pi}{2}}}$. By Cauchy-Schwarz

$\displaystyle \int_{\mathbb{H}^{+}}{|G \overline{G'}V'\Phi'|y}\;dx\wedge dy \leq \left(\int_{\mathbb{H}^{+}}{|G'|^2|\Phi'|y}\;dx\wedge dy\right)^{1/2}\left(\int_{\mathbb{H}^{+}}{|G V'(\Phi')^{1/2}|^2y}\;dx\wedge dy\right)^{1/2},$

and we recognize the first term on the rhs as the ${B^{1/2}}$ in ${A\lesssim B+A^{1/2}B^{1/2}}$. As for the second term,

$\displaystyle |G V'(\Phi')^{1/2}|=|G e^{-iV} V'e^{iV}(\Phi')^{1/2}|\leq e^{\frac{\pi}{2}}|\left(e^{iV}\right)' G(\Phi')^{1/2}|.$

We can therefore apply (4) with ${f=G(\Phi')^{1/2}}$, ${g = e^{iV}}$, ${C=e^{\frac{\pi}{2}}}$, and obtain

$\displaystyle \left(\int_{\mathbb{H}^{+}}{|G V'(\Phi')^{1/2}|^2y}\;dx\wedge dy\right)^{1/2} \leq e^{\frac{3\pi}{4}} \left(\int_{\mathbb{H}^{+}}{|G|^2|\Phi'|}\;dx\right)^{1/2},$

in which we recognize term ${A^{1/2}}$, and we’re done. $\Box$
Proof: (of Lemma 3)

The result will be a consequence of Lemma 2 and Schur’s test. First notice that for ${f(z)}$ to be in ${\Omega_{+}^2}$ is equivalent to ${F(z):=f(z) d(z)^{1/2}}$ to be in ${L^2(\Omega_{+})}$. Moreover, since ${\Gamma}$ is the common boundary of ${\Omega_{+}}$ and ${\Omega_{-}}$, Lemma 2 is true with ${\Omega_{-}}$ in place of ${\Omega_{+}}$. Then, applying it to ${Tf}$,

$\displaystyle \|Tf\|_{L^2(\Gamma)} \lesssim \|(Tf)'\|_{\Omega_{-}} = \sup_{\|g\|_{\Omega_{-}}\leq 1} \left\langle(Tf)',g\right\rangle_{\Omega_{-}},$

and

$\displaystyle (Tf)'(z) = -2\int_{\Omega_{+}}{\frac{f(\zeta)d(\zeta)}{(z-\zeta)^3}}\;\frac{d\zeta\wedge d\overline{\zeta}}{2i}.$

Thus

$\displaystyle \left\langle(Tf)',g\right\rangle_{\Omega_{-}} = -2 \int_{\Omega_{-}}{\int_{\Omega_{+}}{\frac{f(\zeta)\overline{g(z)}d(z)d(\zeta)}{(z-\zeta)^3}}\;\frac{d\zeta\wedge d\overline{\zeta}}{2i}}\;\frac{dz\wedge d\overline{z}}{2i}$

$\displaystyle = -2 \int_{\Omega_{-}}{\int_{\Omega_{+}}{\frac{F(\zeta)\overline{G(z)}d(z)^{1/2}d(\zeta)^{1/2}}{(z-\zeta)^3}}\;\frac{d\zeta\wedge d\overline{\zeta}}{2i}}\;\frac{dz\wedge d\overline{z}}{2i},$

where ${F(\zeta)=f(\zeta)d(\zeta)^{1/2}}$, ${G(z)=g(z)d(z)^{1/2}}$. Now we introduce the linear operator ${S\,:\, L^2(\Omega_{+})\rightarrow L^2(\Omega_{-})}$ given by

$\displaystyle SF(z):= -2\int_{\Omega_{+}}{F(\zeta)\frac{d(z)^{1/2}d(\zeta)^{1/2}}{(z-\zeta)^3}}\;\frac{d\zeta\wedge d\overline{\zeta}}{2i},$

that is the operator given by integration against kernel ${K\,:\, \Omega_{-}\times \Omega_{+} \rightarrow \mathbb{C}}$,

$\displaystyle K(z,w):=-2\frac{d(z)^{1/2}d(w)^{1/2}}{(z-w)^3},$

so that one has

$\displaystyle \left\langle(Tf)',g\right\rangle_{\Omega_{-}} = \int_{\Omega_{-}}{SF(z) \overline{G(z)}}\;\frac{dz\wedge d\overline{z}}{2i}.$

To conclude ${\|Tf\|_{L^2(\Gamma)} \lesssim \|f\|_{\Omega_{+}}}$ from ${\|Tf\|_{L^2(\Gamma)} \lesssim \|(Tf)'\|_{\Omega_{-}}}$ we need to prove ${\|(Tf)'\|_{\Omega_{-}}\lesssim \|f\|_{\Omega_{+}}}$, with constants depending on ${A}$ only. From the above discussion, this latest statement is implied (and equivalent to) by

$\displaystyle \|SF\|_{L^2(\Omega_{-})}\lesssim \|F\|_{L^2(\Omega_{+})},$

and we claim that this is indeed true. To see this it is sufficient to apply Schur’s test to the kernel ${K}$. So,

$\displaystyle \int_{\Omega_{+}}{|K(z,w)|}\;\frac{dw\wedge d\overline{w}}{2i} = 2 d(z)^{1/2}\int_{\Omega_{+}}{\frac{d(w)^{1/2}}{|z-w|^3}}\;\frac{dw\wedge d\overline{w}}{2i},$

and since ${d(z)\leq |z-w|}$ (they belong to opposite sides of ${\Gamma}$)

$\displaystyle d(z)^{1/2}\int_{\Omega_{+}}{\frac{d(w)^{1/2}}{|z-w|^3}}\;\frac{dw\wedge d\overline{w}}{2i} \leq d(z)^{1/2}\int_{\{|z-w|>d(z)\}}{\frac{1}{|z-w|^{5/2}}}\;\frac{dw\wedge d\overline{w}}{2i}$

$\displaystyle = d(z)^{1/2} \int_{d(z)}^{+\infty}{\int_{0}^{2\pi}{\frac{1}{r^{3/2}}}\,d\theta}\;dr$

$\displaystyle = 2\pi d(z)^{1/2}d(z)^{-1/2} = 2\pi.$

By symmetry the integral in ${z}$ is bounded by the same quantity, and thus by Schur’s lemma ${S}$ is ${L^2(\Omega_{+})\rightarrow L^2(\Omega_{-})}$ bounded. $\Box$
The alternative proof will come soon, maybe with some related facts about ${T(1)}$, ${T(b)}$ theorems.

[1] Conformal is equivalent to ${|\Phi'|\neq 0}$ everywhere.

[2] Such functions are often referred to as univalent, in the literature.

[3] Ball and disk are synonyms here.

[4] Schwarz’s lemma states that for a holomorphic function ${f\,:\, \delta\mathbb{D}\rightarrow r\mathbb{D}}$ it holds ${|f'(0)|\leq \frac{r}{\delta}}$. Since ${\Phi}$ is conformal, ${\Phi^{-1}}$ is holomorphic. Put ${\delta=d(\Phi(z))}$ and the inequality follows applying Schwarz’s lemma to ${\Phi^{-1}\,:\, \Phi(z)+\delta\mathbb{D} \rightarrow B(z,y)}$.

[5] Here ${\partial}$ is the ordinary differentiation in ${\mathbb{C}}$, i.e. ${\partial = \frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)}$, and ${\overline{\partial}}$ is the formal conjugate of ${\partial}$. It is ${\overline{\partial}\; \overline{f} = \overline{\partial f}}$. For an holomorphic function ${\overline{\partial} f = 0}$.