TITLE: Using IVT prove that a polynomial of even degree has atleast two real roots if $a_n a_0 \lt 0$
QUESTION [2 upvotes]: $P(x) = a_n x^{n} + a_{n-1}x^{n-1} + \cdots +a_1x+ a_0$
$n = 2k$
Show that $P(x)$ has at least two real roots if $a_na_0 \lt 0$

I think I need to find some interval of length $|N|$ in which the graph has a different sign compared to the sign everywhere else. Any ideas/help on how to start  ? Thanks in advance!

REPLY [4 votes]: Without loss of generality we may assume that $a_n\gt 0$. For if it is not, we can multiply $P(x)$ by $-1$ wiithout changing the roots. Then $a_0\lt 0$.
Note that $P(0)\lt 0$. If we can show that $P(a)\gt 0$ for some positive $a$, it will follow by the Intermediate Value Theorem that $P(x)=0$ for some $x$ between $0$ and $a$, that is, for some positive $x$.
By dividing top and bottom of
$$\frac{a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0}{a_nx^n}\tag{1}$$ by $a_nx^n$ we can show that
$$\lim_{x\to\infty} \frac{P(x)}{a_nx^n}=1.\tag{1}$$
In particular, if $x$ is large enough positive, $P(x)$ is positive, so there is a positive $a$ such that $P(a)\gt 0$.
We now show that there is a negative number $-b$ such that $P(-b)\gt 0$. That will imply that $P(x)=0$ has a root between $-b$ and $0$, that is, a negative root.
An argument essentially identical to the argument we used for (1) shows that
$$\lim_{x\to-\infty} \frac{P(x)}{a_nx^n}=1.\tag{2}$$
Since $n$ is even, for negative $x$ we have that $a_nx^n$ is positive. So if $x$ is large enough negative, by (2) $P(x)$ is positive. This completes the proof.

REPLY [3 votes]: Suppose $a_0 < 0$ and $a_n >0$
Firstly we have $P(0) = a_0 < 0$.
$P$ is of even degree, which means $\lim_{x \to +\infty} P(x) = \lim_{x\to -\infty}P(x) = +\infty$, since $a_n > 0$
Apply IVT on $[0, +\infty)$ and $(-\infty, 0]$, we find two roots for $P(x)$.
If $a_0 > 0$ and $a_n <0$ we can argue in a similar way.

REPLY [3 votes]: Hints
The polynomial is of even degree.
What does this means about the limits at $x\rightarrow-\infty$ and $+\infty$?
Now what is $P(0)$? Using the fact that $a_na_0<0$ means that $a_n$ and $a_0$ are of opposite signs, you should be able to finish.


TITLE: Why does Abel's identity imply either $W = 0$ or $W \neq 0$ everywhere?
QUESTION [4 upvotes]: Let $y_1$ and $y_2$ be solutions to the linear differential equation $A(x)y'' + B(x)y' + C(x)y = 0 $ and let $W = W(y_1, y_2)$ be the Wronskian of the solutions. Why does Abel's identity
$\displaystyle W(y_1, y_2)(x) = W(y_1, y_2)(x_0)\cdot exp\left(-\int{\frac{B(x)}{A(x)}dx}\right) $
imply the Wronskian is either zero everywhere or nonzero everywhere?
I know that $W \neq 0$ if the solutions are linearly independent and $W=0$ if the solutions are linearly dependent. But how does Abel's identity show this for all $x$ in the domain?
Thanks for the help.

REPLY [5 votes]: Suppose that $W$ were ever zero. Since the $\exp$ term is never zero, that means the $W(y_1, y_2)(x_0)$ term must be zero. But that means that $W$ is zero everywhere.


TITLE: solve $y'=ay+b$
QUESTION [5 upvotes]: I have this differential equation which I want to solve
$\displaystyle\frac{dT_i}{dt}=\frac{1}{RC}(T_a-T_i)+\frac{1}{C} \Phi_h$
I know it is in the form
$y'=ay+b$
But how can I solve it ?

REPLY [0 votes]: $$\dfrac{1}{a}\int \dfrac{dy}{y+\dfrac{b}{a}}=\dfrac{1}{a}\int\dfrac{d\left(y+\dfrac{b}{a}\right)}{y+\dfrac{b}{a}}=\int dx$$


text:
\begin{document}

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\thanks{Research supported in part by NSF grant DMS-0805852, a U.S.-China Collaboration in Mathematical Research supplementary to NSF grant DMS-0500909, and by Fok Ying Tung Education Foundation Grant 111004.}

\address{Department of Applied Mathematics, Tongji University, Shanghai, 200092, China}
\address{Department of Mathematical Sciences,
Rutgers University, Camden, NJ 08102, U.S.A.}
\email{boychen@tongji.edu.cn}
\email{sfu@camden.rutgers.edu}

\title{Comparison of the Bergman and Szeg\"{o} kernels}
\author{Bo-Yong Chen and Siqi Fu}
\date{}
\maketitle

\bigskip

\begin{abstract}
The quotient of the Szeg\"{o} and Bergman kernels for a smooth bounded pseudoconvex
domains in ${\mathbb C}^n$ is bounded from above by $\delta|\log\delta|^p$ for any $p>n$, where $\delta$ is the distance to the boundary. For a class of domains that
includes  those of D'Angelo finite type and those with plurisubharmonic defining functions, the quotient is also bounded from below by $\delta|\log\delta|^p$ for any $p<-1$. Moreover, for convex domains, the quotient is bounded from above and below by constant multiples of $\delta$.

\bigskip

\noindent{{\sc Mathematics Subject Classification} (2000): 32A25, 32W05, 32U35.}

\smallskip

\noindent{{\sc Keywords}: Bergman kernel, Szeg\"{o} kernel, $\bar\partial$-operator, $L^2$-estimate, Diederich-Forn{\ae}ss exponent, pluricomplex Green function.}
\end{abstract}

\bigskip

\bigskip

\section{Introduction}

The Bergman and Szeg\"o kernels are two important reproducing kernels in complex analysis. They are related yet distinct. Whereas the Bergman kernel $K$ (as a measure) is  biholomorphically invariant, the Szeg\"o kernel $S$ is not. The former is connected to the $\dbar$-problem and the $\dbar$-Neumann Laplacian and the latter the $\bar{\partial}_b$-problem and the Kohn Laplacian. In his book published in 1972, Stein posted the following problem: What are the relations between $K$ and $S$? He further noted that the relation between $K$ and $S$ was known only in very special circumstances (\cite[p.~20]{Stein72}).

There has been an extensive literature that connects the $\dbar$-Neumann Laplacian to boundary pseudo-differential operators associated with the Kohn Laplacian (cf.~\cite{GreinerStein77, NRSW89, Kohn02}) and mapping properties of the Szeg\"{o} projection to that of the Bergman projection (cf.~\cite{Boas87, BonamiCharpentier90, Machedon88, PhongStein77}). However, there are few results, as far as we know, that directly relate these two kernels.

In this paper, we study boundary behavior of the quotient $S(z, z)/K(z, z)$ of the Szeg\"{o} and Bergman kernels for a smooth bounded pseudoconvex domain $\Omega$ in $\C^n$.
When $\Omega$ is strictly pseudoconvex, boundary limiting behavior of the Bergman kernel
$K(z, z)$ was obtained by H\"{o}rmander \cite{Hormander65} and asymptotic expansions for the Bergman and Szeg\"{o} kernels were established by Fefferman \cite{Fefferman74} and
Boutet-Sj\"{o}strand \cite{BoutetSjostrand76}. As a result, $S(z, z)/K(z, z)$ is asymptotically $\delta(z)/n$ near the boundary, where $\delta(z)$ is the Euclidean distance to the boundary. When $\Omega$ is a pseudoconvex domain of finite type in $\C^2$ or a convex domain of finite type in $\C^n$, estimates of the Bergman kernel on diagonal from above and below were obtained by Catlin \cite{Catlin89} and J. Chen \cite{ChenJH89}. Estimates for the Bergman and Szeg\"{o} kernels (on and off diagonal) and their derivatives from above were established by McNeal\cite{McNeal89, McNeal94}, Nagel et al \cite{NRSW89}, and McNeal-Stein \cite{McNealStein97}. It follows that on
these domains, $S(z, z)/K(z, z)\le C\delta(z)$ for some positive constant $C$.

Our main result can be stated as follows:

\begin{theorem}\label{th:main}
Let $\Omega\subset\subset\C^n$ be a pseudoconvex domain with $C^2$-smooth boundary.
\begin{enumerate}
\item For any $a\in (0,\ 1)$, there exists a constant $C>0$ such that
$$
\frac{S(z,z)}{K(z, z)}\le  C \delta(z)|\log(\delta(z))|^{n/a}.
$$

\item If there exist a neighborhood $U$ of $b\Omega$, a bounded continuous plurisubharmonic function $\varphi$ on $U\cap\Omega$, and a defining function $\rho$ of $\Omega$ satisfying $i\partial\dbar\varphi \ge i\rho^{-1} \partial\dbar\rho$ on $U\cap\Omega$ as currents, then there
exist constants $a\in (0,\ 1]$ and $C>0$ such that
$$
\frac{S(z,z)}{K(z, z)}\ge  C \delta(z)|\log(\delta(z))|^{-1/a}.
$$
\end{enumerate}
\end{theorem}

The constant $a$ in the second part of the theorem is a Diederich-Forn{\ae}ss exponent (\cite{DiederichFornaess77}): Namely, there exists a negative plurisubharmonic function $\varphi$ on $\Omega$ such that $C_1 \delta^a(z)\le -\varphi(z)\le C_2 \delta^{a}(z)$ for some positive constants $C_1$ and $C_2$.
It was shown by Catlin \cite{Catlin84, Catlin87} that any smooth bounded pseudoconvex domain of D'Angelo finite type satisfies Property ($P$). Sibony further showed that
for a smooth bounded pseudoconvex domain satisfying Property ($P$), the Diederich-Forn{\ae}ss index, the supremum of the Diederich-Forn{\ae}ss exponents,  is one (see \cite[Theorem~2.4]{Sibony91}).  More recently, Forn{\ae}ss and Herbig \cite{FornaessHerbig08} showed that a smooth bounded domain with a defining function that is plurisubharmonic on the boundary also has Diederich-Forn{\ae}ss index one.

For the convenience of the discussion, a bounded domain that satisfies the condition in (2) will be called {\it $\delta$-regular}. As we will show in Section~\ref{sec:lower}, such a domain is necessarily hyperconvex with a positive Diederich-Forn{\ae}ss index. It is easy to see that the class of $\delta$-regular domains includes smooth bounded pseudoconvex domains with a defining function that is plurisubharmonic on $b\Omega$, and
it is a consequence of the above-mentioned work of Catlin \cite{Catlin87} that this class of domains also includes pseudoconvex domains of D'Angelo finite type (see Proposition~\ref{pro:c} below). Therefore, in light of these and Theorem~\ref{th:main}, we have:

\begin{theorem}
Let $\Omega$ be a smooth bounded pseudoconvex domain in $\C^n$. Suppose that $b\Omega$ is
either of D'Angelo finite type or has a defining function that is plurisubharmonic on $b\Omega$. Then for any constant $a\in (0,\ 1)$, there exist positive
constants $C_1$ and $C_2$ such that
\begin{equation}\label{eq:cor1}
C_1 \delta(z)|\log(\delta(z))|^{-1/a}\le\frac{S(z,z)}{K(z, z)}\le  C_2 \delta(z)|\log(\delta(z))|^{n/a}.
\end{equation}
\end{theorem}

The logarithmic terms in the above theorems do not materialize when the domain is convex. More precisely, we have:

\begin{theorem}\label{th:convex}
Let $\Omega\subset\subset\C^n$ be a bounded convex domain with $C^2$-smooth boundary. Then there exist positive constants $C_1$ and $C_2$ such that
\begin{equation}\label{eq:convex}
C_1 \delta(z)\le S(z,z)/K(z, z)\le  C_2\delta(z).
\end{equation}
\end{theorem}

Our analysis depends on the $L^2$-estimates for the $\bar\partial$-operator by H\"{o}rmander \cite{Hormander65}, Demailly \cite{Demailly82}, and Berndtsson \cite{Berndtsson01}. We also make essential use of Blocki's estimates for the pluricomplex Green function on hyperconvex domains \cite{Blocki04}.

This paper is organized as follows: In Section~\ref{sec:prelim}, we
establish necessary background for the Hardy spaces, the Bergman and Szeg\"{o} kernels. In
Section~\ref{sec:weighted}, we review the relevant $L^2$-estimates of the $\bar\partial$-operator by H\"{o}rmander \cite{Hormander65}, Demailly \cite{Demailly82}, and Berndtsson \cite{Berndtsson01}. The first part of Theorem~\ref{th:main} is proved in Section~\ref{sec:upper} and the second part in Section~\ref{sec:lower}.

\section{Preliminaries}\label{sec:prelim}

We first establish necessary harmonic analysis background. We refer the reader to \cite{Stein72, Stein93} for an extensive treatise on the subject. Let $D$ be a bounded domain in $\R^N$ with $C^2$-smooth boundary.  Let $D_\eps=\{x\in D \mid \delta_D(x)>\eps\}$, where $\delta_D(x)$ denotes the Euclidean distance to the boundary $bD$.  For $1<p<\infty$, the harmonic Hardy space $h^p(D)$ is the space of harmonic functions $f$ such that
\[
\|f\|^p_{h^p}=\limsup_{\eps\to 0^+} \int_{bD_\eps} |f(z)|^p \, dS <\infty.
\]
The level sets $bD_\eps$ in the above definition can be replaced by those of any defining function of $D$ (see \cite{Stein72}). A classical result says that the non-tangential limit $f^*(y)$ of $f$ exists for almost every point $y$ on $bD$. Furthermore, $f^*\in L^p(bD)$, $\|f\|_{h^p}=\|f^*\|_{L^p(bD)}$, and
\[
f(x)=\int_{bD} P(x, y) f^*(y)\, dS(y),
\]
where $P(x, y)$ is the Poisson kernel of $D$.

Throughout the paper, we will use $C$, together with subscripts, to denote positive constants which could be different in different appearances. We will need the following two simple lemmas.

\begin{lemma}\label{lm:hhardy}
Let $D_1\subset D_2$ be bounded domains in $\R^N$ with $C^2$-smooth boundaries. There exists a positive constant $C$ such that
\begin{equation}\label{eq:hhardy}
\|f\|_{h^p(D_1)}\le C\|f\|_{h^p(D_2)}
\end{equation}
for any $f\in h^p(D_2)$.
\end{lemma}

\begin{proof} The proof follows the same lines of arguments as in the proof of Theorem 1 in \cite{Stein72}.  We provide the detail below. Let
\[
g(x)=\int_{bD_2} P_2(x, y)|f^*(y)|^p\, dS(y),
\]
where $P_2(x, y)$ is the Poisson kernel of $D_2$. Then for any $x\in D_2$,
\[
|f(x)|^p=\left|\int_{bD_2} P_2(x, y) f^*(y)\, dS(y)\right|^p\le \int_{bD_2}P_2(x, y)|f^*(y)|^p\, dS(y)=g(x).
\]
Thus $g(x)$ is a harmonic majorant of $|f|^p$ on $D_2$. Now fix a point $x_0$ in $D_1$. Let $G_1(x, y)$ be the Green function of $D_1$.  Let $D_1^\eps=\{x\in D_1 \mid G_1(x, y)>\eps\}$.  Then $-\partial G_1(x, y)/\partial \nu_y  =P_\eps (x, y)$ is the
Poisson kernel of $D_1^\eps$, where $\nu_y$ is the outward normal direction on $bD_1^\eps$.
Let $\pi_\eps\colon bD_1^\eps\to bD_1$ be the projection along the normal direction. Since $P_\eps (x_0, \pi^{-1}_\eps(y))$ converges uniformly on $bD_1$ to $P_1(x_0, y)$ and $C_1=\min\{P_1(x_0, y) \mid y\in bD_1\}>0$, we have
\[
g(x_0)=\int_{bD_1^\eps} P_\eps(x_0, y) g(y)\, dS \ge \frac{C_1}{2} \int_{bD_1^\eps} g(y)\, dS
\]
for sufficiently small $\eps>0$.  It follows that
\[
\begin{aligned}
\int_{b D_1^\eps} |f(x)|^p \, dS &\le \int_{bD_1^\eps} g(x)\, dS \le \frac{2}{C_1} g(x_0)
=\frac{2}{C_1} \int_{b D_2} P_2(x_0, y) |f^*(y)|^p \, dS \\
&\le \frac{2C_2}{C_1} \int_{bD_2} |f^*(y)|^p \, dS =\frac{2C_2}{C_1}\|f\|^p_{h^p(D_2)},
\end{aligned}
\]
where $C_2=\max\{P_2(x_0, y)\mid y\in bD_2\}<\infty$.  Thus \eqref{eq:hhardy} holds with $C=(2C_2/C_1)^{1/p}$.
\end{proof}

In what follows, we will also use $f$ to denote the boundary values $f^*$ for
$f\in h^p(D)$.

\begin{lemma}\label{lm:hardy} Let $D$ be a bounded domain in $\R^N$ with $C^2$-smooth boundary.
For any harmonic function $f$ on $D$,
\begin{equation}\label{eq:hardy1}
\limsup_{\eps\to 0^+} \int_{bD_\eps} |f|^p\, dS
=\limsup_{r\to 1^-} (1-r)\int_D |f(x)|^p \delta^{-r}(x)\, dV.
\end{equation}
Furthermore, when the above limits are finite, then $f\in h^p(D)$ and
\begin{equation}\label{eq:hardy2}
\int_{bD} |f|^p\, dS=\lim_{r\to 1^-} (1-r)\int_D |f(x)|^p \delta^{-r}(x)\, dV.
\end{equation}
\end{lemma}

\begin{proof}  If the limit on the left hand side of \eqref{eq:hardy1} is finite,  then $f\in h^p(D)$.
Hence
\[
\lim_{\eps\to 0^+} \int_{bD_\eps} |f|^p\, dS=\int_{bD} |f|^p\, dS.
\]
Write
\[
\lambda(\eps)=\int_{bD_\eps} |f|^p\, dS.
\]
Then $\lambda(\eps)$ is continuous on $[0, a]$ for any sufficiently small $a>0$. Therefore,
\[
\lim_{r\to 1^-} (1-r)\int_{D} |f|^p \delta^{-r}\, dV=\lim_{r\to 1^-} (1-r)\int_0^{a}\eps^{-r}\lambda(\eps)\, d\eps=\lambda(0)=\int_{bD} |f|^p\, dS.
\]
Now suppose the limit on the right hand side of \eqref{eq:hardy1} is finite. For any sufficiently small $0<\eps_1<\eps_2$, we assume that $\lambda(\eps)$ takes its minimum on $[\eps_1, \eps_2]$ at $\eps_0$. Then
\[
(1-r)\int_{D} |f|^p \delta^{-r}\, dV\ge (1-r)\int_{\eps_1\le\delta\le\eps_2} |f|^p \delta^{-r}\, dV\ge(\eps_2^{1-r}-\eps_1^{1-r}) \lambda(\eps_0).
\]
Taking $\liminf_{\eps_1\to 0^+}$ and then $\limsup_{r\to 1^-}$, we then have
\[
\infty>\limsup_{r\to 1^-} (1-r)\int_D |f(x)|^p \delta^{-r}(x)\, dV\ge \liminf_{\eps_1\to 0^+} \lambda(\eps_0).
\]
It follows that there exists a sequence $\eps_j\to 0^+$ such that $\lambda(\eps_j)$ is bounded. Let $\pi_\eps\colon bD_\eps\to bD$ be the projection along the outward
normal direction. Then $f_j(x)=f(\pi^{-1}_{\eps_j}(x))$ is a bounded sequence in $L^p(bD)$. By Alaoglu's theorem, it has a subsequence that converges to some $\tilde f\in L^p(bD)$ in the weak* topology. It follows that
\[
f(x)=\int_{bD} P(x, y)\tilde f(y)\, dS(y).
\]
Hence $f\in h^p(D)$ and we can refer back to the first part of the proof.
\end{proof}

We now review the rudiments on the Bergman and Szeg\"{o} kernels. Let $\Omega$ be a bounded domain in $\C^n$ and let $A^2(\Omega)$ be the Bergman space, the space of square integrable holomorphic functions on $\Omega$. The Bergman kernel $K_\Omega(z, w)$ is the reproducing kernel of $A^2(\Omega)$:
\[
f(z)=\int_\Omega K_\Omega(z, w) f(w)\, dV, \qquad \forall f\in A^2(\Omega), \ \forall z\in\Omega.
\]
Assume that $b\Omega$ is of class $C^2$. The Hardy space $H^2(\Omega)$ is the space of holomorphic functions on $\Omega$ that are also in  $h^2(\Omega)$.  The Szeg\"{o} kernel is
the reproducing kernel of $H^2(\Omega)$:
\[
f(z)=\int_{b\Omega} S_\Omega(z, w) f(w)\, dS, \qquad \forall f\in H^2(\Omega), \ \forall z\in\Omega.
\]
It follows from these reproducing properties that
\begin{equation}\label{eq:bergman}
K_\Omega(z, z)=\sup\{ |f(z)|^2; \  f\in A^2(\Omega), \|f\|_\Omega\le 1\}
\end{equation}
and
\begin{equation}\label{eq:szego}
\quad S_\Omega(z, z)=\sup\{|f(z)|^2; \  f\in H^2(\Omega), \|f\|_{b\Omega}\le 1\}.
\end{equation}
From \eqref{eq:bergman}, we know that the Bergman kernel has the decreasing property: if $\Omega_1\subset\Omega_2$, then $K_{\Omega_1}(z, z)\ge K_{\Omega_2}(z, z)$.  Combining Lemma~\ref{lm:hhardy} with \eqref{eq:szego}, we have:

\begin{lemma}\label{lm:lszego}
Let $\Omega_1\subset\Omega_2$ be bounded domains in $\C^n$ with $C^2$-smooth boundaries. Then there exists a constant $C>0$ such that
\begin{equation}\label{eq:lszego}
S_{\Omega_2}(z, z)\le C S_{\Omega_1}(z, z)
\end{equation}
for all $z\in\Omega_1$.
\end{lemma}

\section{Weighted $L^2$-estimates for the $\bar\partial$-operator}\label{sec:weighted}

In this section, we review relevant weighted $L^2$-estimates for the $\dbar$-operator of H\"{o}rmander, Demailly,  and Berndtsson. We will only state their results for $(0, 1)$-forms, which are what we will need later in this paper. Let $\Omega$ be a bounded pseudoconvex domain in $\C^n$ and let $\psi$ be a plurisubharmonic function on $\Omega$. Let $L^2(\Omega,e^{-\psi})$ be the Hilbert space of all measurable functions satisfying
$$
\|f\|^2_\psi=\int_\Omega |f|^2 e^{-\psi}\, dV<\infty
$$
and let $L^2_{(0, 1)}(\Omega, e^{-\psi})$ be the space of $(0, 1)$-forms with coefficients in $L^2(\Omega, e^{-\psi})$.  Suppose $\partial\dbar\psi \ge c \partial\dbar |z|^2$ as currents where $c$ is a positive continuous function. (Here and in what follows, we will drop the letter $i$ from the real $(1, 1)$-form $i\partial\dbar\psi$.) H\"ormander's theorem says that for any $\dbar$-closed $(0, 1)$-form $f$, one can solve the equation
\begin{equation}\label{eq:d-bar}
\bar{\partial}u=f
\end{equation}
in the sense of distribution, together with the estimate
\begin{equation}\label{eq:hormander}
\int_\Omega |u|^2 e^{-\psi}\, dV\le 2\int_\Omega |f|^2 e^{-\psi}/c\, dV,
\end{equation}
provided the right hand side is finite (\cite[Theorem 2.2.1$'$ ]{Hormander65}; see also \cite[Lemma4.4.1]{Hormander91}).  Suppose $\psi\in C^2(\Omega)$. For any $(0, 1)$-form $f$, let
\[
|f|_{\partial\dbar\psi}=\sup\{|\langle f, X\rangle|; \ \ X\in T^{0, 1}(\Omega),\ |X|_{\partial\dbar\psi} \le 1\}
\]
be the norm induced by the $(1, 1)$-form $\partial\dbar\psi$, where $\langle \cdot, \cdot\rangle$
denotes the pairing of a form and a vector and $|X|_{\partial\dbar\psi}=\partial\dbar\psi(\overline{X}, X)$ is the length of $X$ with respect to $\partial\dbar\psi$. According to Demailly's reformulation of H\"{o}rmander's theorem, one can solve $\dbar u=f$ with the following estimate

\begin{equation}\label{eq:demailly1}
\int_\Omega |u|^2 e^{-\psi}\, dV\le \int_\Omega |f|_{\partial\bar{\partial}\psi}^2 e^{-\psi}\, dV,
\end{equation}
provided the right hand side is finite (see \cite[Theorem~4.1]{Demailly82}). It follows that for any function $u$ in the orthogonal complement of the nullspace $\scriptn(\dbar)$
in $L^2(\Omega, e^{-\psi})$, we have
\begin{equation}\label{eq:demailly2}
\int_\Omega |u|^2 e^{-\psi}\, dV\le \int_\Omega |\dbar u|_{\partial\bar{\partial}\psi}^2 e^{-\psi}\, dV.
\end{equation}

The following theorem is a slight reformulation of a result due to Berndtsson (\cite[Theorem~2.8]{Berndtsson01}). Berndtsson's proof uses an integration by parts formula related to the $\partial\dbar$-Bochner-Kodaira technique of Siu (see Section 3 in \cite{Siu82}). We provide a proof here as a simple application of \eqref{eq:demailly2}.
Similar approach was used in \cite{BerndtssonCharpentier00} to prove an estimate of Donnelly-Fefferman \cite{DonnellyFefferman83}.

\begin{theorem}\label{prop:bo}
Let $\Omega\subset\subset\C^n$ be a bounded pseudoconvex domain.  Let $\rho\in C^2(\Omega)$ with $\rho<0$.  Suppose that there exists a plurisubhamornic function $\psi\in C^2(\Omega)$ such that
$$
\Theta:=(-\rho)\partial\bar{\partial} \psi+\partial\bar{\partial}\rho
$$
is positive. Let $u$ be the solution to \eqref{eq:d-bar} that is orthogonal to $\scriptn(\dbar)$ in $L^2(\Omega, e^{-\psi})$. Then for any $0<r<1$,
\begin{equation}\label{eq:bo}
(1-r)\int_\Omega |u|^2 (-\rho)^{-r}e^{-\psi}\, dV\le \frac1r \int_\Omega |f|^2_{\Theta}(-\rho)^{1-r} e^{-\psi}\, dV.
\end{equation}
\end{theorem}

\begin{proof} Let $\phi=-r\log(-\rho)$ and $\varphi=\phi+\psi$. Then $ue^\phi \perp \scriptn(\dbar)$ in $L^2(\Omega, e^{-\varphi})$.  Applying \eqref{eq:demailly2} to
$ue^{\phi}$ with weight $e^{-\varphi}$, we have
\[
\int_\Omega |u|^2 e^{\phi-\psi} \, dV\le \int_\Omega |\dbar u+ u\dbar\phi|^2_{\partial\dbar\varphi} e^{\phi-\psi}\, dV.
\]
It remains to show that
\begin{equation}\label{eq:d1}
|\dbar u+u\dbar\phi|^2_{\partial\dbar\varphi}\le r|u|^2+\frac{1}{r}|\dbar u|^2_{\Theta}(-\rho).
\end{equation}
Notice that
\[
\begin{aligned}
\partial\dbar\varphi&=\frac{r}{-\rho}\Theta+\frac{1}{r}\partial\phi\wedge\dbar\phi+
(1-r)\partial\dbar\psi \\
&\ge \frac{r}{-\rho}\Theta+\frac{1}{r}\partial\phi\wedge\dbar\phi =: \widetilde\Theta.
\end{aligned}
\]
Thus
\begin{equation}\label{eq:d2}
|\dbar u+u\dbar\phi |^2_{\partial\dbar\varphi}\le |\dbar u+u\dbar\phi|^2_{\widetilde\Theta}
=\sup\{\frac{|\langle\dbar u+u\dbar\phi, X\rangle|^2}{\frac{r}{-\rho}|X|^2_\Theta+\frac{1}{r}|\langle\dbar\phi, X\rangle|^2};\ \  X\in T^{0, 1}(\Omega)\}.
\end{equation}
Inequality \eqref{eq:d1} then follows from \eqref{eq:d2} and the inequalities $|\langle \dbar u, X\rangle|\le |\dbar u|_\Theta |X|_\Theta$ and
\[
2|u\langle \dbar\phi, X\rangle\overline{\langle\dbar u, X\rangle}|\le 2|u||X|_\Theta |\dbar u|_\Theta |\langle \dbar\phi, X\rangle|\le\frac{r^2}{-\rho}|u|^2|X|^2_\Theta+\frac{-\rho}{r^2}|\dbar u|^2_\Theta|\langle\dbar \phi, X\rangle|^2.
\]

\end{proof}

\section{Upper bound estimates}\label{sec:upper}

We prove the first part of Theorem~\ref{th:main} in this section. For a bounded domain $\Omega$ in $\C^n$, the {\it pluricomplex Green function} with a pole at $w\in \Omega$ is defined by
$$
g_\Omega(z,w)=\sup\left\{u(z); \ u\in PSH(\Omega),\,u<0,\,\lim\sup_{z\rightarrow w}(u(z)-\log|z-w|)<\infty\right\}.
$$
It is known that for any bounded hyperconvex domain $\Omega$, the pluricomplex Green function $g_\Omega(\cdot, w)\colon \Omega\to [-\infty, 0)$ is a continuous plurisubharmonic function such that $\lim_{z\to b\Omega} g_\Omega(z, w)=0$ (\cite{Demailly87}; see also Chapter 5 in \cite{Klimek91}).

Recall that a constant $a\in (0,\, 1]$ is said to be a {\it Diederich-Forn{\ae}ss exponent} for a bounded pseudoconvex domain $\Omega$ if there exist a negative plurisubharmonic function $\varphi$ on $\Omega$ and positive constants $C_1$ and $C_2$ such that
\begin{equation}\label{eq:df}
C_1 \delta^a(z)\le -\varphi(z) \le C_2 \delta^a(z).
\end{equation}
The supremum of all Diederich-Forn{\ae}ss exponents is called the {\it Diederich-Forn{\ae}ss index} of $\Omega$. It follows from the work of Diederich and Forn{\ae}ss that any bounded pseudoconvex domain with $C^2$-smooth boundary has a positive
Diederich-Forn{\ae}ss index, which can be arbitrarily small (\cite{DiederichFornaess77,
DiederichFornaess77b}). It was proved by Demailly \cite{Demailly87} that  bounded pseudoconvex Lipschitz domains are hyperconvex. More recently, Harrington \cite{Harrington07} showed that bounded pseudoconvex Lipchitz domains have indeed
positive Diederich-Forn{\ae}ss indices.

We will make essential use of the following quantitative estimate for the pluricomplex Green function due to Blocki (\cite[Theorem~5.2]{Blocki04};  see also \cite{Herbort00} for prior related results). We provide a proof below, following mostly Blocki's arguments\footnote{There are slight inaccuracies in the proof of Theorem 5.2 in \cite{Blocki04}: The inequality (5.6) and the choice of $\eps$ there seem to be incorrect.},
because we will need to refer back to it.

\begin{theorem}\label{th:blocki}
Let $\Omega$ be a bounded pseudoconvex domain in ${\mathbb C}^n$. Suppose there exists a negative plurisubharmonic function $\varphi$ on $\Omega$ such that
\begin{equation}\label{eq:est}
C_1\delta^a(z)\le -\varphi(z)\le C_2\delta^b(z), \qquad z\in\Omega
\end{equation}
for some positive constants $C_1, C_2$, and $a\ge b$. Then there exists positive constants $\delta_0$ and $C$ such that
\begin{equation}\label{eq:blocki}
\{z\in\Omega;\ g_\Omega(z,w)\le -1\}\subset \left\{ C^{-1}\delta^{\frac{a}b}(w)|\log{\delta(w)}|^{-\frac1b}\le \delta(z)\le C\delta^{\frac{b}a} (w)|\log{\delta(w)}|^{\frac{n}a}\right\},
\end{equation}
for any $w\in \Omega$ with $\delta(w)\le \delta_0$.
\end{theorem}

\begin{proof} Assume that
$\Omega$ has diameter $R$. Let $w\in\Omega$ with $r=\delta(w)\le e^{-2}$. Let $z\in\Omega$. Suppose that $\delta=\delta(z)\le r/2$. It follows from comparison with the pluricomplex Green function of $B(w, R)$ that
\begin{equation}\label{eq:est1}
g_{\Omega}(\zeta, w)\ge \log(|\zeta-w|/R)
\end{equation}
for all $\zeta\in\Omega$. By the maximal property of the pluricomplex Green function, we have
\begin{equation}\label{eq:fu1}
g_\Omega(\zeta, w)\ge \frac{\log(2R/r)}{\inf\{|\varphi(\zeta)|; \ \zeta\in \overline{B(w, \frac{r}2)}\}} \varphi(\zeta)
\end{equation}
on $\Omega\setminus B(w, r/2)$ because the same inequality holds on the boundary.  By \eqref{eq:est},
\begin{equation}\label{eq:est2}
\inf\{|\varphi(\zeta)|; \ \zeta\in \overline{B(w, \frac{r}2)}\}\ge C (r/2)^a.
\end{equation}
Therefore,
\begin{equation}\label{eq:b1}
g_\Omega(z, w)\ge -C\frac{\delta^b}{r^a}\log\frac{1}{r}.
\end{equation}
It follows that
\begin{align}
\{z\in\Omega; \ g_\Omega(z, w)\le -1\}&\subset\left\{\delta(z)> \frac{r}2 \text{ or }
\delta(z)\ge C r^{\frac{a}b}(\log(1/r))^{-\frac1b} \right\}\notag\\
&\subset \left\{\delta(z)\ge C^{-1}\delta^{\frac{a}b}(w)|\log{\delta(w)}|^{-\frac1b}\right\},\label{eq:bb}
\end{align}
provided the last constant $C$ is sufficiently large.

Now suppose that $e^{-2}\ge\delta(z)\ge 2r$. It follows from \eqref{eq:b1} that for any $0<\eps<r/2$,
\begin{equation}\label{eq:fu2}
\inf_{\delta(\zeta)=\eps} g_\Omega(\zeta, w) \ge -C\frac{\eps^b}{r^a} \log\frac{1}r.
\end{equation}
We also obtain from \eqref{eq:b1} that
\begin{equation}\label{eq:fu3}
g_\Omega(w, z)\ge -C \frac{r^b}{\delta^a}\log \frac{1}{\delta},
\end{equation}
by reversing the r\^{o}les of $z$ and $w$. By Theorem 5.1 in \cite{Blocki04}, we have
\begin{equation}\label{eq:b2}
g_\Omega(z, w)\ge -C \frac{\log\frac{1}{\eps}}{\log\frac{r}{2\eps}}\left(\frac{\eps^b}{r^a}\log\frac{1}{r}+
\frac{r^{\frac{b}{n}}}{\delta^{\frac{a}{n}}}
(\log\frac1\eps)^{1-\frac{1}{n}}(\log\frac{1}{\delta})^{\frac{1}{n}}\right).
\end{equation}
We have followed closely Blocki's proof thus far. Here is where we start to deviate. Suppose further that
\begin{equation}\label{eq:est3}
\delta(z)\ge r^{\frac{b}{a}}\left(\log\frac{1}{r}\right)^{\frac{n}{a}}.
\end{equation}
Set
\begin{equation}\label{eq:est4}
\eps=\frac{1}{2}\frac{r^{\frac{1}{n}+\frac{a}{b}}}{\delta^{\frac{a}{bn}}}
\left(\frac{\log\log\frac{1}{r}}{\log\frac{1}{r}}\right)^\frac{1}{b}.
\end{equation}
Since $\delta\ge 2r$, $\log\log\frac{1}{r}\le \log\frac{1}{r}$, and
\begin{equation}\label{eq:est5}
\frac{1}{n}+\frac{a}{b}-\frac{a}{bn}\ge 1,
\end{equation}
we have
\[
\eps\le \frac{r^{\frac{1}{n}+\frac{a}{b}-\frac{a}{bn}}}{2^{1+\frac{a}{bn}}}<\frac{1}{2} r
\]
as required.
From \eqref{eq:est3} and \eqref{eq:est4}, we know that
\[
\frac{r}{2\eps}\ge r^{1-\frac{a}{b}} \left(\log\frac{1}{r}\right)^\frac{1}{b}\left(\frac{\log\frac{1}{r}}{\log\log\frac{1}{r}}
\right)^\frac{1}{b}\ge C \left(\log\frac{1}{r}\right)^\frac{1}{b},
\]
provided $r=\delta(w)$ is sufficiently small. (Recall that $b\le a$.)
Therefore,
\begin{equation}\label{eq:est6}
\log\frac{r}{2\eps}\ge C\log\log\frac{1}{r}.
\end{equation}
It is easy to see that
\begin{equation}\label{eq:est7}
\log\frac{1}{\eps}\le C \log \frac{1}{r} \quad\text{and}\quad
\log\frac{1}{\delta}\le C\log\frac{1}{r}.
\end{equation}
Combining \eqref{eq:est4}, \eqref{eq:est6}, and \eqref{eq:est7} with \eqref{eq:b2}, we then obtain
\[
g_\Omega(z, w)\ge -C\frac{r^{\frac{b}{n}}}{\delta^{\frac{a}{n}}} \log\frac{1}{r}.
\]
Therefore,
\[
\{g_\Omega(z, w)<-1\}\subset\{z\in\Omega; \ \delta(z)\ge e^{-2} \text{ or } \delta(z)\le C\delta^{b/a}(w)|\log \delta(w)|^{n/a}\}.
\]
Together with \eqref{eq:bb}, we then obtain \eqref{eq:blocki} by choosing a sufficiently small $\delta_0$ and a sufficiently large $C$. \end{proof}

We also need the following localization of the Bergman kernel (\cite[Lemma~4.2]{Chen99}; also \cite[Proposition~3.6]{Herbort99}).

\begin{proposition}\label{prop:c}
Let $\Omega$ be a bounded pseudoconvex domain in $\C^n$. Then there exists a positive constant $C$ such that for any $w\in \Omega$,
\[
K_\Omega(w,w)\ge C K_{\{g_\Omega(\cdot,w)< -1\}}(w,w).
\]
\end{proposition}

To illustrate the idea of the proof, we first prove the following weaker version of Theorem~\ref{th:main}~(1):

\begin{proposition}\label{prop:weak}
Let $\Omega\subset\subset\C^n$ be a pseudoconvex domain with $C^2$-smooth boundary. Suppose that the Diederich-Forn{\ae}ss index of $\Omega$ is $\beta$.
Then for any $a\in (0,\ \beta)$, there exists a constant $C>0$ such that
\begin{equation}\label{eq:weak}
\frac{S(z,z)}{K(z, z)}\le  C \delta(z)|\log(\delta(z))|^{n/a}.
\end{equation}
\end{proposition}

\begin{proof} By the definition of the Diederich-Forn{\ae}ss index, there exists a negative plurisubharmonic function $\varphi$ satisfying \eqref{eq:df}.  By Theorem~\ref{th:blocki}, there exists a positive constant $C$ such that
\begin{equation}\label{eq:b7}
\{g_\Omega(\cdot, w)<-1\}\subset \{\delta(\cdot)<C\delta(w)|\log\delta(w)|^{n/a}\}
\end{equation}
for any $w\in\Omega$ sufficiently closed to the boundary.  Therefore, for any $f\in H^2(\Omega)$,
\[
\int_{\{g_\Omega(\cdot, w)<-1\}} |f|^2\, dV\le \int_{0}^{C\delta(w)|\log\delta(w)|^{n/a}} \, d\eps
\int_{b\Omega_\eps} |f|^2 \, dS\le C\|f\|^2_{b\Omega}\delta(w)|\log\delta(w)|^{n/a}.
\]
It then follows from the extremal properties \eqref{eq:bergman} and \eqref{eq:szego} that
\[
S(w, w)\le C \delta(w)|\log\delta(w)|^{n/a}K_{\{g(\cdot, w)<-1\}}(w, w).
\]
Applying Proposition~\ref{prop:c}, we then conclude the proof of the proposition.
\end{proof}

To get from Proposition~\ref{prop:weak} to the first statement of Theorem~\ref{th:main}, we use Lemma~\ref{lm:lszego} to localize the Szeg\"{o} kernel and then apply the following fact: For any $z_0\in b\Omega$ and $a\in (0,\ 1)$, there exist a defining function $r$ of $\Omega$ and a neighborhood $U$ of $z_0$ such that $\varphi_2=-(-r)^a$ is strictly plurisubharmonic on $U\cap\Omega$ (\cite{DiederichFornaess77}, Remark on p.~133).  The problem is that this function $\varphi_2$ is not an exhaustion function of $U\cap\Omega$ and thus one cannot directly apply Theorem~\ref{th:blocki}. We now show how to overcome this difficulty and prove Theorem~\ref{th:main}~(1).

Let $\chi(t)$ be a smooth function such that $\chi(t)=0$ when $t\le 1$, $\chi(t)>0$ is strictly increasing and convex when $t>1$. We may further assume that $\chi(t)=\exp(-1/(t-1))$ when $t\in (1,\ 5/4)$ so that $(\chi(t))^b\in C^\infty(\R)$ for any positive number $b$. Let
\[
\varphi_1(z)=-(-r)^a+M\chi(|z-z_0|^2/m^2).
\]
Then $\varphi_1$ is strictly plurisubharmonic on $U\cap\Omega$.  Write $g(z)=M\chi(|z-z_0|^2/m^2)$. Let
\[
\widetilde\Omega=\{z\in\Omega; \ \ \varphi_1(z)=-(-r)^a+g<0\}.
\]
By choosing $m$ sufficiently small and $M$ sufficiently large, we know that
$B(z_0, m)\cap \Omega\subset\widetilde\Omega$ and
$\widetilde\Omega\subset B(z_0, 2m)$. Furthermore, $\widetilde\Omega$ is pseudoconvex with a $C^2$-smooth defining function
\[
\tilde r=r+g^{1/a}
\]
(see, for example, \cite[pp.~470--471]{Bell86}).

Evidently, $\varphi_1$ is a plurisubharmonic exhaustion function for $\widetilde\Omega$. However, $\varphi_1$ does not satisfy \eqref{eq:df}. In fact, it is easy to show that
there exists a positive constant $C_1$ such that
\begin{equation}\label{eq:loc1}
C_1|\tilde r|\le -\varphi_1\le |\tilde r|^a, \quad z\in\widetilde\Omega.
\end{equation}
If we directly invoke Theorem~\ref{th:blocki} with \eqref{eq:loc1}, we obtain
\[
\{g_{\widetilde\Omega}(\cdot, w)<-1\}\subset \{\delta(\cdot)<C\delta^a(w)|\log\delta(w)|^{n}\}.
\]
Consequently, we have as in the proof of Proposition~\ref{prop:weak} that
\[
S(z, z)/K(z, z)\le C \delta^{a}(z)|\log\delta(z)|^{n},
\]
which is even weaker than \eqref{eq:weak}.

Instead of directly appealing to Theorem~\ref{th:blocki}, we proceed
as follows. We follow the proof of Theorem~\ref{th:blocki} with $\Omega$ replaced by $\widetilde\Omega$, and with $\delta(z)=\delta_{\widetilde\Omega}(z)$ now denoting the Euclidean distance to $b\widetilde\Omega$. Notice that $C^{-1}\delta\le |\tilde r|\le C\delta$ on $\widetilde\Omega$. Assume that $|w-z_0|<m$. Applying \eqref{eq:fu1} to the function $\varphi_1$, we have
\begin{equation}\label{eq:fu2b}
\inf_{\delta(\zeta)=\eps} g_{\widetilde\Omega}(\zeta, w)\ge \frac{\log(2R/r)}{\inf\{|\varphi_1(\zeta)|; \ \zeta\in \overline{B(w, \frac{r}2)}\}} \inf_{\delta(\zeta)=\eps}\varphi_1(\zeta)\ge -C\frac{\eps^a}{r^a} \log\frac{1}r,
\end{equation}
which is our analogue in this case to \eqref{eq:fu2}. (Here we have $a=b$.)  Now applying \eqref{eq:fu1} to the function $\varphi_2=-(-r)^a$ with the r\^{o}le of $z$ and $w$ reversed, we have
\[
g_{\widetilde\Omega}(\zeta, z)\ge \frac{\log(2R/\delta)}{\inf\{|\varphi_2(\zeta)|; \ \zeta\in \overline{B(z, \frac{\delta}2)}\}} \varphi_2(\zeta).
\]
on $\widetilde\Omega\setminus B(z, \delta/2)$. It follows that
\begin{equation}\label{eq:fu3b}
g_{\widetilde\Omega}(w, z)\ge -C\frac{r^a}{\delta^a}\log\frac{1}{\delta},
\end{equation}
which plays the r\^{o}le of \eqref{eq:fu3} in this case. Following exactly
the same lines for the rest of the proof of Theorem~\ref{th:blocki}, we then
obtain
\[
\{z\in\widetilde\Omega; \ g_{\widetilde\Omega}(z, w)<-1\}\subset \{z\in\widetilde\Omega; \ \delta(z)\le C\delta(w)|\log\delta(w)|^{n/a}\}.
\]
From the proof of Proposition~\ref{prop:weak}, we then have
\[
\frac{S_{\widetilde\Omega}(w,w)}{K_{\widetilde\Omega}(w, w)}\le  C \delta(w)|\log(\delta(w))|^{n/a},
\]
when $w$ is sufficiently close to $z_0$.  By the localization property of the Bergman kernel (see the proof of Theorem 1 in \cite{Ohsawa81}; also \cite[Proposition 1]{DFH84}), $K_\Omega(w, w)\ge CK_{\widetilde\Omega}(w, w)$. Together with Lemma~\ref{lm:lszego}, we then conclude the proof of the first statement in Theorem~\ref{th:main}.

\section{Lower bound estimates}\label{sec:lower}

Recall that a continuous function $\rho$ is said to be a defining function of a domain $\Omega\subset\C^n$ if $\Omega=\{z\in \C^n; \ \rho(z)<0\}$ and $C^{-1}\delta\le\rho\le C\delta$ for a constant $C>0$. We also assume the defining function $\rho$ to be in the same smoothness class as that of the boundary $b\Omega$. A bounded domain $\Omega\subset\C^n$ is {\it $\delta$-regular} if there exist a neighborhood $U$ of $b\Omega$, a bounded continuous plurisubharmonic function $\varphi$ on $U\cap\Omega$, and a defining function $\rho$ of
$\Omega$ such that
\begin{equation}\label{eq:delta}
\partial\dbar\varphi \ge \rho^{-1} \partial\dbar\rho
\end{equation}
on $U\cap\Omega$ as currents. By adding $|z|^2$ to $\varphi$, we may assume that it is strictly plurisubharmonic. By Richberg's approximation theorem (\cite[Satz~4.3]{Richberg68}), we may further assume that $\varphi\in C^\infty(\Omega)$.

\begin{proposition}\label{prop:h} Let $\Omega\subset\subset\C^n$ be a $\delta$-regular domain. Then $\Omega$ is hyperconvex with a positive Diederich-Forn{\ae}ss index.
\end{proposition}

\begin{proof} Let $\varphi$ and $\rho$ be the functions that satisfy \eqref{eq:delta}. Assume $0\le\varphi\le M$ for some positive constant $M$. Let $\psi=e^\varphi$ and $K=e^M$. Then
\begin{equation}\label{eq:s}
1\le\psi\le K,  \quad \partial\psi\wedge\dbar\psi\le K \partial\dbar\psi, \quad
\text{ and } \quad \partial\dbar\psi\ge \frac{\partial\dbar\rho}{\rho}+K^{-1}\partial\psi\wedge\dbar\psi,
\end{equation}
on $U\cap\Omega$. Let
\[
\tilde\rho=\rho e^{-\psi} \quad \text{and} \quad r=-(-\tilde\rho)^\eta.
\]
It follows from a simple (formal) computation and \eqref{eq:s} that
\[
\begin{aligned}
\partial\dbar r=&\eta(-\tilde\rho)^\eta\left(\partial\dbar -\log(-\tilde\rho)-\eta\frac{\partial\tilde\rho\wedge\dbar\tilde\rho}{\tilde\rho^2}\right)\\
&\ge \eta(-\tilde\rho)^\eta\left(\frac{1}{K}\partial\psi\wedge\dbar\psi+\frac{\partial\rho
\wedge\dbar\rho}{\rho^2}-
\eta\frac{\partial\tilde\rho\wedge\dbar\tilde\rho}{\tilde\rho^2}\right)\\
&\ge\eta(-\tilde\rho)^\eta\left((1-\eta)\frac{\partial\rho
\wedge\dbar\rho}{\rho^2}+\eta\frac{\partial\rho}{\rho}\wedge\dbar\psi+\eta\partial\psi\wedge
\frac{\dbar\rho}{\rho}+(\frac{1}{K}-\eta)\partial\psi\wedge\dbar\psi\right).
\end{aligned}
\]
We then obtain from the Schwarz inequality that $\partial\dbar r$ is a positive current on $U\cap\Omega$, provided $\eta$ is sufficiently small. The extension of $r$ to the whole domain $\Omega$ is standard (see \cite[p.~133]{DiederichFornaess77}).
\end{proof}

\begin{proposition}\label{pro:c}
Let $\Omega$ be a smooth pseudoconvex bounded domain in $\C^n$. If $\Omega$ has a defining
function that is plurisubharmonic on $b\Omega$ or if $b\Omega$ is of D'Angelo finite type, then $\Omega$ is $\delta$-regular.
\end{proposition}

\begin{proof} If $\Omega$ has a defining function $\rho$ that is plurisubharmonic on $b\Omega$. Then
$\partial\dbar\rho\ge  C\rho \partial\dbar |z|^2$. Therefore in this case, we
can choose $\varphi(z)=C|z|^2$ with a sufficiently large $C$.

The case when $\Omega$ is of D'Angelo finite type is a consequence of Catlin's construction of bounded plurisubharmonic function \cite{Catlin87} (see \cite[p.~464]{Straube97} for a related discussion): There exist positive constants $\tau<1$, $C>0$, and a smooth bounded plurisubharmonic function $\lambda$ on $\Omega$ such that
\begin{equation}
\partial\bar{\partial}\lambda\ge C \frac{\partial\bar{\partial}|z|^2}{|\rho|^{\tau}}.
\end{equation}
By Oka's lemma, we can choose a defining function $\rho$ such that $\partial\dbar(-\log(-\rho))\ge \partial\dbar |z|^2$. It then follows from a theorem
of Diederich-Fornaess that for any sufficiently small $\eta$,
\begin{equation}
\partial\bar{\partial}( -(-\rho)^\eta )\ge C \eta |\rho|^{\eta}\left(\partial\dbar |z|^2+|\rho|^{-2}\partial\rho\wedge\dbar\rho\right)
\end{equation}
for some positive constant $C$ (\cite{DiederichFornaess77}, Theorem 1 and its proof; compare also \cite[Lemma 2.2]{CSW04}).

Now we fix an $\eta\in (0,\ \tau)$. Write $N=|\partial\rho|^{-1}\sum \rho_{\bar z_j}{\partial/\partial z_j}$.
For any $(1, 0)$-vector $X$, write $X_N=\langle X, N\rangle N$ and
$X_T=X-X_N$.  By the pseudoconvexity of $b\Omega$, we have
\[
\rho^{-1}\partial\dbar\rho (X, \ov{X}) \le C(|X|^2+|\rho|^{-1}|X||X_N|)\le C(|\rho|^{-\eta} |X|^2+ |\rho|^{-2+\eta}|X_N|^2)
\]
The desirable function is then given by
\[
\varphi=C(\lambda-(-\rho)^\eta)
\]
for any sufficiently large $C>0$.
\end{proof}

We are now in position to prove the second part of Theorem~\ref{th:main}.  Let $\kappa$ be
a standard Friedrichs mollifier. Let $\eps_j$ be a decreasing sequence of positive number tending to $0$.  Let $w$ be a point in $\Omega$, sufficiently closed to the boundary $b\Omega$. Let $g_j=g_\Omega(\cdot, w)*\kappa_{\eps_j}$. Then $g_j$ is a decreasing sequence of plurisubharmonic functions on $\Omega_j=\{z\in\Omega; \ \delta(z)<\eps_j\}$ with limit $g_\Omega(\cdot, w)$. By Oka's lemma, $\Omega_j$ is pseudoconvex.  Let
$$
\psi=2ng_\Omega(\cdot, w)-\log(-g_\Omega(\cdot, w)+1)+\varphi  \quad \text{and}\quad
\psi_j=2n g_j-\log(-g_j+1)+\varphi,
$$
where $\varphi$ is the smooth bounded strictly plurisubharmonic function, obtained from the $\delta$-regularity assumption, such that $\partial\dbar\varphi \ge \rho^{-1}\partial\dbar\rho$ for a defining function $\rho$ of $\Omega$. Clearly $\psi_j$ is a plurisubharmonic function on $\Omega_j$. Moreover,
\begin{equation}\label{eq:f1}
\partial\bar{\partial}\psi_j\ge \partial\log(-g_j+1)\wedge\bar{\partial}\log(-g_j+1)
\end{equation}
and $\Theta_j=(-\rho)\partial\bar{\partial}\psi_j+\partial\bar{\partial}\rho$ is positive
on $\Omega_j$. Let $\chi:{\mathbb R}\rightarrow [0,1]$ be a $C^\infty$ cut-off function such that $\chi|_{(-\infty,-1)}=1$ and $\chi|_{(0,\infty)}=0$. Put
$$
v_j=\bar{\partial}\chi(-\log(-g_j)) \frac{K_\Omega(\cdot,w)}{\sqrt{K_\Omega(w,w)}}.
$$
Let $u_j$ be the solution to $\dbar u_j=v_j$ that is in the orthogonal complement of $\scriptn(\dbar)$ in $L^2(\Omega_j, e^{-\psi_j})$. By Demailly's estimate \eqref{eq:demailly1},
$$
\int_{\Omega_j} |u_j|^2 e^{-\psi_j}\, dV\le \int_{\Omega_j} |v_j|^2_{\partial\bar{\partial}\psi_j}e^{-\psi_j}\, dV.
$$
It follows from \eqref{eq:f1} that the right hand side is uniformly bounded from above, independent of $j$. Passing to a subsequence, we may assume that $u_j$ converges to $u\in L^2(\Omega, e^{-\psi})$ in the weak* topology. (We extend $u_j=0$ on $\Omega\setminus\Omega_j$.) Let
$$
f=\chi(-\log(-g_\Omega(\cdot,w)))K_\Omega(\cdot,w)/K_\Omega(w,w)^{1/2}-u.
$$
Then $f$ is holomorphic on $\Omega$. Since $u$ is holomorphic in a neighborhood of $w$ and $g_\Omega(z,w)=\log|z-w|+O(1)$ near $w$,
\[
u(w)=0.
\]
By Theorem~\ref{prop:bo}, for any $0<r<1$, we have
\begin{align}
(1-r)\int_{\Omega_j}|u_j|^2 (-\rho)^{-r} e^{-\psi_j}\, dV  &\le  \frac1r \int_{\Omega_j} |v_j|^2_{\Theta_j}(-\rho)^{1-r}e^{-\psi_j}\, dV\nonumber\\
& \le \frac{C}r \int_{{\rm supp\,}\bar{\partial}\chi(\cdot)}\frac{|K_\Omega(\cdot,w)|^2}{K_\Omega(w,w)}(-\rho)^{-r}\, dV.
\end{align}
By Theorem~\ref{th:blocki},
\[
\begin{aligned}
{\rm supp\,}\bar{\partial}\chi(\cdot)&\subset\{-e\le g_j(\cdot,w)\le -1\}\subset\{g_\Omega(\cdot, w)\le -1\}\\
&\subset \{C^{-1}|\rho(w)||\log(-\rho(w))|^{-1/a}\le |\rho|\},
\end{aligned}
\]
where $a$ is a Diederich-Forn{\ae}ss exponent for $\Omega$. Therefore, passing to the limit, we have
\begin{equation}\label{eq:lower}
(1-r)\int_\Omega|u|^2 (-\rho)^{-r} e^{-\psi}\, dV\le \frac{C}r\cdot \frac{|\log(-\rho(w))|^{r/a}}{|\rho(w)|^r}.
\end{equation}
Notice that $f$ is a holomorphic function on $\Omega$ such that $f(w)=K_\Omega(w,w)^{1/2}$ and $f=-u$ near $b\Omega$. Since $e^{-\psi}\ge e^{-\varphi}\ge C>0$,  by Lemma~\ref{lm:hardy} (and its proof), $f\in H^2(\Omega)$. Combining with \eqref{eq:lower}, we have
\begin{align*}
\int_{b\Omega} |f|^2\, dS&\le C\lim_{r\to 1^-} (1-r)\int_\Omega|f|^2 (-\rho)^{-r} e^{-\psi}\, dV\\
&=C\lim_{r\to 1^-} (1-r)\int_\Omega|u|^2 (-\rho)^{-r} e^{-\psi}\, dV\\
&\le C \frac{|\log(-\rho(w))|^{1/a}}{|\rho(w)|}.
\end{align*}
It then follows from the extremal property \eqref{eq:szego} of the Szeg\"{o} kernel that
\[
S_\Omega(w,w)\ge C\frac{|\rho(w)|}{|\log(-\rho(w))|^{1/a}}\cdot K_\Omega(w,w).
\]
This concludes the proof of Theorem~\ref{th:main}.

The proof of Theorem~\ref{th:convex} is similar that of Theorem~\ref{th:main}. The only difference is that, instead of Theorem~\ref{th:blocki}, we use the following well known estimate for the pluricomplex Green function on convex domains:
$$
\{g_\Omega(\cdot,w)\le -1\}\subset \{C^{-1}\delta(w)\le \delta(\cdot)\le C\delta(w)\}
$$
(see \cite{Blocki04}, Theorem 5.4).

\begin{remark}
It follows from the Ohsawa-Takegoshi extension theorem \cite{OhsawaTakegoshi87} that for any bounded pseudoconvex Lipschitz domain $\Omega$ in $\C^n$, $K(z, z)\ge C\delta^{-2}(z)$
for some constant $C>0$. Ohsawa  \cite{Ohsawa03} conjectured that the analogue estimate $S(z, z)\ge C\delta^{-1}(z)$ holds. Theorem~\ref{th:convex} confirms this conjecture for convex domains. Moreover, Theorem~\ref{th:main} implies that for a bounded $\delta$-regular domain $\Omega$ with $C^2$-smooth boundary,
\begin{equation}
S(z, z)\ge C\delta^{-1}(z)|\log\delta(z)|^{-1/a},
\end{equation}
where $a$ is any Diederich-Forn{\ae}ss exponent of $\Omega$.
\end{remark}

\bibliography{survey}
\providecommand{\bysame}{\leavevmode\hbox
to3em{\hrulefill}\thinspace}

config: arxiv


TITLE: Why is sound reflected and not transmitted through a wall?
QUESTION [22 upvotes]: We know that speed of sound is higher in a denser medium. So when a sound wave strikes a wall, why does it echo instead of passing through it with a speed faster than the speed of sound in air?

REPLY [24 votes]: Whenever a wave reaches a boundary between two mediums some of the energy is reflected and some is transmitted.
The important parameter is not just the speed of the wave on either side of the boundary but what is called the acoustic impedance (= density $\times$ speed).
If there is a large difference between the acoustic impedances then you will get most of the wave reflected.
So if you have an air (speed of sound 330 m/s, density 1.2 kg/m$^3$)  brick wall (4200 m/s, 1850 kg/m$^3$) boundary a lot of the sound will be reflected.
So when you walk down a tunnel the sound that you emit gets reflected off the walls and even off the open end of the tunnel and comes back to you as an echo.

REPLY [13 votes]: Sound follows like everything else in nature. Minimize energy. So while sound can travel faster in a denser medium, the fact that sound is produced by mechanical waves mean that more energy is required to transmit the sound 100% through the wall with the same intensity.
And just like any other wave, when a sound wave hits a solid object like a wall, it will undergo reflection, dissipation and transmission. This transmission usually reduces the intensity of the travelling sound wave which is why when you are in an adjacent room, you can hear a fain part of the conversation from the next room. But at the same time, if the wall is made from a very dense material, then the sound waves don't have enough energy in creating mechanical waves in the material of the wall and thus you usually end up with only reflection (echo) and some dissipation.


TITLE: What does Noether's Normalization lemma even mean?
QUESTION [10 upvotes]: Noether's Normalization lemma states that

For any field $k$, and any finitely generated commutative $k$-algebra $A$, there exists a non-negative integer $d$ and algebraically independent elements $y_1,y_2,\dots,y_d\in A$ such that $A$ is a finitely generated module over the polynomial ring $k[y_1,y_2,\dots,y_d]$.

I'm trying to understand the statement of this lemma. A finitely generated commutative $k$-algebra would be $k[x_1,x_2,\dots,x_n]$. Clearly, $x_1,x_2,\dots,x_n$ are algebraically independent. Aren't we essentially saying that $k[x_1,x_2,\dots,x_n]$ is a finitely generated module over $k[x_1,x_2,\dots,x_n]$? What is there to prove in that?

REPLY [13 votes]: It is important to know that finitely generated $k$-algebras are not just of the form $k[x_1,...,x_n]$. They can also be quotients of these rings by ideals. For instance $A=\mathbb{C}[x,y]/(x^2-y^3)$. This is a $\mathbb{C}$-algebra, but will not be isomorphic to any polynomial ring over $\mathbb{C}$. One way to see this is that its localization at the maximal ideal $(x,y)$ will not be integrally closed in the fraction field of $A$. The element $x/y$, for instance, will be integral over $A_{(x,y)}$ but not actually in $A$.
However, if we localize any polynomial ring $P$ over $\mathbb{C}$ at a maximal ideal $m$, we find that $P_m$ is integrally closed in the fraction field of $P$.
This may not be the best way to see that $A$ is not a polynomial ring, but it is very geometric. The ring $A$ is the coordinate ring of the plane curve given by $x^2 = y^3$. This curve has a cusp at the origin and is quite singular. This geometric fact is somehow wrapped up in the algebra of the local ring $A_{(x,y)}$. Any polynomial ring is the coordinate ring of a certain affine space. Affine spaces are nice and smooth, and this fact is again reflected somehow in the algebra of its localizations at maximal ideals.
What normalization tells us is that finitely $k$-algebras are reasonable objects. They are integral extensions (very nice extensions) of polynomial rings (very nice rings).
One geometric consequence (at least over an algebraically closed field) that normalization gives us is that any algebraic set is a cover of an affine space. Normalization gives us some inclusion of a polynomial ring $P=k[y_1,...,y_d]$ into the ring $A$. If we think of this geometrically, $A$ is the coordinate ring of some variety, and $P=k[y_1,...,y_d]$ is the coordinate ring of some $d$-dimensional linear subspace. The geometric picture is the reversal of the algebraic picture, in that an inclusion of $P$ into $A$ means the variety associated with $A$ projects (surjectively) onto the linear space associated with $P$. An important consequence of this is the Nullstellensatz.


text:
\begin{document}

\title{On the maximum rank of a real binary form}
\author{A.~Causa  \and R.~Re}
\date{}
\maketitle
\begin{abstract} We show that a real binary form $f$ of degree
$n\geq 3$ has $n$ distinct real roots if and only if for any
$(\alpha,\beta)\in\mathbb{R}^2\setminus\{0\}$ all the forms
$\alpha f_x+\beta f_y$ have $n-1$ distinct real roots. This
answers to a question of P. Comon and G. Ottaviani in \cite{coot},
and allows to complete their argument to show that $f$ has
symmetric rank $n$ if and only if it has $n$ distinct real roots.
\end{abstract}
\section{Introduction}
This paper deals with the following problem: Given a degree $n$ polynomial $f\in\mathbb
K[x_1,\ldots,x_m]$ find the rank (or Waring rank) of $f$ i.e.~the minimum number of summands
which achieve the following decomposition:\[f=\lambda_1l_1^n+\cdots
+\lambda_rl_r^n\quad\mbox{with }\lambda_i\in\mathbb K\mbox{ and }l_i\mbox{ linear forms.}\]
 For $\mathbb K=\mathbb C$ and $f$ generic the answer has been given
 (see \cite{alhi,otbr}), nevertheless some questions remain unsolved, e.g.~it's not yet
 known which is the stratification of the set of complex polynomials by the rank. However one
 can see \cite{cose} for an answer in the binary case.\\
 In the real case, i.e.~$\mathbb K=\mathbb R$, the situation becomes more complicated. In
 contrast to the complex case which has a generic rank, in the real case the generic rank is
 substituted by the concept of typical rank.
A rank $k$ is said \emph{typical} for a given degree $n$ if there exists a euclidean open set in
the space of real degree $n$ polynomials such that any $f$ in such open et has rank $k$.
 We will prove the following theorem, posed as a question in \cite{coot}.
\begin{theorem}\label{thm:main} Let $f(x,y)$ be a real homogeneous polynomial of degree $n\geq 3$ without multiple roots in $\mathbb{C}$.
 Then $f$ has all real roots if and only if for any $(0,0)\not=(\alpha,\beta)\in\mathbb{R}^2$ the polynomial $\alpha f_x+\beta f_y$ has
 $n-1$ distinct real roots.\end{theorem}
Notice that the ``only if" part of the theorem is easy. Indeed given
any $(\alpha,\beta)\not=(0,0)$ one may consider a new coordinate
system $l,$ $m$  on the projective line, such that $x=\alpha
l+\alpha' m$, and $y=\beta l+\beta' m$, so that
$\partial_l=\alpha\partial_x+\beta\partial_y$. Writing $f$ as a
function of $l,m$ and de-homogenizing by setting $m=1$ one sees
that $f_l$ has $n-1$ distinct roots by the theorem of Rolle.

In \cite{coot} the result above has been considered in connection
with the problem of determining the \emph{rank} of a real binary
form, that is the minimum number $r$ such that $f(x,y)=\lambda_1
l_1^n+\cdots+\lambda_r l_r^n$, with $\lambda_i\in\mathbb{R}$ and
$l_i=\alpha_ix+\beta_i y\in \mathbb{R}[x,y]$ for $i=1,\ldots r$.
Using the arguments already given in \cite{coot} and applying
Theorem \ref{thm:main}, one gets the following result.
\begin{corollary}\label{cor:rank} A real binary form $f(x,y)$ of degree $n\geq 3$ without multiple roots in $\mathbb{C}$ has rank $n$ if and only
if it has $n$ distinct real roots.\end{corollary}

We leave the following question open for further
investigations. Partial evidence for it has been given from the results in \cite{coot}, where it
has given a positive answer for $n\leq 5$, and where the reader can find references for the
existing literature on rank problems for real tensors.
\begin{question} Are all the ranks $\lfloor n/2\rfloor +1\leq k\leq n$ typical for real binary forms of degree $n$?\end{question}
\section{Main Theorem}
Let $f(x,y)$ be a real homogeneous polynomial of degree $n\geq 3$ without multiple roots in $\mathbb{C}$. Then $\nabla f(x,y)\not=(0,0)$ for any $(x,y)\not=(0,0)$ and one can define the maps $\bar{\phi}:S^1\to S^1$ and $\bar{\psi}:S^1\to S^1$ setting, for any $(x,y)$ with $x^2+y^2=1$, $\bar{\phi}(x,y)=|\nabla f|^{-1}(f_x,f_y)$ and
$\bar{\psi}(x,y)=|\nabla f|^{-1}(xf_x+yf_y, -y f_x+xf_y)$, with $|\nabla f|=(f_x^2+f_y^2)^{1/2}$. Setting
$(x,y)=(\cos\theta,\sin\theta)$, one can also write $\bar{\phi}$ and $\bar{\psi}$ as functions of $\theta$. \vskip1mm \noindent
{\bf Notation.} We denote $\partial_\theta=-y\partial_x+x\partial_y$ the basis tangent
vector to $S^1$ at the point $(x,y)$. Given any differentiable map $\phi:S^1\to M$ to a differentiable manifold $M$, we denote $\phi_\ast:T_\theta S^1\to T_{\phi(\theta)}M$ the associated tangent map. If $M=S^1$, and the map $\phi$ is defined in terms of angular coordinates by the function $\theta_1(\theta)$, we recall that the \emph{degree}, or \emph{winding number}, of $\phi$ is the number
\[\deg \phi=\frac{1}{2\pi}\int_0^{2\pi} \theta_1'(\theta)d\theta.\]
This is always an integer number, and for any $z\in S^1$ one has
$\# \phi^{-1}(z)\geq |\deg \phi\, |$. \vskip1mm \noindent
The following lemmas are straightforward calculations and their proofs are omitted.
\begin{lemma}\label{lm:winding} Assume that $\theta_1'(\theta)$ never vanishes. Then $\# \phi^{-1}(z)=|\deg \phi\,|$ for any $z\in S^1$.\end{lemma}

We assume that for any $(\alpha ,\beta)\in\mathbb{P}^1(\mathbb{R})$ the
polynomial $\alpha f_x+\beta f_y$ has $n-1$ distinct roots in $\mathbb{R}$. Under this
assumption, we want to show that the absolute value of the degree of $\bar{\psi}$ is $n$. Since
$f(x,y)=0$ iff $\bar{\psi}(x,y)=(0,\pm 1)$ this implies that $f$ has all its roots in
$\mathbb{R}$. Indeed $\bar{\psi}(-x,-y)=(-1)^n\bar{\psi}(x,y)$, henceforth when $n$ is even
$n/2$ real roots of $f(x,y)=0$ are in $\bar{\psi}^{-1}(0,1)$ and the other $n/2$ roots are in
$\bar{\psi}^{-1}(0,-1)$; otherwise when $n$ is odd one gets $\bar{\psi}^{-1}(0,1)=
\bar{\psi}^{-1}(0,-1)$, hence $\bar{\psi}^{-1}(0,1)$ is the set of the $n$ real roots of
$f(x,y)=0$.
\begin{lemma}\label{lm:partialtheta} Let $F:S^1\to \mathbb{R}^2$ be a differentiable function defined by
$F(x,y)=(F_1(x,y),F_2(x,y))=(a(\theta),b(\theta))$. Then $F_\ast(\partial_\theta)=A\partial_x+B\partial_y$ with
$$\begin{array}{l}A=-y F_{1x}+xF_{1y}=a'(\theta)\\ B=-yF_{2x}+xF_{2y}=b'(\theta).\end{array}$$\end{lemma}
\noindent
{\bf Notation.} Given a map $f:S^1\to \mathbb{R}^2$,
which one can write $f(\theta)=(a(\theta),b(\theta))$, we denote
with $(f,f_\theta)$ the matrix
$$ \left(\begin{array}{ll}a(\theta)&
b(\theta)\\a'(\theta)& b'(\theta)\end{array}\right).$$
Notice that the sign of the determinant of this matrix expresses\\
if $f_\ast$ is orientation-preserving at the point $f(\theta)$.
\begin{lemma}\label{lm:normaliz}
Let $g:S^1\to \mathbb{R}^2$ and $\rho:S^1\to \mathbb{R}_+$ be differentiable functions. Then
$\det(g,g_\theta)=\rho^{-2}\det(\rho g,(\rho g)_\theta)$.\end{lemma}
Notice that if $\bar{g}:S^1\to S^1$ is the map $\bar{\phi}(x,y)=|\nabla f|^{-1}(f_x,f_y)$ then one may calculate
the sign of  $\det(\bar{\phi},\bar{\phi}_\theta)$ by reducing to the simpler map $\phi=(f_x,f_y):S^1\to \mathbb{R}^2$.
\vskip1mm\noindent {\bf Notation.} We denote by $H(f)=\det\left(\begin{array}{ll} f_{xx}& f_{xy}\\ f_{xy}&
f_{yy}\end{array}\right)$, the {\em hessian} of $f$.
\begin{proposition}\label{prop:hessian}Let $\phi:S^1\to \mathbb{R}^2$ be the map defined above. Then
$$\det(\phi,\phi_\theta)=(n-1)^{-1}H(f).$$ \end{proposition}
\begin{proof} We have $\phi_\ast(\partial_\theta)=A \partial_x+B\partial_y$ with $A$ and $B$ determined as in Lemma \ref{lm:partialtheta}, hence
\begin{eqnarray*} \det(\phi,\phi_\theta)&=&\det\left(\begin{array}{ll}f_x& f_y\\-yf_{xx}+xf_{xy}&-yf_{yx}+xf_{yy}\end{array}\right)\\
&=&\frac{1}{n-1}\det
\left(\begin{array}{cc}xf_{xx}+yf_{xy}& xf_{yx}+yf_{yy}\\-yf_{xx}+xf_{xy}&-yf_{yx}+xf_{yy}\end{array}\right)\\
&=&\frac{1}{n-1}\det\left(\begin{array}{cc}x&y\\
-y&x\end{array}\right)
\det\left(\begin{array}{cc} f_{xx}& f_{xy}\\ f_{xy}& f_{yy}\end{array}\right)\\
&=&\frac{1}{n-1} H(f)\end{eqnarray*}
\end{proof}
\begin{proposition}\label{prop:rotation} Let $\phi:S^1\to\mathbb{R}^2\cong\mathbb{C}$ defined by $\phi(\theta)=a(\theta)+ib(\theta)$ and $\psi:S^1\to\mathbb{C}$ defined by $\psi(\theta)=e^{-i\theta}\phi(\theta)$.  Then $\det(\psi,\psi_\theta)=\det(\phi,\phi_\theta)-a^2-b^2$.
\end{proposition}
\begin{proof} We calculate $\psi'(\theta)=(a'+b+i(b'-a))e^{-i\theta}$. It follows that
\[
\det(\psi,\psi_\theta)=\det\left(\begin{array}{ll}a&b\\a'+b&b'-a\end{array}\right)=
\det\left(\begin{array}{ll}a&b\\a'&b'\end{array}\right)-a^2-b^2.\]\qed\end{proof}
Notice that by Lemma \ref{lm:normaliz}, if $\phi:S^1\to \mathbb{R}^2$ is given in polar coordinates by
$\phi(\theta)=\rho(\theta)\bar{\phi}(\theta)$, with $\bar{\phi}:S^1\to S^1$, then
$\det(\phi,\phi_\theta)=\rho^2\det(\bar{\phi},\bar{\phi}_\theta)$. Moreover, expressing $\bar{\phi}$  in terms of angular coordinates by means of a function $\theta_1(\theta)$,

one sees easily that $\det(\bar{\phi},\bar{\phi}_\theta)=\theta_1'(\theta)$. We are
interested in $\bar{\phi}=|\nabla f|^{-1}(f_x,f_y)$ and
$\bar{\psi}=|\nabla f|^{-1}(xf_x+yf_y,-yf_x+xf_y)$. In this case
we get the following result.
\begin{corollary}\label{cor:theta} In the notations above, the following statements hold.\begin{enumerate}
\item
$\det(\bar{\phi},\bar{\phi}_\theta)=\theta_1'(\theta)=(n-1)^{-1}|\nabla
f|^{-2}H(f)$. \item $\deg \bar{\psi}=\deg
\bar{\phi}-1$.\end{enumerate}\end{corollary}
\begin{proof} The first statement follows form Proposition \ref{prop:hessian} and Lemma \ref{lm:normaliz}.
The second one follows from Proposition \ref{prop:rotation} and Lemma \ref{lm:normaliz} since\[
\deg\bar{\psi}=\frac{1}{2\pi}\int\det(\bar{\psi},\bar{\psi}_\theta)=\frac{1}{2\pi}\int\det(\bar{\phi},\bar{\phi}_\theta)-1
=\deg\bar{\phi}-1.\]\end{proof} Now we are ready to complete the proof of Theorem
\ref{thm:main}.
\begin{proof}[Proof of Theorem \ref{thm:main}]
Since for any
$(\alpha,\beta)\in\mathbb{R}^2\setminus\{0\}$ the polynomial $\alpha f_x+\beta f_y$ has $n-1$
distinct real roots, then the map $(f_x,f_y):\mathbb{P}^1_\mathbb{R}\to \mathbb{P}^1_\mathbb{R}$
has no ramification at any real point of $\mathbb{P}^1_\mathbb{R}$. Equivalently, the jacobian
of $\bar{\phi}$ which is equal to the hessian $H(f)$ is always non-zero at the real points of
$\mathbb{P}^1_\mathbb{R}$. We call the map $\bar{\phi}:S^1\to S^1$ defined by
$\bar{\phi}=|\nabla f|^{-1}(f_x,f_y)$ and we also express it as
$\theta_1=\theta_1(\theta)$ in angular coordinates. By the observation above
 and Corollary \ref{cor:theta}
it follows that the derivative $\theta_1'(\theta)$ is non vanishing at any $\theta\in S^1$. Hence $\theta_1'(\theta)$ is either always positive or always negative.

\noindent {\em Claim:} $\theta_1'(\theta)<0$ for any $\theta$.

The sign of $\theta_1'(\theta)$ is the same than the sign of $H(f)$. Since we already know that it is constant it will be sufficient to evaluate it at a single point $(x,y)\in S^1$. We choose to examine the point $(1,0)$. One observes that for any binary form $g(x,y)={m\choose 0}a_0x^m+{m\choose 1}a_1 x^{m-1}y+\cdots+{m\choose m}a_my^m$ of degree $m\geq 3$, the Hessian $H(g)$ calculated at $(1,0)$ is equal to
\[m(m-1)\det\left(\begin{array}{ll}a_0&a_1\\a_1& a_2\end{array}\right).\] Similarly the Hessian of its derivative $g_x$ at $(1,0)$ is given by \[m(m-1)(m-2)\det\left(
\begin{array}{ll}a_0&a_1\\a_1&a_2\end{array}\right).\]
Therefore we find that $$H(g)(1,0)=(m-2)^{-1}H(g_x)(1,0).$$ Applying this result to $g=f$, we are reduced to compute the sign of $H(f_x)$. We know that $f_x$ has $n-1$ distinct real roots, so all of its
derivatives $\partial_x^i(f_x)$ have all their roots real and distinct, up to $i=n-3$.
The last of these derivatives is $h=\partial_x^{n-2}f$, and its Hessian is a constant equal to
$-\Delta(h)$,  hence $H(h)<0$. Applying recursively the reduction step, we find that
$H(f)(1,0)=(n-2)^{-1}H(f_x)(1,0)=((n-2)!)^{-1}H(h)<0,$ proving the claim.

By Corollary \ref{cor:theta}({\em 1.}), Lemma \ref{lm:winding} and applying the claim above, we
get that $\deg\bar{\phi}<0$ and $\#\phi^{-1}(z)=|\deg \phi\,|$ for any $z\in S^1$, hence
$\deg\bar{\phi}=-n+1$. Moreover, by Corollary \ref{cor:theta}({\em 2.}), we have
$\deg\bar{\psi}=\deg\bar{\phi} -1=-n$, hence $\#\mbox{real roots}(f)\geq |\deg\bar{\psi}\,|=n$.
This completes the proof of the Theorem.\end{proof}
We conclude giving a self-contained proof of the result on the rank of a real binary form
mentioned in the introduction. The arguments given are all already in \cite{coot}.
\begin{proof}[Proof of Corollary \ref{cor:rank}.]
 The statement holds for $n=3$, as shown in
\cite{coot}, Proposition 2.2. Assuming $n>3$, suppose the statement holds in degree $n-1$.
Assume rank$(f)=r$, so one can write $f=\lambda_1l_1^n+\cdots+\lambda_rl_r^n$, with $r$ minimal.
Then one can consider $l=l_1$ and $m=l_r$ and $g(t)=m^{-n}f$, with $t=l/m$. One sees that
$m^{-n+1}f_l=g'(t)$ can be expressed as a sum of at most $r-1$ $n$-th powers of linear forms in
$t$. If $f$ has $n$ distinct real roots then, by induction hypothesis $f_l$ has $n-1$ distinct
real roots and we find $r-1\geq n-1$, i.e. $r\geq n$. Since the inequality $r\leq n$ always
holds, as shown in \cite{coot} Proposition 2.1, we have $r=n$. Conversely, if the rank of $f$ is
$n$ then take $r=n$ and consider any derivative $\alpha f_x+\beta f_y=f_l$, after defining a
suitable coordinate system $l,m$, as explained in the introduction. We can consider the
polynomial $g'(t)=m^{-n+1}f_l$. If it has rank $<n-1$ then by indefinite integration over $t$ one
sees easily that $f$ has rank$<n$, contrary to the assumption. So rank$(f_l)=n-1$ and, by
induction hypothesis, it also holds that $f_l$ has $n-1$ distinct real roots. By the
arbitrariness of $l$ and by Theorem \ref{thm:main}, we conclude that $f$ has $n$ distinct
roots.
\end{proof}
\noindent
\textbf{Acknowledgements} The second author thanks G.~Ottaviani for posing the question giving rise to this paper and for stimulating discussions.