Nevanlinna classes for non radial weights in the unit disc. Applications

TL;DR

This paper introduces Nevanlinna classes for non-radial weights in the unit disc, providing Blaschke type theorems and applications to eigenvalues of non-self-adjoint Schrödinger operators, offering new proofs and improvements.

Contribution

It develops a new framework of Nevanlinna classes for non-radial weights and applies complex variables methods to derive Blaschke type theorems, enhancing previous results.

Findings

Established Nevanlinna classes for non-radial weights

Derived Blaschke type theorems using several complex variables

Improved understanding of eigenvalues of non-self-adjoint Schrödinger operators

Abstract

We introduce Nevanlinna classes associated to non radial weights in the unit disc in the complex plane and we get Blaschke type theorems relative to these classes by use of several complex variables methods. This gives alternative proofs and improve some results of Boritchev, Golinski and Kupin useful, in particular, for the study of eigenvalues of non self adjoint Schr\"odinger operators.

Equations

$${\int }_{\mathbb{D}}(g△\log \left|f\right|-\log \left|f\right|△g)={\int }_{\mathbb{T}}(g{\partial }_n\log \left|f\right|-\log \left|f\right|{\partial }_{n}g)$$

$$\sum _{a\in Z(f)}g(a)={\int }_{\mathbb{D}}\log \left|f\right|△g-{\int }_{\mathbb{T}}\log \left|f\right|{\partial }_{n}g.$$

$$A_3=\sum _{j=1}^n{B}_j\le s\left|q\right|(p+1)(1-{\left|z\right|}^2{)}^{p-1/2}{\left|R(sz)\right|}^2{\log }^{-}\left|f(sz)\right|.$$

$$C(s\rho )\le M(s\rho ){\int }_{\mathbb{T}}{\log }^{+}\left|f(s\rho e^{i\theta })\right|.$$

$$C(s\rho )\le M(s\rho )m(s\rho {)}^{-1}{\int }_{\mathbb{T}}{\left|R(s\rho e^{i\theta })\right|}^2{\log }^{+}\left|f(s\rho e^{i\theta })\right|.$$

$$B\le {\int }_0^u(1-{\rho }^2{)}^{p-1}(1-{\rho }^2{)}^{\delta }C(s\rho )\rho d\rho ,$$

Full text

1. 1 Introduction.
2. 2 Basic notations and results.
3. 3 Case $\displaystyle p>0.$
4. 4 Case $\displaystyle p=0.$
5. 5 Application : $\displaystyle L^{\infty}$ bounds.
6. 6 Case of a closed set in $\displaystyle{\mathbb{T}}.$
7. 7 The mixed case.
8. 8 Mixed cases with $\displaystyle L^{\infty}$ bounds.
9. 9 Appendix.

1 Introduction.

We shall work with classes of holomorphic functions whose zeroes may appear as eigenvalues of Schrödinger operators with complex valued potential. So having information on these zeroes gives information on the operator.

Let $\displaystyle F:=\{\eta_{j},\ j=1,...,n\}\subset{\mathbb{T}}\ ;$ we associate to $F$ the rational function with $\displaystyle q_{j}\in{\mathbb{R}},\ R(z):=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}}$ and we set, as a clearly non radial weight, $\displaystyle\varphi(z)=\left|{R(z)}\right|^{2}\ ;$ we also need to set $\displaystyle\ \gamma(z):=\left|{\sum_{j=1}^{n}{q_{j}(z-\eta_{j})^{-1}}}\right|.$

Definition 1.1

We shall say that the holomorphic function $f$ is in the generalised Nevanlinna class with weight $\displaystyle\varphi,\ {\mathcal{N}}_{\varphi,p}({\mathbb{D}}),$ if there is $\displaystyle 0<\delta<1$ such that, for $\displaystyle p>0:$

$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ {\left\|{f}\right\|}_{{\mathcal{N}}_{\varphi,p}}:=\sup_{1-\delta\leq s<1}\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\varphi(sz)\log^{+}\left|{f(sz)}\right|}<\infty.$

For $\displaystyle p=0\ :$

$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ {\left\|{f}\right\|}_{{\mathcal{N}}_{\varphi,0}}:=\sup_{1-\delta\leq s<1}\int_{{\mathbb{T}}}{\varphi(se^{i\theta})\log^{+}\left|{f(se^{i\theta})}\right|d\theta}+$

$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\sup_{1-\delta\leq s<1}\int_{{\mathbb{D}}}{\varphi(sz)\gamma(sz)\log^{+}\left|{f(sz)}\right|}<\infty.$**

In order to state the results we get, we set, for $\displaystyle p>0:$

if $\displaystyle q_{j}>-p/2,\ \tilde{q}_{j}:=q_{j}\ ;$ else we choose any $\displaystyle\tilde{q}_{j}>-p/2\ ;$ for $\displaystyle p=0:\ \tilde{q}_{j}:=(q_{j})_{+}\ ;$ then we set

$\displaystyle\tilde{\varphi}(z):=\left|{\prod_{j=1}^{n}{(z-\eta_{j})}^{\tilde{q}_{j}}}\right|.$

We get the following Blaschke type theorem:

Theorem 1.2

Suppose $\displaystyle f\in{\mathcal{N}}_{\varphi,p}({\mathbb{D}})$ is such that $\displaystyle\ \left|{f(0)}\right|=1,$ then we have:

$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})^{p+1}\tilde{\varphi}(a)}\leq c(\tilde{\varphi}){\left\|{f}\right\|}_{{\mathcal{N}}_{\varphi,p}},$

the constant $\displaystyle c(\tilde{\varphi})$ depending only on $\displaystyle\tilde{\varphi}.$

We can apply these theorems to the case of $\displaystyle L^{\infty}$ bounds.

With $\displaystyle R(z):=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}},\ \eta_{j}\in{\mathbb{T}},\ q_{j}\in{\mathbb{R}},$ we set $\displaystyle\forall\epsilon>0,\ R_{\epsilon}(z):=\prod_{j=1}^{n}{(z-\eta_{j})^{(q_{j}-1+\epsilon)_{+}}}.$ We define, $\displaystyle\forall j=1,...,n,$ if $\displaystyle q_{j}-1>-p/2,\ \tilde{q}_{j}=q_{j}$ else we choose $\displaystyle\tilde{q}_{j}>1-p/2,$ and we set $\displaystyle\tilde{R}_{0}(z):=\prod_{j=1}^{n}{(z-\eta_{j})}^{\tilde{q}_{j}-1}.$

We get as a corollary of our results:

Theorem 1.3

Suppose the holomorphic function $f$ in $\displaystyle{\mathbb{D}}$ verifies $\displaystyle\ \left|{f(0)}\right|=1$ and $\displaystyle\ \left|{f(z)}\right|\leq\exp\frac{D}{(1-\left|{z}\right|^{2})^{p}\left|{R(z)}\right|}$ with $\displaystyle R(z):=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}},\ \eta_{j}\in{\mathbb{T}},\ q_{j}\in{\mathbb{R}},$ then we have:

for $\displaystyle p=0,$

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|)\left|{R_{\epsilon}(a)}\right|}\leq Dc(R).$

For $\displaystyle p>0$

$\displaystyle\ \forall\epsilon>0,\ \sum_{a\in Z(f)}{(1-\left|{a}\right|)^{1+p+\epsilon}\left|{\tilde{R}_{0}(a)}\right|}\leq Dc(\epsilon,R).$**

Now recall that Boritchev, Golinskii and Kupin [4] proved, in particular:

Theorem 1.4

Let $\displaystyle f\in{\mathcal{H}}({\mathbb{D}}),\ \left|{f(0)}\right|=1$ and $\displaystyle\zeta_{j},\ \xi_{k}\in{\mathbb{T}},$ satisfy the growth condition :

$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \log^{+}\left|{f(z)}\right|\leq\frac{K}{(1-\left|{z}\right|)^{p}}\frac{\prod_{j=1}^{n}{\left|{z-\zeta_{j}}\right|^{r_{j}}}}{\prod_{k=1}^{m}{\left|{z-\xi_{k}}\right|^{q_{k}}}},\ \ \ z\in{\mathbb{D}},\ p,\ q_{k},\ r_{j}\geq 0.$

Then for every $\displaystyle\epsilon>0,$ there is a positive number $\displaystyle C_{3}=C_{3}(E,F,p,\{q_{k}\},\{r_{j}\},\epsilon)$ such that the following Blaschke condition holds:

$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \sum_{\zeta\in Z(f)}{(1-\left|{\zeta}\right|)^{p+1+\epsilon}}\frac{\prod_{k=1}^{m}{\left|{\zeta-\xi_{k}}\right|^{(q_{k}-1+\epsilon)_{+}}}}{\prod_{j=1}^{n}{\left|{\zeta-\zeta_{j}}\right|^{\min(p,r_{j})}}}\leq C_{3}\cdot K.$

If $\displaystyle p=0,$ the factor $\displaystyle(1-\left|{\zeta}\right|)^{1+\epsilon}$ can be replaced by $\displaystyle(1-\left|{\zeta}\right|).$

Comparing our result with the previous one, we get:

$\bullet$ for $\displaystyle p>0$ and $\displaystyle q\leq-p/2$ their result is better ;

$\bullet$ for $\displaystyle p>0$ and $\displaystyle q>-p/2$ our is better ;

$\bullet$ for $\displaystyle p=0$ the two results are identical.

The reason is that they have a threshold of $\displaystyle-p$ and our is $\displaystyle-p/2.$

As we shall see our results are based only on:

$\bullet$ the green formula ;

$\bullet$ the "zeroes" formula (see the next section) ;

which are the tools we use in several complex variables when dealing with problems on zeroes of holomorphic functions.

The methods used in several complex variables already proved their usefulness in the one variable case. For instance:

$\bullet$ the corona theorem of Carleson [6] is easier to prove and to understand thanks to the proof of T. Wolff based on L. Hörmander [7] ;

$\displaystyle\bullet$ the characterization of interpolating sequences by Carleson for $\displaystyle H^{\infty}$ and by Shapiro & Shields for $\displaystyle H^{p}$ are also easier to prove by these methods (see [1], last section, where they allow me to get the bounded linear extension property for the case $\displaystyle H^{p}$ ; the $\displaystyle H^{\infty}$ case being done by Pehr Beurling [3]).

So it is not surprising that in the case of zero set, they can also be useful.

In this paper all the computations are completely elementary: derivations of usual functions and straightforward estimates.

This work was already presented in an international workshop in November 2016 in Toulouse, France and in May 2017 in Bedlewo, Poland, during the conference on : "Hilbert spaces of entire functions and their applications".

2 Basic notations and results.

Let $f$ be an holomorphic function in the unit disk ${\mathbb{D}}$ of the complex plane, $\displaystyle{\mathcal{C}}^{\infty}(\bar{\mathbb{D}}),$ and $g$ a $\displaystyle{\mathcal{C}}^{\infty}$ smooth function in the closed unit disk $\displaystyle\bar{\mathbb{D}}$ such that $\displaystyle g=0$ on $\displaystyle{\mathbb{T}}.$

The only measures we shall deal with are the Lebesgue measures: of the plane when we integrate in the unit disc $\displaystyle{\mathbb{D}}$ or of the torus when we integrate on $\displaystyle{\mathbb{T}}:=\partial{\mathbb{D}}.$ So usually I shall not write explicitly the measure.

The Green formula gives:

[TABLE]

where $\displaystyle\partial_{n}$ is the normal derivative. With the "zero" formula: $\Delta\log\left|{f}\right|=\sum_{a\in Z(f)}{\delta_{a}}$ we get

$\displaystyle\ \sum_{a\in Z(f)}{g(a)}=\int_{{\mathbb{D}}}{\log\left|{f}\right|\triangle g}+\int_{{\mathbb{T}}}{(g\partial_{n}\log\left|{f}\right|-\log\left|{f}\right|\partial_{n}g)}.$

Because $\displaystyle g=0$ on $\displaystyle{\mathbb{T}},$

[TABLE]

So, in order to get estimates on $\displaystyle\ \sum_{a\in Z(f)}{g(a)},$ we have to compute $\partial_{n}g$ and $\Delta g.$ In this work, $g$ will always be of the form

$\displaystyle g_{s}(z)=(1-\left|{z}\right|^{2})^{1+p}\varphi(sz),$

where $\varphi(z)$ will be smooth and positive in $\displaystyle{\mathbb{D}}.$

We get a Blaschke type theorem if we can control

$\displaystyle\ \int_{{\mathbb{D}}}{\log\left|{f}\right|\triangle g}-\int_{{\mathbb{T}}}{\log\left|{f}\right|\partial_{n}g}\leq c{\left\|{f}\right\|}$

because then we get

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})^{p+1}\varphi(sa)}\leq c{\left\|{f}\right\|},$

where $\displaystyle\ {\left\|{f}\right\|}$ is a "norm" linked to the function $\displaystyle f.$ To get an idea of what happens here, suppose first that $\displaystyle p>0,$ and we set $\displaystyle f_{s}(z):=f(sz)\ ;$ so the equation (2.2) simplifies to

$\displaystyle\ \sum_{a\in Z(f_{s})}{g_{s}(a)}=\int_{{\mathbb{D}}}{\log\left|{f(sz)}\right|\triangle g_{s}(z)}=\int_{{\mathbb{D}}}{\log^{+}\left|{f(sz)}\right|\triangle g_{s}(z)}-\int_{{\mathbb{D}}}{\log^{-}\left|{f(sz)}\right|\triangle g_{s}(z)}.$

The strategy is quite obvious: we compute $\Delta g_{s}$ and we estimate the two quantities

$\displaystyle A_{+}(s):=\int_{{\mathbb{D}}}{\log^{+}\left|{f(sz)}\right|\triangle g_{s}(z)}$ and $\displaystyle A_{-}(s):=-\int_{{\mathbb{D}}}{\log^{-}\left|{f(sz)}\right|\triangle g_{s}(z)}.$

Because $\displaystyle\log^{+}\left|{f(sz)}\right|$ is directly related to the size of $f,$ we just take the sum of the absolute value of the terms in $\Delta g_{s}$ to estimate $\displaystyle A_{+}.$

For $A_{-}$ we have to be more careful because we want to control terms containing $\displaystyle\log^{-}\left|{f(sz)}\right|$ by terms containing only $\displaystyle\log^{+}\left|{f(sz)}\right|.$

This work is presented the following way.

$\bullet$ In the next section we study the case of $\varphi(z)=\left|{R(z)}\right|^{2}$ with $\displaystyle R(z)=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}},\ \eta_{j}\in{\mathbb{T}},\ q_{j}\in{\mathbb{R}}$ and $\displaystyle p>0.$ This is the easiest case but the problematic is already here.

$\bullet$ In section 4 we study, with the same $\varphi,$ the case $\displaystyle p=0.$

$\bullet$ In section 5 we get the $\displaystyle L^{\infty}$ bounds and we retrieve some results of Boritchev, Golinskii and Kupin [4].

$\bullet$ In section 6 we recall the case of a weight which is a power of the distance to a closed set $E$ in $\displaystyle{\mathbb{T}}.$

$\bullet$ in section 7 we study the mixed case associated to a closed set $E$ in $\displaystyle{\mathbb{T}}$ and a finite set $F.$

$\bullet$ Finally in the appendix we prove technical, but important, lemmas.

3 Case $\displaystyle p>0.$

Let $\displaystyle F:=\{\eta_{1},...,\eta_{n}\}\subset{\mathbb{T}}$ be a finite sequence of points on $\displaystyle{\mathbb{T}}.$ We shall work with the rational function $\displaystyle R(z)=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}},\ q_{j}\in{\mathbb{R}}$ and we set $\varphi(z):=\left|{R(z)}\right|^{2}.$ In order to have a smooth function in the disc we set $\displaystyle g_{s}(z):=(1-\left|{z}\right|^{2})^{1+p}\left|{R(sz)}\right|^{2},$ with $\displaystyle 0\leq s<1,$ and:

$\displaystyle\Delta g_{s}=4\partial\bar{\partial}g_{s}=4\partial\bar{\partial}[(1-\left|{z}\right|^{2})^{1+p}\left|{R(sz)}\right|^{2}]=\Delta[(1-\left|{z}\right|^{2})^{p+1}]\left|{R(sz)}\right|^{2}+(1-\left|{z}\right|^{2})^{p+1}\Delta[\left|{R(sz)}\right|^{2}]+$

$\displaystyle+8\Re[\partial((1-\left|{z}\right|^{2})^{p+1})\bar{\partial}(\left|{R(sz)}\right|^{2})].$

Straightforward computations give the following lemma, which separates the positive terms, the negative terms and the terms with no fixed sign:

Lemma 3.1

We have

$\Delta g_{s}(z)=\Delta_{+}-\Delta_{-}+\Delta_{\mp}$

with

$\displaystyle\Delta_{+}:=4(1-\left|{z}\right|^{2})^{p-1}[p(p+1)\left|{z}\right|^{2}+s^{2}(1-\left|{z}\right|^{2})^{2}\left|{\sum_{j=1}^{n}{q_{j}(sz-\eta_{j})^{-1}}}\right|^{2}]\left|{R(sz)}\right|^{2}$

$\displaystyle\Delta_{-}:=4(p+1)(1-\left|{z}\right|^{2})^{p}\left|{R(sz)}\right|^{2}$

$\displaystyle\Delta_{\mp}:=8s\Re[(-(r+1)(1-\left|{z}\right|^{2})^{r}\bar{z}\ )(\sum_{j=1}^{n}{q_{j}(s\bar{z}-\bar{\eta}_{j})^{-1}})]\left|{R(sz)}\right|^{2}.$**

Because $p>0\Rightarrow\partial_{n}g_{s}=0$ on $\displaystyle{\mathbb{T}},$ and formula (2.2), with $\displaystyle f_{s}(z):=f(sz),$ reduces to:

$\displaystyle\ \sum_{a\in Z(f_{s})}{g_{s}(a)}=\int_{{\mathbb{D}}}{\log\left|{f(sz)}\right|\triangle g_{s}(z)}.$

We have to estimate $\displaystyle\ \int_{{\mathbb{D}}}{\log\left|{f(sz)}\right|\triangle g_{s}(z)}$ and for it, we decompose:

$\displaystyle\log\left|{f(sz)}\right|\triangle g_{s}(z)=\log^{+}\left|{f(sz)}\right|\triangle g_{s}(z)-\log^{-}\left|{f(sz)}\right|\triangle g_{s}(z).$

We shall first group the terms containing $\displaystyle\log^{+}\left|{f(sz)}\right|.$ We set

$\displaystyle A_{+}(s):=\Delta_{+}\log^{+}\left|{f(sz)}\right|-\Delta_{-}\log^{+}\left|{f(sz)}\right|+\Delta_{\mp}\log^{+}\left|{f(sz)}\right|.$

And $\displaystyle T_{+}(s):=\int_{{\mathbb{D}}}{A_{+}(s)dm(z)}.$ We set also $\displaystyle P_{{\mathbb{D}},+}(s):=\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\left|{R(sz)}\right|^{2}\log^{+}\left|{f(sz)}\right|}.$

Proposition 3.2

We have, with $\displaystyle\ \left|{q}\right|:=\sum_{j=1}^{n}{\left|{q_{j}}\right|},\ T_{+}(s)\leq 4[p(p+1)\left|{z}\right|^{2}+4\left|{q}\right|^{2}+2\left|{q}\right|]P_{{\mathbb{D}},+}(s).$

Proof.

We have $\displaystyle A_{+}\leq\Delta_{+}\log^{+}\left|{f(sz)}\right|+\Delta_{\mp}\log^{+}\left|{f(sz)}\right|$ because $-\Delta_{-}$ is negative. We use that $\displaystyle(1-\left|{z}\right|^{2})\leq 2\left|{sz-\eta_{j}}\right|$ then elementary estimates on the modulus of the reminding terms end the proof. $\hfill\blacksquare$

We shall now group the terms containing $\displaystyle\log^{-}\left|{f(sz)}\right|.$ We set

$\displaystyle A_{-}(s,z):=-\Delta_{+}\log^{-}\left|{f(sz)}\right|+\Delta_{-}\log^{-}\left|{f(sz)}\right|-\Delta_{\mp}\log^{-}\left|{f(sz)}\right|$

and $\displaystyle P_{{\mathbb{D}},-}(s):=\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\left|{z}\right|^{2}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}$ and $\displaystyle T_{-}(s):=\int_{{\mathbb{D}}}{A_{-}(s,z)}.$

Proposition 3.3

Suppose that $\displaystyle\forall j=1,...,n,\ q_{j}\geq 0,$ then

$\displaystyle T_{-}(s)\leq(p+1)[4c(1,u)+s\left|{q}\right|c(1/2,u)]P_{{\mathbb{D}},+}(s).$**

Proof.

Set

$\displaystyle A_{2}:=\Delta_{-}\log^{-}\left|{f(sz)}\right|=4(p+1)(1-\left|{z}\right|^{2})^{p}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|.$

We apply the "substitution" lemma 9.1 from the appendix with $\delta=1,$ to get

$\displaystyle\ \int_{{\mathbb{D}}}{A_{2}}\leq 4(p+1)(1-u^{2})\frac{1}{u^{2}}P_{{\mathbb{D}},-}(s)+4(p+1)c(1,u)P_{{\mathbb{D}},+}(s).$

Now set

$\displaystyle B_{j}:=8q_{j}(p+1)(1-\left|{z}\right|^{2})^{p}\Re[\bar{z}(\bar{z}-\bar{\eta}_{j})^{-1}]\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|,$

and

$\displaystyle A_{3}:=-\Delta_{\mp}\log^{-}\left|{f(sz)}\right|=$

$\displaystyle=-8\Re[(-(p+1)(1-\left|{z}\right|^{2})^{p}\bar{z}\ )(\sum_{j=1}^{n}{q_{j}(\bar{z}-\bar{\eta}_{j})^{-1}})]\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|\ ;$

we get $\displaystyle A_{3}=\sum_{j=1}^{n}{B_{j}}.$ But

$\displaystyle\Re[\bar{z}(s\bar{z}-\bar{\eta}_{j})^{-1}]=\frac{1}{\left|{sz-\eta_{j}}\right|^{2}}\Re[\bar{z}(sz-\eta_{j})],$

hence by lemma 9.2 from the appendix, we have $\displaystyle\Re(\bar{z}(z-\eta))\leq 0$ iff $\displaystyle z\in{\mathbb{D}}\cap D(\frac{\eta_{j}}{2},\ \frac{1}{2}).$ So, with $\displaystyle q_{j}\geq 0,$ the part in $\displaystyle{\mathbb{D}}\cap D(\frac{\eta_{j}}{2},\ \frac{1}{2})$ is negative and can be ignored. It remains

$\displaystyle B_{j}\leq(p+1)s(1-\left|{z}\right|^{2})^{p}\left|{R(sz)}\right|^{2}{{\rm\kern 1.99997pt\vrule width=0.20004pt,height=6.32915pt,depth=0.0pt\kern-3.69995pt1}}_{D(\frac{\eta_{j}}{2},\frac{1}{2})^{c}}(z)\Re[q_{j}\bar{z}(\bar{z}-\bar{\eta}_{j})^{-1}]\log^{-}\left|{f(sz)}\right|.$

But for $\displaystyle z\in D(\frac{\eta_{j}}{2},\frac{1}{2})^{c},\ (1-\left|{z}\right|^{2})\leq 2\left|{z-\eta_{j}}\right|^{2}$ hence,

$\displaystyle{{\rm\kern 1.99997pt\vrule width=0.20004pt,height=6.32915pt,depth=0.0pt\kern-3.69995pt1}}_{D(\frac{\eta_{j}}{2},\frac{1}{2})^{c}}(z)\Re[\bar{z}(\bar{z}-\bar{\eta}_{j})^{-1}]\leq 2(1-\left|{z}\right|^{2})^{-1/2}{{\rm\kern 1.99997pt\vrule width=0.20004pt,height=6.32915pt,depth=0.0pt\kern-3.69995pt1}}_{D(\frac{\eta_{j}}{2},\frac{1}{2})^{c}}(z)\leq 2(1-\left|{z}\right|^{2})^{-1/2}.$

So we get

$\displaystyle B_{j}\leq sq_{j}(p+1)(1-\left|{z}\right|^{2})^{p-1/2}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|$

and, provided that $\displaystyle q_{j}\geq 0,$

[TABLE]

We can again apply the "substitution" lemma 9.1 with $\delta=1/2,$ this time and we get

$\displaystyle\ \int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1/2}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(z)}\right|}\leq(1-u^{2})^{1/2}\frac{1}{u^{2}}P_{{\mathbb{D}},-}(s)+c(1/2,u)P_{{\mathbb{D}},+}(s).$

So finally

$\displaystyle\ \int_{{\mathbb{D}}}{A_{3}}\leq s\left|{q}\right|(p+1)(1-u^{2})^{1/2}\frac{1}{u^{2}}P_{{\mathbb{D}},-}(s)+s\left|{q}\right|(p+1)c(1/2,u)P_{{\mathbb{D}},+}(s).$

Integrating $\displaystyle A_{-}(s,z)$ over $\displaystyle{\mathbb{D}}$ and adding, we get, with $\displaystyle A_{1}:=-\Delta_{+}\log^{-}\left|{f(sz)}\right|,$

$\displaystyle T_{-}(s)\leq\int_{{\mathbb{D}}}{(A_{1}+A_{2}+A_{3})}\leq-4p(p+1)P_{{\mathbb{D}},-}(s)+4(p+1)(1-u^{2})\frac{1}{u^{2}}P_{{\mathbb{D}},-}(s)+$

$\displaystyle+4(p+1)c(1,u)P_{{\mathbb{D}},+}(s)+s\left|{q}\right|(p+1)(1-u^{2})^{1/2}\frac{1}{u^{2}}P_{{\mathbb{D}},-}(s)+$

$\displaystyle+s\left|{q}\right|(p+1)c(1/2,u)P_{{\mathbb{D}},+}(s).$

The key point here is that the "bad terms" in $\displaystyle\log^{-}\left|{f(z)}\right|$ can be controlled by the "good" one: $\displaystyle A_{1}:=-\Delta_{+}\log^{-}\left|{f(sz)}\right|.$

We can choose $\displaystyle 0<u<1$ such that

$\displaystyle-4p(p+1)+4(p+1)(1-u^{2})\frac{1}{u^{2}}+s\left|{q}\right|(p+1)(1-u^{2})^{1/2}\frac{1}{u^{2}}\leq 0$

just taking, because $\displaystyle p>0,\ \ {\sqrt{1-u^{2}}}\leq\frac{4p}{4+s\left|{q}\right|}.$ Hence we get, provided that $\displaystyle\forall j=1,...,n,\ q_{j}\geq 0,$

$\displaystyle T_{-}(s)\leq(p+1)[4c(1,u)+s\left|{q}\right|c(1/2,u)]P_{{\mathbb{D}},+}(s).$ $\hfill\blacksquare$

We can also get results for $\displaystyle q_{j}<0$ the following way. We cut the disc in disjoint sectors around the points $\displaystyle\eta_{j}:\ {\mathbb{D}}=\Gamma_{0}\cup\bigcup_{j=1}^{n}{\Gamma_{j}}$ with

$\displaystyle\forall j=1,...,n,\ \Gamma_{j}:=\{z\in{\mathbb{D}}::\left|{\frac{z}{\left|{z}\right|}-\eta_{j}}\right|<\alpha\},\ \Gamma_{0}:={\mathbb{D}}\backslash\bigcup_{j=1}^{n}{\Gamma_{j}}.$

This is possible because the points $\eta_{j}$ are in finite number so $\alpha>0$ exists.

Proposition 3.4

Set $\displaystyle\ \left|{q}\right|_{\infty}:=\max_{k=1,...,n}\left|{q_{k}}\right|$ and suppose $\displaystyle\ \left|{q}\right|_{\infty}<p/4,$ then there exist $\displaystyle u<1,\ \gamma<1$ such that:

$\displaystyle T_{-}(s)\leq 4(p+1)[c(1,u)+2\frac{\left|{q}\right|}{\alpha}c(1,u)+2\left|{q}\right|_{\infty}(1-\gamma)^{-1}c(1,\gamma)]P_{{\mathbb{D}},+}(s).$**

Proof.

We have

$\displaystyle\ \left|{-\Delta_{\mp}}\right|=\left|{-8s\Re[(-(p+1)(1-\left|{z}\right|^{2})^{p}\bar{z}\ )(\sum_{j=1}^{n}{q_{j}(s\bar{z}-\bar{\eta}_{j})^{-1}})]}\right|\left|{R(sz)}\right|^{2}\leq$

$\displaystyle\leq 8(p+1)(1-\left|{z}\right|^{2})^{p}\sum_{j=1}^{n}{\left|{q_{j}}\right|\left|{sz-\eta_{j}}\right|^{-1}}\left|{R(sz)}\right|^{2}.$

Now we set

$\displaystyle A^{\prime}_{3}:=\left|{-\Delta_{\mp}\log^{-}\left|{f(sz)}\right|}\right|\leq 8(p+1)(1-\left|{z}\right|^{2})^{p}\sum_{j=1}^{n}{\left|{q_{j}}\right|\left|{sz-\eta_{j}}\right|^{-1}}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|$

and

$\displaystyle\forall k=0,1,...,n,\ f_{k}(z):=8(p+1)(1-\left|{z}\right|^{2})^{p}\sum_{j=1,j\neq k}^{n}{\left|{q_{j}}\right|\left|{sz-\eta_{j}}\right|^{-1}}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|$

and on $\Gamma_{k},$ including $\displaystyle k=0,$ we get

$\displaystyle\forall z\in\Gamma_{k},\ f_{k}(z)\leq 8(p+1)\frac{\left|{q}\right|}{\alpha}(1-\left|{z}\right|^{2})^{p}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|.$

Hence we have

$\displaystyle\forall k=0,...,n,\ \forall z\in\Gamma_{k},\ A^{\prime}_{3}\leq$

$\displaystyle\leq 8(p+1)\frac{\left|{q}\right|}{\alpha}(1-\left|{z}\right|^{2})^{p}+8(p+1)(1-\left|{z}\right|^{2})^{p}\left|{q_{k}}\right|\left|{sz-\eta_{k}}\right|^{-1}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|.$

Now we integrate in the disc and we get

$\displaystyle\ \int_{{\mathbb{D}}}{A^{\prime}_{3}}\leq 8(p+1)\frac{\left|{q}\right|}{\alpha}\sum_{k=0}^{n}{\int_{\Gamma_{k}}{(1-\left|{z}\right|^{2})^{p}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}+}$

$\displaystyle+8(p+1)\sum_{k=0}^{n}{\left|{q_{k}}\right|\int_{\Gamma_{k}}{(1-\left|{z}\right|^{2})^{p}\left|{sz-\eta_{k}}\right|^{-1}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}}=:B_{1}+B_{2}.$

But

$\displaystyle\ \int_{\Gamma_{k}}{(1-\left|{z}\right|^{2})^{p}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}\leq\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}$

and we can apply the "substitution" lemma 9.1, with $\delta=1,$ to get

$\displaystyle\ \int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}\leq(1-u^{2})\frac{1}{u^{2}}P_{{\mathbb{D}},-}(s)+c(1,u)P_{{\mathbb{D}},+}(s).$

So the first term in $\displaystyle\ \int_{{\mathbb{D}}}{A^{\prime}_{3}}$ is controlled by

$\displaystyle B_{1}\leq 8(p+1)\frac{\left|{q}\right|}{\alpha}(1-u^{2})\frac{1}{u^{2}}P_{{\mathbb{D}},-}(s)+8(p+1)\frac{\left|{q}\right|}{\delta}c(1,u)P_{{\mathbb{D}},+}(s).$

For the second one we first localise near the boundary:

$\displaystyle B_{2}:=8(p+1)\sum_{k=0}^{n}{\left|{q_{k}}\right|\int_{\Gamma_{k}}{(1-\left|{z}\right|^{2})^{p}\left|{sz-\eta_{k}}\right|^{-1}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}}=$

$\displaystyle=8(p+1)\sum_{k=0}^{n}{\left|{q_{k}}\right|\int_{D(0,\gamma)\cap\Gamma_{k}}{(1-\left|{z}\right|^{2})^{p}\left|{sz-\eta_{k}}\right|^{-1}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}}+$

$\displaystyle+8(p+1)\sum_{k=0}^{n}{\left|{q_{k}}\right|\int_{\Gamma_{k}\backslash D(0,\gamma)}{(1-\left|{z}\right|^{2})^{p}\left|{sz-\eta_{k}}\right|^{-1}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}}=:$

$\displaystyle=:C_{1}+C_{2}.$

We get

$\displaystyle C_{1}\leq 8(p+1)\left|{q}\right|_{\infty}(1-\gamma)^{-1}\int_{D(0,\gamma)}{(1-\left|{z}\right|^{2})^{p}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}.$

The proof of the "substitution" lemma 9.1, gives with $\gamma$ in place of $u,$

$\displaystyle C_{1}\leq 8(p+1)\left|{q}\right|_{\infty}(1-\gamma)^{-1}c(1,\gamma)P_{{\mathbb{D}},+}(s).$

Now for $\displaystyle C_{2}$ we have

$\displaystyle C_{2}:=8(p+1)\sum_{k=0}^{n}{\left|{q_{k}}\right|\int_{\Gamma_{k}\backslash D(0,\gamma)}{(1-\left|{z}\right|^{2})^{p}\left|{sz-\eta_{k}}\right|^{-1}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}}\leq$

$\displaystyle\leq 8(p+1)\sum_{k=0}^{n}{\left|{q_{k}}\right|\frac{1}{\gamma^{2}}\int_{\Gamma_{k}\backslash D(0,\gamma)}{(1-\left|{z}\right|^{2})^{p}\left|{z}\right|^{2}\left|{sz-\eta_{k}}\right|^{-1}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}}.$

We use $\displaystyle(1-\left|{z}\right|^{2})\leq 2\left|{sz-\eta_{k}}\right|$ to get

$\displaystyle C_{2}\leq 16(p+1)\frac{1}{\gamma^{2}}\sum_{k=0}^{n}{\left|{q_{k}}\right|\int_{\Gamma_{k}}{(1-\left|{z}\right|^{2})^{p-1}\left|{z}\right|^{2}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}}\leq 16(p+1)\left|{q}\right|_{\infty}\frac{1}{\gamma^{2}}P_{{\mathbb{D}},-}(s).$

We have, with the notations of proposition 3.3, replacing $\displaystyle A_{3}$ by $\displaystyle A^{\prime}_{3},$

$\displaystyle T_{-}(s)\leq\int_{{\mathbb{D}}}{(A_{1}+A_{2}+A^{\prime}_{3})}\leq$

$\displaystyle-4p(p+1)P_{{\mathbb{D}},-}(s)+4(p+1)(1-u^{2})\frac{1}{u^{2}}P_{{\mathbb{D}},-}(s)+4(p+1)c_{3}(1,u)P_{{\mathbb{D}},+}(s)+$

$\displaystyle+8(p+1)\frac{\left|{q}\right|}{\alpha}(1-u^{2})\frac{1}{u^{2}}P_{{\mathbb{D}},-}(s)+8(p+1)\frac{\left|{q}\right|}{\alpha}c_{3}(1,u)P_{{\mathbb{D}},+}(s)+$

$\displaystyle+8(p+1)\left|{q}\right|_{\infty}(1-\gamma)^{-1}c(1,\gamma)P_{{\mathbb{D}},+}(s)+16(p+1)\left|{q}\right|_{\infty}\frac{1}{\gamma^{2}}P_{{\mathbb{D}},-}(s).$

Let us see the terms containing $\displaystyle\log^{-}\left|{f(sz)}\right|,$ we set:

$\displaystyle D(s,\gamma,u):=[-4p(p+1)+8(p+1)\frac{\left|{q}\right|}{\alpha}(1-u^{2})\frac{1}{u^{2}}+16(p+1)\left|{q}\right|_{\infty}\frac{1}{\gamma^{2}}]P_{{\mathbb{D}},-}(s).$

So

$\displaystyle D(s,\gamma,u)=16(-\frac{p}{4}+\frac{\left|{q}\right|_{\infty}}{\gamma^{2}}+\frac{\left|{q}\right|}{2\alpha}\frac{1-u^{2}}{u^{2}})(p+1)P_{{\mathbb{D}},-}(s).$

Now suppose that $\displaystyle\ \left|{q}\right|_{\infty}<p/4$ and first choose $\gamma<1$ big enough to have $\displaystyle-\frac{p}{4}+\frac{\left|{q}\right|_{\infty}}{\gamma^{2}}=:-\epsilon<0$ which is clearly possible, then choose $\displaystyle u<1$ such that $\displaystyle\ \frac{\left|{q}\right|}{2\alpha}\frac{1-u^{2}}{u^{2}}-\epsilon\leq 0$ which is also clearly possible because $\displaystyle\epsilon>0.$ So we get with these choices of $u$ and $\gamma,$

$\displaystyle T_{-}(s)\leq[4(p+1)c(1,u)+8(p+1)\frac{\left|{q}\right|}{\alpha}c(1,u)+8(p+1)\left|{q}\right|_{\infty}(1-\gamma)^{-1}c(1,\gamma)]P_{{\mathbb{D}},+}(s).$ $\hfill\blacksquare$

As a corollary of these two propositions, we get

Corollary 3.5

Suppose $\displaystyle\ \forall j,\ q_{j}>-p/4,$ then there is a constant $\displaystyle c(p,R)$ such that:

$\displaystyle T_{-}(s)\leq c(p,R)P_{{\mathbb{D}},+}(s).$**

Proof.

As above we can separate the points $\eta_{j}$ where $\displaystyle-p/4<q_{j}<0$ from the points $\eta_{j}$ with $\displaystyle q_{j}\geq 0.$ Then we apply the relevant proof to each case. $\hfill\blacksquare$

We are lead to the following definition:

Definition 3.6

Let $\displaystyle R(z)=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}},\ q_{j}\in{\mathbb{R}}.$ We say that an holomorphic function $f$ is in the generalised Nevanlinna class $\displaystyle{\mathcal{N}}_{\left|{R}\right|^{2},p}({\mathbb{D}})$ for $\displaystyle p>0,$ if $\displaystyle\exists\delta>0,\ \delta<1$ such that

$\displaystyle\ {\left\|{f}\right\|}_{{\mathcal{N}}_{\left|{R}\right|^{2},p}}:=\sup_{1-\delta<s<1}\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\left|{R(sz)}\right|^{2}\log^{+}\left|{f(sz)}\right|}<\infty.$**

And we get the Blaschke type condition:

Theorem 3.7

Let $\displaystyle R(z)=\prod_{j=1}^{n}{(z-\eta_{j})}^{q_{j}},\ q_{j}\in{\mathbb{R}}.$ Suppose $\displaystyle p>0,\ j=1,...,n,\ q_{j}>-p/4$ and $\displaystyle f\in{\mathcal{N}}_{\left|{R}\right|^{2},p}({\mathbb{D}})$ with $\displaystyle\ \left|{f(0)}\right|=1,$ then

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})^{1+p}\left|{R(a)}\right|^{2}}\leq c(p,R){\left\|{f}\right\|}_{{\mathcal{N}}_{\left|{R}\right|^{2},p}}.$**

Proof.

We apply the formula (2.2), to get, with $\displaystyle g_{s}(z)=(1-\left|{z}\right|^{2})^{1+p}\left|{R(sz)}\right|^{2},$

$\displaystyle\forall s<1,\ \ \sum_{a\in Z(f_{s})}{(1-\left|{a}\right|^{2})^{1+p}\left|{R(sa)}\right|^{2}}=\int_{{\mathbb{D}}}{\log\left|{f(sz)}\right|\triangle g_{s}(z)}$

because with $\displaystyle p>0,\ \partial_{n}g_{s}=0$ on $\displaystyle{\mathbb{T}}.$

Now we use Proposition 3.2 to get that

$\displaystyle\ \int_{{\mathbb{D}}}{\log^{+}\left|{f(sz)}\right|\triangle g}_{s}(z)\leq 4[p(p+1)\left|{z}\right|^{2}+4\left|{q}\right|^{2}+2\left|{q}\right|]P_{{\mathbb{D}},+}(s),$

and corollary 3.5 to get

$\displaystyle-\int_{{\mathbb{D}}}{\log^{-}\left|{f(sz)}\right|\triangle g_{s}(z)}\leq c(p,R)P_{{\mathbb{D}},+}(s).$

So adding we get

$\displaystyle\forall s<1,\ \ \sum_{a\in Z(f_{s})}{(1-\left|{a}\right|^{2})^{1+p}\left|{R(sa)}\right|^{2}}\leq c(p,R)P_{{\mathbb{D}},+}(s).$

We are in position to apply lemma 9.5 from the appendix, with $\varphi(z)=\left|{R(z)}\right|^{2},$ to get

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})^{1+p}\left|{R(a)}\right|^{2}}\leq c(p,R)\sup_{1-\delta<s<1}P_{{\mathbb{D}},+}(s),$

because $\displaystyle\ \left|{R(z)}\right|^{2}$ is positive. $\hfill\blacksquare$

Corollary 3.8

Let $\displaystyle R(z)=\prod_{j=1}^{n}{(z-\eta_{j})}^{q_{j}},\ q_{j}\in{\mathbb{R}}.$ Suppose $\displaystyle p>0$ and $\displaystyle f\in{\mathcal{N}}_{\left|{R}\right|,p}({\mathbb{D}})$ with $\displaystyle\ \left|{f(0)}\right|=1,$ and let $\displaystyle\forall j=1,...,n,$ if $\displaystyle q_{j}>-p/2,\ \tilde{q}_{j}=q_{j}$ else choose $\displaystyle\tilde{q}_{j}>-p/2,$ and set $\displaystyle\tilde{R}(z):=\prod_{j=1}^{n}{(z-\eta_{j})}^{\tilde{q}_{j}},$ then

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})^{1+p}\left|{\tilde{R}(a)}\right|}\leq c(p,\tilde{q},R){\left\|{f}\right\|}_{{\mathcal{N}}_{\left|{R}\right|,p}}.$**

Proof.

In order to apply theorem 3.7 to $\displaystyle\tilde{R}$ we have to show that $\displaystyle f\in{\mathcal{N}}_{\left|{R}\right|,p}({\mathbb{D}})\Rightarrow f\in{\mathcal{N}}_{\left|{\tilde{R}}\right|,p}({\mathbb{D}}).$

But

$\displaystyle\tilde{R}(sz):=\prod_{j=1}^{n}{(sz-\eta_{j})}^{\tilde{q}_{j}}=\prod_{j=1}^{n}{(sz-\eta_{j})}^{q_{j}}{\times}\prod_{j=1}^{n}{(sz-\eta_{j})}^{\tilde{q}_{j}-q_{j}},$

and the only point is for the $j$ such that $\displaystyle q_{j}\leq-p/2.$ So set $\displaystyle r_{j}:=\tilde{q}_{j}-q_{j}\geq 0,$ we have $\displaystyle\ \left|{sz-\eta_{j}}\right|\leq 2$ hence $\displaystyle\ \left|{sz-\eta_{j}}\right|^{r_{j}}\leq 2^{r_{j}}$ so $\displaystyle\ \left|{\tilde{R}(sz)}\right|\leq 2^{\left|{r}\right|}\left|{R(sz)}\right|$ with $\displaystyle\ \left|{r}\right|:=\sum_{j=1}^{n}{r_{j}}.$

Putting it in $\displaystyle\ {\left\|{f}\right\|}_{{\mathcal{N}}_{\left|{\tilde{R}}\right|,p}}$ we get

$\displaystyle\ {\left\|{f}\right\|}_{{\mathcal{N}}_{\left|{\tilde{R}}\right|,p}}:=\sup_{1-\delta<s<1}\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\left|{\tilde{R}(sz)}\right|\log^{+}\left|{f(sz)}\right|}\leq$

$\displaystyle\leq 2^{\left|{r}\right|}\sup_{1-\delta<s<1}\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\left|{R(sz)}\right|\log^{+}\left|{f(sz)}\right|}=2^{\left|{r}\right|}{\left\|{f}\right\|}_{{\mathcal{N}}_{\left|{R}\right|,p}}.$

So we are done. $\hfill\blacksquare$

4 Case $\displaystyle p=0.$

Now we set: $\displaystyle g_{s}(z)=(1-\left|{z}\right|^{2})\left|{R(sz)}\right|^{2}$ and we have that

$\displaystyle\partial_{n}g_{s}(z)=-2\left|{z}\right|\left|{R(sz)}\right|^{2}+(1-\left|{z}\right|^{2})\partial_{n}(\left|{R(sz)}\right|^{2})$

which is not [math] on $\displaystyle{\mathbb{T}},$ so we have to add the boundary term:

$\displaystyle B(s):=-\int_{{\mathbb{T}}}{\log\left|{f(sz)}\right|\partial_{n}g_{s}}=2\int_{{\mathbb{T}}}{\left|{R(sz)}\right|^{2}\log^{+}\left|{f(sz)}\right|}-2\int_{{\mathbb{T}}}{\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}=:$

$\displaystyle=:B_{+}(s)-B_{-}(s).$

We shall use as above, for $\displaystyle t_{0}\in[0,1[,$

$\displaystyle P_{{\mathbb{T}},-}(t_{0}):=\sup_{0\leq s\leq t_{0}}\int_{{\mathbb{T}}}{\left|{R(se^{i\theta})}\right|^{2}\log^{-}\left|{f(se^{i\theta})}\right|}$

and

$\displaystyle P_{{\mathbb{T}},+}(t_{0}):=\sup_{0\leq s\leq t_{0}}\int_{{\mathbb{T}}}{\left|{R(se^{i\theta})}\right|^{2}\log^{+}\left|{f(se^{i\theta})}\right|}.$

Now we set

$\displaystyle A_{+}(s):=4s^{2}(1-\left|{z}\right|^{2})[\left|{\sum_{j=1}^{n}{q_{j}(sz-\eta_{j})^{-1}}}\right|^{2}]\left|{R(sz)}\right|^{2}\log^{+}\left|{f(sz)}\right|-4\left|{R(sz)}\right|^{2}\log^{+}\left|{f(sz)}\right|+$

$\displaystyle+8s\Re[(-\bar{z}\ )(\sum_{j=1}^{n}{q_{j}(s\bar{z}-\bar{\eta}_{j})^{-1}})]\left|{R(sz)}\right|^{2}\log^{+}\left|{f(sz)}\right|+B_{+}(s).$

Set also $\displaystyle T_{+}(s):=\int_{{\mathbb{D}}}{A_{+}(s)},$ and with $\displaystyle\gamma(z):=\sum_{j=1}^{n}{\left|{q_{j}}\right|\left|{z-\eta_{j}}\right|^{-1}},$

$\displaystyle\ P_{\gamma,+}(s):=\int_{{\mathbb{D}}}{\gamma(sz)\left|{R(sz)}\right|^{2}\log^{+}\left|{f(sz)}\right|}.$

Proposition 4.1

We have

$\displaystyle T_{+}(s)\leq 8(\left|{q}\right|+1)P_{\gamma,+}(s)+B_{+}(s).$**

Proof.

Set

$\displaystyle A_{1}:=4s^{2}\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})[\left|{\sum_{j=1}^{n}{q_{j}(sz-\eta_{j})^{-1}}}\right|^{2}]\left|{R(sz)}\right|^{2}\log^{+}\left|{f(sz)}\right|}.$

Using $\displaystyle(1-\left|{z}\right|^{2})\leq 2\left|{sz-\eta_{j}}\right|,$ we get $\displaystyle A_{1}\leq 8\left|{q}\right|P_{\gamma,+}(s).$

Set $\displaystyle A_{2}:=-\int_{{\mathbb{D}}}{4\left|{R(sz)}\right|^{2}\log^{+}\left|{f(sz)}\right|.}$ Then $\displaystyle A_{2}\leq 0$ and it can be forgotten.

Finally set

$\displaystyle A_{3}:=\int_{{\mathbb{D}}}{8s\Re[(-\bar{z}\ )(\sum_{j=1}^{n}{q_{j}(s\bar{z}-\bar{\eta}_{j})^{-1}})]\left|{R(sz)}\right|^{2}\log^{+}\left|{f(sz)}\right|.}$

Again we get $\displaystyle A_{3}\leq 8sP_{\gamma,+}(s).$

Summing the $\displaystyle A_{j}$ we get

$\displaystyle T_{+}(s)\leq 8(\left|{q}\right|+1)P_{\gamma,+}(s)+B_{+}(s).$ $\hfill\blacksquare$

We shall now group the terms containing $\displaystyle\log^{-}\left|{f(sz)}\right|.$ We set

$\displaystyle-A_{-}(s,z):=-4\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|+(1-\left|{z}\right|^{2})\Delta(\left|{R(sz)}\right|^{2})(sz)\log^{-}\left|{f(sz)}\right|+$

$\displaystyle+8s\Re[(-\bar{z}\ )(\sum_{j=1}^{n}{q_{j}(s\bar{z}-\bar{\eta}_{j})^{-1}})]\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|+B_{-}(s).$

and $\displaystyle T_{-}(s):=\int_{{\mathbb{D}}}{A(s,z)}.$

Proposition 4.2

We have

$\displaystyle T_{-}(s)\leq 2[2c^{\prime}_{3}(1,u)+2\left|{q}\right|c^{\prime}_{3}(1/2,u)]P_{{\mathbb{T}},+}(t_{0})+$

$\displaystyle+2(1-u^{2})^{1/2}[2(1-u^{2})^{1/2}+2\left|{q}\right|]P_{{\mathbb{T}},-}(t_{0})-B_{-}(s).$**

Proof.

We have $\displaystyle\Delta[(1-\left|{z}\right|^{2})]=-4$ so

$\displaystyle A_{1}(s):=-\int_{{\mathbb{D}}}{\Delta((1-\left|{z}\right|^{2}))\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}=4\int_{{\mathbb{D}}}{\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}.$

We can apply the second part of the substitution lemma 9.1 with $\displaystyle\delta=1,$ we get for any $\displaystyle u<1,$

$\displaystyle\forall s\leq t_{0},\ \int_{{\mathbb{D}}}{\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}\leq c(1,u)P_{{\mathbb{T}},+}(t_{0})+\frac{1}{2}(1-u^{2})P_{{\mathbb{T}},-}(t_{0}).$

So we get

$\displaystyle A_{1}(s)\leq 4c(1,u)P_{{\mathbb{T}},+}(t_{0})+2(1-u^{2})P_{{\mathbb{T}},-}(t_{0}).$

For

$A_{2}:=-\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})\Delta(\left|{R(sz)}\right|^{2})(sz)\log^{-}\left|{f(sz)}\right|}=$

$\displaystyle=-4s^{2}\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})\left|{R^{\prime}(sz)}\right|^{2}(sz)\log^{-}\left|{f(sz)}\right|}\leq 0,$

so we can forget it.

Now we arrive at the "bad term"

$\displaystyle A_{3}:=-\int_{{\mathbb{D}}}{8\Re[\partial((1-\left|{z}\right|^{2}))\bar{\partial}(\left|{R(sz)}\right|^{2})]\log^{-}\left|{f(sz)}\right|}.$

Copying the proof done in the case $\displaystyle p>0,$ we use again lemma 9.2 and we integrate inequality (3.3) with $\displaystyle p=0:$

$\displaystyle A_{3}\leq s\left|{q}\right|\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{-1/2}\left|{R(sz)}\right|^{2}}.$

Now we are in position to apply the second part of lemma 9.1 with $\displaystyle\delta=1/2,$ so we get

$\displaystyle\forall s\leq t_{0},\ \int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{-1/2}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}\leq 2c(1/2,u)P_{{\mathbb{T}},+}(t_{0})+(1-u^{2})^{1/2}P_{{\mathbb{T}},-}(t_{0}),$

and

$\displaystyle A_{3}\leq 2s\left|{q}\right|c(1/2,u)P_{{\mathbb{T}},+}(t_{0})+2s\left|{q}\right|(1-u^{2})^{1/2}P_{{\mathbb{T}},-}(t_{0}).$

Summing all, we get

$\displaystyle T_{-}(s)\leq 4c(1,u)P_{{\mathbb{T}},+}(t_{0})+2(1-u^{2})P_{{\mathbb{T}},-}(t_{0})+2s\left|{q}\right|c(1/2,u)P_{{\mathbb{T}},+}(t_{0})+$

$\displaystyle+2s\left|{q}\right|(1-u^{2})^{1/2}P_{{\mathbb{T}},-}(t_{0})-B_{-}(s).$

Hence

$\displaystyle T_{-}(s)\leq 2[2c(1,u)+2\left|{q}\right|c(1/2,u)]P_{{\mathbb{T}},+}(t_{0})+2(1-u^{2})^{1/2}[2(1-u^{2})^{1/2}+2\left|{q}\right|]P_{{\mathbb{T}},-}(t_{0})-B_{-}(s).$ $\hfill\blacksquare$

Definition 4.3

Let $\displaystyle R(z)=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}},\ q_{j}\in{\mathbb{R}}.$ We say that an holomorphic function $f$ is in the generalised Nevanlinna class $\displaystyle{\mathcal{N}}_{\left|{R}\right|^{2},0}({\mathbb{D}})$ if $\displaystyle\exists\delta>0,\ \delta<1$ such that

$\displaystyle\ {\left\|{f}\right\|}_{{\mathcal{N}}_{\left|{R}\right|^{2},0}}:=\sup_{1-\delta<s<1}\int_{{\mathbb{T}}}{\left|{R(se^{i\theta})}\right|^{2}\log^{+}\left|{f(se^{i\theta})}\right|}+$

$\displaystyle+\sup_{1-\delta<s<1}\int_{{\mathbb{D}}}{\gamma(sz)\left|{R(sz)}\right|^{2}\log^{+}\left|{f(sz)}\right|}<\infty,$

with $\displaystyle\gamma(z):=\sum_{j=1}^{n}{\left|{q_{j}}\right|\left|{z-\eta_{j}}\right|^{-1}}.$

We get the Blaschke type condition:

Theorem 4.4

Let $\displaystyle R(z)=\prod_{j=1}^{n}{(z-\eta_{j})}^{q_{j}},\ q_{j}\in{\mathbb{R}}.$ Suppose $\displaystyle\forall j=1,...,n,\ q_{j}\geq 0$ and $\displaystyle f\in{\mathcal{N}}_{\left|{R}\right|^{2},0}({\mathbb{D}})$ with $\displaystyle\ \left|{f(0)}\right|=1,$ then there exists a constant $\displaystyle c(R)$ depending only on $R$ such that

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})\left|{R(a)}\right|^{2}}\leq c(R){\left\|{f}\right\|}_{{\mathcal{N}}_{\left|{R}\right|^{2},0}}.$**

Proof.

Fix $\displaystyle t_{0}\in[0,1[,$ by lemma 9.3 in the appendix, we have that

$\displaystyle h(s):=\int_{{\mathbb{T}}}{\left|{R(se^{i\theta})}\right|^{2}\log^{-}\left|{f(se^{i\theta})}\right|}$

is a continuous function of $s\in[0,t_{0}]$ hence its supremum is achieved at $\displaystyle s_{0}=s(t_{0})\in[0,t_{0}],$ i.e.

$\displaystyle P_{{\mathbb{T}},-}(t_{0})=B_{-}(s_{0}):=\int_{{\mathbb{T}}}{\left|{R(s_{0}e^{i\theta})}\right|^{2}\log^{-}\left|{f(s_{0}e^{i\theta})}\right|}.$

Let us consider, for any $\displaystyle t\in[0,t_{0}],$

$\displaystyle\Sigma(t,s_{0}):=\sum_{a\in Z(f_{t})}{g_{t}(a)}+\sum_{a\in Z(f_{s_{0}})}{g_{s_{0}}(a)}.$

We have, by (2.2),

$\displaystyle\Sigma(t,s_{0})\leq T_{+}(t)+T_{+}(s_{0})+T_{-}(t)+T_{-}(s_{0}).$

By use of proposition 4.1 we get

$\displaystyle T_{+}(s)\leq 8(\left|{q}\right|+1)\int_{{\mathbb{D}}}{\gamma(z)\left|{R(sz)}\right|^{2}\log^{+}\left|{f(sz)}\right|}+B_{+}(s),$

and by use of proposition 4.2 we get for $\displaystyle s\in[0,t_{0}],$

$\displaystyle T_{-}(s)\leq 2[2c(1,u)+2\left|{q}\right|c(1/2,u)]P_{{\mathbb{T}},+}(t_{0})+2(1-u^{2})^{1/2}[2(1-u^{2})^{1/2}+2\left|{q}\right|]P_{{\mathbb{T}},-}(t_{0})-B_{-}(s).$

Hence

$\displaystyle\Sigma(t,s_{0})\leq T_{+}(t)+T_{+}(s_{0})+T_{-}(t)+T_{-}(s_{0})\leq 8(\left|{q}\right|+1)\int_{{\mathbb{D}}}{\gamma(z)\left|{R(tz)}\right|^{2}\log^{+}\left|{f(tz)}\right|}+B_{+}(t)+$

$\displaystyle+8(\left|{q}\right|+1)\int_{{\mathbb{D}}}{\gamma(z)\left|{R(s_{0}z)}\right|^{2}\log^{+}\left|{f(s_{0}z)}\right|}+B_{+}(s_{0})+$

$\displaystyle+4[2c(1,u)+2s\left|{q}\right|c(1/2,u)]P_{{\mathbb{T}},+}(t_{0})+$

$\displaystyle+4(1-u^{2})^{1/2}[2(1-u^{2})^{1/2}+2\left|{q}\right|]P_{{\mathbb{T}},-}(t_{0})-B_{-}(t)-B_{-}(s_{0}).$

We forget the negative term $\displaystyle-B_{-}(t):=-\int_{{\mathbb{T}}}{2\left|{R(tz)}\right|^{2}\log^{-}\left|{f}\right|}\leq 0$ and we recall that

$\displaystyle P_{{\mathbb{T}},-}(t_{0})=B_{-}(s_{0}):=\int_{{\mathbb{T}}}{\left|{R(s_{0}z)}\right|^{2}\log^{-}\left|{f}\right|}.$

Now choose a suitable $u<1$ such that

$\displaystyle 4(1-u^{2})^{1/2}[2(1-u^{2})^{1/2}+2\left|{q}\right|]\ -1\leq 0$

i.e. $\displaystyle(1-u^{2})^{1/2}\leq\frac{1}{8(\left|{q}\right|+1)},$ which is independent of $\displaystyle t_{0}.$ It remains

$\displaystyle\Sigma(t,s_{0})\leq 8(\left|{q}\right|+1)\int_{{\mathbb{D}}}{\gamma(z)\left|{R(tz)}\right|^{2}\log^{+}\left|{f(tz)}\right|}+B_{+}(t)+$

$\displaystyle+8(\left|{q}\right|+1)\int_{{\mathbb{D}}}{\gamma(z)\left|{R(s_{0}z)}\right|^{2}\log^{+}\left|{f(s_{0}z)}\right|}+B_{+}(s_{0})+$

$\displaystyle+4[2c(1,u)+2s\left|{q}\right|c(1/2,u)]P_{{\mathbb{T}},+}(t_{0}).$

Then, because $\displaystyle t\in[0,t_{0}],\ s_{0}\in[0,t_{0}],$ we get \displaystyle B_{+}(t)\leq P_{{\mathbb{T}},+}(t_{0})\ ;\ B_{+}(s_{0})\leq P_{{\mathbb{T}},+}(t_{0})\ ;\hence

$\displaystyle\Sigma(t,s_{0})\leq 16(\left|{q}\right|+1)P_{\gamma,+}(t_{0})+2P_{{\mathbb{T}},+}(t_{0})+4[2c(1,u)+2\left|{q}\right|c(1/2,u)]P_{{\mathbb{T}},+}(t_{0}).$

So finally

$\displaystyle\Sigma(t,s_{0})\leq 16(\left|{q}\right|+1)P_{\gamma,+}(t_{0})+2[1+2(2c(1,u)+2\left|{q}\right|c(1/2,u))]P_{{\mathbb{T}},+}(t_{0}).$

We get, taking $\displaystyle t=t_{0}<1$ and the suitable $u,$ independent of $\displaystyle t_{0},$

$\displaystyle\ \sum_{a\in Z(f_{t_{0}})}{g_{t_{0}}(a)}\leq\Sigma(t,s_{0})\leq 16(\left|{q}\right|+1)P_{\gamma,+}(t_{0})+2[1+2(2c(1,u)+2\left|{q}\right|c(1/2,u))]P_{{\mathbb{T}},+}(t_{0}).$

Setting

$\displaystyle c(R):=\max(16(\left|{q}\right|+1),\ 2[1+2(2c(1,u)+2\left|{q}\right|c(1/2,u))]),$

which is still independent of $\displaystyle t_{0},$ we get

$\displaystyle\forall t_{0}\in[0,1[,\ \ \sum_{a\in Z(f_{t_{0}})}{(1-\left|{a}\right|^{2})\left|{R(t_{0}a)}\right|^{2}}\leq c(R){\left\|{f}\right\|}_{{\mathcal{N}}_{\left|{R}\right|^{2},0}}$

hence using the second part of lemma 9.5 from the appendix, with $\varphi(z)=\gamma(z)\left|{R(z)}\right|^{2},\ \psi(z)=\left|{R(z)}\right|^{2},$ we get

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})\left|{R(a)}\right|^{2}}\leq c(R){\left\|{f}\right\|}_{{\mathcal{N}}_{\left|{R}\right|^{2},0}}.$ $\hfill\blacksquare$

Corollary 4.5

Let $\displaystyle R(z)=\prod_{j=1}^{n}{(z-\eta_{j})}^{q_{j}},\ q_{j}\in{\mathbb{R}}.$ Suppose $\displaystyle f\in{\mathcal{N}}_{\left|{R}\right|,0}({\mathbb{D}})$ with $\displaystyle\ \left|{f(0)}\right|=1,$ and set $\displaystyle\tilde{R}(z):=\prod_{j=1}^{n}{(z-\eta_{j})}^{(q_{j})_{+}},$ then there exists a constant $\displaystyle c(R)$ depending only on $R$ such that

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})\left|{\tilde{R}(a)}\right|}\leq c(R){\left\|{f}\right\|}_{{\mathcal{N}}_{\left|{R}\right|,0}}.$**

Proof.

We have to prove that $\displaystyle f\in{\mathcal{N}}_{\left|{R}\right|,0}\Rightarrow f\in{\mathcal{N}}_{\left|{\tilde{R}}\right|,0}.$ But if $\displaystyle q<0$ then:

$\displaystyle\ \left|{z-\eta}\right|\leq 2\Rightarrow\left|{z-\eta}\right|^{q}\geq 2^{q}\Rightarrow 1=\left|{z-\eta}\right|^{(q)_{+}}\leq 2^{-q}\left|{z-\eta}\right|^{q}.$

Putting it in the definition of $\displaystyle\ {\left\|{f}\right\|}_{{\mathcal{N}}_{\left|{R}\right|,0}}$ we are done. $\hfill\blacksquare$

5 Application : $\displaystyle L^{\infty}$ bounds.

We shall retrieve some of the results of Boritchev, Golinskii and Kupin [4], [5].

Suppose the function $f$ verifies $\displaystyle\ \left|{f(z)}\right|\leq\exp\frac{D}{\left|{R(z)}\right|}$ with $\displaystyle R(z):=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}}.$

We deduce that $\displaystyle\ \left|{R(z)}\right|\log\left|{f(z)}\right|$ is in $\displaystyle L^{1}({\mathbb{T}})$ with a better exponent of almost 1 over the rational function $R.$ Precisely set

$\displaystyle\forall\epsilon\geq 0,\ R_{\epsilon}(z):=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}-1+\epsilon}},$

we have:

Lemma 5.1

If the function $f$ verifies $\displaystyle\ \left|{f(z)}\right|\leq\exp\frac{D}{\left|{R(z)}\right|}$ with $\displaystyle R(z):=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}},$ we have

$\displaystyle\ \forall\epsilon>0,\ \int_{{\mathbb{T}}}{\left|{R_{\epsilon}(e^{i\theta})}\right|\log^{+}\left|{f(e^{i\theta})}\right|}\leq DC(\delta,\epsilon).$**

Proof.

The hypothesis gives $\displaystyle\ \left|{R(z)}\right|\log^{+}\left|{f(z)}\right|\leq D$ and

$\displaystyle\ \frac{R_{\epsilon}(z)}{R(z)}=\prod_{j=1}^{n}{\frac{(z-\eta_{j})^{q_{j}-1+\epsilon}}{(z-\eta_{j})^{q_{j}}}}=\prod_{j=1}^{n}{(z-\eta_{j})^{-1+\epsilon}},$

so

$\displaystyle\ \left|{R_{\epsilon}(z)}\right|\log^{+}\left|{f(z)}\right|\leq\frac{\left|{R_{\epsilon}(z)}\right|}{\left|{R(z)}\right|}D\leq D\prod_{j=1}^{n}{(z-\eta_{j})^{-1+\epsilon}}.$

Because the points $\displaystyle\{\eta_{k}\}$ are separated on the torus $\displaystyle{\mathbb{T}}$ by $\alpha>0$ say and $\displaystyle\ \left|{z-\eta_{j}}\right|^{-1+\epsilon}$ is integrable for the Lebesgue measure on the torus $\displaystyle{\mathbb{T}}$ because $\displaystyle\epsilon>0,$ we get:

$\displaystyle\ \int_{{\mathbb{T}}}{\frac{\left|{R_{\epsilon}(e^{i\theta})}\right|}{\left|{R(e^{i\theta})}\right|}\left|{R(e^{i\theta})}\right|\log^{+}\left|{f(e^{i\theta})}\right|}\leq D\int_{{\mathbb{T}}}{\prod_{j=1}^{n}{\left|{e^{i\theta}-\eta_{j}}\right|^{-1+\epsilon}}}\leq DC(\alpha,\epsilon).$ $\hfill\blacksquare$

Theorem 5.2

Suppose the holomorphic function $f$ verifies $\displaystyle\ \left|{f(0)}\right|=1$ and $\displaystyle\ \left|{f(z)}\right|\leq\exp\frac{D}{(1-\left|{z}\right|^{2})^{p}\left|{R(z)}\right|}$ with $\displaystyle R(z):=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}},\ q_{j}\in{\mathbb{R}}.$ For $\displaystyle p=0,$ we set $\displaystyle\tilde{R}_{\epsilon}(z):=\prod_{j=1}^{n}{(z-\eta_{j})}^{(q_{j}-1+\epsilon)_{+}}$ and we get:

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|)\left|{\tilde{R}_{\epsilon}(a)}\right|}\leq Dc(\epsilon,p,R).$

For $\displaystyle p>0,\ \forall j=1,...,n,$ if $\displaystyle q_{j}-1>-p/2$ set $\displaystyle\tilde{q}_{j}=q_{j}$ else choose $\displaystyle\tilde{q}_{j}>1-p/2,$ and set $\displaystyle\tilde{R}_{0}(z):=\prod_{j=1}^{n}{(z-\eta_{j})}^{\tilde{q}_{j}-1},$ then:

$\displaystyle\ \forall\epsilon>0,\ \sum_{a\in Z(f)}{(1-\left|{a}\right|)^{1+p+\epsilon}\left|{\tilde{R}_{0}(a)}\right|}\leq Dc(\epsilon,R).$**

Proof.

$\bullet$ Case $\displaystyle p=0.$

We shall apply the corollary 4.5 with $\displaystyle R_{\epsilon}$ instead of $\displaystyle R.$

To apply corollary 4.5 we have to show that

$\displaystyle\sup_{s<1}\int_{{\mathbb{D}}}{\left|{R_{\epsilon}(sz)}\right|s\left|{\sum_{j=1}^{n}{q_{j}(z-\eta_{j})^{-1}}}\right|\log^{+}\left|{f(sz)}\right|}<\infty$

and

$\displaystyle\sup_{s<1}\int_{{\mathbb{T}}}{\left|{R_{\epsilon}(se^{i\theta})}\right|\log^{+}\left|{f(se^{i\theta})}\right|}<\infty.$

The hypothesis gives $\displaystyle\ \left|{R(z)}\right|\log^{+}\left|{f(z)}\right|\leq D$ so we get

$\displaystyle\ \left|{R_{\epsilon}(sz)}\right|\log^{+}\left|{f(sz)}\right|\leq D\prod_{j=1}^{n}{\left|{1-s\bar{\eta}_{j}z}\right|^{-1+\epsilon}},$

because, as already seen, $\displaystyle\ \frac{R_{\epsilon}(sz)}{R(sz)}=\prod_{j=1}^{n}{(1-s\bar{\eta}_{j}z)^{-1+\epsilon}},$

so we get:

$\displaystyle\ \left|{R_{\epsilon}(sz)}\right|\sum_{k=1}^{n}{\left|{1-s\bar{\eta}_{k}z}\right|^{-1+\epsilon}}\log^{+}\left|{f(z)}\right|\leq 2D\left|{q}\right|\sum_{k=1}^{n}{\prod_{j\neq k}{(\left|{1-s\bar{\eta}_{j}z}\right|^{-1+\epsilon}})\left|{1-s\bar{\eta}_{k}z}\right|}^{-2+\epsilon}.$

Because the points $\displaystyle\{\eta_{k}\}$ are separated by an $\alpha>0$ and $\displaystyle\ \left|{1-\bar{\eta}_{j}z}\right|^{-2+\epsilon}$ is integrable for the Lebesgue measure on the disc $\displaystyle{\mathbb{D}}$ because $\displaystyle\epsilon>0,$ we get:

$\displaystyle\ \sup_{s<1}\int_{{\mathbb{D}}}{\left|{R_{\epsilon}(sz)}\right|s\sum_{j=1}^{n}{\left|{q_{j}}\right|\left|{1-s\bar{\eta}_{j}z}\right|^{-1}}\log^{+}\left|{f_{s}}\right|dm(z)}\leq 2D\left|{q}\right|c(\alpha,\epsilon).$

Now to apply corollary 4.5 we need also to compute

$\displaystyle\ \int_{{\mathbb{T}}}{\left|{R_{\epsilon}(se^{i\theta})}\right|\log^{+}\left|{f(se^{i\theta})}\right|}\leq\int_{{\mathbb{T}}}{\frac{\left|{R_{\epsilon}(se^{i\theta})}\right|}{\left|{R(e^{i\theta})}\right|}\left|{R(se^{i\theta})}\right|\log^{+}\left|{f(se^{i\theta})}\right|}\leq$

$\displaystyle\leq D\int_{{\mathbb{T}}}{\left|{\prod_{j=1}^{n}{(1-s\bar{\eta}_{j}e^{i\theta})^{-1+\epsilon}}}\right|}.$

Again the points $\displaystyle\{\eta_{k}\}$ are separated by $\alpha$ and $\displaystyle\ \left|{1-\bar{\eta}_{j}e^{i\theta}}\right|^{-1+\epsilon}$ is integrable for the Lebesgue measure on the torus $\displaystyle{\mathbb{T}}$ because $\displaystyle\epsilon>0.$ So we get:

$\displaystyle\ \sup_{s<1}\int_{{\mathbb{T}}}{\left|{R_{\epsilon}(se^{i\theta})}\right|\log^{+}\left|{f(se^{i\theta})}\right|}\leq c(\alpha,\epsilon),$

which ends the proof of the case $\displaystyle p=0.$

$\bullet$ Case $\displaystyle p>0.$

We shall show that $\displaystyle\forall\epsilon>0,\ f\in{\mathcal{N}}_{R_{0},p+\epsilon}({\mathbb{D}}).$ For this we have to prove:

$\displaystyle\ {\left\|{f}\right\|}_{R_{0},p+\epsilon}:=\sup_{s<1}(\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p+\epsilon-1}\left|{R_{0}(sz)}\right|\log^{+}\left|{f(sz)}\right|})<\infty.$

Because $\displaystyle\ \left|{f(sz)}\right|\leq\exp\frac{D}{(1-\left|{sz}\right|^{2})^{p}\left|{R(sz)}\right|}$ we get

$\displaystyle I(s,\epsilon):=\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p+\epsilon-1}\left|{R_{0}(sz)}\right|\log^{+}\left|{f(sz)}\right|})\leq\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p+\epsilon-1}\frac{\left|{R_{0}(sz)}\right|}{\left|{R(sz)}\right|}\left|{R(sz)}\right|\log^{+}\left|{f}\right|}\leq$

$\displaystyle\leq\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p+\epsilon-1}\frac{\left|{R_{0}(sz)}\right|}{\left|{R(sz)}\right|}\frac{D}{(1-\left|{sz}\right|^{2})^{p}}}.$

Now, as already seen, $\displaystyle\ \frac{R_{0}(sz)}{R(sz)}=\prod_{j=1}^{n}{(1-s\bar{\eta}_{j}z)^{-1}},$ so we get, because $\displaystyle\forall s\leq 1,\ (1-\left|{z}\right|^{2})\leq(1-\left|{sz}\right|^{2}),$

$\displaystyle I(s,\epsilon)\leq D\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{\epsilon-1}\prod_{j=1}^{n}{(1-s\bar{\eta}_{j}z)^{-1}}}.$

Now we apply lemma 9.4 with $\displaystyle p=\epsilon$ to get

$\displaystyle\sup_{s<1}\int_{{\mathbb{D}}}{(1-\left|{sz}\right|^{2})^{-1+\epsilon}\prod_{j=1}^{n}{(1-s\bar{\eta}_{j}z)^{-1}}}\leq c(\epsilon,\alpha).$

Hence

$\displaystyle\ {\left\|{f}\right\|}_{R_{0},p+\epsilon}\leq Dc(\epsilon,\delta)\Rightarrow f\in{\mathcal{N}}_{R_{0},p+\epsilon}({\mathbb{D}}).$

But then corollary 3.8 gives that

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|)^{1+p+\epsilon}\left|{\tilde{R}_{0}(a)}\right|}\leq C{\left\|{f}\right\|}_{R_{0},p+\epsilon}\leq CDc(\epsilon,\alpha),$

which ends the proof of the theorem. $\hfill\blacksquare$

6 Case of a closed set in $\displaystyle{\mathbb{T}}.$

Let $\displaystyle E=\bar{E}\subset{\mathbb{T}}$ be a closed set in $\displaystyle{\mathbb{T}}\ ;$ in [2], we associate to it a $\displaystyle{\mathcal{C}}^{\infty}({\mathbb{D}})$ function $\displaystyle h(z)$ (called $\varphi(z)$ in [2]) such that $\displaystyle h(z)\simeq d(z,E)$ and setting $\displaystyle g_{s}(z):=(1-\left|{z}\right|^{2})^{p+1}h(sz)^{q}\in{\mathcal{C}}^{\infty}(\bar{\mathbb{D}}),$ with $\displaystyle 0<s<1$ and $\displaystyle q>0,$ we proved there:

Theorem 6.1

We have:

$\displaystyle\ \int_{{\mathbb{D}}}{\triangle g_{s}(z)\log\left|{f(sz)}\right|}\lesssim\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}h(sz)^{q}\log^{+}\left|{fsz}\right|}.$**

This lead to the definition:

Definition 6.2

Let $\displaystyle E=\bar{E}\subset{\mathbb{T}}.$ We say that an holomorphic function $f$ is in the generalised Nevanlinna class $\displaystyle{\mathcal{N}}_{h^{q},p}({\mathbb{D}})$ for $\displaystyle p>0$ if $\displaystyle\exists\delta>0,\ \delta<1$ such that

$\displaystyle\ {\left\|{f}\right\|}_{{\mathcal{N}}_{h^{q},p}}:=\sup_{1-\delta<s<1}\int_{{\mathbb{D}}}{(1-\left|{z}\right|)^{p-1}h(sz)^{q}\log^{+}\left|{f(sz)}\right|}<\infty.$**

And we proved the Blaschke type condition:

Theorem 6.3

Let $\displaystyle E=\bar{E}\subset{\mathbb{T}}.$ Suppose $\displaystyle q>0$ and $\displaystyle f\in{\mathcal{N}}_{h^{q},p}({\mathbb{D}})$ with $\displaystyle\ \left|{f(0)}\right|=1,$ then

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})^{1+p}h(a)^{q}}\leq c{\left\|{f}\right\|}_{{\mathcal{N}}_{h^{q},p}}.$**

Corollary 6.4

Let $\displaystyle E=\bar{E}\subset{\mathbb{T}}.$ Suppose $\displaystyle q\in{\mathbb{R}}$ and $\displaystyle f\in{\mathcal{N}}_{d(\cdot,E)^{q},p}({\mathbb{D}})$ with $\displaystyle\ \left|{f(0)}\right|=1,$ then

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})^{1+p}d(a,E)^{q}}\leq c{\left\|{f}\right\|}_{{\mathcal{N}}_{d(\cdot,E)^{q},p}}.$**

7 The mixed case.

We shall combine the case of the rational function $\displaystyle R(z)=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}},\ q_{j}\in{\mathbb{R}}$ with the case of the closed set $\displaystyle E\subset{\mathbb{T}}$ treated in [2]. For this we shall consider $\varphi(z):=\left|{R(sz)}\right|^{2}h(sz)^{q}$ and $\displaystyle g_{s}(z):=(1-\left|{z}\right|^{2})^{1+p}\varphi(sz).$

We make the hypothesis that $\displaystyle\forall j=1,...,n,\ \eta_{j}\notin E.$ We set $2\mu:=\min_{j=1,...,n}d(\eta_{j},E)$ then we have that $\mu>0.$

Because

$\displaystyle\Delta g_{s}(z)=\Delta[(1-\left|{z}\right|^{2})^{p+1}]\varphi(sz)+(1-\left|{z}\right|^{2})^{p+1}\Delta[\varphi(sz)]+8\Re[\partial((1-\left|{z}\right|^{2})^{p+1})\bar{\partial}(\varphi(sz))],$

and

$\Delta[\varphi(sz)]=s^{2}h(sz)^{q}\Delta[\left|{R(sz)}\right|^{2}]h(sz)^{q}+s^{2}\left|{R(sz)}\right|^{2}\Delta[h(sz)^{q}]+8s^{2}\Re[\bar{\partial}\left|{R(sz)}\right|^{2}{\times}\partial(h(sz)^{q})],$

we are lead to set:

$\displaystyle A_{1}:=\frac{1}{2}\left|{R(sz)}\right|^{2}\Delta[(1-\left|{z}\right|^{2})^{p+1}]h(sz)^{q},\ A_{2}:=\frac{1}{2}h(sz)^{q}\Delta[(1-\left|{z}\right|^{2})^{p+1}]\left|{R(sz)}\right|^{2}$

so

$\displaystyle\Delta[(1-\left|{z}\right|^{2})^{p+1}]\varphi(sz)=A_{1}+A_{2}.$

And

$\displaystyle A_{3}:=(1-\left|{z}\right|^{2})^{p+1}s^{2}h(sz)^{q}\Delta[\left|{R(sz)}\right|^{2}]h(sz)^{q}$

$\displaystyle A_{4}:=s^{2}(1-\left|{z}\right|^{2})^{p+1}\left|{R(sz)}\right|^{2}\Delta[h(sz)^{q}]$

$\displaystyle A_{5}:=8s^{2}(1-\left|{z}\right|^{2})^{p+1}\Re[\bar{\partial}\left|{R(sz)}\right|^{2}{\times}\partial(h(sz)^{q})]$

$\displaystyle A_{6}:=8h(sz)^{q}\Re[\partial((1-\left|{z}\right|^{2})^{p+1})\bar{\partial}(\left|{R(sz)}\right|^{2})]$

$\displaystyle A_{7}:=8\left|{R(sz)}\right|^{2}\Re[\partial((1-\left|{z}\right|^{2})^{p+1})\bar{\partial}(h(sz)^{q})]\ ;$

and we get

$\displaystyle\Delta g_{s}(z)=A_{1}+A_{2}+A_{3}+A_{4}+A_{5}+A_{6}+A_{7}.$

It remains to see that grouping these terms in the right way, this was already treated by the $F$ case or by the $E$ one.

Theorem 7.1

We have, for $\displaystyle p>0:$

$\displaystyle\ \int_{{\mathbb{D}}}{\triangle g_{s}(z)\log\left|{f(sz)}\right|}\lesssim\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\left|{R(sz)}\right|^{2}h(sz)^{q}\log^{+}\left|{fsz}\right|}.$**

Proof.

We first group the terms

$\displaystyle B_{1}:=A_{1}\log\left|{f(sz)}\right|+A_{4}\log\left|{f(sz)}\right|+A_{7}\log\left|{f(sz)}\right|,$

these terms contain no derivatives of $\displaystyle\ \left|{R(sz)}\right|^{2}$ and so verify theorem 6.1 with $h^{q}$ replaced by $\displaystyle\ \left|{R(sz)}\right|^{2}h(sz)^{q}$ i.e.

$\displaystyle\ \int_{{\mathbb{D}}}{B_{1}(s,z)}\lesssim\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\ \left|{R(sz)}\right|^{2}h(sz)^{q}\log^{+}\left|{fsz}\right|}.$

Now we group the terms

$\displaystyle B_{2}:=A_{2}\log\left|{f(sz)}\right|+A_{3}\log\left|{f(sz)}\right|+A_{6}\log\left|{f(sz)}\right|,$

these terms contain no derivatives of $\displaystyle h(sz)$ and so verify also

$\displaystyle\ \int_{{\mathbb{D}}}{B_{2}(s,z)}\lesssim\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\ \left|{R(sz)}\right|^{2}h(sz)^{q}\log^{+}\left|{fsz}\right|}.$

It remains $\displaystyle A_{5}\log\left|{f(sz)}\right|$ but again the homogeneity is the right one and we get

$\displaystyle\ \int_{{\mathbb{D}}}{A_{5}(s,z)\log^{+}\left|{fsz}\right|}\lesssim\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\ \left|{R(sz)}\right|^{2}h(sz)^{q}\log^{+}\left|{fsz}\right|}.$

So it remains $\displaystyle A_{5}\log^{-}\left|{f(sz)}\right|,$ and, in order to separate the points, we consider:

$\displaystyle\forall j=1,...,n,\ G_{j}:=\{z\in\bar{\mathbb{D}}::\left|{\frac{z}{\left|{z}\right|}-\eta_{j}}\right|<\delta\}\ ;\ G:=\bigcup_{j=1}^{n}{G_{j}}.$

Then we need:

Lemma 7.2

There are two constants $\displaystyle a(\mu),\ b(\mu),$ just depending on $\mu,$ such that:

$\displaystyle\forall z\in G,\ \partial h(sz)\simeq a(\mu).$

And

$\displaystyle\forall z\notin G,\bar{\partial}\ \left|{R(sz)}\right|^{2}\simeq b(\mu).$**

Proof.

Recall that we have $\displaystyle{\mathbb{T}}\backslash E=\bigcup_{j\in{\mathbb{N}}}{(\alpha_{j},\beta_{j})}$ where the $\displaystyle F_{j}:=(\alpha_{j},\beta_{j})$ are the contiguous intervals to $\displaystyle E$ and $\displaystyle\Gamma_{j}:=\{z=re^{i\psi}\in{\mathbb{D}}::\psi\in(\alpha_{j},\beta_{j})\}.$ We set:

$\displaystyle\forall z\in\Gamma_{j},\ h(z):=\eta_{j}(z)\psi_{j}(z)^{q}+(1-\left|{z}\right|^{2})^{2q},\ \forall z\in\Gamma_{E},\ h_{E}(z):=(1-\left|{z}\right|^{2})^{2q}$

with $\displaystyle\chi\in{\mathcal{C}}^{\infty}({\mathbb{R}}),\ t\leq 2\Rightarrow\chi(t)=0,\ t\geq 3\Rightarrow\chi(t)=1$ and

$\displaystyle\forall z\in\Gamma_{j},\ \psi_{j}(z):=\frac{\left|{z-\alpha_{j}}\right|^{2}\left|{z-\beta_{j}}\right|^{2}}{\delta_{j}^{2}},\ \eta_{j}(z):=\chi(\frac{\left|{z-\alpha_{j}}\right|^{2}}{(1-\left|{z}\right|^{2})^{2}})\chi(\frac{\left|{z-\beta_{j}}\right|^{2}}{(1-\left|{z}\right|^{2})^{2}}).$

An easy computation using the first lemma in the appendix of [2] gives $\displaystyle\forall z\in G,\ \partial h(sz)\simeq a(\mu)$ because $z$ is far from $\displaystyle E.$

And with $\displaystyle R(z)=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}},$ again an easy computation gives $\displaystyle\forall z\notin G,\bar{\partial}\ \left|{R(sz)}\right|^{2}\simeq b(\mu)$ because $z$ is far from $\displaystyle\ \bigcup_{j=1}^{n}{\{\eta_{j}\}}.$ $\hfill\blacksquare$

We can treat the $\displaystyle A_{5}\log^{-}\left|{f(sz)}\right|$ term easily now ; recall

$\displaystyle A_{5}\log^{-}\left|{f(sz)}\right|:=8s^{2}(1-\left|{z}\right|^{2})^{p+1}\Re[\bar{\partial}\left|{R(sz)}\right|^{2}{\times}\partial(h(sz)^{q})]\log^{-}\left|{f(sz)}\right|\ ;$

cut the disc $\displaystyle{\mathbb{D}}=G\cup({\mathbb{D}}\backslash G),$ so

$\displaystyle\ \int_{{\mathbb{D}}}{A_{5}\log^{-}\left|{f(sz)}\right|}=\int_{G}{A_{5}\log^{-}\left|{f(sz)}\right|}+\int_{{\mathbb{D}}\backslash G}{A_{5}\log^{-}\left|{f(sz)}\right|}.$

On $G$ we have, by lemma 7.2, $\displaystyle\partial h(sz)\simeq a(\mu)$ and we win a $\displaystyle(1-\left|{z}\right|^{2})$ so we can apply the substitution lemma 9.1 to get

$\displaystyle\ \int_{G}{A_{5}\log^{-}\left|{f(sz)}\right|}\leq c_{5}P_{{\mathbb{D}},+}(s).$

On $\displaystyle{\mathbb{D}}\backslash G$ we have, by lemma 7.2, $\displaystyle\bar{\partial}\ \left|{R(sz)}\right|^{2}\simeq b(\mu)$ and we win again a $\displaystyle(1-\left|{z}\right|^{2})$ so we can apply the substitution lemma 9.1 to get

$\displaystyle\ \int_{{\mathbb{D}}\backslash G}{A_{5}\log^{-}\left|{f(sz)}\right|}\leq c^{\prime}_{5}P_{{\mathbb{D}},+}(s),$

so finally we get

$\displaystyle\ \int_{{\mathbb{D}}}{A_{-}(s,z)}\leq c_{6}P_{{\mathbb{D}},+}(s),$

which ends the proof of the theorem. $\hfill\blacksquare$

So we are lead to

Definition 7.3

Let $\displaystyle E=\bar{E}\subset{\mathbb{T}}$ and $\displaystyle R(z)=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}},\ q_{j}\in{\mathbb{R}}$ with $\displaystyle\forall j=1,...,n,\ \eta_{j}\notin E.$ Set $\displaystyle\varphi(z)=\left|{R(z)}\right|^{2}h(z)^{q}.$ We say that an holomorphic function $f$ is in the generalised Nevanlinna class $\displaystyle{\mathcal{N}}_{\varphi,p}({\mathbb{D}})$ if $\displaystyle\exists\delta>0,\ \delta<1$ such that

$\displaystyle\ {\left\|{f}\right\|}_{{\mathcal{N}}_{\varphi,p}}:=\sup_{1-\delta<s<1}\int_{{\mathbb{D}}}{(1-\left|{z}\right|)^{p-1}\varphi(sz)\log^{+}\left|{f(sz)}\right|}.$**

And we have the Blaschke type condition, still using lemma 9.5 from the appendix, with $\displaystyle\varphi(z)=\left|{R(z)}\right|^{2}h(z)^{q}:$

Theorem 7.4

Let $\displaystyle E=\bar{E}\subset{\mathbb{T}}$ and $\displaystyle R(z)=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}},\ q_{j}\in{\mathbb{R}},\ q_{j}>p/4,$ with $\displaystyle\forall j=1,...,n,\ \eta_{j}\notin E.$ Suppose $\displaystyle q>0$ and $\displaystyle f\in{\mathcal{N}}_{\varphi,p}({\mathbb{D}})$ with $\displaystyle\ \left|{f(0)}\right|=1,$ then

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})^{1+p}\varphi(a)\left|{R(a)}\right|^{2}}\leq c{\left\|{f}\right\|}_{{\mathcal{N}}_{\varphi,p}}.$**

As for the case of the rational function $R$ only, we get the

Corollary 7.5

Let $\displaystyle E=\bar{E}\subset{\mathbb{T}}$ and $\displaystyle R(z)=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}},\ q_{j}\in{\mathbb{R}},$ with $\displaystyle\forall j=1,...,n,\ \eta_{j}\notin E.$ Let $\displaystyle\forall j=1,...,n,$ if $\displaystyle q_{j}>-p/2,\ \tilde{q}_{j}=q_{j}$ else choose $\displaystyle\tilde{q}_{j}>-p/2$ and set $\displaystyle\tilde{R}(z):=\prod_{j=1}^{n}{(z-\eta_{j})}^{\tilde{q}_{j}},$ and $\displaystyle\varphi(z)=\left|{R(z)}\right|h(z)^{q},\ \tilde{\varphi}(z)=\left|{\tilde{R}(z)}\right|h(z)^{q}.$ Suppose $\displaystyle q>0$ and $\displaystyle f\in{\mathcal{N}}_{\varphi,p}({\mathbb{D}})$ with $\displaystyle\ \left|{f(0)}\right|=1,$ then

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})^{1+p}\tilde{\varphi}(a)}\leq c(\varphi){\left\|{f}\right\|}_{{\mathcal{N}}_{\varphi,p}}.$

Corollary 7.6

Let $\displaystyle E=\bar{E}\subset{\mathbb{T}}$ and $\displaystyle R(z)=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}},\ q_{j}\in{\mathbb{R}},$ with $\displaystyle\forall j=1,...,n,\ \eta_{j}\notin E.$ Let $\displaystyle\forall j=1,...,n,$ if $\displaystyle q_{j}>-p/2,\ \tilde{q}_{j}=q_{j}$ else choose $\displaystyle\tilde{q}_{j}>-p/2$ and set $\displaystyle\tilde{R}(z):=\prod_{j=1}^{n}{(z-\eta_{j})}^{\tilde{q}_{j}},$ and $\displaystyle\varphi(z)=\left|{R(z)}\right|d(z,E)^{q},\ \tilde{\varphi}(z)=\left|{\tilde{R}(z)}\right|d(z,E)^{(q)_{+}}.$ Suppose $\displaystyle f\in{\mathcal{N}}_{\varphi,p}({\mathbb{D}})$ with $\displaystyle\ \left|{f(0)}\right|=1,$ then

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})^{1+p}\tilde{\varphi}(a)}\leq c(\varphi){\left\|{f}\right\|}_{{\mathcal{N}}_{\varphi,p}}.$**

Proof.

Still using that $\displaystyle h(z)\simeq d(z,E)$ and copying the proof of corollary 4.5 we are done. $\hfill\blacksquare$

We proceed exactly the same way for the case $\displaystyle p=0$ to set, with $\displaystyle\gamma(z):=\sum_{j=1}^{n}{\left|{q_{j}}\right|\left|{z-\eta_{j}}\right|^{-1}}:$

Definition 7.7

Let $\displaystyle E=\bar{E}\subset{\mathbb{T}}$ and $\displaystyle R(z)=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}},\ q_{j}\in{\mathbb{R}}$ with $\displaystyle\forall j=1,...,n,\ \eta_{j}\notin E.$ Set $\displaystyle\varphi(z)=\left|{R(z)}\right|^{2}h(z)^{q}.$ We say that an holomorphic function $f$ is in the generalised Nevanlinna class $\displaystyle{\mathcal{N}}_{\varphi,0}({\mathbb{D}})$ if $\displaystyle\exists\delta>0,\ \delta<1$ such that

$\displaystyle\ {\left\|{f}\right\|}_{{\mathcal{N}}_{\varphi,0}}:=\sup_{1-\delta<s<1}\int_{{\mathbb{T}}}{\varphi(se^{i\theta})\log^{+}\left|{f(se^{i\theta})}\right|}+\sup_{1-\delta<s<1}\int_{{\mathbb{D}}}{\varphi(z)\gamma(z)h(z)^{-1}\log^{+}\left|{f(z)}\right|}.$**

And we have the Blaschke type condition, still using lemma 9.5 from the appendix,

Theorem 7.8

Let $\displaystyle E=\bar{E}\subset{\mathbb{T}}$ and $\displaystyle\varphi$ as above. Suppose $\displaystyle q>0$ and $\displaystyle f\in{\mathcal{N}}_{\varphi,0}({\mathbb{D}})$ with $\displaystyle\ \left|{f(0)}\right|=1,$ then

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})\varphi(a)}\leq c{\left\|{f}\right\|}_{{\mathcal{N}}_{\varphi,0}}.$**

Corollary 7.9

Let $\displaystyle E=\bar{E}\subset{\mathbb{T}}$ and $\displaystyle R(z)=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}},\ q_{j}\in{\mathbb{R}},$ with $\displaystyle\forall j=1,...,n,\ \eta_{j}\notin E.$ Suppose $\varphi(z):=\left|{R(z)}\right|d(z,E)^{q}$ and $\displaystyle f\in{\mathcal{N}}_{\varphi,0}({\mathbb{D}})$ with $\displaystyle\ \left|{f(0)}\right|=1,$ and set $\displaystyle\tilde{R}(z):=\prod_{j=1}^{n}{(z-\eta_{j})}^{(q_{j})_{+}},$ then

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})d(a,E)^{(q)_{+}}\left|{\tilde{R}(a)}\right|^{2}}\leq c{\left\|{f}\right\|}_{{\mathcal{N}}_{\varphi,0}}.$**

Proof.

Again using that $\displaystyle h(z)\simeq d(z,E)$ and copying the proof of corollary 4.5 we are done. $\hfill\blacksquare$

8 Mixed cases with $\displaystyle L^{\infty}$ bounds.

As in section 7 we can mixed the two previous cases and we get, by a straightforward adaptation of the previous proofs,

Theorem 8.1

Suppose that $\displaystyle f\in{\mathcal{H}}({\mathbb{D}}),\ \left|{f(0)}\right|=1$ and

$\displaystyle\forall z\in{\mathbb{D}},\ \log^{+}\left|{f(z)}\right|\leq\frac{K}{(1-\left|{z}\right|^{2})^{p}}\frac{1}{\left|{R(z)}\right|d(z,E)^{q}},$

with $\displaystyle p>0,$ and $\displaystyle R(z):=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}},\ q_{j}\in{\mathbb{R}},$ if $\displaystyle q_{j}-1>-p/2$ set $\displaystyle\tilde{q}_{j}=q_{j}$ else choose $\displaystyle\tilde{q}_{j}>1-p/2,$ and set $\displaystyle\tilde{R}_{0}(z):=\prod_{j=1}^{n}{(z-\eta_{j})}^{\tilde{q}_{j}-1},$ then we have, with $\displaystyle\epsilon>0,$

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})^{1+p+\epsilon}\left|{\tilde{R}_{0}(a)}\right|d(a,E)^{(q-\alpha(E)+\epsilon)_{+}}}\leq c(p,q,R,E,\epsilon)K.$**

And

Theorem 8.2

Suppose that $\displaystyle f\in{\mathcal{H}}({\mathbb{D}}),\ \left|{f(0)}\right|=1$ and

$\displaystyle\forall z\in{\mathbb{D}},\ \log^{+}\left|{f(z)}\right|\leq K\frac{1}{\left|{R(z)}\right|d(z,E)^{q}},$

with $\displaystyle p=0,$ and $\displaystyle R(z):=\prod_{j=1}^{n}{(z-\eta_{j})^{q_{j}}},\ q_{j}\in{\mathbb{R}},$ set $\displaystyle\tilde{R}_{\epsilon}(z):=\prod_{j=1}^{n}{(z-\eta_{j})}^{(q_{j}-1+\epsilon)_{+}}$

then, with $\displaystyle\epsilon>0,$

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})\left|{\tilde{R}_{\epsilon}(a)}\right|d(a,E)^{(q-\alpha(E)+\epsilon)_{+}}}\leq c(q,R,E,\epsilon)K.$**

9 Appendix.

Lemma 9.1

(Substitution)  Suppose $\delta>0,\ 0<u<1$ and $\displaystyle\ \left|{f(0)}\right|=1,$ then

$\displaystyle\ \int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1+\delta}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}\leq(1-u^{2})^{\delta}\frac{1}{u^{2}}P_{{\mathbb{D}},-}(s)+c(\delta,u)P_{{\mathbb{D}},+}(s),$

with $\displaystyle c(\delta,u):=2{\times}4^{\left|{q}\right|}(1-u)^{\delta-\alpha-\beta},\ \alpha:=-2\max_{j=1,...,n}(0,-q_{j}),\ \beta:=2\max_{j=1,...,n}(q_{j}),$

and $\displaystyle P_{{\mathbb{D}},-}(s):=\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\left|{z}\right|^{2}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|},\ P_{{\mathbb{D}},+}(s):=\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\left|{R(sz)}\right|^{2}\log^{+}\left|{f(sz)}\right|}.$

We also have:

$\displaystyle\forall s\leq t_{0},\ \int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{\delta-1}\left|{R(s\rho e^{i\theta})}\right|^{2}\log^{-}\left|{f(sz)}\right|}\leq c(\delta,u)P_{{\mathbb{T}},+}(t_{0})+\frac{1}{2\delta}(1-u^{2})^{\delta}P_{{\mathbb{T}},-}(t_{0}),$

with

$\displaystyle P_{{\mathbb{T}},+}(t_{0}):=\sup_{0\leq s\leq t_{0}}\int_{{\mathbb{T}}}{\left|{R(se^{i\theta})}\right|^{2}\log^{+}\left|{f(se^{i\theta})}\right|d\theta}$

and

$\displaystyle P_{{\mathbb{T}},-}(t_{0}):=\sup_{0\leq s\leq t_{0}}\int_{{\mathbb{T}}}{\left|{R(se^{i\theta})}\right|^{2}\log^{-}\left|{f(se^{i\theta})}\right|d\theta}.$**

Proof.

Because this lemma is a key one for us, we shall give a detailed proof of it. We have

$\displaystyle A:=\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1+\delta}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}=\int_{D(0,u)}{(1-\left|{z}\right|^{2})^{p-1+\delta}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(z)}\right|}+$

$\displaystyle\ +\int_{{\mathbb{D}}\backslash D(0,u)}{(1-\left|{z}\right|^{2})^{p-1+\delta}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(z)}\right|}=:B+C.$

Clearly for the second term we have

$\displaystyle\ C:=\int_{{\mathbb{D}}\backslash D(0,u)}{(1-\left|{z}\right|^{2})^{p-1+\delta}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}\leq$

$\displaystyle(1-u^{2})^{\delta}\frac{1}{u^{2}}\int_{{\mathbb{D}}\backslash D(0,u)}{(1-\left|{z}\right|^{2})^{p-1}\left|{z}\right|^{2}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}.$

For the first one, we have

$\displaystyle B:=\int_{D(0,u)}{(1-\left|{z}\right|^{2})^{p-1+\delta}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}$

and, changing to polar coordinates,

$\displaystyle\ B=\int_{0}^{u}{(1-\rho^{2})^{p-1+\delta}\{\int_{{\mathbb{T}}}{\left|{R(s\rho e^{i\theta})}\right|^{2}\log^{-}\left|{f(s\rho e^{i\theta})}\right|d\theta}\}\rho d\rho}.$

We set

$\displaystyle M(\rho):=\sup_{\theta\in{\mathbb{T}}}\left|{R(\rho e^{i\theta})}\right|^{2}\leq 4^{\left|{q}\right|}(1-\rho)^{-2\max_{j=1,...,n}(0,-q_{j})},$

because we have $\displaystyle\ \left|{z-\eta_{j}}\right|\leq 2$ and $\displaystyle\ \left|{\rho e^{i\theta}-\eta_{j}}\right|\geq(1-\rho).$

So we get

$\displaystyle\ C(s\rho):=\int_{{\mathbb{T}}}{\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}\leq M(s\rho)\int_{{\mathbb{T}}}{\log^{-}\left|{f(s\rho e^{i\theta})}\right|}.$

Because $\displaystyle\log\left|{f(z)}\right|$ is subharmonic, we get

$\displaystyle 0=\log\left|{f(0)}\right|\leq\int_{{\mathbb{T}}}{\log\left|{f(s\rho e^{i\theta})}\right|}=\int_{{\mathbb{T}}}{\log^{+}\left|{f(s\rho e^{i\theta})}\right|}-\int_{{\mathbb{T}}}{\log^{-}\left|{f(s\rho e^{i\theta})}\right|}.$

So we have

[TABLE]

Now we set $\displaystyle m(\rho):=\inf_{\theta\in{\mathbb{T}}}\left|{R(\rho e^{i\theta})}\right|^{2}$ and the same way as for $\displaystyle M(\rho),$ we get $\displaystyle m(\rho)\geq(1-\rho)^{2\max_{j=1,...,n}(q_{j})}.$

Putting it in (9.4), we get

[TABLE]

We notice that $\displaystyle\sup_{s<1}\sup_{\rho<u}\frac{M(s\rho)}{m(s\rho)}=\sup_{\rho<u}\frac{M(\rho)}{m(\rho)}$ hence, setting

$\displaystyle c(\delta,u):=\sup_{s<1}\sup_{\rho<u}\frac{M(s\rho)}{m(s\rho)}(1-\rho^{2})^{\delta},$

we get

$\displaystyle c(\delta,u)\leq 2{\times}4^{\left|{q}\right|}(1-u)^{\delta-\alpha-\beta},$

with

$\displaystyle\alpha:=-2\max_{j=1,...,n}(0,-q_{j}),\ \beta:=2\max_{j=1,...,n}(q_{j}).$

Now we have

[TABLE]

hence $\displaystyle B\leq c(\delta,u)P_{{\mathbb{D}},+}(s).$

Adding $B$ and $C$ gives the first part of the lemma.

For the second one, from the definition of $C$ with $\displaystyle p=0,$

$\displaystyle C:=\int_{{\mathbb{D}}\backslash D(0,u)}{(1-\left|{z}\right|^{2})^{-1+\delta}\left|{R(sz)}\right|^{2}\log^{-}\left|{f(sz)}\right|}$

we get passing in polar coordinates and with $\displaystyle 0\leq s\leq t_{0}<1,$

$\displaystyle C=\int_{u}^{1}{(1-\rho^{2})^{\delta-1}\int_{{\mathbb{T}}}{\left|{R(s\rho e^{i\theta})}\right|^{2}\log^{-}\left|{f(s\rho e^{i\theta})}\right|d\theta}\rho d\rho}$

$\displaystyle\leq P_{{\mathbb{T}},-}(t_{0})\int_{u}^{1}{(1-\rho^{2})^{\delta-1}\rho d\rho}\leq\frac{1}{2\delta}(1-u^{2})^{\delta}P_{{\mathbb{T}},-}(t_{0}).$

Now from (9.5) and (9.6) we get

$\displaystyle B\leq P_{{\mathbb{T}},+}(t_{0})c(\delta,u)\int_{0}^{u}{(1-\rho^{2})^{\delta-1}\rho d\rho}\leq P_{{\mathbb{T}},+}(t_{0})c(\delta,u).$

Adding $C$ with $B$ we get the second part of the lemma. $\hfill\blacksquare$

Lemma 9.2

Let $\displaystyle\eta\in{\mathbb{T}},$ then we have $\displaystyle\Re(\bar{z}(z-\eta))\leq 0$ iff $\displaystyle z\in{\mathbb{D}}\cap D(\frac{\eta}{2},\ \frac{1}{2}).$

Proof.

We set $\displaystyle z=\eta t,$ then we have

$\displaystyle\bar{z}(z-\eta)=\bar{\eta}\bar{t}(\eta t-\eta)=\bar{t}(t-1).$

Hence

$\displaystyle\Re(\bar{z}(z-\eta))=\Re(\bar{t}(t-1))=\Re(r^{2}-re^{i\theta})=r^{2}-r\cos\theta.$

Hence with $\displaystyle t=x+iy=re^{i\theta,\ }x=r\cos\theta,\ y=r\sin\theta,$ we get

$\displaystyle\Re(\bar{t}(t-1))\leq 0\iff x^{2}+y^{2}-x\leq 0$

which means $\displaystyle(x,y)\in D(\frac{1}{2},\ \frac{1}{2})$ hence $\displaystyle z\in{\mathbb{D}}\cap D(\frac{\eta}{2},\ \frac{1}{2}).$ $\hfill\blacksquare$

Lemma 9.3

Let $\varphi$ be a continuous function in the unit disc $\displaystyle{\mathbb{D}}.$ We have that:

$\displaystyle s\leq t\in]0,1[\rightarrow\gamma(s):=\int_{{\mathbb{T}}}{\varphi(se^{i\theta})\log^{-}\left|{f(se^{i\theta})}\right|d\theta}$

is a continuous function of $\displaystyle s\in[0,t].$

Proof.

Because $\displaystyle s\leq t<1,$ the holomorphic function in the unit disc $\displaystyle f(se^{i\theta})$ has only a finite number of zeroes say $\displaystyle N(t).$ As usual we can factor out the zeros of $f$ to get

$\displaystyle f(z)=\prod_{j=1}^{N}{(z-a_{j})}g(z)$

where $\displaystyle g(z)$ has no zeros in the disc $\displaystyle\bar{D}(0,t).$ Hence we get

$\displaystyle\log\left|{f(z)}\right|=\sum_{j=1}^{N}{\log\left|{z-a_{j}}\right|}+\log\left|{g(z)}\right|.$

Let $\displaystyle a_{j}=r_{j}e^{\alpha_{j}},\ r_{j}>0$ because $\displaystyle\ \left|{f(0)}\right|=1,$ then it suffices to show that

$\displaystyle\gamma(s):=\int_{{\mathbb{T}}}{\varphi(se^{i\theta})\log^{-}\left|{se^{i\theta}-re^{i\alpha}}\right|d\theta}$

is continuous in $s$ near $\displaystyle s=r,$ because $\displaystyle\ \int_{{\mathbb{T}}}{\varphi(se^{i\theta})\log^{-}\left|{g(se^{i\theta})}\right|d\theta}$ is clearly continuous.

To see that $\gamma(s)$ is continuous at $\displaystyle s=r,$ it suffices to show

$\displaystyle\gamma(s_{n})\rightarrow\gamma(r)$ when $\displaystyle s_{n}\rightarrow r.$

But

$\displaystyle\forall\theta\neq 0,\ \varphi(se^{i\theta})\log\left|{se^{i\theta}-r}\right|\rightarrow\varphi(re^{i\theta})\log\left|{re^{i\theta}-r}\right|$

and $\displaystyle\log\frac{1}{\left|{se^{i\theta}-r}\right|}\leq c_{\epsilon}\left|{se^{i\theta}-r}\right|^{-\epsilon}$ with $\displaystyle\epsilon>0.$ So choosing $\displaystyle\epsilon<1,$ we get that $\displaystyle\log\frac{1}{\left|{se^{i\theta}-r}\right|}\in L^{1}({\mathbb{T}})$ uniformly in $s.$ Because $\displaystyle\varphi(se^{i\theta})$ is continuous uniformly in $\displaystyle s\in[0,t]$ we get also $\displaystyle\varphi(se^{i\theta})\log\frac{1}{\left|{se^{i\theta}-r}\right|}\in L^{1}({\mathbb{T}})$ uniformly in $s.$ So we can apply the dominated convergence theorem of Lebesgue to get the result. $\hfill\blacksquare$

Lemma 9.4

The function $\displaystyle(1-\left|{z}\right|^{2})^{p-1}\prod_{j=1}^{n}{\left|{z-\eta_{k}}\right|^{-1}},$ with $\displaystyle p>0,$ is integrable for the Lebesgue measure in the disc $\displaystyle{\mathbb{D}}$ and we have the estimate

$\displaystyle\ \int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\prod_{j=1}^{n}{\left|{z-\eta_{k}}\right|^{-1}}}\leq c(p,\alpha)<\infty,$

where the constant $\displaystyle\alpha$ is twice the length of the minimal arc between the points $\displaystyle\{\eta_{j}\}_{j=1,...,n}\subset{\mathbb{T}}.$

Proof.

Because the points $\displaystyle\eta_{k}$ are separated on the torus $\displaystyle{\mathbb{T}}$ we can assume that we have disjoint sectors $\displaystyle\Gamma_{j}$ based on the arcs $\displaystyle\{\eta_{j}-\alpha,\eta_{j}+\alpha\}_{j=1,...,n}\subset{\mathbb{T}}$ for a $\alpha>0.$ Let $\displaystyle\Gamma_{0}:={\mathbb{D}}\backslash\bigcup_{j=1}^{n}{\Gamma_{j}.}$ We have

$\displaystyle A:=\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\prod_{j=1}^{n}{\left|{z-\eta_{k}}\right|^{-1}}dm(z)}=\sum_{j=0}^{n}{\int_{\Gamma_{j}}{(1-\left|{z}\right|^{2})^{p-1}\prod_{k=1}^{n}{\left|{z-\eta_{k}}\right|^{-1}}dm(z)}}.$

We set

$\displaystyle A_{0}:=\int_{\Gamma_{0}}{(1-\left|{z}\right|^{2})^{p-1}\prod_{k=1}^{n}{\left|{z-\eta_{k}}\right|^{-1}}dm(z)},$

and we get

$\displaystyle\forall z\in\Gamma_{0},\ \forall k=1,...,n,\ \left|{z-\eta_{k}}\right|\geq\alpha\Rightarrow\prod_{k=1}^{n}{\left|{z-\eta_{k}}\right|^{-1}}\leq\alpha^{-n}.$

So

$\displaystyle A_{0}\leq\alpha^{-n}\int_{\Gamma_{0}}{(1-\left|{z}\right|^{2})^{p-1}dm(z)}\leq\alpha^{-n}\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}dm(z)}\leq 2\pi\alpha^{-n}.$

For computing $\displaystyle A_{j}$ we can assume that $\displaystyle\eta_{j}=1$ by rotation and $\displaystyle\Gamma_{j}$ based on the arc $\displaystyle(-\alpha,\ \alpha)\ ;$ so we have, because $\displaystyle\ \prod_{k=1}^{n}{\left|{z-\eta_{k}}\right|^{-1}}\leq\alpha^{-(n-1)}\left|{1-z}\right|,$

$\displaystyle A_{j}:=\int_{\Gamma_{j}}{(1-\left|{z}\right|^{2})^{p-1}\prod_{k=1}^{n}{\left|{z-\eta_{k}}\right|^{-1}}dm(z)}\leq\alpha^{-(n-1)}\int_{\Gamma_{j}}{(1-\left|{z}\right|^{2})^{p-1}\left|{1-z}\right|^{-1}dm(z)}.$

Set $\displaystyle\beta:=\frac{p}{2}>0,$ then we have $\displaystyle(1-\left|{z}\right|^{2})^{\beta}<2^{\beta}\left|{1-z}\right|^{\beta}$ hence

$\displaystyle A_{j}\leq\alpha^{-(n-1)}2^{\beta}\int_{\Gamma_{j}}{(1-\left|{z}\right|^{2})^{\beta-1}\left|{1-z}\right|^{\beta-1}dm(z)}.$

Changing to polar coordinates, we get

$\displaystyle A_{j}\leq\alpha^{-(n-1)}2^{\beta}\int_{0}^{1}{(1-\rho^{2})^{\beta-1}\rho\{\int_{-\delta}^{\delta}{\left|{1-\rho e^{i\theta}}\right|^{\beta-1}d\theta}\}d\rho}.$

Because $\displaystyle\beta>0,$ we get

$\displaystyle\forall\rho\leq 1,\ \int_{-\alpha}^{\alpha}{\left|{1-\rho e^{i\theta}}\right|^{\beta-1}d\theta}\leq c(\alpha,\beta)$

and

$\displaystyle\ \int_{0}^{1}{(1-\rho^{2})^{\beta-1}\rho d\rho}\leq c(\beta).$

So adding the $\displaystyle A_{j},$ we end the proof of the lemma. $\hfill\blacksquare$

Lemma 9.5

Let $\varphi(z)$ be a positive function in $\displaystyle{\mathbb{D}}$ and $\displaystyle f\in{\mathcal{H}}({\mathbb{D}})\ ;$ set $\displaystyle f_{s}(z):=f(sz)$ and suppose that:

$\displaystyle\forall s<1,\ \sum_{a\in Z(f_{s})}{(1-\left|{a}\right|^{2})^{p+1}\varphi(sa)}\leq\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\varphi(sz)\log^{+}\left|{f(sz)}\right|},$

then, for any $\displaystyle 1>\delta>0$ we have

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})^{p+1}\varphi(a)}\leq\sup_{1-\delta<s<1}\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\varphi(sz)\log^{+}\left|{f(sz)}\right|}.$

We have also:

let $\varphi(z),\ \psi(z)$ be positive continuous functions in $\displaystyle{\mathbb{D}}$ and $\displaystyle f\in{\mathcal{H}}({\mathbb{D}})$ such that:

$\displaystyle\forall s<1,\ \sum_{a\in Z(f)\cap D(0,s)}{(1-\left|{a}\right|^{2})\varphi(sa)}\leq\int_{{\mathbb{D}}}{\varphi(sz)\log^{+}\left|{f(sz)}\right|}+\int_{{\mathbb{T}}}{\psi(se^{i\theta})\log^{+}\left|{f(se^{i\theta})}\right|}$

then, for any $\displaystyle 1>\delta>0$ we have

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})\varphi(a)}\leq\sup_{1-\delta<s<1}\int_{{\mathbb{D}}}{\varphi(sz)\log^{+}\left|{f(sz)}\right|}+\sup_{1-\delta<s<1}\int_{{\mathbb{T}}}{\psi(se^{i\theta})\log^{+}\left|{f(sz)}\right|}.$**

Proof.

We have $\displaystyle a\in Z(f_{s})\iff f(sa)=0,$ i.e. $\displaystyle b:=sa\in Z(f)\cap D(0,s).$ Hence the hypothesis is

$\displaystyle\forall s<1,\ \sum_{a\in Z(f)\cap D(0,s)}{(1-\left|{\frac{a}{s}}\right|^{2})^{p+1}\varphi(a)}\leq\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\varphi(sz)\log^{+}\left|{f(sz)}\right|}.$

We fix $\displaystyle 1-\delta<r<1,\ r<s<1,$ then, because $\displaystyle Z(f)\cap D(0,r)\subset Z(f)\cap D(0,s)$ and $\varphi\geq 0,$ we have

$\displaystyle\ \sum_{a\in Z(f)\cap D(0,r)}{(1-\left|{\frac{a}{s}}\right|^{2})^{p+1}\varphi(a)}\leq\sum_{a\in Z(f)\cap D(0,s)}{(1-\left|{\frac{a}{s}}\right|^{2})^{p+1}\varphi(a)}\leq$

$\displaystyle\leq\sup_{1-\delta<s<1}\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\varphi(z)\log^{+}\left|{f(z)}\right|}.$

In $\displaystyle D(0,r)$ we have a finite fixed number of zeroes of $f,$ and, because $(1-\left|{\frac{a}{s}}\right|^{2})^{p+1}$ is continuous in $\displaystyle s\leq 1$ for $\displaystyle a\in{\mathbb{D}},$ we have

$\displaystyle\forall a\in Z(f)\cap D(0,r),\ \lim_{s\rightarrow 1}(1-\left|{\frac{a}{s}}\right|^{2})^{p+1}=(1-\left|{a}\right|^{2})^{p+1}.$

Hence

$\displaystyle\ \sum_{a\in Z(f)\cap D(0,r)}{(1-\left|{a}\right|^{2})^{p+1}\varphi(a)}\leq\sup_{1-\delta<s<1}\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\varphi(sz)\log^{+}\left|{f(sz)}\right|}.$

Because the right hand side is independent of $r<1$ and $\varphi$ is positive in $\displaystyle{\mathbb{D}}$ so the sequence

$\displaystyle S(r):=\sum_{a\in Z(f)\cap D(0,r)}{(1-\left|{a}\right|^{2})^{p+1}\varphi(a)}$

is increasing with $r,$ we get

$\displaystyle\ \sum_{a\in Z(f)}{(1-\left|{a}\right|^{2})^{p+1}\varphi(a)}\leq\sup_{1-\delta<s<1}\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\varphi(sz)\log^{+}\left|{f(sz)}\right|}.$

This proves the first part. The proof of the second one is just identical. $\hfill\blacksquare$

Remark 9.6

(i) As can be easily seen by the change of variables $\displaystyle u=sz,$ if $\displaystyle p\geq 1$ we have:

$\displaystyle\sup_{1-\delta<s<1}\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\varphi(sz)\log^{+}\left|{f(sz)}\right|}\lesssim\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\varphi(z)\log^{+}\left|{f(z)}\right|}.$

(ii) We also have that if $\displaystyle\varphi(z)\log^{+}\left|{f(z)}\right|$ is subharmonic, then:

$\displaystyle\sup_{1-\delta<s<1}\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\varphi(sz)\log^{+}\left|{f(sz)}\right|}\leq\int_{{\mathbb{D}}}{(1-\left|{z}\right|^{2})^{p-1}\varphi(z)\log^{+}\left|{f(z)}\right|}.$

But (ii) is not the case in general in our setting.