TITLE: Forced oscillations
QUESTION [0 upvotes]: Suppose we are using a pendulum.
From one extremity we just leave it, it reaches the other extremity and again reaches us at a certain time period. We will be able to apply the next force to it when it reaches us after a certain time period and hence we will be able to account for the damping force which it experiences.
This simply means that we are applying a force at a frequency which is equal to the natural frequency of the vibrating body.
But if we now apply a force at a frequency which is other than the natural frequency of the vibrating body, it is written in our book that its amplitude decreases. I searched for this doubt and I got that as the force is not in the same frequency as the natural frequency of the vibrating body the force is not applied at the time when the body is at the vicinity of the force. And so due to damping force its amplitude decreases, but we were said that a vibrating body acquires the frequency of the external periodic force, so the force should act at the correct point of time when the body is in its vicinity; as a result its amplitude should increase rather than decrease, but we do not find it so. Why?
I request you give a lucid explanation to this because I am a tenth grade student. Please don't use any complicated terminologies.

REPLY [0 votes]: A resonant driving force will cause the amplitude to increase until some damping force (such as friction) removes energy as fast as it is being added.  There is an infamous example:  On a very windy day a long suspension bridge near Tacoma started to oscillate (standing torsional waves).  The changes in shape caused the forces from the wind to get into resonance.  The damping forces were inadequate, and the bridge came apart.


TITLE: Understanding the solution of a question about sufficient statistics
QUESTION [0 upvotes]: So this is a question and solution to an old exam I was practicing for my statistics course.In the solution for 1a), they say that $L(\mathbf X;a,k)=k^ne^{\sum_{i=1}^nX_i}e^{nka} \mathbf 1(X_{(1)} \geq a)$. Why do they use the $\mathbf 1$ there, and what does that $X_{(1)}$ mean?
I calculated $L(\mathbf X;a,k)=k^ne^{\sum_{i=1}^nX_i}e^{nka}(X_i \geq a)$, but than you get in trouble for the sufficient statistic for $a$.
Thanks in advance!

REPLY [1 votes]: $X_{(1)}$ means the first order statistic of the $X_i$, in other words the minimum of the observations; similarly $X_{(n)}$ would the $n$th order statistic i.e. the $n$ lowest value
$\mathbf 1(X_{(1)} \geq a)$ is an indicator function taking the value $1$ when the event in the bracket occurs and the value $0$ when it does not, so here is $1$ when the minimum (and so all) $X_i$ are greater than equal to $a$ but $0$ when any of the $X_i$ are less than $a$


TITLE: On the Banach–Alaoglu theorem: is the unit ball of an equivalent norm also weak-* compact?
QUESTION [1 upvotes]: Suppose that $E$ is a Banach space and let $E^*$ denote its dual space with canonical norm $\lVert\bullet\rVert_{E^*}$. Suppose that $\lvert\bullet\rvert_{E^*}$ is an equivalent norm on $E^*$. The Banach–Alaoglu theorem tells us that the unit ball with respect to the norm $\lVert\bullet\rVert_{E^*}$ is weak-* compact. Is it true that the unit ball with respect to the norm $\lvert\bullet\rvert_{E^*}$ is also weak-* compact?
More generally, if $E$ is a locally convex topological vector space, and the polar topology of $E^*$ is induced by some norm $\lvert\bullet\rvert_{E^*}$. Is it true that the unit ball is weak-* compact?
If $E$ is reflexive, then the unit ball of $E^*$ with respect to the norm $\lvert\bullet\rvert_{E^*}$ is strongly closed and convex, hence weakly closed and therefore weak-* closed. I don't know in general, whether it's true.
Any ideas? Thanks!

REPLY [2 votes]: No. If this were true then the isomorphic Banach spaces $L_\infty$ and $\ell_\infty$ would have isomorphic preduals, but $L_1$ is not isomorphic to $\ell_1$. Indeed, you are asking whether all bounded isomorphisms $X^*\to X^*$ are weakly*-to-weakly* continuous (in which case they would have pre-adoints being isomorphisms on their own). This is not the case.
The drastic case of $\ell_1$ shows that that you can have uncountably many balls (closed, symmetric around the origin, convex sets with non-empty interior) in $\ell_1$ that are not homeomorphic to in the whatever weak* topology.


text:
\newcommand{\overmat}[2]{
	\makebox[0pt][l]{
		$\smash{
			\color{white}
			\overbrace{
				\phantom{
					\begin{matrix}#2\end{matrix}
				}
			}^{\text{\color{black}#1}}
		}$
	}#2
}
\newcommand\bovermat[2]{
	\makebox[0pt][l]{
		$\smash{
			\overbrace{
				\phantom{
					\begin{matrix}#2\end{matrix}
				}
			}^{\text{#1}}
		}$
	}#2
}

In this chapter we use the results from \cref{bkk-chapter} to prove B\'ezout's theorem (\cref{bezout-1}) and two of its classical generalizations: the weighted homogeneous version (\cref{wt-bezout-1}) and the weighted multi-homogeneous version (\cref{multi-bezout}). The weighted degrees considered in these results have the property that the weight of each variable is {\em positive}. In \cref{weighted-application,weighted-multapplication} we establish more general versions of these results involving arbitrary weighted degrees as special cases of the extension of Bernstein's theorem to $\kk^n$. We continue to assume that $\kk$ is an algebraically closed field.

\section{Weighted degree}
Let $\omega$ be an integral element of $\rnstar$. The {\em weighted degree} corresponding to $\omega$, which by an abuse of notation we also denote by $\omega$, is the map $\kk[x_1, \ldots, x_n] \to \zz \cup \{-\infty\}$ given by
\begin{align*}
f \mapsto \max_{\supp(f)}(\omega)
\end{align*}
Given $f = \sum_\alpha c_\alpha x^\alpha \in \kk[x_1, \ldots, x_n]$, the {\em leading form} of $f$ with respect to $\omega$ is $\ld_\omega(f) := \sum_{ \langle \omega, \alpha \rangle = \omega(f) }c_\alpha x^{\alpha}$. We say that $f$ is {\em weighted homogeneous} with respect to $\omega$ (or in short, {\em $\omega$-homogeneous}) if $\omega (x^\alpha)$ are equal for all $\alpha$ such that $c_\alpha \neq 0$, or equivalently, if $\supp(\ld_\omega(f)) = \supp(f)$.

\section{$\wpn$ as a compactification of $\kk^n$ when the $\omega_j$ are positive and $\omega_0 = 1$}
Assume $\omega$ is an integral element of $\rnnstar{n+1}$ with coordinates $(\omega_0, \ldots, \omega_n)$ with respect to the basis dual to the standard basis of $\rr^{n+1}$ such that each $\omega_j$ is {\em positive}. The {\em weighted projective space} $\wpn$ corresponding to $\omega$ was constructed in \cref{weighted-toric-section}. In this section we treat the case that $\omega_0 = 1$. For each $j = 0, \ldots, n$, let $U_j := \wpn \setminus V(x_j)$, be the ``coordinate charts'' of $\wpn$ considered in \cref{wt-identification-section}. Since $\omega_0 = 1$, \cref{toric-k^n} implies that the map $(a_1, \ldots, a_n) \mapsto [1:a_1: \cdots : a_n]$ induces an isomorphism between $\kk^n$ and $U_0$, and therefore $\wpn$ is a compactification of $\kk^n$. The set of points at infinity (with respect to $U_0 \cong \kk^n$) on $\wpn$ is $V(x_0) = \{[0: a_1: \cdots :a_n] : (a_1, \ldots, a_n) \in \kk^n \setminus \{0\}\}
\cong \wwpnn{\omega_1, \ldots, \omega_n}{n-1}$. Since $ \omega_0 = 1$, for each polynomial $f \in \kk[x_1, \ldots, x_n]$, we can define its {\em weighted homogenization} with respect to $\omega$ as $\tilde f := x_0^{\omega(f)}f(x_1/x_0^{\omega_1}, \ldots, x_n/x_0^{\omega_n})$. It is straightforward to check that $\tilde f$ is $\omega$-homogeneous and $\ld_\omega(\tilde f) = \ld_\omega(f)$.

\begin{prop} \label{wpn-closure}
Identify $\kk^n$ with $U_0$.
\begin{enumerate}
\item \label{wpn-closure-Z} Let $f \in \kk[x_1, \ldots, x_n]$. Then $V(\tilde f)$ is the Zariski closure in $\wpn$ of $V(f) \subset \kk^n$.
\item \label{wpn-closure-infty} If $f_1, \ldots, f_k \in \kk[x_1, \ldots, x_n]$, then the following are equivalent:
\begin{enumerate}
\item $\bigcap_{j=1}^k V(\tilde f_j) \setminus \kk^n = \emptyset$.
\item there is no common zero of $\ld_\omega(f_1), \ldots, \ld_\omega(f_k)$ on $\kk^n \setminus \{\origin\}$.
\end{enumerate}
\end{enumerate}
\end{prop}

\begin{proof}
At first we prove assertion \eqref{wpn-closure-Z}. If $f$ is a nonzero constant, then both $V(f)$ and $V(\tilde f)$ are empty. So assume $f$ is a non-constant polynomial. If $n = 1$, then $V(f)$ consists of finitely many points and it is straightforward to check that $V(f) = V(\tilde f) \subset \wwpnn{\omega_0, \omega_1}{1}$. So assume $n \geq 2$. Since $V(\tilde f) \cap \kk^n = V(f)$, \cref{thm:pure-dimension} implies that each irreducible component of $\dim(V(\tilde f))$ has dimension $n-1$. On the other hand,
\begin{align}
V(\tilde f) \cap (\pp^n \setminus \kk^n) = V(\tilde f) \cap V(x_0) = V(\ld_\omega(f)) \cap V(x_0) \label{wpn-infty-eqn}
\end{align}
Since $\ld_\omega(f)$ is a nonzero polynomial and $V(x_0) \cong \wwpnn{\omega_1, \ldots, \omega_n}{n-1}$, \cref{thm:pure-dimension} implies that each irreducible component of $V(\tilde f) \setminus \kk^n$ has dimension $n-2$, and therefore can not be an irreducible component of $V(\tilde f)$. Consequently every irreducible component of $V(\tilde f)$ intersects $\kk^n$ and therefore contains an irreducible component of $V(f)$. Since $V(\tilde f)$ is Zariski closed in $\wpn$, this completes the proof of assertion \eqref{wpn-closure-Z}. Assertion \eqref{wpn-closure-infty} follows from identity \eqref{wpn-infty-eqn}.
\end{proof}

\section{Weighted B\'ezout theorem} \label{wt-bezout-proof-section}
We use the following notation throughout the rest of the book: for a nonnegative integer $n$ we denote by $[n]$ the set $\{1, \ldots, n\}$; if $I \subseteq [n]$ and $k$ is a field (in most cases $k$ will be either $\kk$ or $\rr$), we write
\begin{align}
\ki & := \{(x_1, \ldots, x_n) \in k^n: x_i =  0\ \text{if}\ i \not\in I\}  \cong \kii{|I|}, \label{ki}\\
\kstari &:=  \{\prod_{i\in I} x_i \neq 0\} \cap \ki \cong \kstarii{|I|} \label{kstari}
\end{align}
where $k^* := k \setminus \{0\}$. Note that $\kii{\emptyset} = \kstarii{\emptyset} = \{0\}$.

\begin{thm}[Weighted B\'ezout theorem (\cref{wt-bezout})] \label{wt-bezout-1}
Let $\omega$ be a weighted degree on $\kk[x_1, \ldots, x_n]$ with positive weights $\omega_j$ for $x_j$, $j = 1, \ldots, n$. Then the number of isolated solutions of polynomials $f_1, \ldots, f_n$ on $\kk^n$ is bounded above by $(\prod_j \omega(f_j))/(\prod_j \omega_j)$. This bound is exact if and only if the leading weighted homogeneous forms of $f_1, \ldots, f_n$ have no common solution other than $(0, \ldots, 0)$.
\end{thm}

\begin{proof}
Due to multiadditivity of intersection multiplicities (assertion \eqref{mult-additive} of \cref{int-mult-curve}), we may replace each $f_i$ by $f_i^m$ for an appropriate positive integer $m$ and assume that $m_{i,j} := \omega(f_i)/\omega_j$ is an integer for each $i,j$. Let $\scrP_i$ be the simplex in $\rr^n$ with vertices at the origin and at $m_{i,j}e_j$, $j = 1, \ldots, n$, where $e_1, \ldots, e_n$ are the standard unit vectors in $\rr^n$. Analogous to the definition of $\multPntorusiso$ in \eqref{multPntorusiso}, define
\begin{align*}
\multPkniso := \max\{\multkniso{p_1}{p_n}: \supp(p_j) \subseteq \scrP_j,\ j =1 , \ldots, n\}
\end{align*}

It suffices to show that
\begin{prooflist}
\item \label{weighted-multpknsio} $\multPkniso = (\prod_j \omega(f_j))/(\prod_j \omega_j)$, and
\item \label{weighted-non-degeneracy} $\multfkniso = \multPkniso$ if and only if there is no common zero of $\ld_\omega(f_1), \ldots, \ld_\omega(f_n)$ on $\kk^n \setminus \{0\}$.
\end{prooflist}
Assertion \ref{weighted-multpknsio} follows from Bernstein's theorem (\cref{bkk-bound-thm}) and the following claim.

\begin{proclaim} \label{wt-bezout-claim}
\begin{enumerate}
\item $\mv(\scrP_1, \ldots, \scrP_n) = (\prod_j \omega(f_j))/(\prod_j \omega_j)$.
\item \label{wt-equal-mult} $\multPkniso = \multPntorusiso$.
\end{enumerate}
\end{proclaim}

\begin{proof}
Since $\scrP_j = \frac{\omega(f_j)}{\omega(f_1)}\scrP_1$ for each $j = 1, \ldots, n$, the properties of mixed volume from \cref{mixed-unique} imply that
\begin{align*}
\mv(\scrP_1, \ldots, \scrP_n)
	&= \mv(\scrP_1, \frac{\omega(f_2)}{\omega(f_1)}\scrP_1, \ldots, \frac{\omega(f_n)}{\omega(f_1)}\scrP_1)
	= \prod_{j=2}^n \frac{\omega(f_j)}{\omega(f_1)} \mv(\scrP_1, \ldots, \scrP_1)\\
	&= \prod_{j=2}^n \frac{\omega(f_j)}{\omega(f_1)} n!\vol_n(\scrP_1)
	=  \prod_{j=2}^n \frac{\omega(f_j)}{\omega(f_1)}   \prod_{j=1}^n m_{1,j}
	=  \prod_{j=1}^n \frac{\omega(f_j)}{\omega_j}
\end{align*}
This proves the first assertion of the claim.

For the second assertion, let $p_j$ be an arbitrary polynomial supported at $\scrP_j$, $j = 1, \ldots, n$. Since it is clear that $\multPkniso \geq \multPntorusiso$, it suffices to show that there are $p'_j$ supported at $\scrP_j$ such that $\multkniso{p_1}{p_n} \leq \multntorusiso{p'_1}{p'_n}$. Write $\scrA_j := \scrP_j \cap \zz^n$, $j = 1, \ldots, n$. Pick $(\scrA_1, \ldots, \scrA_n)$-non-degenerate $(q_1, \ldots, q_n) \in \scrL(\scrA_1, \ldots, \scrA_n)$, and set $r_i(x,t) := (1-t)p_i + tq_i$. For each $\epsilon \in \kk$, write $r_{\epsilon,i} := r_i|_{t = \epsilon}$. Note that $r_{1,i} = q_i$ and $r_{0,i} = p_i$. Pick a generic $\epsilon \in \kk$. Due to \cref{bkk-0} we may assume that $(r_{\epsilon, 1}, \ldots, r_{\epsilon,n})$ is also $(\scrA_1, \ldots, \scrA_n)$-non-degenerate. Since the intersection of an $\scrA_j$ with a coordinate subspace is a (nonempty) face of $\scrA_j$, it is then straightforward to check (e.g.\ using \cref{finite-remark}) from the definition of $(\scrA_1, \ldots, \scrA_n)$-non-degeneracy that $r_{\epsilon, 1}, \ldots, r_{\epsilon,n}$ do not have any common zero on $\kk^n \setminus \nktorus$. Assertion \eqref{openly-max} of \cref{mult-deformation-global} then implies that $\multkniso{p_1}{p_n} \leq \multkniso{r_{\epsilon, 1}}{r_{\epsilon, n}} = \multntorusiso{r_{\epsilon, 1}}{r_{\epsilon, n}}$, as required.
\end{proof}

Now we prove assertion \ref{weighted-non-degeneracy}. At first assume $\multfkniso < \multPkniso$. We will show that the leading weighted homogeneous forms $\ld_\omega(f_i)$ of $f_i$ have a common zero on $\kk^n \setminus \{0\}$. Let $\omega' := (1, \omega_1, \ldots, \omega_n)$. Embed $\kk^n$ into $\wpnprime$ via the map $(x_1, \ldots, x_n) \mapsto [1: x_1: \cdots : x_n]$. Set $p_i := f_i$, $i = 1, \ldots, n$. Let $q_i$ and $r_i := (1-t)p_i + tq_i$ be as in the proof of \cref{wt-bezout-claim}. Let $V$ be the (finite) set of common zeroes of $q_1, \ldots, q_n$ on $\kk^n$ and $C \subseteq \kk^{n+1}$ be the union of irreducible components of $V(r_1, \ldots, r_n)$ which intersect $V \times \{1\}$. Assertion \eqref{onely-isolated} of \cref{mult-deformation-global} implies that $C$ is a curve. Since $\multfkniso < \multkniso{q_1}{q_n}$, assertion \eqref{non-max-condition} of \cref{mult-deformation-global} implies that one of the following holds:
\begin{prooflist}[resume]
\item \label{possibility-non-isolated} there is a positive dimensional component of $V(f_1, \ldots, f_n) \subset \kk^n$, or
\item \label{possibility-at-infinity} $C$ ``has a point at infinity at $t = 0$'', i.e.\ if $\bar C$ is the closure of $C$ in $\wpnprime \times \pp^1$, then $\bar C \cap ((\wpnprime \setminus X) \times \{0\}) \neq \emptyset$.
\end{prooflist}
Denote the weighted homogeneous coordinates on $\wpnprime$ by $[x_0:  \cdots : x_n]$. Let $\tilde f_i$ and $\tilde q_i$ be the {\em weighted homogenization} with respect to $\omega'$ respectively of $f_i$ and $q_i$. \Cref{wpn-closure} implies that the closures of $V(f_i)$ and $V(q_i)$ in $\wpnprime$ are respectively $V(\tilde f_i)$ and $V(\tilde q_i)$. Since $\omega(f_i) = \omega(q_i)$ for each $i$, it follows that the closure of $V(r_i)$ in $\wpnprime \times \kk$ is $V(\tilde r_i)$, where $\tilde r_i := (1-t)\tilde f_i + t \tilde q_i$. If \ref{possibility-non-isolated} holds, then the closure of $V(f_1, \ldots, f_n)$ in $\wpnprime$ contains a point $a' \in \wpnprime \setminus \kk^n$. The weighted homogeneous coordinates of $a'$ are of the form $[0: a_1: \cdots : a_n]$ with $a = (a_1, \ldots, a_n) \in \kk^n \setminus \{0\}$. Then $\tilde f_i(a')  = 0$, and therefore $\ld_\omega(f_i)(a) = 0$ for each $i$. On the other hand, if \ref{possibility-at-infinity} holds, then let $(a',0) \in \bar C \cap ((\wpnprime \setminus X) \times \{0\})$. Since $\tilde r_i(a',0) = 0$ for each $i$, this again yields a common zero of the $\ld_\omega(f_i)$ on $\kk^n\setminus \{0\}$, as required. \\

It remains to show the necessity of the non-degeneracy condition. Assume the leading weighted homogeneous forms $\ld_\omega(f_i)$ of $f_i$ have a common solution $a \in \kk^n \setminus \{\origin\}$. As in \cref{bkk-necessary-section}, pick BKK non-degenerate $g_1, \ldots, g_n$ with a common zero $b \in \kk^n$ such that for each $i$, $\np(g_i) = \scrP_i$ and $\ld_\omega(g_i)(a) \neq 0 \neq f_i(b)$. Define a rational curve $C$ on $\kk^n$ via the parametrization $c(t):= (c_1(t), \ldots, c_n(t)) :\kk \to \kk^n$ from \eqref{trick-curve} with $\nu := -\omega$, i.e.
\begin{align*}
c_j(t) := a_jt^{-\omega_j} + (b_j - a_j)t^{-\omega_j + k_j},\ j = 1, \ldots, n.
\end{align*}
where each $k_j$ is a positive integer. Define $h_1, \ldots, h_n$ as in \eqref{trick-deformation} with $m_j = -\omega(f_j) = -\omega(g_j)$, i.e.\
\begin{align*}
h_j &:=
	t^{\omega(f_j)}f_j(c(t))g_j - t^{\omega(g_j)}g_j(c(t))f_j
\end{align*}
Note that $t^{\omega(f_j)}f_j(c(t))$ and $t^{\omega(g_j)}g_j(c(t))$ are {\em polynomials} in $t$. The same arguments as in the alternate proof of necessity of BKK non-degeneracy in \cref{modified-proof} show that $\multfkniso < \multgkniso$, so that the weighted homogeneous bound is {\em not} exact.
\end{proof}

\begin{cor}[B\'ezout's theorem (\cref{bezout})] \label{bezout-1}
The number of isolated solutions of polynomials $f_1, \ldots, f_n$ on $\kk^n$ is bounded above by $\prod_j \deg(f_j)$. This bound is exact if and only if the leading homogeneous forms of $f_1, \ldots, f_n$ have no common solution other than $(0, \ldots, 0)$.
\end{cor}

\begin{proof}
In \cref{wt-bezout-1} take $\omega$ to be the usual degree of polynomials.
\end{proof}

Combining the arguments of the proof of assertion \eqref{wt-equal-mult} of \cref{wt-bezout-claim} with \cref{thm:pure-dimension,bkk-bound-thm,positively-mixed} gives the following characterization of polytopes $\scrP_1, \ldots, \scrP_n$ such that $\multPkniso = \multPntorusiso$. Given a coordinate subspace $H$ of $\rr^n$, we write $\thp :=  \{j: \scrP_j \cap H \neq \emptyset\} \subset [n]$.

\begin{lemma} \label{equal-multpn}
Let $\scrP_1, \ldots, \scrP_n$ be convex integral polytopes in $\rzeroo{n}$. Then the following are equivalent:
\begin{enumerate}
\item $\multPkniso = \multPntorusiso$.
\item For each proper coordinate subspace $H$ of $\rr^n$, one of the following is true:
\begin{enumerate}
\item either there is a coordinate subspace $H'$ of $\rr^n$ such that $H' \supset H$ and $|\thprimep| < \dim(H')$, or
\item $\thp$ is nonempty and $\{\scrP_j \cap H: j \in \thp\}$ is a {\em dependent} collection of polytopes. \qed
\end{enumerate}
\end{enumerate}
\end{lemma}

\section{Products of weighted projective spaces}
In this section we examine the closures of the hypersurfaces in a product of weighted projective spaces. This would be useful in the proof of the weighted multi-homogeneous B\'ezout bound (\cref{multi-bezout}). Let $\bar X := \prod_{j=1}^s \wwpnn{\omega_j}{n_j}$, where each $\omega_j$ is a weighted degree on $A_j := \kk[x_{j,0}, \ldots, x_{j,n_j}]$ such that the weight $\omega_{j,k}$ of $x_{j,k}$ is positive for each $j,k$. Let $A := \kk[x_{j,k}: 1 \leq j \leq s,\ 0 \leq k \leq n_j]$, and for each $j$, let $\tilde \omega_j$ be the trivial extension of $\omega_j$ to $A$, i.e.\
\begin{align*}
\tilde \omega_j(x_{k,l})
	&=
	\begin{cases}
	\omega_{k,l} & \text{if}\ k = j,\\
	0 & \text{otherwise.}
	\end{cases}
\end{align*}
We say that a polynomial $h\in A$ is {\em (weighted multi-) homogeneous with respect to $\Omega := \{\omega_1, \ldots, \omega_s\}$}, or in short, $f$ is {\em $\Omega$-homogeneous}, if it is $\tilde \omega_j$-homogeneous for each $j = 1, \ldots, s$. If $h$ is $\Omega$-homogeneous, then $V(h) := \{a \in \bar X: h(a) = 0\}$ is a well defined Zariski closed subset of $\bar X$.

\subsection{The case that \bm{$\omega_{j,0} = 1$} for each \bm{$j = 1, \ldots, s$}} \label{multi-compactisection}
In this section we consider the case that $\omega_{j,0} = 1$ for each $j$. In this case each $\wwpnn{\omega_j}{n_j}$ is a compactification of $\kk^{n_j}$, and therefore $\bar X$ is a compactification of $\kk^n$, where $n := \sum_{j=1}^s n_j$. More precisely, $\kk^n$ can be identified with $\bar X \setminus V( \prod_{j=1}^s x_{j,0})$, and we may treat $B := \kk[x_{j,k}: 1 \leq j \leq s,\ 1 \leq k \leq n_j] \subset A$ as the coordinate ring of $\kk^n$. Let $f \in B$. Then {\em $\Omega$-homogenization} $\tilde f$ of $f$ is formed by substituting $x_{j,k}/x_{j,0}$ for $x_{j,k}$ in $f$ for each $j,k$, and then clearing out the denominator. The following result is the analogue of assertion \eqref{wpn-closure-Z} of \cref{wpn-closure}, and follows from the same arguments.

\begin{prop} \label{multi-closure-Z}
Let $f \in B$. If $\kk$ is algebraically closed, then $V(\tilde f)$ is the Zariski closure in $\bar X$ of $V(f)\subset \kk^n$. \qed
\end{prop}

The set of points at infinity on $\bar X$ is $\bar X \setminus \kk^n = V(\prod_{j=1}^s x_{j,0}) = \bigcup_{J} Y_J$, where the union is over all {\em nonempty} subsets $J$ of $[s] := \{1, \ldots, s\}$, and $Y_J$ are defined as follows:
\begin{align*}

Y_J
	&=  V(x_{j,0} : j \in J) \setminus V(\prod_{j \not\in J} x_{j,0})
	\cong \prod_{j \in J} \wwpnn{\omega'_j}{n_j-1} \times \prod_{j \not\in J} \kk^{n_j}
\end{align*}
where $\omega'_j$ are the restriction of $\omega_j$ to $\kk[x_{j,1}, \ldots, x_{j,n_j}]$ for each $j$. Fix a nonempty subset $J$ of $[s]$.  Given $f \in B$, we would like to compute the points at infinity on $V(\tilde f)$ which belong to $Y_J$. If $f = \sum_\alpha c_\alpha x^\alpha$, we write $\ld_{\Omega, J}(f)$ for be sum of $c_\alpha x^\alpha$ over all $\alpha$ such that $\tilde \omega_j(x^\alpha) = \tilde \omega_j(f)$ for each $j \in J$. In other words, $\ld_{\Omega, J}(f)$ is obtained from $\tilde f$ by substituting $x_{j,0} = 0$ for each $j \in J$ and $x_{j,0} = 1$ for each $j \not\in J$.

\begin{example}
Let $s = 3$, $n_1 = n_2 = 1$ and $n_3 = 2$ (so that $n = 4$). Let $\omega_1, \omega_2$ be the usual degree in respectively $x_{1,1}$ and $x_{2,1}$ coordinates, and $\omega_3$ be the weighted degree in $(x_{3,1}, x_{3,2})$ coordinates corresponding to weights $2$ for $x_{3,1}$ and $3$ for $x_{3,2}$. Let $f = x_{1,1}^5 +  x_{2,1}^7 + x_{1,1}^5x_{2,1}^7 + x_{3,1}^3 +  x_{1,1}^5x_{3,2}^2 $. Then $\omega_1(f) = 5$, $\omega_2(f) = 7$, $\omega_3(f) = 6$, and
\begin{align*}
\ld_{\Omega, J}(f)
	&=
	\begin{cases}
		x_{1,1}^5 + x_{1,1}^5x_{2,1}^7 +  x_{1,1}^5x_{3,2}^2 & \text{if}\ J = \{1\},\\
		x_{2,1}^7  + x_{1,1}^5x_{2,1}^7 & \text{if}\ J = \{2\},\\
		x_{3,1}^3 +  x_{1,1}^5x_{3,2}^2 & \text{if}\ J = \{3\},\\
		x_{1,1}^5x_{2,1}^7 	 & \text{if}\ J = \{1,2\},\\
		x_{1,1}^5x_{3,2}^2 	& \text{if}\ J = \{1,3\},\\
		0 & \text{if $J = \{2, 3\}$ or $J = \{1,2,3\}$.}\\
	\end{cases}
\end{align*}
Note the following difference from the weighted homogeneous case: if $|J| \geq 2$, then it might happen that $\ld_{\Omega, J}(f) = 0$ even if $f$ is a nonzero polynomial.
\end{example}

Since $\ld_{\Omega, J}(f)$ is $\tilde \omega_j$-homogeneous for each $j \in J$, it defines a well defined Zariski closed subset of $Y_J$. It is straightforward to check that this set is precisely the intersection of $V(\tilde f)$ and $Y_J$, which is the content of the next result.

\begin{prop} \label{multi-closure-infty}
Assume $\kk$ is algebraically closed. If $f \in B$, then $V(\tilde f) \cap Y_J = V(\ld_{\Omega, J}(f)) \cap Y_J$. If $f_1, \ldots, f_k \in B$, then the following are equivalent
\begin{enumerate}
\item $\bigcap_{j=1}^k V(\tilde f_j) \setminus \kk^n = \emptyset$.
\item For each nonempty subset $J$ of $[s]$, there is no common zero of $\ld_{\Omega, J}(f_1), \ldots, \ld_{\Omega, J}(f_k)$ on $\prod_{j \in J} (\kk^{n_j} \setminus \{\origin\}) \times \prod_{j \not\in J} \kk^{n_j}$. \qed
\end{enumerate}
\end{prop}

\section{Weighted multi-homogeneous B\'ezout theorem}

We now generalize the weighted B\'ezout bound to the multi-projective setting. Let $\mscrI := (I_1, \ldots, I_s)$ be an ordered partition of $[n] := \{1, \ldots, n\}$, i.e.\ $[n] = \bigcup_j I_j$ and $\sum_j |I_j| = n$. For each $j = 1, \ldots, s$, let $\omega_j$ be a weighted degree on $\kk[x_k: k \in I_j]$ with positive weights $\omega_{j,k}$ for $x_k$, $k  \in I_j$. Let $f_1, \ldots, f_n$ be polynomials on $\kk^n$. Given $d_{i,j} := \omega_j(f_i)$, we would like to compute a (sharp) upper bound of the number of isolated solutions of $f_1, \ldots, f_n$. Let $n_j := |I_j|$ and $l_j$ be the least common multiple of $\omega_{j,1}, \ldots, \omega_{j,n_j}$ and $\scrS_j$ be the simplex in $\rzeroo{n_j}$ defined by $\{ \alpha: \langle \omega_j , \alpha \rangle \leq l_j\}$. Note that $\supp(f_i) \subseteq \scrP_i := \prod_{j=1}^s (d_{i,j}/l_j) \scrS_j \subset \rr^n$. By definition $\mv(\scrP_1, \ldots, \scrP_n)$ is the coefficient of $\lambda_1 \cdots \lambda_n$ in the polynomial
\begin{align*}
\vol_n(\sum_{i=1}^n \lambda_i \scrP_i)
	&= \vol_n(\sum_{i=1}^n \lambda_i \prod_{j=1}^s \frac{d_{i,j}}{l_j}  \scrS_j)
	= \vol_n(\prod_{j=1}^s (\sum_{i=1}^n \lambda_i \frac{d_{i,j}}{l_j}  \scrS_j))
	= \prod_{j=1}^s (\sum_{i=1}^n \lambda_i \frac{d_{i,j}}{l_j} )^{n_j} \vol_n(\prod_{j=1}^s \scrS_j) \\
	&= \frac{\prod_{j=1}^s\prod_{k=1}^{n_j}(l_j/\omega_{j,k})} {\prod_{j=1}^s (n_j! l_j^{n_j})}
	\prod_{j=1}^s (\sum_{i=1}^n \lambda_i d_{i,j})^{n_j}
	= \frac{ \prod_{j=1}^s (\sum_{i=1}^n \lambda_i d_{i,j})^{n_j}}
	 {\prod_{j=1}^s (n_j! \prod_{k=1}^{n_j}\omega_{j,k})}
\end{align*}
The coefficient of $\lambda_1 \cdots \lambda_n$ in $\prod_{j=1}^s (\sum_{i=1}^n \lambda_i d_{i,j})^{n_j}$ is the {\em permanent} of the following $n \times n$ matrix:
\vspace{2mm}

\begin{align*}
\begin{matrix}
D(\mscrI, \vec d)
	&:=
	  \begin{pmatrix}
	  \bovermat{$n_1$ times}{ d_{1,1}  & \cdots & d_{1,1}}
	  	&
	  \overmat{$\mkern-3.5mu\cdots$}{\cdots & \cdots}
	  	&
	  \bovermat{$n_s$ times}{ d_{1,s}  & \cdots & d_{1,s}}
	   	\\[0.5em]
	   	\vdots & & \vdots & & & \vdots & & \vdots \\
	   	d_{n, 1}  & \cdots & d_{n,1} & \cdots & \cdots & d_{n,s}  & \cdots & d_{n,s}
	   \end{pmatrix}
\end{matrix}
\end{align*}
The preceding observations together with \cref{equal-multpn} imply that
\begin{align}
\multfkniso
    \leq \multPkniso
    = \mv(\scrP_1, \ldots, \scrP_n)
	= \frac{ \perm(D(\mscrI, \vec d))}
		 {(\prod_j n_j!) (\prod_{j,k}\omega_{j,k})}
		 \label{wt-multi-homogeneous-bound}
\end{align}
Note that when $s = 1$, then $\Omega$ consists of only one weighted degree $\omega$ and $\perm(D(\mscrI, \vec d)) = n! \omega(f_1) \cdots \omega(f_n)$, so that the bound from \eqref{wt-multi-homogeneous-bound} is precisely the weighted B\'ezout bound $\prod_{j=1}^n(\omega(f_j)/\omega_j)$. Now we determine the condition for the attainment of this bound. For each $j = 1, \ldots, s$, fix a new indeterminate $u_j$. Let $B'_j$ be the ring of polynomials (over $\kk$) in $u_j$ and $x_k$, $k \in I_j$. Note that each $\xpp{\scrP_i}$ is isomorphic to $\prod_{j=1}^s \wwpnn{\omega'_j}{n_j}$, where $\omega'_j$ is the weighted degree on $B'_j$ such that $\omega'_j(u_j) = 1$ and $\omega'_j(x_k) = \omega_{j,k}$, $k \in I_j$. For each $g \in \kk[x_1, \ldots, x_n]$ and each nonempty subset $J$ of $[s]$, define $\ld_{\Omega, J}(g)$ as in \cref{multi-compactisection}. Finally, note that since $\mscrI$ is a partition of $[n]$, we may identify $\kk^n$ with $\prod_{j=1}^s \Kii{I_j}$.

\begin{thm}[Weighted multi-homogeneous B\'ezout theorem] \label{multi-bezout}
The number of isolated solutions of polynomials $f_1, \ldots, f_n$ on $\kk^n$ is bounded by \eqref{wt-multi-homogeneous-bound}. This bound is exact if and only if the following holds: for each nonempty subset $J$ of $[s]$, there is no common zero of $\ld_{\Omega, J}(f_1), \ldots, \ld_{\Omega, J}(f_n)$ on $\prod_{j \in J} (\Kii{I_j} \setminus \{\origin\}) \times \prod_{j \not\in J} \Kii{I_j}$ (upon identification of $\kk^n$ with $\prod_{j=1}^s \Kii{I_j}$).
\end{thm}

\begin{proof}
We have already proved that \eqref{wt-multi-homogeneous-bound} holds. Let $\tilde f_i$ be the {\em $\Omega$-homogenization} of $f_i$ defined as in \cref{multi-compactisection}. If the bound is not exact, then it follows by the same arguments as in the proof of \cref{wt-bezout-1} that $\bigcap_{j=1}^s V(\tilde f_j) \setminus \kk^n \neq \emptyset$, and then \cref{multi-closure-infty} implies that the condition in the second assertion of \cref{multi-bezout} is violated. Now assume there is a nonempty subset $J$ of $[s]$ such that $\ld_{\Omega, J}(f_1), \ldots, \ld_{\Omega, J}(f_n)$ have a common zero $a = (a_1, \ldots, a_n) \in \prod_{j \in J} (\Kii{I_j} \setminus \{\origin\}) \times \prod_{j \not\in J} \Kii{I_j}$. Replacing each $f_i$ by $f_i^m$ for some appropriate positive integer $m$, we may assume that the vertices of each $\scrP_i$ have integer coordinates. Let $I := \{i: a_i \neq 0\} \subseteq [n]$; in other words, $I$ is the smallest subset of $[n]$ such that $a \in \Kstari$. Since $J$ is nonempty, it follows that $I$ is also nonempty; in fact $I_j \cap I \neq \emptyset$ for each $j \in J$. For each $j \in J$, let $\scrT_j$ be the facet of $\scrS_j$ determined by $\langle \omega_j, \alpha \rangle = l_j$. Fix $i \in [n]$. Define
\begin{align*}
\scrQ_i := \ri \cap \left (\prod_{j \in J}(d_{i,j}/l_j)\scrT_j \times \prod_{j \not\in J} (d_{i,j}/l_j)\scrS_j \right)
\end{align*}
Then $\scrQ_i$ is a proper (nonempty) face of $\scrP_i$. Let $f_{i, \scrQ_i}$ be the component of $f_i$ supported at $\scrQ_i$, i.e.\ if $f_i = \sum_\alpha c_\alpha x^\alpha$, then $f_{i, \scrQ_i} = \sum_{\alpha \in \scrQ_i} c_\alpha x^\alpha$. It is straightforward to check that $f_{i,\scrQ_i}(a) = 0$ for each $i$. Now choose an integral element $\nu \in \rnstar$ such that $\scrQ_i = \In_\nu(\scrP_i)$ for each $i$ [you have to check that such $\nu$ exists!]. As in \cref{bkk-necessary-section}, pick BKK non-degenerate $g_1, \ldots, g_n$ with a common zero $b \in \kk^n$ such that for each $i$, $f_i(b) \neq 0$, $\np(g_i) = \scrP_i$ and $\In_\nu(g_i)(a) \neq 0$. Define a rational curve $C$ on $\kk^n$ via the parametrization from \eqref{trick-curve} and define $h_1, \ldots, h_n$ as in \eqref{trick-deformation} with $m_i = \nu(g_i)$ for each $i$. Then $t^{-m_j}f_j(c(t))$ and $t^{-m_j}g_j(c(t))$ are polynomials in $t$ and the same arguments as in \cref{bkk-necessary-section} show that $\multfkniso < \multgkniso$, as required.
\end{proof}

\section{Notes}
A version of the weighted B\'ezout theorem appears in \cite{damon}. I.\ Shafarevich gave the bound for the ``multi-homogeneous'' case  (i.e.\ the case that all weights are $1$) of the weighted multi-homogeneous B\'ezout theorem in the first edition of \cite{shaf1} in the 1970s; see also \cite{morgan-sommese}. We could not locate any past reference for the non-degeneracy condition or the estimate for the general weighted multi-homogeneous version.

config: arxiv


TITLE: Covariance dependent Binomial variables
QUESTION [1 upvotes]: Suppose $Y \sim \mathrm{Bin}(n,p)$ given $\Theta=(p,q)$ and $Z \sim \mathrm{Bin}(y,q)$ given $Y=y$ and $\Theta=(p,q)$.
Now I want to determine the variance of $Z-Y$, but I don't know how. I know $\operatorname{Cov}(Z,Y)=E(ZY)-E(Z)E(Y)$, but I'm stuck on the first term.

REPLY [1 votes]: We'll use the result $\mathbb{E}[X] = \mathbb{E}[ \mathbb{E}[X\mid Y] ]$
It's easy to see that
$$\mathbb{E}[Z\mid Y] = Y q $$
Now,
\begin{align}
\mathbb{E}[ZY ] &= \mathbb{E} [ \mathbb{E}[ZY \mid  Y] ] \\
&= \mathbb{E}[ Y \mathbb{E}[Z\mid Y] ] \\
&= \mathbb{E}[ q Y^2 ]\\
&= npq ( np + 1 - p)
\end{align}
You'll also need the second moment of $Z$, can you figure out how to calculate it now?


text:
\begin{document}
\null
\vspace{1cm}
\begin{center}
{\LARGE Local calibrations for minimizers of the  Mumford-Shah

\vspace{.5cm}

 functional with rectilinear discontinuity sets}
\vspace{1cm}

 {\normalsize Gianni \sc Dal Maso}

\vspace{.25cm}
 {\normalsize Maria Giovanna \sc Mora}

\vspace{.25cm}
{\normalsize Massimiliano \sc Morini}

\vspace{1cm}

{\normalsize S.I.S.S.A.}\\
{\normalsize Via Beirut 2-4, 34014 Trieste, Italy}\\
{\normalsize e-mail: \tt dalmaso@sissa.it, mora@sissa.it,
morini@sissa.it}

\vspace{3cm}

\begin{minipage}[t]{11.5cm}
{\normalsize
\centerline{ {\bf Abstract}}
\noindent Using a calibration method,
we prove that, if $w$ is a function which satisfies
all Euler conditions for the Mumford-Shah functional on a two-dimensional
open set $\Omega$, and the discontinuity set of $w$ is a segment
connecting two boundary points, then for every point $(x_0, y_0)$ of
$\Om$ there exists a neighbourhood $U$ of $(x_0, y_0)$ such that
$w$ is a minimizer of the Mumford-Shah functional on $U$ with respect
to its own boundary values on $\partial U$.}
\vspace{.5cm}\\

\noindent{\normalsize {\bf AMS (MOS) subject classifications:}
49K10 (primary),
49Q20 (secondary\-)
}

\vspace{.5cm}
\noindent{\normalsize {\bf Key words:} free-discontinuity problems,
calibration method}
\end{minipage}
\end{center}

\setcounter{page}{0}
\thispagestyle{empty}

\vfill

\pagebreak

\clearpage
\section{Introduction}

This paper deals with Dirichlet minimizers of the Mumford-Shah
functional (see \cite{Mum-Sha1} and \cite{Mum-Sha2})
\begin{equation} \label{g1}
\int_{\Om} |\nabla u(x,y)|^{2} dx\,dy +  \haus(S_{u}) \,,
\end{equation}
where $\Omega$ is a bounded open subset of $\rdue$ with a Lipschitz
boundary, $\haus$ is the one-dimensional Hausdorff measure,
$S_{u}$ is the set of essential
discontinuity points of the unknown function $u$, while $\nabla u$
denotes its approximate gradient (see \cite{Amb} or \cite{Amb-Fus-Pal}).

A {\it Dirichlet minimizer\/} of (\ref{g1}) in $\Om$  is a function $w$ which
belongs to the space $SBV(\Om)$ of special functions of bounded
variation in $\Om$ (see
\cite{Amb} or \cite{Amb-Fus-Pal}) and satisfies the inequality
$$
\int_{\Om} |\nabla w(x,y)|^{2} dx \,dy + \haus(S_{w}) \,\le\,
\int_{\Om} |\nabla u(x,y)|^{2} dx \,dy + \haus(S_{u})
$$
for every function $u\in SBV(\Om)$ with the same trace as $w$ on $\partial\Om$.

Suppose that $w$ is a Dirichlet minimizer
of (\ref{g1}) in $\Om$ and that $S_{w}$ is a regular curve. Then the following
equilibrium conditions are satisfied (see \cite{Mum-Sha1} and
\cite{Mum-Sha2}):
\begin{description}
\item[(a)] $w$ is harmonic on $\Om\setminus S_{w}$;
\item[(b)] the normal derivative of $w$ vanishes on both sides of $S_{w}$;
\item[(c)] the curvature of $S_{w}$ is equal to the
difference of the squares of the tangential derivatives of $w$
on both sides of $S_{w}$.
\end{description}
Elementary examples show that conditions (a), (b), and (c) are not
sufficient for the Dirichlet minimality of $w$.

In this paper we prove that, if
$S_{w}$ is a straight line segment connecting two
points of $\partial \Om$, and  the tangential derivatives
$\diff_{\tau} w$ and $\diff_{\tau}^2 w$
of $w$ do not vanish on both sides of $S_{w}$, then (a), (b), and (c)
imply that every point $(x_{0}, y_{0})$ in $\Om$ has
an open neighbourhood $U$
such that $w$ is a Dirichlet minimizer of (\ref{g1}) in $U$. In
other words, under our assumptions, conditions (a), (b), and (c) are
also sufficient for the Dirichlet minimality in small domains.
We hope that our proof will be useful in the future to achieve the same
result without our special assumptions on $S_{w}$.

The proof is obtained by using the calibration method
adapted in
\cite{Alb-Bou-DM}
to the functional (\ref{g1}).
We construct an explicit calibration for $w$
in the cylinder $U{ \times }\erre$, where $U$ is a suitable
neighbourhood of $(x_{0}, y_{0})$. This construction is elementary
when $(x_{0}, y_{0}) \notin S_{w}$ (see \cite{Alb-Bou-DM}), so we
consider only the case $(x_{0}, y_{0})\in S_{w}$.

The plan of the paper is the following.
In Section 2 we fix the notations and we
recall the main result of \cite{Alb-Bou-DM}.
In Theorem \ref{x-x} we consider
the special case of the function
$$w(x,y) := \begin{cases}
\;\;\; x & \text{if $y>0$}, \\
-x & \text{if $y<0$},
\end{cases}$$
and give in full detail the first
example of a calibration for a
discontinuous function
which is not locally constant.
In Theorem \ref{x-x+1} we adapt the same construction to the function
$$w(x,y) := \begin{cases}
x+ 1 & \text{if $y>0$}, \\
x & \text{if $y<0$}.
\end{cases}$$
In Section 4
we consider the general case of a function $w$ satisfying (a),
(b), and (c) and with $S_{w}=\{ (x,y) \in \Omega : y=0 \}$.
If $S_w$ is connected, only two situations are possible:
\begin{eqnarray}
& \partial_{x} w(x,0+) = - \partial_{x} w(x,0-) \quad  \hbox{on} \ \, S_{w}, &
\label{opp} \\
& \partial_{x} w(x,0+) = \partial_{x} w(x,0-) \quad  \hbox{on} \ \, S_{w}. &
\label{conc}
\end{eqnarray}
The former case (\ref{opp}) is studied in Theorem \ref{harm} by a
suitable change of variables and by adding two new parameters to the
construction used in Theorem \ref{x-x}.
The minor changes for the case (\ref{conc})
are considered in Theorem \ref{harm2}.

\section{Preliminary results}

Let $\Omega$ be a bounded open subset of $\rdue$ with a Lipschitz
boundary and let
$$\Omega_{0} = \{ (x,y) \in \Omega : y \neq 0 \}, \qquad
S = \{ (x,y)\in \Omega : y=0 \}.$$
For every vector field $\varphi :
\Om { \times } \erre \to \rdue { \times } \erre$ we define the maps
$\varphi^{x}, \; \varphi^{y}, \; \varphi^{z} : \Om { \times } \erre \to
\erre$ by
$$\varphi (x,y,z) = (\varphi^{x}(x,y,z), \varphi^{y}(x,y,z),
\varphi^{z}(x,y,z)).$$
We shall consider the collection ${\cal F}$ of all piecewise $C^{0}$
vector fields $\varphi :
\Om { \times } \erre \to \rdue { \times } \erre$
with the following property: there exists a finite
number $g_{1}, \ldots, g_{k}$ of functions in $C^{1}(\overline{\Om})$
such that the sets
$$A_{i} : = \{ (x,y,z): (x,y)\in \Om, \ g_{i}(x,y) < z < g_{i+1}(x,y) \}$$
are nonempty and $\varphi \in C^{1}(\overline{A_{i}}, \rdue {
\times } \erre)$ for $i=0, \ldots, k$, where we put $g_{0}=-\infty$
and $g_{k+1} = + \infty$.
Therefore, the discontinuity set of a vector field in ${\cal F}$ is contained in
a finite number of regular surfaces.

Let $w \in C^{1}(\Om_{0})$ be a function such that
$\int_{\Om_{0}} |\nabla w|^{2} dx\, dy < + \infty$.
The upper trace of $w$ on $S$ is denoted by $w(x,0+)$, and the lower
trace by $w(x,0-)$. Therefore, the approximate upper and lower limits
$w^{+}(x,0)$ and $w^{-}(x,0)$
are given by
$$w^{+}(x,0) = \max \{ w(x,0+), w(x,0-) \} \qquad \hbox{and}
\qquad w^{-}(x,0) = \min \{ w(x,0+), w(x,0-) \}.$$
A {\it calibration \/} for $w$ is a bounded vector field $\varphi
\in {\cal F}$ which is continuous on the graph of $w$ and
satisfies the following properties:
\begin{description}
\item[(a)] $\dive \varphi = 0$ in the sense of distributions in
$\Om { \times } \erre$;
\item[(b)] $(\varphi^{x}(x,y,z))^{2} + (\varphi^{y}(x,y,z))^{2} \leq
4 \varphi^{z}(x,y,z)$ at every continuity point $(x,y,z)$ of $\varphi$;
\item[(c)]  $(\varphi^{x}, \varphi^{y})(x,y,w(x,y)) = 2 \nabla  w(x,y)$
and
$\varphi^{z}(x,y,w(x,y)) = |\nabla w(x,y)|^{2}$ for every $(x,y)\in
\Om_{0}$;
\item[(d)] $\displaystyle \left( \int_{t_{1}}^{t_{2}} \varphi^{x}(x,y,z)\, dz
\right)^{2} + \left( \int_{t_{1}}^{t_{2}} \varphi^{y}(x,y,z)\, dz
\right)^{2} \leq 1$ for every $(x,y)\in \Om$ and for every $t_{1},
t_{2}\in \erre$;
\item[(e)] $\displaystyle \int_{w^{-}(x,0)}^{w^{+}(x,0)} \varphi^{x}(x,0,z)\, dz =0$
and
$\displaystyle \int_{w^{-}(x,0)}^{w^{+}(x,0)} \varphi^{y}(x,0,z)\, dz =1$ for every
$(x,0)\in S$.
\end{description}

The following theorem is proved in \cite{Alb-Bou-DM}.

\begin{theorem}
If there exists a calibration $\varphi$ for $w$, then $w$ is a
Dirichlet minimizer of the Mumford-Shah functional (\ref{g1}) in $\Om$.
\end{theorem}

If $\Om$ is a circle with centre on the $x$-axis, and
$w\in C^{1}(\Om_{0})$ with $\int_{\Om_{0}} |\nabla w|^{2} dx\, dy < + \infty$,
then $w$ satisfies the Euler conditions (a), (b), and (c)
if and only if $w$ has one of the following forms:
\begin{equation}\label{caso1}
w(x,y) =
\begin{cases}
\;\;\; u(x,y) & \text{if $y>0$}, \\
-u(x, y) + c_1 & \text{if $y<0$},
\end{cases}
\end{equation}
or
\begin{equation}\label{caso2}
w(x,y) =
\begin{cases}
u(x,y) + c_2 & \text{if $y>0$}, \\
u(x, y)  & \text{if $y<0$},
\end{cases}
\end{equation}
where $u\in C^1(\Om)$ is harmonic with normal derivative
vanishing on $S$ and $c_1$, $c_2$ are real constants.
For our purposes, it is enough to consider the case $c_1=0$ in (\ref{caso1})
and $c_2 =1$ in (\ref{caso2}).

\section{A model case}

In this section we consider in (\ref{caso1}) and in
(\ref{caso2}) the particular function $u(x,y)=x$ and we
deal with the minimality of the functions
\begin{equation}\label{ics}
w (x,y) :=
\begin{cases}
\;\;\; x & \text{if $y>0$}, \\
-x & \text{if $y<0$},
\end{cases}
\end{equation}
and
\begin{equation}\label{costante}
w (x,y) :=
\begin{cases}
x+ 1 & \text{if $y>0$}, \\
x & \text{if $y<0$}.
\end{cases}
\end{equation}
The aim of the study of these simpler cases (but we will see
that they involve the main difficulties) is to clarify the ideas of the general construction.

\begin{theorem}\label{x-x}
Let $w: \rdue \to \erre$ be the
function defined by
$$w(x,y) := \begin{cases} \;\;\;  x & \text{if $y > 0$}, \\
-x & \text{if $y < 0$}.
\end{cases}$$
Then every point $(x_{0}, y_{0}) \neq (0,0)$ has an open
neighbourhood $U$ such that $w$ is a Dirichlet minimizer
in $U$ of
the Mumford-Shah functional (\ref{g1}).
\end{theorem}

{\bf Proof.}
The result follows by Theorem 4.1 of \cite{Alb-Bou-DM} if $y_{0}\neq 0$.
We consider now the case $y_{0}=0$, assuming for simplicity
that $x_0 >0$.
We will construct a local calibration of $w$ near $(x_{0}, 0)$.
Let us fix  $\varepsilon >0$ such that
\begin{equation}\label{130}
0 < \varepsilon < \frac{x_{0}}{10}, \qquad 0< \varepsilon <
\frac{1}{32}.
\end{equation}
For $0<\delta<\varepsilon$ we
consider the open rectangle
$$U := \{ (x,y)\in \rdue: |x-x_0| <\varepsilon, |y| < \delta \}$$
and the following subsets of $U{ \times } \erre$ (see Fig.~1)
\begin{figure}[p]
\begin{center}
  \includegraphics[height=0.9\textheight]{Fig1.eps}
\caption{Section of the sets
$A_{1},\ldots,A_{5}$ at $x={\rm constant}$.}
\end{center}
\end{figure}
\begin{eqnarray*}
A_1 & := & \{(x,y,z)\in U{ \times } \erre : x-\alpha(y) < z < x+\alpha(y) \}, \\
A_2 & := & \left\{(x,y,z)\in U{ \times } \erre : b +\coeff \, y< z < b + \coeff \, y
+ h \right\}, \\
A_3 & := & \{(x,y,z)\in U{ \times } \erre : -h < z< h \}, \\
A_4 & := & \left\{(x,y,z)\in U{ \times } \erre : -b + \coeff \, y -h < z < -b
+ \coeff \, y \right\}, \\
A_5 & := & \{(x,y,z)\in U{ \times } \erre : -x-\alpha(-y) < z <
-x+\alpha(-y) \}, \end{eqnarray*}
where
$$\alpha(y) :=  \sqrt{4\varepsilon^2 - (\varepsilon-y)^2},$$
$$h:=\frac{x_0 -3\varepsilon}{4}, \qquad \coeff :=
\frac{\lambda}{4} - \frac{1}{\lambda},  \qquad
b := 2h+ \coeff \, \delta, \qquad \lambda :=
\frac{1-4\varepsilon}{2h}.$$
We will assume that
\begin{equation}\label{120}
\delta < \frac{x_0 -3\varepsilon}{8 \mcoeff},
\end{equation}
so that the sets $A_1, \dots, A_5$ are pairwise disjoint.

For every $(x,y,z) \in U{ \times } \erre $, let us
define the vector
$\varphi(x,y,z) = (\varphi^x, \varphi^y, \varphi^z)(x,y,z)\in \rtre$ as
follows:
$$\begin{cases}
\displaystyle \left(\frac{2(\varepsilon-y)}{\sqrt{(\varepsilon-y)^2 +
(z-x)^2}}, \frac{-2(z-x)}{\sqrt{(\varepsilon-y)^2 + (z-x)^2}}, 1
\right) & \text{if $(x,y,z)\in A_1$}, \\
\\
\displaystyle \left( 0, \lambda, \frac{\lambda^2}{4} \right) & \text{if $(x,y,z)\in
A_2$}, \\
\\
\displaystyle (f(y),0,1) & \text{if $(x,y,z)\in A_3$}, \\
\\
\displaystyle \left( 0, \lambda, \frac{\lambda^2}{4} \right) & \text{if $(x,y,z)\in
A_4$}, \\
\\
\displaystyle \left( \frac{-2(\varepsilon+y)}{\sqrt{(\varepsilon+y)^2 + (z+x)^2}},
\frac{2(z+x)}{\sqrt{(\varepsilon+y)^2
+ (z+x)^2}}, 1 \right) & \text{if $(x,y,z)\in A_5$}, \\
\\
\displaystyle (0,0,1) & \text{otherwise},
\end{cases}$$
where
$$f(y) := - \frac{1}{h}
\left( \int_{0}^{\alpha(y)} \frac{\varepsilon-y}{\sqrt{t^2 + (\varepsilon-y)^2}}\, dt -
\int_0^{\alpha(-y)} \frac{\varepsilon+y}{\sqrt{t^2 + (\varepsilon+y)^2}}\, dt
\right).$$

Note that $A_1 \cup A_5$ is an open neighbourhood of $\graph
(w)\cap ( U { \times } \erre )$.
The purpose of the definition of $\varphi$ in $A_1$ and $A_5$
(see Fig.~2)
\begin{figure}[ht]

\begin{center}
  \includegraphics[width=14cm]{Fig2.eps}
\caption{Section of the set $A_{1}$ at $z={\rm constant}$.}
\end{center}
\end{figure}
is to provide a divergence free vector field satisfying
condition (c) of Section~2 and such that
\begin{eqnarray*}
\varphi^y (x,0,z) > 0 & & \hbox{for} \;\; |z|<x, \\
\varphi^y (x,0,z) < 0 & & \hbox{for} \;\; |z|>x.
\end{eqnarray*}
These properties are crucial in order to obtain (d) and (e) simultaneously.

The r\^ole of $A_2$ and $A_4$ is to
give the main contribution to the integral in (e).
To explain this fact, suppose, for a moment, that $\varepsilon = 0$;
in this case we would have $A_1= A_5 = \emptyset$ and
$$\int_{-x}^x \varphi^y (x,0,z)\, dz = 1,$$
so that the $y$-component of equality (e) would be satisfied.

The purpose of the definition of $\varphi$ in $A_3$ is to correct the
$x$-component of $\varphi$, in order to obtain (d).

\

We shall prove that, for a suitable choice of
$\delta$, the vector
field $\varphi$ is a calibration for $w$ in the rectangle $U$.

Note that for a given $z\in \erre$ we have
\begin{equation}\label{div2}
\partial_x \varphi^x (x,y,z) + \partial_y \varphi^y (x,y,z)=0
\end{equation}
for every $(x,y)$ such that $(x,y,z)\in A_1 \cup A_5$. This implies
$\varphi$ is divergence free in $A_1 \cup A_5$.
Moreover $\dive \varphi =0$ in the other sets $A_i$,
and the normal component  of
$\varphi$  is continuous across $\partial A_i$:
the choice of $\coeff$ ensures that this property holds for $\partial
A_{2}$ and $\partial A_{4}$ (see Fig.~3).
\begin{figure}[hp]
\begin{center}
  \includegraphics[height=0.9\textheight]{Fig3.eps}
  \caption{Section of the set $A_{2}$ at $x={\rm constant}$.}
\end{center}
\end{figure}
Therefore
$\varphi$ is divergence free in the sense of distributions
in $U{ \times } \erre$.

On the graph of $w$ we have
$$\varphi(x,y,w(x,y)) = \begin{cases}
(\;\;\; 2,0,1) & \text{if $y>0$}, \\
(-2,0,1) & \text{if $y<0$},
\end{cases}$$
so condition (c) is satisfied.

Inequality (b) is clearly satisfied in all regions: the only non
trivial case is $A_3$,
where we have, using (\ref{130}),
$$|f(y)|  \leq \frac{ 4\left( \alpha(y) +
\alpha(-y)
\right) }{x_0 -3\varepsilon}
 \leq  \frac{8\sqrt{3}\varepsilon }{x_0 - 3\varepsilon}
<2.$$

We now compute
\begin{equation}
\int_{-x}^{x} \varphi^y(x,y,z)\, dz .
\end{equation}
Let us fix $y$ with $|y|< \delta$. Since $\varphi^y(x,y,z)$ depends
on $z-x$, we have
\begin{equation}\label{01}
\int_{x-\alpha(y)}^x \varphi^y(x,y,z)\, dz = \int_x^{x+\alpha(y)}
\varphi^y(\xi, y, x)\, d\xi.
\end{equation}
Using (\ref{div2}) and applying the divergence
theorem to the curvilinear triangle
$$T=\{ (\xi, \eta) \in \rdue: \xi> x, \; \eta< y, \; (\varepsilon-\eta)^2 +
(x-\xi)^2<
4\varepsilon^{2} \}$$
(see Fig.~4),
we obtain
\begin{equation}\label{02}
\int_x^{x+\alpha(y)}
\varphi^y(\xi, y, x)\, d\xi= \int_{-\varepsilon}^y
\varphi^x(x, \eta, x)\, d\eta = 2(y+\varepsilon).
\end{equation}
\begin{figure}[t]

\begin{center}
  \includegraphics[width=14cm]{Fig4.eps}
  \caption{The curvilinear triangle $T$.}
\end{center}
\end{figure}
{}From (\ref{01}) and (\ref{02}), we get
\begin{equation}\label{03}
\int_{x-\alpha(y)}^x \varphi^y(x,y,z)\, dz =
2(y+\varepsilon).
\end{equation}
Similarly we can prove that
\begin{equation}\label{04}
\int_{-x}^{-x +\alpha(-y)}\varphi^y(x,y,z)\, dz =
2(-y+\varepsilon).
\end{equation}
Using the definition of $\varphi$ in $A_2$, $A_3$, $A_4$, we obtain
\begin{equation}\label{phiy}
\int_{-x}^x \varphi^y(x,y,z)\, dz = 1.
\end{equation}
On the other hand, by the definition of $f$, we have immediately that
\begin{equation}\label{phix}
\int_{-x}^x \varphi^x(x,y,z)\, dz = 0.
\end{equation}
{}From these equalities it follows in particular
that condition (e) is satisfied on the jump set
$S_{w} \cap U = \{ (x,y)\in U: y=0\}$.

Let us begin now the proof of (d).
Let us fix $(x,y)\in U$.
For every $t_{1}< t_{2}$ we set
$$\I{t_{1}}{t_{2}} := \int_{t_{1}}^{t_{2}} (\varphi^x,\varphi^y)(x,y,z)\, dz.$$
It is enough to consider the case $-x-\alpha(-y) \leq t_{1} \leq t_{2}
\leq x - \alpha(y)$. We can write
\begin{eqnarray*}
\I{t_{1}}{t_{2}} & = & \I{t_{1}}{-x} + \I{-x}{x} + \I{x}{t_{2}}, \\
\I{t_{1}}{-x} & = & \I{t_{1} \land (-x +\alpha(-y))}{-x} + \I{t_{1}
\lor (-x +\alpha(-y))}{-x +\alpha(-y)}, \\
\I{x}{t_{2}} & = & \I{x}{t_{2} \lor (x-\alpha(y))} + \I{x-\alpha(y)}{
t_{2} \land (x -\alpha(y))}.
\end{eqnarray*}
Therefore
$$\I{t_{1}}{t_{2}}  =  \I{-x}{x} + \I{t_{1} \land (-x
+\alpha(-y))}{-x} + \I{x}{t_{2} \lor (x-\alpha(y))}$$
\begin{equation}\label{111}
\mbox{} + \I{t_{1}
\lor (-x +\alpha(-y))}{t_{2} \land (x -\alpha(y))} -
\I{-x+\alpha(-y)}{x-\alpha(y)}.
\end{equation}

Let $B$ be the ball of radius $4\varepsilon$
centred at $(0,-4\varepsilon)$. We want to prove that
\begin{equation}\label{109}
\I{x}{t} \in \overline{B}
\end{equation}
for every
$t$ with $x-\alpha(y) \leq t \leq x+\alpha(y)$.
Let us denote the components of $\I{x}{t}$ by
$a^x$ and  $a^y$.
Arguing as in the proof of (\ref{03}), we get the identity
\begin{equation}
a^{y} = 2(\varepsilon-y) - 2 \sqrt{(t-x)^2 + (\varepsilon-y)^2} \leq 0.
\end{equation}
As $|\varphi^x|\leq 2$, we have also
$$( a^{x} )^2 \leq 4(t-x)^2
= ( 2(\varepsilon -y) - a^{y}
)^2 - 4(\varepsilon -y)^2.$$
{}From these estimates it follows that
$$( a^{x} )^2 +
( a^{y} + 4\varepsilon
)^2 \leq 16 \varepsilon^2,$$
which proves (\ref{109}).
In the same way we can prove that
\begin{equation}\label{110}
\I{t}{-x} \in \overline{B}
\end{equation}
for every $t$ with $-x-\alpha(-y) \leq t \leq -x +\alpha(-y).$

If $f(y)\geq 0$, we define
$$C : = ([ 0, 2h f(y) ] { \times } [0,
{\textstyle \frac{1}{2}}
-2\varepsilon]) \cup (\{ 2hf(y) \} {\times }  [0, 1-4\varepsilon]);$$
if $f(y)\leq 0$, we simply replace $[ 0, 2h f(y) ]$ by $[ 2h f(y),
0 ]$.
From the definition of $\varphi$ in $A_{2}$, $A_{3}$, $A_{4}$, it
follows that
\begin{equation}\label{112}
\I{-x+\alpha(-y)}{x-\alpha(y)} = (2h f(y), 1-4\varepsilon)
\end{equation}
and
\begin{equation}\label{114}
\I{s_{1}}{s_{2}} \in C
\end{equation}
for $-x+\alpha(-y) \leq s_{1} \leq s_{2} \leq x - \alpha(y)$.
Let $D := C - (2h f(y), 1-4\varepsilon)$, i.e.,
$$D = ([ -2h f(y) , 0 ] { \times } [-1 +
4\varepsilon,
-{\textstyle \frac{1}{2}}
+2\varepsilon]) \cup (\{ 0 \} {\times }  [-1+4\varepsilon , 0]),$$
for $f(y)\geq 0$; the interval $[ -2h f(y) , 0 ]$ is replaced by
$[ 0, -2h f(y) ]$ when $f(y)\leq 0$.
From (\ref{111}), (\ref{phiy}), (\ref{phix}),
(\ref{109}), (\ref{110}), (\ref{112}) and (\ref{114})
we obtain
\begin{equation}\label{122}
\I{t_{1}}{t_{2}} \in (0,1) + 2\overline{B} + D.
\end{equation}
As $f(0)=0$, we can choose $\delta$ so that (\ref{120}) is satisfied
and
\begin{equation}\label{121}
|2h f(y)| = \frac{x_{0} -3\varepsilon}{2} |f(y)| \leq \varepsilon
\end{equation}
for $| y | < \delta$.
It is then easy to see that, by (\ref{130}),
the set $(0,1) + 2\overline{B} + D$ is contained in the unit ball centred at
$(0,0)$.
So that (\ref{122}) implies (d). \qed

\

{\bf Remark.} The assumption $(x_0 , y_0)\neq (0,0)$
in Theorem \ref{x-x} cannot be dropped.
Indeed, there is no
neighbourhood $U$ of $(0,0)$ such that $w$ is a Dirichlet minimizer of
the Mumford-Shah
functional in $U$.

To see this fact, let $\psi$ be a function defined on the square
$Q = (- 1 ,1) { \times } (-1 ,1)$ satisfying the boundary condition
$\psi = w$ on $\partial Q$
and such that $S_{\psi} = ((-1, -1/2) \cup (1/2, 1)) { \times } \{ 0 \}$.
For every $\varepsilon$, let $\psi_{\varepsilon }$ be the function
defined on $Q_{\varepsilon} = \varepsilon Q$ by
$\psi_{\varepsilon}(x, y) := \varepsilon \psi (x / \varepsilon,
y / \varepsilon)$.
Note that $\psi_{\varepsilon}$ satisfies the boundary condition
$\psi_{\varepsilon}= w$ on $\partial Q_{\varepsilon}$.
Let us
compute the Mumford-Shah functional for $\psi_{\varepsilon}$ on $Q_{\varepsilon}$:
$$\int_{Q_{\varepsilon}} |\nabla \psi_{\varepsilon}|^2 dx\, dy +
\haus(S_{\psi_{\varepsilon}}) =
\varepsilon^2 \int_{Q} |\nabla \psi |^2 dx\, dy + \varepsilon.$$
Since
$$\int_{Q_{\varepsilon}} |\nabla w|^2 dx\, dy + \haus(S_{w}) = 4\varepsilon^2 +
2\varepsilon,$$
we have
$$\int_{Q_{\varepsilon}} |\nabla \psi_{\varepsilon}|^2 dx\, dy +
\haus(S_{\psi_{\varepsilon}})
< \int_{Q_{\varepsilon}} |\nabla w|^2 dx\, dy + \haus(S_{w})$$
for $\varepsilon$
sufficiently small.

\

The construction shown in the proof of Theorem \ref{x-x} can be easily adapted to
define a calibration for the function $w$ in (\ref{costante}).

\begin{theorem}\label{x-x+1}
Let $w: \rdue \to \erre$ be the
function defined by
$$w(x,y) : = \begin{cases}
x+1  & \text{if $y > 0$}, \\
x & \text{if $y < 0$}.
\end{cases}$$
Then every point $(x_{0}, y_{0})\in \rdue$ has an open
neighbourhood $U$ such that $w$ is a Dirichlet minimizer in $U$ of
the Mumford-Shah functional (\ref{g1}).
\end{theorem}

{\bf Proof.}
The result follows by Theorem 4.1 of \cite{Alb-Bou-DM} if $y_{0}\neq 0$.
We consider now the case $y_{0}=0$;
we will construct a local calibration of $w$ near $(x_{0}, 0)$,
using the same technique as in Theorem \ref{x-x}.
We give only the new definitions of the sets $A_1, \ldots, A_5$ and of the
function $\varphi$, and leave to the reader the verification of the fact that
this function is a calibration for suitable values of the involved
parameters.

Let us fix  $\varepsilon >0$ such that
\begin{equation}\label{130b}
0 < \varepsilon < \frac{1}{24}, \qquad 0< \varepsilon <
\frac{1}{32}.
\end{equation}
For $0<\delta<\varepsilon$ we
consider the open rectangle
$$U := \{ (x,y)\in \rdue: |x-x_0| <\varepsilon, |y| < \delta \}$$
and the following subsets of $U { \times } \erre$
\begin{eqnarray*}
A_1 & := & \{(x,y,z)\in U{ \times } \erre  : x+1 - \alpha(y) < z < x+1 +\alpha(y) \}, \\
A_2 & := & \{(x,y,z)\in U {\times } \erre : b +  \kappa (\lambda ) \, y + 3h < z
< b + \kappa (\lambda ) \, y + 4h  \}, \\
A_3 & := & \{(x,y,z)\in U { \times } \erre : x_0 + 3 \varepsilon + 2h <z < x_0 +
3 \varepsilon +3 h \}, \\
A_4 & := & \{ (x,y,z)\in U { \times } \erre : b + \kappa (\lambda ) \, y < z <
b + \kappa (\lambda ) \, y + h \}, \\
A_5 & := & \{(x,y,z)\in U { \times } \erre: x - \alpha (-y) <z < x+ \alpha (-y)   \}, \\
\end{eqnarray*}
where
$$\alpha(y) :=  \sqrt{4\varepsilon^2 - (\varepsilon-y)^2},$$
$$h :=\frac{1 -6 \varepsilon}{5}, \qquad \kappa ( \lambda ) :=
\frac{\lambda}{4} - \frac{1}{\lambda},  \qquad
b : = x_0 + 3 \varepsilon + \kappa (\lambda ) \, \delta, \qquad
\lambda :=
\frac{1- 4 \varepsilon }{2h}.$$
We will assume that
\begin{equation}\label{120b}
\delta < \frac{1 -6 \varepsilon}{10 | \kappa (\lambda ) | },
\end{equation}
so that the sets $A_1, \ldots, A_5$ are pairwise disjoint.

For every $(x,y,z) \in U { \times } \erre$, let us define the vector
$\varphi ( x,y,z) \in \rtre$ as follows:
$$\begin{cases}
\displaystyle \left(\frac{2( \varepsilon -y)}{\sqrt{( \varepsilon -y)^2 + (z-x-1)^2}},
\frac{-2(z-x-1)}{\sqrt{(\varepsilon -y)^2 + (z-x-1)^2}}, 1
\right) & \text{if $(x,y,z)\in A_1$}, \\
\\
\displaystyle \left( 0, \lambda, \frac{\lambda^2}{4} \right) & \text{if $(x,y,z)\in A_2$}, \\
\\
\displaystyle (f(y),0,1) & \text{if $(x,y,z)\in A_3$}, \\
\\
\displaystyle \left( 0, \lambda, \frac{\lambda^2}{4} \right) & \text{if $(x,y,z)\in A_4$}, \\
\\
\displaystyle \left( \frac{2(\varepsilon +y)}{\sqrt{(\varepsilon +y)^2 + (z-x)^2}}, \frac{2(z-x)}{\sqrt{(
\varepsilon +y)^2 + (z-x)^2}}, 1 \right) &
\text{if $(x,y,z)\in A_5$}, \\
\\
\displaystyle (0,0,1) & \text{otherwise},
\end{cases}$$
where
$$f(y) := - \frac{2}{h}\left( \int_{0}^{
\alpha(y)} \frac{\varepsilon -y}{\sqrt{t^2 + (\varepsilon -y)^2}} \, dt
+ \int_{0}^{
\alpha(-y)} \frac{\varepsilon +y}{\sqrt{t^2 + (\varepsilon +y)^2}}\, dt
\right)$$
for every $|y| <  \delta$.
\qed

\section{The general case}

In this section we denote by $\Omega$ a ball in $\rdue$ centred at $(0,0)$
and we consider as $u$ in (\ref{caso1}) and in (\ref{caso2}) a generic
harmonic function with normal derivative vanishing on $S$.
We add the technical assumption that the
first and second order tangential derivatives of
$u$ are not zero on $S$.

\begin{theorem}\label{harm}
Let $u: \Omega \to \erre$ be a harmonic function such that
$\partial_{y}u(x,0)=0$ for $(x,0)\in\Omega$, and let
$w: \Omega \to \erre$ be the function defined by
$$
w(x,y) := \begin{cases}\;\;\;u(x,y) & \text{for $y>0$},
\\
-u(x, y) & \text{for $y<0$}.
\end{cases}$$
Assume that $u_{0}:=u(0,0)\neq 0$, $\partial_{x}u(0,0)\neq 0$, and
$\partial_{x}^2 u(0,0)\neq 0$.
Then there
exists an open
neighbourhood $U$ of $(0,0)$ such that $w$ is a Dirichlet minimizer in $U$ of
the Mumford-Shah functional (\ref{g1}).
\end{theorem}

{\bf Proof.} We may assume $u(0, 0)>0$ and $\partial_{x}u(0,0) >0$.
We shall give the proof only for $\partial_{x}^2 u(0,0)> 0$,
and we shall explain at the end the modification needed for
$\partial_{x}^2u(0,0)< 0$.
Let $v:\Omega\to \erre$ be the harmonic conjugate of $u$ that
vanishes on $y=0$, i.e., the function satisfying
$\partial_{x}v(x,y) = -\partial_{y}u(x,y)$,
$\partial_{y}v (x,y) = \partial_{x}u(x,y)$, and
$v(x,0)=0$.

Consider a small neighbourhood $U$ of $(0,0)$ such that the map
$\Phi(x,y):= (u(x,y), v(x,y))$ is invertible on $U$ and $\diff_x u
>0$ on $U$. We call
$\Psi$ the inverse function $(u,v) \mapsto (\xi(u,v), \eta(u,v))$,
which is defined
in the
neighbourhood $V:= \Phi (U)$ of $(u_{0},0)$.
Note that, if $U$ is small enough, then $\eta (u,v) =0$
if and only if $v=0$. Moreover,
\begin{equation}\label{psijacob}
D\Psi = \left( \begin{array}{cc}
\diff_u \xi & \diff_v \xi  \\
\diff_u \eta  & \diff_v \eta
\end{array} \right) =
\frac{1}{|\nabla u|^2} \left(
\begin{array}{cc}
\diff_x u & \diff_x v  \\
\diff_y u  & \diff_y v
\end{array}
\right),
\end{equation}
where, in the last formula, all functions are computed at $(x,y)= \Psi(u,v)$,
and so
$\diff_u \xi = \diff_v \eta$, $\diff_v \xi = - \diff_u \eta$ and
$\diff_u \eta (u,0)=0$, $\diff_v \eta (u,0) > 0$. In particular, $\xi$
and $\eta$ are harmonic, and
\begin{equation}\label{nulla}
\diff_{u}^{2}\eta (u,0) =0, \qquad \diff_{v}^{2}\eta (u,0) =0.
\end{equation}
On $U$ we will use the coordinate system $(u,v)$ given by $\Phi$.
By (\ref{psijacob}) the canonical
basis of the tangent space to $U$ at a point $(x,y)$
is given by
\begin{equation}\label{diff}
\ddu = \frac{\nabla u}{|\nabla u|^2}, \qquad \ddv =\frac{\nabla v}{|\nabla v|^2}.
\end{equation}
For every $(u,v)\in V$,
let $G(u,v)$ be the matrix associated with the first fundamental form
of $U$ in the coordinate system $(u,v)$,
and let $g(u,v)$  be its determinant.
By (\ref{psijacob}) and (\ref{diff}),
\begin{equation}\label{gi}
g = ((\diff_u \eta)^2 + (\diff_v \eta)^2)^{2} = \frac{1}{|\nabla u
(\Psi) |^4}.
\end{equation}
We set $\gamma(u,v) := \sqrt[4]{g(u,v)}$.

The calibration $\varphi(x,y,z)$ on $U { \times } \erre$
will be written as
\begin{equation}\label{calibra2}
\varphi(x,y,z) = \frac{1}{\gamma^2 (u(x,y),
v(x,y))}\phi(u(x,y),
v(x,y), z).
\end{equation}
We will adopt the following representation for $\phi : V { \times } \erre \to \rtre$:
\begin{equation}\label{calibra}
\phi (u,v,z)=
\phi^u (u,v,z) \ddu +  \phi^v (u,v,z) \ddv +  \phi^z (u,v,z)
\ddz,
\end{equation}
where $\ddz$ is the third vector of the canonical basis of $\rtre$,
and $\ddu$, $\ddv$ are computed at the point $\Psi (u,v)$.
We now reformulate the conditions of Section 2
in this new coordinate system.
It is known from Differential Geometry (see, e.g., \cite[Proposition
3.5]{Cha}) that,
if $X = X^u \ddu + X^v \ddv$
is a vector field on $U$,
then the divergence of $X$ is given by
\begin{equation}\label{div}
\dive X = \frac{1}{\gamma^2} (\diff_{u}(\gamma^2 X^u) +
\diff_{v} (\gamma^2 X^v)).
\end{equation}
Using (\ref{diff}), (\ref{gi}), (\ref{calibra2}), (\ref{calibra}), and
(\ref{div})
it turns out
that $\varphi$ is a calibration if
the following conditions are satisfied:
\begin{description}
\item[(a)] $\displaystyle
     \diff_{u} \phi^u +
	\diff_{v} \phi^v +
	\diff_{z} \phi^z = 0$ for every $(u,v,z) \in V { \times } \erre$;
\item[(b)] $\displaystyle (\phi^u (u,v,z))^2 + (\phi^v(u,v,z))^2 \leq
            4 \phi^z(u,v,z)$
	for every  $(u,v,z) \in V { \times } \erre$;
\item[(c)] $\displaystyle \phi^u (u,v, \pm u)=\pm 2$,
	$\displaystyle \phi^v(u,v, \pm u)=0$, and
	$\displaystyle \phi^z(u,v, \pm u)= 1$
	for every $(u,v)\in V$;
\item[(d)] $\displaystyle \left( \int_{t_1}^{t_2} \phi^u(u,v,z)\,
	dz \right)^2 + \left( \int_{t_1}^{t_2}
	\phi^v(u,v,z)\, dz \right)^2 \leq
	\gamma^2(u,v)$
	for every $(u,v) \in V$, $t_1, t_2 \in \erre$;
\item[(e)] $\displaystyle \int_{-u}^{u} \phi^u(u,0,z)\, dz = 0$ and
	$\displaystyle \int_{-u}^u \phi^v (u,0,z)\, dz = \gamma(u,0)$
	for every $(u,0) \in V$.
\end{description}

Given suitable parameters $\varepsilon >0$, $h>0$, $\lambda >0$, that
will be chosen later,
and assuming
\begin{equation}\label{intorno}
V=
\{ (u,v): |u-u_{0}| < \delta, |v| < \delta \},
\end{equation}
with $\delta < \varepsilon$,
we consider the following subsets of $V{ \times } \erre$
\begin{eqnarray*}
A_1 & := & \{(u,v,z)\in V{ \times } \erre : u
-\alpha(v) < z < u +\alpha(v) \}, \\
A_2 & := & \left\{(u,v,z)\in V{ \times } \erre : 3h +\beta (u,v) < z <
3h +
\beta (u,v)
+ 1/\lambda \right\}, \\
A_3 & := & \{(u,v,z)\in V{ \times } \erre : -h < z< h \}, \\
A_4 & := & \left\{(u,v,z)\in V{ \times } \erre : -3h + \beta (u,v)
-1/\lambda < z < -3h
+ \beta (u,v) \right\}, \\
A_5 & := & \{(u,v,z)\in V{ \times } \erre : -u -\alpha(-v) < z < -
u + \alpha(-v) \},
\end{eqnarray*}
where
$$\alpha(v) :=  \sqrt{4\varepsilon^2 - (\varepsilon -v)^2},$$
and $\beta$ is a suitable smooth function satisfying
$\beta (u,0) =0$, which will be defined later.
It is easy to see that, if $\varepsilon$ and $h$ are sufficiently
small, while $\lambda$ is sufficiently large, then
the sets $A_{1}, \ldots , A_{5}$ are pairwise disjoint,
provided $\delta$ is small enough. Moreover,
since $\gamma(u,0)=
\diff_{v} \eta (u,0)>0$,
by continuity we may assume that
\begin{equation}\label{gamma}
\gamma(u,v) > 128\varepsilon \qquad \hbox{and}
\qquad \diff_{v}\eta (u,v) > 8\varepsilon
\end{equation}
for every $(u,v)\in V$.

For $(u,v) \in V$ and $z\in\erre$ the vector
$\phi(u,v,z)$
introduced in (\ref{calibra2}) is defined
as follows:
$$
\begin{cases}
\displaystyle \frac{2(\varepsilon- v)}{\sqrt{(\varepsilon-v)^2+(z-u)^2}} \ddu
-  \frac{2(z-u)}{\sqrt{(\varepsilon-v)^2
+(z-u)^2}} \ddv + \ddz
& \text{in $A_1$}, \\

\\
\displaystyle -\lambda \sigma (u,v) \frac{v}{\sqrt{(u-a)^2 + v^2}} \ddu +
\lambda \sigma (u,v) \frac{u-a}{\sqrt{(u-a)^2 + v^2}} \ddv +
\mu \ddz
& \text{in $A_2$}, \\

\\
\displaystyle  f(v)\ddu + \ddz  & \text{in
$A_3$}, \\

\\
\displaystyle -\lambda \sigma (u,v) \frac{v}{\sqrt{(u-a)^2 + v^2}} \ddu +
\lambda  \sigma (u,v) \frac{u-a}{\sqrt{(u-a)^2 + v^2}} \ddv +
\mu \ddz
& \text{in $A_4$}, \\

\\
\displaystyle - \frac{2(\varepsilon+ v)}{\sqrt{(\varepsilon+ v)^2 + (z+ u)^2}} \ddu
+ \frac{2(z+ u)}{\sqrt{
(\varepsilon+ v)^2 + (z+ u)^2}} \ddv + \ddz  &
\text{in $A_5$}, \\

\\
\displaystyle \ddz  & \text{\hphantom{in $A_{5}$,}\llap{otherwise,}}
\end{cases}$$
where
\begin{eqnarray}\label{param}
& a < u_{0}- 11 \delta, \qquad \mu >0 & \\
\nonumber \\
& \displaystyle f(v) := - \frac{1}{h} \left( \int_0^{\alpha(v) }
\frac{(\varepsilon - v)}{\sqrt{t^2 + (\varepsilon - v)^2}}\, dt -
\int_0^{\alpha(-v) }
\frac{(\varepsilon + v)}{\sqrt{t^2 + (\varepsilon + v)^2}}\, dt
\right), & \nonumber \\
\nonumber \\
\label{sigma} \displaystyle
 & \sigma (u,v) := \frac{1}{2}
\gamma ( a + \sqrt{(u-a)^2 + v^2},0)
- 2 \varepsilon. &
\end{eqnarray}
We choose $\beta$ as the solution of the
Cauchy problem
\begin{equation}\label{cauchy}
\begin{cases}
\lambda \sigma(u,v) (- v\, \partial_{u} \beta + (u-a)
\partial_{v} \beta )
= (\mu -1) \sqrt{(u-a)^{2} + v^{2}}, & \\
\\
\beta (u,0)=0. &
\end{cases}
\end{equation}
Since the line $v=0$ is not characteristic for the equation near
$(u_{0},0)$,
there exists a unique solution $\beta \in C^{\infty}(V)$, provided
$V$ is small enough.

In the coordinate system $(u,v)$ the definition of the field $\phi$
in $A_{1}$, $A_{3}$, and $A_{5}$
is the same as the definition of $\varphi$ in the proof of Theorem
\ref{x-x}. The crucial difference is in the definition on the sets
$A_{2}$ and $A_{4}$, where now we are forced to introduce two new parameters $a$
and $\mu$. Note that the definition given in Theorem \ref{x-x} can be
regarded as the limiting case as $a$ tends to $+\infty$.

By direct computations it is easy to see that $\phi$ satisfies condition
(a) on $A_{1}$ and $A_{5}$.
Similarly,
the vector field
$$\left( -\frac{v}{\sqrt{(u-a)^2 + v^2}},
\frac{u-a}{\sqrt{(u-a)^2 + v^2}} \right)$$
is divergence free;
since
$(u-a)^{2}+v^{2}$ is constant along the integral curves of this field,
by construction
the same property holds for $\sigma$, so that
$\phi$ satisfies condition (a) in $A_{2}$ and $A_{4}$.

In $A_3$, condition (a) is trivially satisfied.

Note that
the normal component
of $\phi$  is continuous across each $\partial A_i$:
for the region $A_3$ this continuity is guaranteed by our
choice of $\beta$. This implies that (a) is satisfied in the sense of
distributions on $V { \times } \erre$.

In order to satisfy condition (b),
it is enough to take
the parameter $\mu$ such that
$$\frac{\lambda^2}{4}
\sigma^2(u,v)
\leq \mu$$
for every $(u,v)\in V$,
and require that
\begin{equation}\label{effe}
|f(v)|\leq 2.
\end{equation}
Since
\begin{equation}\label{stima4}
|f(v)|  \leq  \frac{\alpha(v) + \alpha(-v)}{h}
\leq \frac{4\, \varepsilon}{h},
\end{equation}
inequality (\ref{effe}) is true if we impose
$$2 \, \varepsilon \leq h.$$

Looking at the definition of $\phi$ on $A_{1}$ and $A_{5}$,
one can check that
condition (c) is satisfied.

Arguing as in the proof of (\ref{03}),
(\ref{04}), (\ref{phix}) in Theorem \ref{x-x},
we find that for every $(u,v) \in V$
$$\int_{-u}^{-u+\alpha(-v)} \phi^u(u,v,z)\, dz
+ \int_{-h}^h \phi^u(u,v,z)\, dz
+ \int_{u-\alpha(v)}^u \phi^u(u,v,z)\, dz =0,$$
$$\int_{-u}^{-u+\alpha(-v)} \phi^v(u,v,z)\, dz
+ \int_{-h}^h \phi^v(u,v,z)\, dz
+ \int_{u-\alpha(v)}^u \phi^v(u,v,z)\, dz = 4\varepsilon.$$
Now, it is easy to see that
\begin{equation}\label{phiu}
\int_{-u}^u \phi^u(u,v,z)\, dz = -2\sigma(u,v)
\frac{v}{\sqrt{(u-a)^2+v^2}},
\end{equation}
\begin{equation}\label{phiv}
\int_{-u}^u \phi^v(u,v,z)\, dz = 4\varepsilon
+ 2 \sigma(u,v)
\frac{u-a}{\sqrt{(u-a)^2+v^2}};
\end{equation}
since for $v=0$ we have
$$\sigma (u,0) = \frac{1}{2} \gamma(u,0)- 2\varepsilon,$$
condition (e) is satisfied.

By continuity, if $\delta$ is small enough,
we have
\begin{equation}\label{stima2}
\int_{-u}^u \phi^v(u,v,z)\, dz > \frac{7}{8} \gamma(u,v)
\end{equation}
for every $(u,v)\in V$.

From now on, we regard the pair $(\phi^u, \phi^v)$ as a vector in $\rdue$.
To prove condition (d) we set
$$I_{\varepsilon,a}(u,v,s,t) :=\int_s^t (\phi^u,\phi^v)(u,v,z)\, dz$$
for every $(u,v)\in V$, and for every $s,t\in \erre$.
We want to compare
the behaviour of the functions $|I_{\varepsilon,a}|^2$ and
$\gamma^2$; to this aim, we define the function
$$\di (u,v,s,t) := |I_{\varepsilon,a}(u,v,s,t)|^2 - \gamma^2(u,v).$$
We have already shown (condition (e)) that
\begin{equation}\label{zero}
\di(u,0,-u,u)=0.
\end{equation}
We start by proving that, if $V$ is sufficiently small, condition (d)
holds for every $(u,v)\in V$, for $t_1$ close to $-u$
and $t_2$ close to $u$.
Using the definition of $\phi(u,v,z)$ on $A_1$ and $A_5$,
one can compute explicitly $\di(u,v,s,t)$
for $|s+u|\leq \alpha(-v)$ and for $|t-u|\leq \alpha(v)$.
By direct computations one obtains
\begin{equation}\label{deriv1}
\nabla_{v,s ,t}\, \di (u,0,-u,u) =0
\end{equation}
for $(u,0)\in V$.

We now want to compute the hessian matrix $\nabla^2_{v,s ,t}\, \di$ at the point
$(u_0, 0, -u_0, u_0)$.
By (\ref{sigma}) and (\ref{gi}), after some easy computations, we get
$$\diff_v^2\sigma (u,0)=
\frac{1}{2(u-a)}\diff_u \gamma(u,0)=
\frac{1}{2(u-a)}
\diff_v\diff_u \eta (u,0).$$
Using this equality and the explicit expression of
$\di$ near $(u_0,0,-u_0,u_0)$, we obtain
\begin{eqnarray*}
\diff^2_v \di (u_0,0,-u_0,u_0) & = & -\frac{8\varepsilon}{(u_0-a)^2}
(\diff_v \eta(u_0,0) - 4\varepsilon) +   \\
& & +\frac{2}{u_0-a} \diff_{v}\eta (u_0,0)\, \diff_v\diff_u \eta(u_0,0)
 - \diff^2_v (\gamma^2)(u_0,0).
\end{eqnarray*}
Since $\eta$ and
$\gamma$ do not depend on $a$ and
$\varepsilon$,  for every $\varepsilon$
satisfying (\ref{gamma}) we can find
$a$ so close to $u_{0}$ that
\begin{equation}\label{det1}
\diff^2_v \di (u_0,0, -u_0, u_0) < 0.
\end{equation}
Moreover, we easily obtain that
\begin{eqnarray*}
&\displaystyle \diff_t^2 \di (u_0,0,-u_0,u_0) =
\diff_s^2 \di (u_0,0,-u_0,u_0) =
8 - \frac{4}{\varepsilon}\diff_v \eta(u_0,0), & \\
&\displaystyle  \diff_v\diff_t \di (u_0,0,-u_0,u_0)  =
\diff_v\diff_s \di (u_0,0,-u_0,u_0) =
-\frac{4}{ u_0-a}(\diff_v \eta(u_0,0)- 4\varepsilon) , & \\
& \diff_t\diff_s \di (u_0,0,-u_0,u_0)  = 8. &
\end{eqnarray*}
By the above expressions,
it follows that
\begin{eqnarray*}
\lefteqn{\det \left( \begin{array}{cc}
\diff_v^2 \di & \diff_v\diff_t \di  \\
& \\
\diff_v\diff_t \di  & \diff^2_t \di
\end{array} \right)(u_0,0,-u_0,u_0) = } \\
 & = & \frac{16}{(u_0-a)^2}\diff_v \eta(u_0,0)
(\diff_v \eta(u_0,0) -4\varepsilon )
+\frac{c_1(\varepsilon)}{u_0-a} + c_2(\varepsilon),
\end{eqnarray*}
where $c_1(\varepsilon)$, $c_2(\varepsilon)$
are two constants depending only on
$\varepsilon$. Then, if $\varepsilon$ satisfies (\ref{gamma}),
$a$ can be chosen so close to $u_0$ that
\begin{equation}\label{det2}
\det \left( \begin{array}{cc}
\diff_v^2 \di & \diff_v\diff_t \di  \\
& \\
\diff_v\diff_t \di  & \diff^2_t \di
\end{array} \right)(u_0,0,-u_0,u_0)>0.
\end{equation}
At last, the determinant of the hessian matrix
of $\di$ at $(u_0,0,-u_0,u_0)$ is given by
$$\det \nabla^2_{v,s ,t}\, \di (u_0,0,-u_0,u_0)=
\frac{1}{u_0-a} (\diff_v \eta(u_0,0))^2
\diff_v \diff_u\eta(u_0,0)
(\diff_v \eta(u_0,0) - 4\varepsilon)
\frac{32}{\varepsilon^2 }  + c_3(\varepsilon),$$
where $c_3(\varepsilon)$
is a constant depending only on
$\varepsilon$.
Since, by (\ref{psijacob}),
$$\diff_v \diff_u\eta(u_0,0) = - \frac{\diff^2_x u(0,0)}{(\diff_x u(0,0))^3},$$
given $\varepsilon$ satisfying (\ref{gamma}), we can choose $a$
so close to $u_0$ that
\begin{equation}\label{det3}
\det \nabla^2_{v,s ,t}\, \di (u_0,0,-u_0,u_0)<0.
\end{equation}
By (\ref{det1}), (\ref{det2}), and (\ref{det3}),
we can conclude that, by a suitable choice of the parameters, the hessian matrix
of $\di$ (with respect to $v,s,t$) at $(u_0,0,-u_0,u_0)$ is negative definite. This fact, with
(\ref{zero}) and (\ref{deriv1}), allows us to state
the existence of a constant $\tau >0$
such that
\begin{equation}\label{stima0}
\di(u,v,s,t)<0
\end{equation}
for $|s+u_0|<\tau$, $|t-u_0|<\tau$,
$(u,v)\in V$, $v\neq 0$,
provided $V$ is sufficiently small.
So, condition (d) is satisfied for $|t_1+u_0|<\tau$ and $|t_2-u_0|<\tau$.
We can assume $\delta < \tau < \alpha (v)$ for every
$(u,v)\in V$.

From now on, since at this point the parameters $\varepsilon$, $a$
have been fixed, we simply write $I$
instead of $I_{\varepsilon, a}$.
We now study the more general case $|t_1+u|<\alpha(-v)$ and $|t_2-u|<\alpha(v)$.

Let us set
$$m_1(u,v):= \max \left\{ | I(u,v,s,t ) | :
\, |s+u|\leq\alpha(-v),\, |t-u|\leq \alpha(v),\, |t-u_0|\geq\tau \right\}.$$
By the definition of $A_1, \ldots, A_5$, for
$\rho= \alpha(\delta) + \delta$ we have $(\phi^u,\phi^v)=0$ on
$(V { \times } [u_0 - \rho, u_0 +\rho])\setminus A_1$
and $(V { \times } [-u_0 - \rho,- u_0 +\rho])\setminus A_5$.
This implies that
$$m_1(u,v):= \max \left\{ | I(u,v,s,t ) | :
\, |s+u_0|\leq\rho,\, \tau \leq |t-u_0|\leq \rho \right\}$$
for $(u,v)\in V$.
The function $m_1$, as supremum of a family of continuous functions, is
lower semicontinuous. Moreover, $m_1$
is also upper semicontinuous; indeed,
suppose, by contradiction, that there exist two sequences
$(u_n)$, $(v_n)$ converging respectively to $u$, $v$,
such that $(m_1(u_n, v_n))$ converges to a limit
$l>m_1(u, v)$; then, there exist $(s_n)$, $(t_n)$
such that
\begin{equation}\label{tnsn}
|s_n+u_n|\leq \alpha(-v_n), \qquad |t_n-u_n|\leq \alpha(v_n),
\qquad |t_n-u_0|\geq\tau,
\end{equation}
and $m_1(u_n, v_n)=| I(u_n,v_n,s_n,t_n ) |$.
Up to subsequences, we can assume that $(s_n)$, $(t_n)$
converge respectively to $s$, $t$ such that,
by (\ref{tnsn}),
$$|s+u|\leq \alpha(-v), \qquad |t-u|\leq \alpha(v),
\qquad |t-u_0|\geq\tau;$$
hence, we have that
$$m_1(u, v)\geq | I(u,v,s,t ) |= \lim_{n\to\infty}
| I(u_n,v_n,s_n,t_n ) | =l>m_1(u, v),$$
which is impossible.
Therefore, $m_1$ is continuous.

Let $B$ be the open ball of radius $4\varepsilon$
centred at $(0,-4\varepsilon)$. Arguing as in (\ref{109}), we can prove that
\begin{equation}\label{901}
I(u,v,u,t) \in B
\end{equation}
whenever $0<|t-u|\leq\alpha(v)$.
In the same way we can prove that
\begin{equation}\label{902}
I(u,v,s,-u) \in B
\end{equation}
for $0<|s+u|\leq\alpha(-v)$.
We can write
\begin{equation}\label{spezza}
I(u,v,s,t) = I(u,v,s,-u) + I(u,v,-u,u) + I(u,v,u,t).
\end{equation}
So, for $|s+u|\leq\alpha(-v)$, $|t-u|\leq \alpha(v)$, and $|t-u_0|\geq\tau$,
by (\ref{902}), (\ref{phiu}), (\ref{phiv}), and (\ref{901}),
we obtain that
$$I(u,0,s,t)\in (0, \gamma(u,0)) + B + \overline{B},$$
hence, by (\ref{gamma}),
$I(u,0,s,t)$ belongs to the open ball of radius $\gamma(u,0)$
centred at $(0,0)$, and so, $m_1(u,0) < \gamma(u,0)$.
By continuity, if $V$ is small enough,
\begin{equation}\label{emme1}
m_1(u,v) < \gamma(u,v)
\end{equation}
for every $(u,v)\in V$.

Analogously, we define
$$m_2(u,v):= \max \left\{ | I(u,v,s,t ) | :
\, |s+u|\leq\alpha(-v), |s+u_0|\geq\tau, |t-u|\leq \alpha(v),  \right\}.$$
Arguing as in the case of $m_1$, we can prove that, if
$V$ is small enough,
\begin{equation}\label{emme2}
m_2(u,v) < \gamma(u,v)
\end{equation}
for every $(u,v)\in V$.

By (\ref{emme1}), (\ref{emme2}), and
(\ref{stima0}), we can conclude that
$I(u,v,t_1,t_2)$ belongs to the ball centred at $(0,0)$ with
radius $\gamma(u,v)$,
for $|t_1 +u|\leq \alpha(-v)$ and $|t_2 -u|\leq \alpha(v)$.
More precisely, let $E(u,v)$ be the intersection of this ball
with the upper half plane
bounded by the horizontal straight line passing through the point
$(0, \frac{3}{4}\gamma(u,v))$:
by (\ref{spezza}), (\ref{stima2}), (\ref{901}), (\ref{902}), and
(\ref{gamma}),
we deduce that
\begin{equation}\label{911}
I(u,v,t_1,t_2)\in E(u,v)
\end{equation}
for $|t_1 +u|\leq \alpha(-v)$ and $|t_2 -u|\leq \alpha(v)$.

We can now conclude the proof of (d).
It is enough to consider the case $-u-\alpha(-v) \leq t_{1} \leq t_{2}
\leq u + \alpha(v)$. We can write
\begin{eqnarray}
I(u,v,t_{1},t_{2}) & = &  I(u,v, t_{1}
\land (-u+\alpha(-v)),
t_{2} \lor (u-\alpha(v))) + \nonumber \\
& & + I(u,v, t_{1} \lor (-u
+\alpha(-v)), t_{2} \land (u -\alpha(v)) )- \label{somma} \\
& & - I(u,v,-u +\alpha(-v), u-\alpha(v)). \nonumber
\end{eqnarray}

By (\ref{911}), it follows that
\begin{equation}\label{inter}
I(u,v, t_{1}
\land (-u+\alpha(-v)),
t_{2} \lor (u-\alpha(v)))\in E(u,v).
\end{equation}

Let $C_{1}(u,v)$ be the parallelogram having three consecutive vertices at
the points
$$(2hf(v), 0), \qquad (0,0), \qquad
\sigma(u,v)\frac{(-v,u-a)}{\sqrt{(u-a)^{2}+v^{2}}},$$
let $C_{2}(u,v)$ be the segment with endpoints
$$(2hf(v),0), \qquad (2hf(v), 0) +
2\sigma(u,v)\frac{(-v,u-a)}{\sqrt{(u-a)^{2}+v^{2}}},$$
and let $C(u,v) :=C_{1}(u,v)\cup C_{2}(u,v)$.

From the definition of $\varphi$ in $A_{2}$, $A_{3}$, $A_{4}$, it
follows that
\begin{equation}\label{922}
I(u,v, -u+\alpha(-v), u-\alpha(v)) = (2hf(v), 0) +
2\sigma(u,v)\frac{(-v,u-a)}{\sqrt{(u-a)^{2}+v^{2}}}
\end{equation}
and
\begin{equation}\label{923}
I(u,v, s_{1}, s_{2}) \in C(u,v)
\end{equation}
for $-u+\alpha(-v) \leq s_{1} \leq s_{2} \leq u- \alpha(v)$.
Let
$$D(u,v) := C(u,v) - (2hf(v), 0) -
2\sigma(u,v)\frac{(-v,u-a)}{\sqrt{(u-a)^{2}+v^{2}}}.$$

From (\ref{somma}), (\ref{inter}),
(\ref{922}), and (\ref{923})
we obtain
\begin{equation}
I(u,v , t_{1}, t_{2}) \in E(u,v) + D(u,v).
\end{equation}
As $|v| < \delta < 10\delta < u-a$ by (\ref{param}),
the angle
that the segment $C_{2}(u,v)$ forms with the vertical is less
than $\arctan (1/10)$.
Moreover, we may assume
that the lenght $2\sigma(u,v)$ of the segment $C_{2}(u,v)$ is
less than $\gamma(u,v)$; indeed, this is true for $v=0$ and, by
continuity, it remains true if $\delta$ is small enough.
By (\ref{gamma}) and (\ref{stima4}), we have also
that $|2hf(v)| \leq \gamma(u,v) /16$.
Using these properties and simple geometric considerations, it is
possible to prove that
$E(u,v) + D(u,v)$
is contained in the ball with centre $(0,0)$ and radius $\gamma(u,v)$.
This concludes the proof of (d).

If $\diff_x^2 u(0,0) <0$,
it is enough to change the definition of $\phi$ in the
sets $A_2$ and $A_4$, as follows:
$$\lambda \sigma(u,v) \frac{v}{\sqrt{(a-u)^2 + v^2}}\ddu
+ \lambda \sigma(u,v) \frac{a-u}{\sqrt{(a-u)^2 + v^2}}\ddv
+ \mu \ddz,$$
where $a> u_0 + 11\delta$ and
$$\sigma(u,v):= \frac{1}{2} \gamma(a-\sqrt{(a-u)^2 + v^2},0)-2\varepsilon.$$
\qed

\

\begin{theorem}\label{harm2}
Let $u: \Omega \to \erre$ be a harmonic function such that $\diff_y u(x,0) = 0$ for
$(x,0) \in \Omega$, and let $w: \Omega \to \erre$ be the function defined by
$$w(x,y) := \begin{cases}
u(x,y) + 1 & \text{for $y>0$}, \\
u(x,y) & \text{for $y<0$}.
\end{cases}$$
Assume that
$\diff_x u(0,0)\neq 0$ and
$\diff_x^2 u(0,0)\neq 0$.
Then there exists an open neighbourhood $U$ of $(0,0)$ such that $w$ is
a Dirichlet minimizer in $U$
of the Mumford-Shah functional (\ref{g1}).
\end{theorem}

{\bf Proof.}
We will write the
calibration $\varphi$ as in (\ref{calibra2}) and we will adopt the representation
(\ref{calibra}) for $\phi$.
We will use the same technique as in Theorem \ref{harm}.
We give only the new definitions of the sets $A_1, \ldots, A_5$ and of the
function $\phi$ when $\diff_x u(0,0)>0$ and $\diff^2_x u(0,0)>0$,
and leave to the reader the verification of the fact that
this function is a calibration for suitable values of the involved
parameters. The case $\diff^2_x u(0,0)<0$ can be treated by the changes
introduced at the end of Theorem \ref{harm}.

Let $u_0 := u(0,0)$.
Given $\varepsilon >0$, $h>0$, $\lambda >0$, and assuming
$$V := \{ (u,v) : |u - u_0 | < \delta, |v| < \delta \},$$
we consider
the following subsets of $V { \times } \erre$
\begin{eqnarray*}
A_1 & : = & \{ (u,v,z) \in V { \times } \erre : u + 1 - \alpha (v) <
z < u + 1 + \alpha(v) \}, \\
A_2 & : = & \{ (u,v,z) \in V { \times } \erre : 5h + \beta(u,v) < z < 5h + \beta(u,v)
+ 1/ \lambda \}, \\
A_3 & : = & \{ (u,v,z) \in  V { \times } \erre : 2h < z < 4h \}, \\
A_4 & : = & \{ (u,v,z) \in  V { \times } \erre : h + \beta(u,v) < z < h + \beta(u,v)
+ 1/ \lambda \}, \\
A_5 & : = & \{ (u,v,z) \in  V { \times } \erre :  u  - \alpha (-v) <
z < u + \alpha(-v) \},
\end{eqnarray*}
where
$$\alpha(v) :=  \sqrt{4\varepsilon^2 - (\varepsilon -v)^2},$$
and $\beta$ is a suitable smooth function satisfying
$\beta (u,0) =0$, which will be defined later.
For $(u,v) \in V$ and $z\in\erre$ the vector
$\phi(u,v,z)$ is defined
as follows:
$$
\begin{cases}
\displaystyle \frac{2(\varepsilon- v)}{\sqrt{(\varepsilon-v)^2+(z-u-1)^2}} \ddu
-  \frac{2(z-u-1)}{\sqrt{(\varepsilon-v)^2
+(z-u-1)^2}} \ddv + \ddz
& \text{in $A_1$}, \\

\\
\displaystyle -\lambda \sigma (u,v) \frac{v}{\sqrt{(u-a)^2 + v^2}} \ddu +
\lambda \sigma (u,v) \frac{u-a}{\sqrt{(u-a)^2 + v^2}} \ddv +
\mu \ddz
& \text{in $A_2$}, \\

\\
\displaystyle  f(v)\ddu + \ddz  & \text{in
$A_3$}, \\

\\
\displaystyle -\lambda \sigma (u,v) \frac{v}{\sqrt{(u-a)^2 + v^2}} \ddu +
\lambda  \sigma (u,v) \frac{u-a}{\sqrt{(u-a)^2 + v^2}} \ddv +
\mu \ddz
& \text{in $A_4$}, \\

\\
\displaystyle  \frac{2(\varepsilon+ v)}{\sqrt{(\varepsilon+ v)^2 + (z- u)^2}} \ddu
+ \frac{2(z- u)}{\sqrt{
(\varepsilon+ v)^2 + (z- u)^2}} \ddv + \ddz  &
\text{in $A_5$}, \\

\\
\displaystyle \ddz  & \text{\hphantom{in $A_{5}$,}\llap{otherwise,}}
\end{cases}$$
where $a < u_{0}- 11 \delta$, $\mu >0$,
\begin{eqnarray*}
& \displaystyle f(v) := - \frac{1}{h} \left( \int_0^{\alpha(v) }
\frac{(\varepsilon - v)}{\sqrt{t^2 + (\varepsilon - v)^2}}\, dt +
\int_0^{\alpha(-v) }
\frac{(\varepsilon + v)}{\sqrt{t^2 + (\varepsilon + v)^2}}\, dt
\right), &  \\
\\
\displaystyle & \sigma (u,v) := \frac{1}{2}
\gamma ( a + \sqrt{(u-a)^2 + v^2},0)
- 2 \varepsilon, &
\end{eqnarray*}
and $\beta$ is the solution of the
Cauchy problem (\ref{cauchy}).
\qed