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\begin{document}

\vspace*{1.0in}

\begin{center} THE INDEX COMPLEX OF A MAXIMAL SUBALGEBRA OF A LIE ALGEBRA
\end{center}
\bigskip

\begin{center} DAVID A. TOWERS
\end{center}
\bigskip

\begin{center} Department of Mathematics and Statistics

Lancaster University

Lancaster LA1 4YF

England

d.towers@lancaster.ac.uk
\end{center}
\bigskip

\begin{abstract}
Let $M$ be a maximal subalgebra of the Lie algebra $L$. A subalgebra $C$ of $L$ is said to be a {\em completion} for $M$ if $C$ is not contained in $M$ but every proper subalgebra of $C$ that is an ideal of $L$ is contained in $M$. The set $I(M)$ of all completions of $M$ is called the {\em index complex} of $M$ in $L$. We use this concept to investigate the influence of the maximal subalgebras on the structure of a Lie algebra, in particular finding new characterisations of solvable and supersolvable Lie algebras.
\par
\noindent {\em Mathematics Subject Classification 2000}: 17B05, 17B20, 17B30, 17B50.
\par
\noindent {\em Key Words and Phrases}: Lie algebras, maximal subalgebra, index complex, ideal index, solvable, supersolvable, Frattini ideal.
\end{abstract}

\section{Introduction}
\medskip

Let $M$ be a maximal subalgebra of the Lie algebra $L$. A subalgebra $C$ of $L$ is said to be a {\em completion} for $M$ if $C$ is not contained in $M$ but every proper subalgebra of $C$ that is an ideal of $L$ is contained in $M$. The set $I(M)$ of all completions of $M$ is called the {\em index complex} of $M$ in $L$. This is analogous to the concept of the index complex of a maximal subgroup of a finite group as introduced by Deskins in \cite{des1}; this concept has since been further studied by a number of authors, including Ballester-Bolinches and Ezquerro (\cite{be}), Beidleman and Spencer (\cite{bs}), Deskins (\cite{des2}), Mukherjee (\cite{muck}), and Mukherjee and Bhattacharya (\cite{mb}).
\par

There are a number of interesting results concerning the question of what certain intrinsic properties of the maximal subalgebras of a Lie algebra $L$ imply about the structure of $L$ itself. For example:
\begin{itemize}
\item[(1)] all maximal subalgebras are ideals of $L$ if and only if $L$ is nilpotent (see \cite{barnes1});
\item[(2)] all maximal subalgebras of $L$ are c-ideals of $L$ if and only if $L$ is solvable (see \cite{cideal});
\item[(3)] if $L$ is solvable then all maximal subalgebras have codimension one in $L$ if and only if $L$ is supersolvable (see \cite{barnes});
\item[(4)] $L$ can be characterised when its maximal subalgebras satisfy certain lattice-theoretic conditions, such as modularity (see \cite{var}).
\end{itemize}
The main purpose of this paper is to seek further such results; in particular, to find a characterisation of supersolvable Lie algebras amongst all Lie algebras, rather than just amongst the solvable ones as in (3) above. We are also seeking a characterisation of solvable Lie algebras in terms of the `size' of their maximal subalgebras. However, since, of course, the maximal subalgebras can have different codimensions in $L$, some other measure of their size is needed. The development of the theory follows closely that of its group-theoretic counterpart, but the proofs are usually different and stronger results can be obtained.
\par

We define the {\em strict core} (resp. {\em core}) of a subalgebra $B \neq 0$ to be the sum of all ideals of $L$ that are proper subalgebras (resp. subalgebras) of $B$, and denote it by $k(B)$ or $k_L(B)$ (resp. $B_L$). Notice that the strict core can differ from the core even when $B$ is an ideal of $L$: for example, if $B$ is a one-dimensional ideal of $L$. It is easy to see that the strict core of any completion $C$ is a proper subalgebra of $C$. The subalgebra $C$ is then a completion of the maximal subalgebra $M$ of $L$ (that is, $C \in I(M)$) if $L = <M, C>$ and $k(C) \subseteq M$. Completions always exist, as the following lemma establishes.
\bigskip

\begin{lemma}\label{l:index} If $M$ is a maximal subalgebra of $L$ then $I(M)$ is non-empty; in fact, $I(M)$ contains an ideal of $L$.
\end{lemma}
\begin{proof} Clearly the set of ideals of $L$ which do not lie in $M$ is a non-empty partially ordered set; choose $C$ to be a minimal element of this set. Then $L = M + C = <M, C>$ and $k(C) \subseteq M$, so $C \in I(M)$.
\end{proof}
\bigskip

In section two we study completions that are ideals and show that if $C$ and $D$ are two such completions of the same maximal subalgebra $M$ then $C/k(C) \cong D/k(D)$. This allows us to define the {\em ideal index} of $M$ in $L$. Next we characterise solvable and supersolvable Lie algebras in terms of the ideal index of their maximal subalgebras. We then consider completions $C$ that are ideals and for which $C/k(C)$ is abelian. A characterisation of solvable Lie algebras and of the solvable radical is given in terms of such completions.
\par

In section three maximal completions are studied. It is shown that over an algebraically closed field a Lie algebra is solvable if and only if every maximal subalgebra of $L$ has an abelian maximal completion. This is analogous to a result of Deskins (see \cite{des2}) for groups. The Lie algebraic result, however, is false if the underlying field is not algebraically closed, even if it has characteristic zero.
\par

The final section is devoted to subideal completions. A characterisation of supersolvable Lie algebras in terms of such completions is found that is analogous to a group-theoretic result of Ballester-Bolinches and Ezquerro (see \cite{be}). The Lie algebraic proof, however, is completely different, as the group theoretic result relies on properties that do not hold in the case of Lie algebras. A counter-example is also given to part of \cite[Corollary 2]{be}.
\par
Throughout, $L$ will denote a finite-dimensional Lie algebra over a field $F$. If $A$ and $B$ are subalgebras of $L$ for which $L = A + B$ and $A \cap B = 0$ we will write $L = A \oplus B$. The ideals $L^{(k)}$ of the derived series are defined inductively by $L^{(0)} = L$, $L^{(k+1)} = [L^{(k)},L^{(k)}]$  for $k \geq 0$; we also write $L^2$ for $L^{(1)}$. If $A$ is a subalgebra of $L$, the {\em centralizer} of $A$ in $L$ is $C_{L}(A) = \{ x \in L : [x, A] = 0\}$.
\bigskip

\section{Ideal completions and the ideal index}
\medskip

If $C$ is an ideal of $L$ and $C \in I(M)$ we call $C$ an {\em ideal completion} of $L$. In this case $C/k(C)$ is a chief factor of $L$ which is avoided by $M$. Up to isomorphism it is uniquely determined by $M$, as is shown by the following result whose proof is based on that of \cite[Lemma 1]{bs}.
\bigskip

\begin{theor}\label{t:isom} Let $C$ and $D$ be ideal completions of the maximal subalgebra $M$ of $L$. Then $C/k(C) \cong D/k(D)$.
\end{theor}
\begin{proof} Let $L$ be a Lie algebra of minimal dimension having a maximal subalgebra $M$ with two ideal completions $C$ and $D$ such that $C/k(C) \not \cong D/k(D)$. If $k(C) \cap k(D) \neq 0$ then by factoring out $k(C) \cap k(D)$ and using the minimality of $L$ we see that $C/k(C) \cong D/k(D)$, which is a contradiction. Hence $k(C) \cap k(D) = 0$. Now $C \cap k(D)$ is an ideal of $L$ and  $C \cap k(D) \subseteq C \cap M$, so $C \cap k(D) \subseteq k(C)$. It follows that $C \cap k(D) \subseteq k(C) \cap k(D) = 0$. Similarly $D \cap k(C) = 0$.
\par
Next we claim that $(C + k(D))/(k(C) + k(D))$ is a ideal completion to $M/(k(C) + k(D))$ in $L/(k(C) + k(D))$. For, certainly it is an ideal supplement. Suppose that $X/(k(C) + k(D))$ is an ideal of $L/(k(C) + k(D))$ properly contained in $(C + k(D))/(k(C) + k(D))$. Then $X \cap C$ is an ideal of $L$ properly contained in $C$, since $X \cap C = C$ implies that $C \subseteq X$ and $X = C + k(D)$. It follows that $X = X \cap (C + k(D)) = X \cap C + k(D) \subseteq M$, proving the claim.
\par
Similarly $(D + k(C))/(k(C) + k(D))$ is an ideal completion to $M/(k(C) + k(D))$ in $L/(k(C) + k(D))$. But now if $k(C) + k(D) \neq 0$ the minimality of $L$ implies that $(C + k(D))/(k(C) + k(D)) \cong (D + k(C))/(k(C) + k(D))$. But $(C + k(D))/(k(C) + k(D)) \cong C/k(C)$ and $(D + k(C))/(k(C) + k(D)) \cong D/k(D)$ so this is a contradiction.
\par
We therefore have $k(C) + k(D) = 0$ and $C, D$ are minimal ideals of $L$. Then $L = M + C = M + D$ and $CD \subseteq C \cap D = 0$. But now $[M \cap C, L] = [M \cap C, M + D] \subseteq M \cap C$ so $M \cap C$ is an ideal of $L$. It follows that $M \cap C = 0$. Similarly $M \cap D = 0$. Thus
$$ C \cong \frac{C + D}{D} = \frac{(C + D) \cap (M + D)}{D} = \frac{(C + D) \cap M + D}{D} \cong (C + D) \cap M.
$$
Similarly, $D \cong (C + D) \cap M$, whence $C \cong D$. This contradiction establishes the result.
\end{proof}
\bigskip

The {\em ideal index} of a maximal subalgebra $M$ in $L$, $\eta(L:M)$, is the dimension of $C/k(C)$, where $C$ is an ideal completion of $M$ in $L$.
\bigskip

\begin{coro}\label{c:unique} If $M$ is a maximal subalgebra of $L$ then $\eta(L:M)$ is well-defined.
\end{coro}
\bigskip

Next we establish how the ideal index behaves with respect to factor algebras. The following result, or rather its corollary, is analogous to \cite[Lemma 2]{bs}, though our proof is somewhat different.
\bigskip

\begin{propo}\label{p:fac} Let $M$ be a maximal subalgebra of $L$ and let $B$ be an ideal of $L$ with $B \subseteq M$. Let $C/B$ be an ideal completion of $M/B$ in $L/B$, put $k(C/B) = K/B$, and let $D$ be an ideal completion of $M$ in $L$. Then $C/K \cong D/k(D)$.
\end{propo}
\begin{proof} In view of Theorem \ref{t:isom} we can assume that $D \subseteq C$. We have that $(k(C) + B)/B$ is an ideal of $L$ which is inside $M/B$ and $C/B$, so $k(C) \subseteq K \cap D$. But also $K \cap D$ is an ideal of $L$ and $K \cap D \subseteq M \cap D$ so $K \cap D \subseteq k(C)$. Hence $k(C) = K \cap D$. But now $D \not \subseteq K$ since $K \subseteq M$, so $K + D = C$. This yields that $C/K = (K + D)/K \cong D/D\cap K = D/k(D)$.
\end{proof}
\bigskip

\begin{coro}\label{c:etafac} Let $M$ be a maximal subalgebra of $L$ and let $B$ be an ideal of $L$ with $B \subseteq M$. Then $\eta(L/B:M/B)= \eta(L:M)$.
\end{coro}
\begin{proof} From Proposition \ref{p:fac} we have that
$$\eta(L/B:M/B)= \dim((C/B)/k(C/B)) = \dim(C/K)$$
$$= \dim(D/k(D)) =\eta(L:M).$$
\end{proof}
\bigskip

A subalgebra $B$ of a Lie algebra $L$ is called a {\em c-ideal} of $L$ if there is an ideal $C$ of $L$ such that $L = B + C$ and $B \cap C \leq B_L$, where $B_L$ is the largest ideal of $L$ contained in $B$. This concept was introduced and studied in \cite{cideal}. We have the following characterisation of maximal subalgebras that are c-ideals in terms of the ideal index.
\bigskip

\begin{theor}\label{t:cideal} Let $M$ be a maximal subalgebra of $L$. Then $M$ is a c-ideal of $L$ if and only if $\eta(L:M) = \dim(L/M)$.
\end{theor}
\begin{proof} ($\Rightarrow$) Let $L$ be a Lie algebra of smallest dimension having a maximal subalgebra $M$ which is a c-ideal but for which $\eta(L:M) \neq \dim(L/M)$. If $M_L \neq 0$ then $M/M_L$ is a c-ideal of $L/M_L$, by \cite[Lemma 2.1]{cideal}, and so $\eta(L/M_L:M/M_L) = \dim((L/M_L)/(M/M_L))$, by the minimality of $L$. But then $\eta(L:M) = \dim(L/M)$, by Corollary \ref{c:etafac}, a contradiction. If $M_L = 0$, by assumption there is an ideal $C$ of $L$ such that $L = M + C$ and $M \cap C \subseteq M_L = 0$. Since $M$ is a maximal subalgebra of $L$, $C$ is a minimal ideal of $L$. But then $\eta(L:M)  = \dim C = \dim(L/M)$. This contradiction yields the required implication.
\par
($\Leftarrow$) Now let $L$ be a Lie algebra of smallest dimension having a maximal subalgebra $M$ for which $\eta(L:M) = \dim(L/M)$ but $M$ is not a c-ideal of $L$. If $L$ is simple then $\eta(L:M) = \dim L$. But then $M = 0$ and $M$ is a c-ideal of $L$, a contradiction. If $M_L \neq 0$ then $M/M_L$ is a c-ideal of $L/M_L$, by the minimality of $L$, giving that $M$ is a c-ideal of $L$, by \cite[Lemma 2.1]{cideal}, a contradiction again. Thus we have that $L$ is a non-simple Lie algebra with $M_L = 0$. Let $B$ be a minimal ideal of $L$. Then $L = M + B$ and $\eta(L:M) = \dim B$. By our assumption then $\dim(L/M) = \dim B$. It follows that $M \cap B = 0$ and $M$ is a c-ideal of $L$. This contradiction again establishes the implication.
\end{proof}
\bigskip

This yields the following characterisation of solvable Lie algebras in terms of the ideal index. The group-theoretic counterpart is \cite[Corollary, page 97]{bs}; its proof uses concepts that have no analogue in Lie algebras.
\bigskip

\begin{coro}\label{c:solv} The Lie algebra $L$ is solvable if and only if $\eta(L:M) = \dim(L/M)$ for all maximal subalgebras $M$ of $L$.
\end{coro}
\begin{proof} Simply combine Theorem \ref{t:cideal} and \cite[Theorem 3.1]{cideal}.
\end{proof}
\bigskip

With some restrictions on the underlying field the existence of a single solvable maximal subalgebra satisfying the above condition is sufficient to ensure that $L$ is solvable. the corresponding result for groups is \cite[Theorem 4]{bs}, though again its proof is completely different.
\bigskip

\begin{coro}\label{c:solv1} Let $L$ be a Lie algebra over a field $F$, where $F$ has characteristic zero or is algebraically closed of characteristic greater than 5. Then $L$ has a solvable maximal subalgebra $M$ with $\eta(L:M) = \dim(L/M)$ if and only if $L$ is solvable.
\end{coro}
\begin{proof} Simply combine Theorem \ref{t:cideal} and \cite[Theorems 3.2 and 3.3]{cideal}.
\end{proof}
\bigskip

We also have the following characterisation of supersolvable Lie algebras in terms of the ideal index.
\bigskip

\begin{coro}\label{c:etass} The Lie algebra $L$ is supersolvable if and only if $\eta(L:M) = 1$ for all maximal subalgebras $M$ of $L$.
\end{coro}
\begin{proof} Suppose first that $L$ is supersolvable and let $M$ be a maximal subalgebra of $L$. Then $L$ has codimension 1 in $L$, by \cite[Theorem 7]{barnes}, so $\eta(L:M) = 1$ by Corollary \ref{c:solv}.
\par
Now suppose that $L$ is any Lie algebra and that $M$ is a maximal subalgebra of $L$ with $\eta(L:M) = 1$. Then there is an ideal completion $C$ of $L$ with $\dim C/k(C) = 1$. Put $C = k(C) + Fx$. Then $L = M + Fx$ and $M$ has codimension 1 in $L$. But now $L$ is solvable, by Corollary \ref{c:solv}, and hence supersolvable, by \cite[Theorem 7]{barnes}.
\end{proof}
\bigskip

The {\em Frattini subalgebra} of $L$, $F(L)$, is the intersection of all of the maximal subalgebras of $L$. The {\em Frattini ideal}, $\phi(L)$, of $L$ is $F(L)_L$. The group-theoretic counterpart of the next result, which has a different proof, is \cite[Theorem 6]{bs}
\bigskip

\begin{theor}\label{t:ss} If $L$ has a supersolvable maximal subalgebra $M$ with $\eta(L:M) = 1$ and $N(L) \not \subseteq M$ then it is supersolvable.
\end{theor}
\begin{proof} Suppose first that $L$ is $\phi$-free. Then $N(L) = Asoc(L)$, by Theorem 7.4 of \cite{frat}, so there is a minimal abelian ideal $A$ of $L$ with $A \not \subseteq M$. Clearly $L = A \oplus M$. Moreover, $\dim A = 1$ as in Corollary \ref{c:etass} above, and so $L$ is supersolvable.
\par
If $L$ is not $\phi$-free then $\eta(L/\phi(L):M/\phi(L)) = 1$, by Corollary \ref{c:etafac}, and $N(L/\phi(L)) = N(L)/\phi(L)$, by \cite[Theorem 6.1]{frat}, so $L/\phi(L)$ satisfies the hypotheses of this theorem. It follows from the paragraph above that $L/\phi(L)$ is supersolvable. But then $L$ is supersolvable, by \cite[Theorem 6]{barnes}.
\end{proof}
\bigskip

We say that the maximal subalgebra $M$ of $L$ has an {\em abelian ideal completion} if it has an ideal completion $C$ such that $C/k(C)$ is abelian. Then we have the following result, which is more straightforward to prove than its analogue in group theory: \cite[Theorem, page 237]{des2}.
\bigskip

\begin{theor}\label{t:ab} The Lie algebra $L$ is solvable if and only if every maximal subalgebra of $L$ has an abelian ideal completion.
\end{theor}
\begin{proof} Suppose first that $L$ is solvable and that $M$ is a maximal subalgebra of $L$. Then there exists $k \geq 2$ such that $L^{(k)} \subseteq M$ but $L^{(k-1)} \not \subseteq M$. Put $C = L^{(k-1)}$. Then $L = M + C$ so $C$ is an ideal completion of $M$. Also, $L^{(k)} \subseteq k(C)$, so $C/k(C)$ is abelian.
\par
Suppose now that every maximal subalgebra of $L$ has an abelian ideal completion, let $M$ be a maximal subalgebra of $L$ and let $C$ be an abelian ideal completion of $M$. Then $[C, M \cap C] \subseteq C^2 \subseteq k(C) \cap C \subseteq M \cap C$, so $M \cap C$ is an ideal of $L$. We infer that $M \cap C \subseteq M_L$ and hence that $M$ is a c-ideal of $L$. The result now follows from \cite[Theorem 3.1]{cideal}.
\end{proof}
\bigskip

Put $F^*(L)$ equal to the intersection of all maximal subalgebras of $L$ which have no abelian ideal completion (with $F^*(L) = L$ if no such maximal subalgebras exist), and let $\phi^*(L) = F^*(L)_L$. Then we have the following characterisation of the radical of a Lie algebra, whose counterpart in group theory is \cite[Theorem B, page 238]{des2}; again our proof is easier.
\bigskip

\begin{theor}\label{t:rad} Let $L$ be any Lie algebra. Then $\phi^*(L) = R(L)$, the solvable radical of $L$.
\end{theor}
\begin{proof} Suppose first that $M$ is a maximal subalgebra of $L$ with $R(L) \not \subseteq M$. Let $C$ be minimal in the set of ideals of $L$ that are inside $R(L)$ but not in $M$. Then $C$ is an ideal completion of $M$ and $C/k(C)$ is a minimal solvable ideal and so abelian. Thus $R(L) \subseteq \phi^*(L)$.
\par
Next suppose that $R(L) = 0$. We wish to show that $\phi^*(L) = 0$. Suppose not, and let $A$ be a minimal ideal of $L$ with $A \subseteq \phi^*(L)$. Since $\phi(L) = 0$ (see \cite[Theorem 6.5]{frat}), there is a maximal subalgebra $M$ of $L$ with $A \not \subseteq M$, so $L = M + A$ and $M_L \cap A = 0$.
\par
Now $F^*(L) \not \subseteq M$, so $M$ has an abelian ideal completion $C$. Clearly $L = M + C$ and $M_L \cap C = k(C)$. If $A \subseteq C$ then $A^2 \subseteq C^2 \cap A \subseteq k(C) \cap A \subseteq M_L \cap A = 0$ and $A$ is abelian, contradicting the fact that $R(L) = 0$. It follows that $A \cap C = 0$ whence $[A, C] = 0$.
\par
Put $K = \{ x \in L : [x,C] \subseteq M_L \}$. Then $M_L + C \subseteq K$. Furthermore, $[M \cap K, C] \subseteq [K, C] \subseteq M_L \cap K \subseteq M \cap K$, so $M \cap K$ is an ideal of $L$. This yields that $M \cap K \subseteq M_L$ from which $K = M \cap K + C \subseteq M_L + C$. We therefore have that $K = M_L + C$ and $M_L + A \subseteq M_L + C$. But now
$$
A \cong \frac{A}{M_L \cap A} \cong \frac{M_L + A}{M_L} \subseteq \frac{M_L + C}{M_L} \cong \frac{C}{M_L \cap C} = \frac{C}{k(C)}
$$
and so $A$ is abelian. This is impossible, since $R(L) = 0$, so $\phi^*(L) = 0$.
\end{proof}
\bigskip

\section{Maximal completions}
\medskip

The set $I(M)$ is partially ordered by set inclusion; call maximal elements of $I(M)$ {\em maximal completions} of $M$ in $L$. Clearly every ideal completion of $M$ in $L$ is a maximal completion of $M$ in $L$, but the converse is not true in general: for example, if $L$ is the two-dimensional non-abelian Lie algebra with basis $x, y$ in which $[x, y] = y$ and $M = F(x + y)$, then $Fx$ is a maximal completion of $M$ in $L$ but is not an ideal of $L$.
\par

Here we are seeking an analogue to a result of Deskins (see \cite{des2}): namely, that if a Lie algebra has a maximal completion $C$ with $C/k(C)$ abelian then it has an abelian ideal completion. As we shall see, this result holds only with conditions on the underlying field. First we consider the structure of Lie algebras with a maximal abelian subalgebra. A Lie algebra $L$ is {\em completely solvable} if $L^{(1)}$ is nilpotent.
\bigskip

\begin{propo}\label{p:compsolv} Let $L$ be a completely solvable Lie algebra. Then $L$ has a maximal subalgebra $M$ that is abelian if and only if either
\begin{itemize}
\item[(i)] $L$ has an abelian ideal of codimension one in $L$; or
\item[(ii)] $L^{(2)} = \phi(L) = Z(L)$, $L^{(1)}/L^{(2)}$ is a chief factor of $L$, and $L$ splits over $L^{(1)}$.
\end{itemize}
\end{propo}
\begin{proof} Suppose first that $L$ has a maximal subalgebra $M$ that is abelian. If $M$ is an ideal of $L$ we have case (i). So suppose that $M$ is self-idealising, in which case it is a Cartan subalgebra of $L$. Now $L^{(2)} \subseteq \phi(L) \subseteq M$, by \cite[Theorem 6.5]{frat}. If $S$ is a subalgebra of $L$ denote by $\bar{S}$ its image under the canonical homomorphism onto $L/L^{(2)}$. Then $\bar{M}$ is a Cartan subalgebra of $\bar{L}$ and $\bar{L}$ has a Fitting decomposition $\bar{L} = \bar{M} \oplus \bar{L}_1$. Now $\bar{L}_1 \subseteq \bar{L}^{(1)} = \overline{L^{(1)}}$ which is abelian, so $\bar{L}_1$ is an ideal of $\bar{L}$. Moreover, since $\bar{M}$ is a maximal subalgebra of $\bar{L}$, $\bar{L}_1$ is a minimal ideal of $\bar{L}$ and $\overline{L^{(1)}} = \bar{L}_1$. It follows that $L^{(1)}/L^{(2)}$ is a chief factor of $L$. Clearly $\phi(\bar{L}) = 0$, whence $\phi(L) = L^{(2)}$. Also $L = M + L^{(1)}$, so letting $C$ be a subspace of $M$ such that $M = C \oplus (M \cap L^{(1)})$ we see that $C$ is a subalgebra of $L$ and $L$ splits over $L^{(1)}$. Finally, $[M,L^{(2)}] \subseteq M^{(1)} = 0$, so $M \subseteq C_L(L^{(2)})$. Since $M$ is a self-idealising maximal subalgebra of $L$ and $C_L(L^{(2)})$ is an ideal of $L$, we have $C_L(L^{(2)}) = L$, whence $L^{(2)} = Z(L)$.
\par
Consider now the converse. If (i) holds the converse is clear. So suppose that (ii) holds. Then $L = C \oplus L^{(1)}$ where $C$ is an abelian subalgebra of $L$. Put $M = C + L^{(2)}$, so $M$ is clearly abelian. Let $M \subseteq T \subseteq L$. Then $\bar{M} \subseteq \bar{T} \subseteq \bar{L}$. But $\bar{L} = \bar{M} \oplus \overline{L^{(1)}}$ and $\overline{L^{(1)}}$ is a minimal abelian ideal of $\bar{L}$. So $\bar{M} \neq \bar{T}$ implies that $\bar{T} = \bar{L}$. It follows that $M$ is a maximal subalgebra of $L$.
\end{proof}
\bigskip

\begin{propo}\label{p:maxab} Let $L$ be a Lie algebra over an algebraically closed field $F$ of any characteristic. Then $L$ has a maximal subalgebra $M$ that is abelian if and only if either
\begin{itemize}
\item[(i)] $L$ has an abelian ideal of codimension one in $L$; or
\item[(ii)] $L^{(1)}$ has dimension one and $\phi(L) = 0$.
\end{itemize}  In either case, $L$ is completely solvable.
\end{propo}
\begin{proof} Suppose first that $L$ has a maximal subalgebra $M$ that is abelian. If $M$ is an ideal of $L$ we have case (i). So suppose that $M$ is self-idealising, in which case it is a Cartan subalgebra of $L$. Let $L = M \oplus L_{1}(M)$ be the Fitting decomposition of $L$ relative to $M$. Then $\{({\rm ad}\,m)\mid_{L_{1}(M)}: m \in M\}$ is a set of simultaneously triangulable linear mappings. So, there exists $0 \neq b \in L_{1}(M)$ such that $[m, b]
= \alpha(m) b$ for every $m \in M$, where $\alpha(m) \in F$. Then we
have that $M + Fb$ is a subalgebra of $L$ strictly containing $M$, whence $M \oplus Fb = L$. But now $L^{(1)} = Fb$, $\phi(L) \subseteq M \cap L^{(1)} = 0$ and we have case (ii).
\par
Consider now the converse. If (i) holds the converse is clear. So suppose that (ii) holds. Then $L = L^{(1)} \oplus M$ for some subalgebra $M$ of $L$, by \cite[Lemma 7.2]{frat}. Clearly $M$ is an abelian maximal subalgebra of $L$.
\end{proof}
\bigskip

\begin{theor}\label{t:maxcomp} Let $L$ be a Lie algebra over an algebraically closed field and let $M$ be a maximal subalgebra of $L$ with a maximal completion $C$ such that $C/k(C)$ is abelian. Then $M$ has an abelian ideal completion in $L$.
\end{theor}
\begin{proof} Let $L$ be a Lie algebra of smallest dimension for which the result is false, and let $M$ be a maximal subalgebra of $L$ with an abelian maximal completion $C$ but no abelian ideal completion. If $L$ is simple then $k(C) = 0$ and the maximality of $C$ in $I(M)$ implies that $C = L$, which is impossible.
\par
If $M$ is an ideal of $L$ choose $A$ to be minimal in the set of ideals of $L$ (not necessarily proper) that are not contained in $M$. Then $A$ is an ideal completion of $M$ in $L$, $A \cap M = k(A)$ and $A/k(A) \cong L/M$ which is one dimensional and hence abelian.
\par
Suppose now that $M_L = 0$. Then $k(C) = 0$ and we can assume that $C$ is not an ideal of $L$. Now $L$ contains a subalgebra $D$ in which $C$ is a maximal subalgebra, and $D$ is solvable, by Proposition \ref{p:maxab}. The result certainly holds if $L$ is solvable, by Theorem \ref{t:ab}, so we can assume that $D \neq L$. It follows from the maximality of $C$ in $I(M)$ that $D \notin I(M)$. Since $<M, D> = L$ we must have that $k(D) \not \subseteq M$. Thus there is a proper subalgebra $K$ of $D$ such that $K$ is an ideal of $L$ and $L = M + K$. Let $A$ be a minimal non-trivial ideal of $L$ inside $D$. Then $M + A = L$, $k(A) = 0$ and $A$ is abelian because $D$ is solvable. This means that $A$ is an abelian ideal completion of $M$.
\par
Suppose next that $M_L = K \neq 0$ and $K \neq k(C)$, and consider the subalgebra $K + C \neq C$. Then $<M, K + C>\,\, = L$, but $K + C \notin I(M)$ because of the maximality of $C$ in $I(M)$. It follows that $k(K + C) \not \subseteq M$, and so the collection of ideals of $L$ inside $K + C$ but not in $M$ is non-empty. Let $A$ be a minimal element of this set. Clearly $A$ is an ideal completion of $M$ in $L$. Also $(K + C)/K \cong C/C \cap K$ is abelian, since $C/k(C)$ is abelian and $k(C) \subseteq C \cap K$, so $A/k(A) = A/K \cap A \cong (K + A)/K$ is abelian, because $A \subseteq K + C$.
\par
Finally suppose that $M_L = K \neq 0$ and $K = k(C)$. If $S$ is a subalgebra of $L$ denote by $\bar{S}$ its image under the canonical homomorphism onto $L/K$. Then $\bar{C}$ is a completion of $\bar{M}$ in $\bar{L}$ and $\bar{M}$ is core-free. If $\bar{C}$ is a maximal element of $I(\bar{M})$, then, by the paragraph above, $\bar{M}$ has an abelian ideal completion $\bar{A}$ in $\bar{L}$. Then $A$ is an ideal completion of $M$ in $L$ and $A/k(A) = A/K \cap A \cong (A + K)/K \cong \bar{A}$ is abelian.
\par
If $\bar{C}$ is not a maximal element of $I(\bar{M})$ then let $\bar{D}$ be minimal amongst those elements of $I(\bar{M})$ which contain $C$ properly. Clearly $C$ is a maximal subalgebra of $D$, so $D$ is not a completion of $M$ in $L$, by the maximality of $C$. Hence $k(D) \not \subseteq M$. Choose $A$ to be a minimal element in the (non-empty) collection of ideals of $L$ lying in $D$ but not in $M$. Then $A$ is an ideal completion of $M$. Also, $D/K$ is solvable, by Proposition \ref{p:maxab}, since $C/K$ is maximal in $L/K$. Moreover, $L/K \supseteq (A + K)/K \cong A/A \cap K$ and $A \cap K = k(A)$, so the chief factor $A/k(A)$ of $L$ is solvable, and thus abelian.
\end{proof}
\bigskip

\begin{coro}\label{c:maxcomp} Let $L$ be a Lie algebra over an algebraically closed field. Then $L$ is solvable if and only if every maximal subalgebra of $L$ has a maximal completion $C$ in $L$ with $C/k(C)$ abelian.
\end{coro}
\begin{proof} This follows from Theorems \ref{t:ab} and \ref{t:maxcomp}.
\end{proof}
\bigskip

Note that Proposition \ref{p:maxab} and Theorem \ref{t:maxcomp} do not hold when the underlying field $F$ is not algebraically closed, even if it has characteristic zero, as the following example shows.
\par

\begin{ex} Let $S$ have basis $e_1, e_2, e_3$ with $[e_1,e_2] = -[e_2,e_1] = e_3$, $[e_2,e_3] = -[e_3,e_2] = -e_1$, $[e_3,e_1] = -[e_1,e_3] = e_2$, all other products being zero (so $L$ is three-dimensional non-split simple), let $\bar{S}$ be an isomorphic copy of $S$ and denote the image of $s \in S$ in $\bar{S}$ by $\bar{s}$. Put $L = S \oplus \bar{S}$ with $[S, \bar{S}] = 0$.
\end{ex}
Every maximal subalgebra of $S$ is one dimensional, and so abelian, showing that Proposition \ref{p:maxab} does not hold. Also, it is easy to check that the diagonal subalgebra $M = \{ x \in L : x = s + \bar{s} \hbox{ for some } s \in S \}$ is maximal in $L$ and that $C = Fe_1 + F\bar{e_1}$ is a maximal abelian completion of $M$ in $L$. However, $M$ has no maximal abelian ideal completion in $L$.
\bigskip

\section{Subideal completions}
\medskip

When the underlying field is algebraically closed of characteristic zero, the following characterisation of supersolvable Lie algebras in terms of maximal completions follows easily from Theorem \ref{t:maxcomp}, Corollary \ref{c:etass} and Lie's Theorem.
\bigskip

\begin{propo}\label{p:supsolv} Let $L$ be a Lie algebra over an algebraically closed field of characteristic zero. Then $L$ is supersolvable if and only if every maximal subalgebra of $L$ has a maximal completion $C$ with $\dim C/k(C) = 1$.
\end{propo}
\bigskip

The above result does not hold, however, over every field of characteristic zero, as the following example shows.
\bigskip

\begin{ex} Let $L = Fa \oplus S$, where $Fa$ is a one-dimensional abelian ideal of $L$ and $S$ is a three-dimensional non-split simple ideal of $L$. Then the maximal subalgebras of $L$ are $S$ and the subalgebras of the form $Fa + Fx$ where $x \in S$. For the former a maximal completion is $C = Fa$; for the latter a maximal completion is $C = Fa + Fy$ where $y \in S \setminus Fx$. In either case $\dim C/k(C) = 1$, but $L$ is not supersolvable.
\end{ex}
\bigskip

So, in order to find another characterisation of supersolvable Lie algebras over a general field in terms of the index complex we follow \cite{be}. If $C$ is a subideal of $L$ and $C \in I(M)$ we call $C$ a {\em subideal completion} of $L$. Unlike ideal completions there is no numerical invariant associated with subideal completions, as the following example shows.
\par

\begin{ex} Let $L$ be the Lie algebra with basis $a, b, c,d$ and products $[a,b] = - [b,a] = c, [b,c] = - [c,b] = d$, all other products being zero. Then $L$ is nilpotent and so all of its subalgebras are subideals of $L$. Put $M = Fb + Fc + Fd$. Then $M$ is a maximal subalgebra of $L$ and it is easy to check that $C_1 = Fa$ and $C_2 = Fa + Fc$ are both completions of $M$ in $L$. However, $k(C_1) = k(C_2) = 0$, so $\dim(C_1/k(C_1)) = 1$, whereas $\dim(C_2/k(C_2)) = 2$.
\end{ex}
\bigskip

In the following we will need subideal completions with an extra property. Let $M$ be a maximal subalgebra of the Lie algebra $L$ and let $S(M) = \{ C \in I(M) : C \hbox{ is a subideal of } L \hbox{ and } L = M + C \}$. Clearly, every ideal completion of $M$ in $L$ is in $S(M)$, so $S(M)$ is non-empty.
\par

Then we have the following characterisation of supersolvable Lie algebras in terms of such completions which is analogous to \cite[Theorem 1]{be}. To use a similar argument to theirs we would require an underlying field of characteristic zero, as it relies on the Baer radical being a nilpotent ideal (see \cite{as}). The argument given below is independent of the underlying field.
\bigskip

\begin{theor}\label{t:subideal} The Lie algebra $L$ is supersolvable if and only if every maximal subalgebra of $L$ has an element $C \in S(M)$ with $\dim C/k(C) = 1$.
\end{theor}
\begin{proof} Suppose first that every maximal subalgebra $M$ of $L$ has an element $C \in S(M)$ with $\dim C/k(C) = 1$. It is clear that $L/\phi(L)$ satisfies the same hypotheses. Moreover, if $L/\phi(L)$ is supersolvable then so is $L$, by \cite[Theorem 6]{barnes}, so we can assume that $L$ is $\phi$-free. Since $k(C) \subseteq M$, we have that $M$ has codimension one in $L$. It follows from \cite{codone} that $L \cong R \oplus S$, where $R$ is a supersolvable ideal of $L$ and $S$ is a three-dimensional simple ideal of $L$.
\par

Suppose that $S \neq 0$, let $M$ be a maximal subalgebra of $L$ containing $R$, let $C \in S(M)$ with $\dim C/k(C) = 1$ and let $\pi : L \rightarrow S$ be the projection homomorphism from $L$ onto $S$. Then $\pi(C)$ is a subideal of $S$, so $\pi(C) = S$ or $0$. Since $\pi(k(C)) = 0$ the former is impossible. The latter implies that $C \subseteq R$ which is also impossible. It follows that $S = 0$ and $L$ is supersolvable.
\par

Conversely, suppose that $L$ is supersolvable. Then $L$ has an ideal completion $C$ (and so $C \in S(M)$) with $\dim C/k(C) = 1$ by Corollary \ref{c:etass}.
\end{proof}
\bigskip

A class ${\mathcal H}$ of finite-dimensional solvable Lie algebras is called a {\em homomorph} if it contains, along with an algebra $L$, all epimorphic images of $L$. The following result is a straightforward adaptation of \cite[Proposition 1]{be}. We include the proof for the convenience of the reader.
\bigskip

\begin{propo}\label{p:hom} Let ${\mathcal H}$ be a homomorph that is closed under ideals, let $M$ be a maximal subalgebra of $L$ and let $N$ be an ideal of $L$ such that $N \subseteq M$. If $C$ is a maximal (respectively, subideal) completion of $M$ in $L$ with $C/k(C) \in {\mathcal H}$, there is a maximal (respectively, subideal) completion $C^*$ of $M$ in $L$ such that $N \subseteq C^*$ and $C^*/k(C^*) \in {\mathcal H}$.
\end{propo}
\begin{proof} Assume that $M$ has a maximal completion $C$ in $L$ with $C/k(C) \in {\mathcal H}$. If $N \subseteq C$ we can take $C^* = C$, so assume that $N \not \subseteq C$. Since $C$ is a maximal completion of $M$ and $C \subset N + C$ we have $k(N + C) \not\subseteq M$. Hence there is a chief factor $C^*/P$ of $L$ such that $N + k(C) \subseteq P \subset C^* \subseteq k(N + C)$ and $L = M + C^*$. Thus $C^*$ is a maximal completion of $M$ and
$$ \frac{k(N + C)}{N + k(C)} \hbox{ is an ideal of } \frac{N + C}{N + k(C)} \in {\mathcal H}, \hbox{ whence } \frac{k(N + C)}{N + k(C)} \in {\mathcal H}.$$
Consequently, $C^*/P \in {\mathcal H}$. Since $k(C^*) = P$, $C^*$ is a maximal completion of $M$ such that $N \subseteq C^*$.
\par

For subideal completions the argument is similar.
\end{proof}
\bigskip

Let ${\mathcal H}$ be as in the above Proposition and let $I({\mathcal H}) = \{L :$ for each maximal subalgebra $M$ of $L$ there exists $C \in I(M)$, maximal in $I(M)$, with $C/k(C) \in {\mathcal H}\}$, $S({\mathcal H}) = \{L :$ for each maximal subalgebra $M$ of $L$ there exists $C \in S(M)$ with $C/k(C) \in {\mathcal H}\}$. Then, as a basis for induction arguments, it is claimed in \cite[Corollary 2]{be} that, in respect of the corresponding concepts for groups, $I({\mathcal H})$ and $S({\mathcal H})$ are saturated homomorphs. However, if $K = \{e,a,b,c\}$ is Klein's 4-group and ${\mathcal H}$ is the homomorph of groups of order one, then it is easy to see that $K \in I({\mathcal H})$, whereas $K/\{e,a\} \not \in I({\mathcal H})$, since it has order two and is the only maximal completion for its trivial maximal subgroup. Less trivial examples are also easy to construct.
\par

In the case of Lie algebras $S({\mathcal H})$ is also a saturated homomorph, but $I({\mathcal H})$ is not. For, if $L$ is as in Example 4.1 above and ${\mathcal H}$ is the homomorph of all abelian Lie algebras, then $L \in I({\mathcal H})$ but $L/Fa \cong S \not \in I({\mathcal H})$, since $S$ itself is the only maximal completion for any of its maximal subalgebras and $S/k(S) \cong S$ is not abelian.
\bigskip


TITLE: How to guarantee the existence of continuity sets?
QUESTION [1 upvotes]: The Skorohod's representation theorem says:
Let $X, X_1, X_2,...$ be S-valued random
elements, where (S, p) is a separable metric space, such that $X_n \xrightarrow{d} X$. Then there exists a common probability space with some random elements $Y, Y_1, Y_2,...$ defined on it such that $Y_n = X_n$ in distribution $\forall n$ and $Y = X$ in distribution, and $Y_n$ converges almost surely to $Y$.
Every proof I found involves the process of the following construction: for all $x \in S$, for any $\epsilon>0$, one can always pick $r$ with $0<r<\epsilon$ such that $P(X\in \partial B(x,r))=0$.
My question might be stupid: why is it always possible to pick such an open ball such that its boundary has measure $0$ in $P_X$? (in other words, an $X$-continuity set, which is defined as: a set $A\in B(S)$ such that $P(X\in \partial A)=0$)
For complete proof, see resource: Kallenberg's Foundations of Modern Probability (2002) p79; or Billingsley's Convergence of Probability measures (1999) p70.

REPLY [1 votes]: $\partial B(x,r)\subset \{y: d(y,x)=r\}$. It is obvious from this that $\partial B(x,r), 0<r<\epsilon$ is a disjoint collection of measurable sets. In a probability space you can only have countably many disjoint sets with positive measure. [If you have not seen this before I will add  a proof]. If $P(\partial B(x,r))>0$ whenever $0<r<\epsilon$ you will uncountably many sets with positive measure, a contradiction. Hence, $P(\partial B(x,r))=0$ for some $r \in (0,\epsilon)$.
Proof of $\partial B(x,r)\subset \{y: d(y,x)=r\}$: Let $y \in \partial B(x,r)$. By definition of boundary there exist sequence $(y_n)$ and $(z_n)$ such that $d(y_n,x) <r, d(z_n,x) \geq r$ and $y_n \to y, z_n \to  y$. By continuity of the distance function we get $d(y,x) \leq r $ and $d(y,x) \geq r$ so we must have $d(y,x)=r$.


text:
\begin{document}

\title{Lipschitz continuous dependence of piecewise constant Lam\'e coefficients {}from boundary data: the case of non flat interfaces
 }

\author{Elena Beretta\thanks{Politecnico di Milano, Italy
(elena.beretta@polimi.it)} \and Elisa Francini\thanks{Universit\`{a} degli Studi di Firenze, Italy (elisa.francini@unifi.it)}\and
Antonino Morassi\thanks{
Universit\`{a} degli Studi di Udine, Italy (antonino.morassi@uniud.it)}\and
Edi Rosset\thanks{
Universit\`{a} degli Studi di Trieste, Italy (rossedi@units.it)}\and
Sergio Vessella\thanks{
Universit\`{a} degli Studi di Firenze, Italy (sergio.vessella@unifi.it)}}

\date{}
\maketitle

\begin{abstract}

We consider the inverse problem of determining the Lam\'e moduli for a piecewise constant elasticity tensor
$\CC= \sum_{j} \CC_j \chi_{D_j}$, where $\{D_j\}$ is a known finite partition of the body $\Omega$, {}from the Dirichlet-to-Neumann map.
We prove that Lipschitz stability estimates can be derived under $C^{1,\alpha}$ regularity assumptions on the interfaces.

\end{abstract}

\noindent
\textbf{MSC: }{Primary 35R30, 35J55 Secondary: 35R05 }

\noindent
\textbf{keywords: }{Inverse boundary value problem, Lam\'{e} system, piecewise constant coefficients, Lipschitz stability}

\section{Introduction\label{sec0}}

An important inverse problem arising {}from engineering
sciences consists in determining the elasticity coefficients
of the material occupying a three dimensional body {}from
measurements of tractions and displacements taken on its
accessible boundary.

The boundary value problem {}from which this inverse problem
originates is as follows. Let $\Omega$ be a bounded domain in
$\RR^3$ made by a linearly elastic isotropic material, with Lam\'{e}
moduli $\mu$ and $\lambda$ satisfying the strong
convexity conditions $\mu(x)\geq \alpha_0 >0$, $2\mu(x) +
3\lambda(x) \geq \beta_0$ in $\Omega$, for some positive constants
$\alpha_0$ and $\beta_0$. For a given $\psi \in H^{
\frac{1}{2}}(\partial \Omega)$, the direct problem consists in
finding the displacement field $u \in H^1(\Omega)$ solution to the
Dirichlet problem

\begin{equation}
    \label{I1.1}
    \left\{\begin{array}{rcl}
             \mbox{div}(\CC  \nabhat u)& = & 0, \quad \mbox{ in }\om, \\
             u & = & \psi, \quad \mbox{ on }\der\om,
           \end{array}\right.
\end{equation}

where $\CC= \lambda(x) I_3 \otimes I_3 + 2\mu(x){\II}_{Sym}$ is
the Lam\'{e} elasticity tensor.

We denote by $\Lambda_{\CC}: H^{ \frac{1}{2}}(\partial \Omega)
\rightarrow H^{ - \frac{1}{2}}(\partial \Omega)$ the
Dirichlet-to-Neumann map associated to the problem \eqref{I1.1},
that is the operator which maps the Dirichlet data $u|_{\partial
\Omega} = \psi$ onto the corresponding Neumann data $(\CC \nabhat
u) \nu|_{\partial \Omega}$, where $\nu$ is the outer unit normal
to $\Omega$.

An interesting inverse problem is the determination of the Lam\'{e} coefficients $\mu$ and $\lambda$ when
$\Lambda_{\CC}$ is known. Most of the results available in the
literature concern the uniqueness issue. A linearized version of
this inverse problem was considered by Ikehata \cite{Ik}. In
\cite{NU1}, Nakamura and Uhlmann established that in two
dimensions the Lam\'{e} moduli are uniquely determined by
$\Lambda_{\CC}$, provided that they are smooth (e.g., $C^\infty(
\overline{\Omega}))$ and sufficiently close to positive constants.
The uniqueness in dimension three was proved in \cite{NU2},
\cite{ER}, \cite{NU3}, assuming that the Lam\'{e} moduli are
$C^\infty( \overline{\Omega})$ and $\mu$ close to a positive
constant. Recent results concern the uniqueness in the case of
partial Cauchy data, see \cite{IUY} for details.

The stability issue for the above inverse problem is expected to
be significantly more difficult than uniqueness and, to our
knowledge, no general result is known. In the simpler context of
an electric conductor, which involves the determination of a
single smooth coefficient in a scalar elliptic equation {}from
boundary measurements, it is well-known that the optimal rate of
continuous dependence is of logarithmic type, see, for instance, \cite{Al1} and \cite{M}. It
follows that logarithmic stability estimates, or even worse ones,
are expected in our case. In addition, the
situation is more complicated because, in several practical
applications, the Lam\'{e} moduli are not smooth and, in some
cases, may also be discontinuous.

In order to have better stability results, a possible way
is based on the
introduction of \emph{suitable} a priori assumptions that are physically
relevant and restore well-posedness. Following the approach suggested by
Alessandrini and Vessella \cite{AV} in the conductivity framework,
in \cite{BFV} the
authors considered a class of piecewise constant elasticity
tensors of the form

\begin{equation}
    \label{I4.1}
    \CC=\sum_{j=1}^N (\lambda_j I_3\otimes I_3
    +2\mu_j{\II}_{Sym})\chi_{D_j}(x),
\end{equation}

where the collection of disjoint Lipschitz domains
$\{D_j\}_{j=1}^N$ forms a known decomposition of the domain
$\Omega$, and $\lambda_j$, $\mu_j$, $j=1,...,N$, are unknown
constants to be determined {}from $\Lambda_\CC$. Assuming that the boundaries of the domains $D_j$
contain \textit{flat portions}, the authors were able to prove a
Lipschitz continuous dependence of the Lam\'{e} moduli {}from the
local Dirichlet-to-Neumann map.

The structure \eqref{I4.1} assumed for $\CC$ fits well in several
problems arising in applications. Polyhedral partitions of
$\Omega$ appear frequently in finite element meshing used for
effective reconstruction of the Lam\'{e} parameters \cite{BJK}. In
identification of material properties of masonry walls or concrete
dams, for example, the actual elasticity coefficients are
approximated by assuming that each finite element or group of
finite elements is made by homogeneous Lam\'{e} material. The
partition of the domain is often suggested by a priori information
on different grades of the material or, in the case of dams, by
the possible presence of natural joints inside the concrete \cite{XJY}.
Obviously, it is not always possible to ensure that the domains
$D_j$ have a flat portion of their boundary in common and,
therefore, in order to address these more general inverse
problems, it is necessary to remove this a priori assumption.

In this paper we prove
a Lipschitz stability estimate assuming
$C^{1,\alpha}$ regularity of some portions $\Sigma_j$ of the interfaces joining contiguous
domains $D_{j-1}$, $D_j$ and on the portion $\Sigma$ of $\partial \Omega$
where the measurements are taken. The precise regularity conditions are
given in Section \ref{subsec2.3} (assumptions \textbf{(A1)}).

Our proof is inspired by the paper \cite{BFV} and is mainly based
on the use of unique continuation properties and on a refined local analysis, near the
$C^{1,\alpha}$ interface $\Sigma_j$, of the behavior of the corresponding biphase fundamental solution
(see Subsection \ref{subsec-sing} for the precise setting). To this aim, a new mathematical tool is
the recent asymptotic approximation of this fundamental solution (see \cite{AdCMR}) in terms
of the biphase fundamental solution associated to a flat interface, which was determined in a close form
by Rongved \cite{R}.

We follow a slightly different procedure to prove the stability estimate. In \cite{BFV} the authors reformulate the direct problem
in terms of the nonlinear forward map $F$ acting on a compact subset $K$ of
$\RR^{2N}$, and use an abstract lemma (see \cite{BaVe}) which ensures that the inverse map
$\left(F|_K\right)^{-1}$ is Lipschitz continuous. Here, instead, we give a more direct proof following the lines
in \cite{AV} for the conductivity framework. As in \cite{AV}, \cite{BFV}, also our proof proceeds
by induction. However, in order to simplify the presentation and to emphasize the crucial points
where new tools are needed, we focus on the first two steps of the induction process. Precisely, the key role
of the asymptotic approximation of the biphase fundamental solution is emphasized in the first step, whereas
the second step explains how to use the transmission conditions at the interface and the stability estimates for
the Cauchy problem to propagate the smallness crossing an interface.

\section{Main result}\label{sec2}
\subsection{Notation and main definitions}\label{subsec2.1}
For every $x\in\RR^3$ we set  $x=(x^\prime,x_3)$ where
$x^\prime\in\RR^{2}$ and $x_3\in \RR$. For every $x\in \RR^3$, $r$
and $L$ positive real numbers we will denote by $B_r(x)$,
$B_r^\prime(x^\prime)$ and $Q_{r,L}(x)$  the open ball in $\RR^3$
centered at $x$ of radius $r$, the open ball in $\RR^2$ centered
at $x^\prime$ of radius $r$ and the cylinder
$B_r^\prime(x^\prime)\times (x_3-Lr,x_3+Lr)$, respectively. In the
sequel $B_r(0)$, $B_r^\prime(0)$ and $Q_{r,L}(0)$ will be denoted
by $B_r$, $B^\prime_r$ and $Q_{r,L}$, respectively.
 We will also denote by $\RR^3_+=\{(x^\prime,x_3)\in\RR^3 | x_3>0\}$,
$\RR^3_-=\{(x^\prime,x_3)\in \RR^3 | x_3<0\}$, $B^+_r=B_r\cap
\RR^3_+$, and $B^-_r=B_r\cap \RR^3_-$.

For any subset $D$ of $\RR^3$ and any $h>0$, we denote by
\[(D)_{h}=\{x \in D| \mbox{ dist}(x,\RR^3\setminus D)>h\}.\]

\begin{definition}
  \label{def:2.1} (${C}^{k,\alpha}$ regularity)
Let $U$ be a bounded domain in $\mathbb{R}^{3}$. Given $k$,
$\alpha$, with $k\in \mathbb{N}$ and $0<\alpha\leq 1$, we say that
$U$ is of class ${C}^{k,\alpha}$ with constants $r_{0}$, $L$, if,
for any $P \in \partial U$, there exists a rigid transformation of
coordinates under which we have $P=0$ and
\begin{equation*}
  U \cap B_{r_{0}}(0)=\{x \in B_{r_{0}}(0) | x_{3}>\varphi(x')
  \},
\end{equation*}

where $\varphi$ is a ${C}^{k,\alpha}$ function on $\RR^2$
satisfying

\begin{equation*}
\varphi(0)=0,
\end{equation*}

\begin{equation*}
\nabla \varphi (0)=0, \quad \hbox{when } k \geq 1,
\end{equation*}

\begin{equation*}
\|\varphi\|_{C^{k,\alpha}(\RR^2)} \leq Lr_{0}.
\end{equation*}

\medskip
\noindent When $k=0$, $\alpha=1$, we also say that $U$ is of
\textit{Lipschitz class with constants $r_{0}$, $L$}.
\end{definition}

\begin{rem}
  \label{rem:2.1}
  We use the convention to normalize all norms in such a way that their
  terms are dimensionally homogeneous and coincide with the
  standard definition when the dimensional parameter equals one.
  For instance, the norm appearing above is meant as follows

\begin{equation*}
  \|\varphi\|_{{C}^{k,\alpha}(\RR^2)} =
  \sum_{i=0}^k r_0^i
  \|\nabla^i\varphi\|_{{L}^{\infty}(\RR^2)}+
  r_0^{k+\alpha}|\nabla^k\varphi|_{\alpha, \RR^2},
\end{equation*}

where
\begin{equation*}
|\nabla^k\varphi|_{\alpha, \RR^2}= \sup_ {\overset{\scriptstyle
x', y'\in \RR^2}{\scriptstyle x'\neq y'}}
\frac{|\nabla^k\varphi(x')-\nabla^k\varphi(y')|} {|x'-y'|^\alpha}.
\end{equation*}

Similarly,

\begin{equation*}
\|u\|_{H^m(\Omega)}=\left(\sum_{i=0}^m
r_0^{2i-3}\int_\Omega|\nabla^i u|^2\right)^{\frac{1}{2}},
 \quad
  \|u\|_{{C}^{k}(\Omega)} =\sum_{i=0}^{k}
  {r_{0}}^{i}\|\nabla^{i} u\|_{{L}^{\infty}(\Omega)},
\end{equation*}
\begin{equation*}
\|u\|_{L^2(\partial\Omega)}=\left(
r_0^{-2}\int_{\partial\Omega}|u|^2\right)^{\frac{1}{2}},
\end{equation*}

where $H^0(\Omega) = L^2(\Omega)$, and so on for trace norms such as
$\|\cdot\|_{H^{\frac{1}{2}}(\partial \Omega)}$,
$\|\cdot\|_{H^{-\frac{1}{2}}(\partial \Omega)}$, where $\Omega$ is
a bounded subset of $\RR^3$ with regular boundary.
\end{rem}

We will also make use of the following notation for matrices and
tensors. Let $\mathcal{L}(\RR^3,\RR^3)$ be the linear space of $3
\times 3$ matrices. For any $A, B \in \mathcal{L}(\RR^3,\RR^3)$,
we set $A : B=\sum_{i,j=1}^3 A_{ij}B_{ij}$, $|A|^2 = A : A$
 and $\hat A=\frac{1}{2}(A+A^T)$. By $I_3$ we denote the $3\times 3$ identity matrix and by ${\II}_{Sym}$ we denote the fourth order tensor such that ${\II}_{Sym}A=\hat A$.

\bigskip

In the whole paper we are going to consider isotropic elastic
materials, hence the fourth order elasticity tensor $\CC$ is given
by
\begin{equation}\label{isotr}
\CC(x)=\lambda(x)I_3\otimes I_3 +2\mu(x){\II}_{Sym}, \quad\hbox{
for  a.e.  }x\mbox{ in } \Omega,
\end{equation}
where $\Omega$ is a bounded domain in $\RR^3$ of Lipschitz class and $(I_3\otimes I_3)A= (I_3 : A) A$ for every
$3 \times 3$ matrix $A$.
The real valued functions $\lambda=\lambda(x)$ and $\mu=\mu(x)\in
L^{\infty}(\om)$ are the Lam\'{e} moduli, and satisfy the strong
convexity condition

\begin{equation}
\label{strong-convexity}
     \alpha_0\leq \mu(x) \leq \alpha_0^{-1}, \quad \lambda(x) \leq
    \alpha_0^{-1}, \quad 2\mu(x) +3\lambda(x) \geq \beta_0, \quad
    \hbox{for a.e. } x \hbox{ in } \Omega,
\end{equation}

where $\alpha_0\in (0,1]$, $\beta_0 \in(0,2]$ are given constants. Let us
notice that the Poisson's ratio $\nu(x) =
\frac{\lambda(x)}{2(\lambda(x)+\mu(x))}$ satisfies

\begin{equation}
\label{Poisson}
    -1 + \frac{\alpha_0 \beta_0}{4} \leq \nu(x) \leq \frac{1}{2} -
    \frac{\alpha_0^2}{4}, \quad
    \hbox{for a.e. } x \hbox{ in } \Omega.
\end{equation}

Under these assumptions, the elasticity tensor  $\CC$ satisfies
the minor and major symmetry conditions

\begin{equation}
\label{symmetry-of-C}
    \CC  A = \widehat{A}, \quad \CC A : B= \CC B : A,
\end{equation}

and the strong convexity condition

\begin{equation}
\label{strong-convexity-of-C}
    \CC  A : A \geq \xi_0 |A|^2,
\end{equation}

where $\xi_0 = \min \{2\alpha_0, \beta_0\}$, for every $A$, $B\in \mathcal{L}(\RR^3,\RR^3)$.

In the sequel we will make use of the following norm in the linear
space of bounded isotropic tensors:
\[\|\CC\|_{\infty}=\max\left\{\|\lambda\|_{L^{\infty}(\om)},\|\mu\|_{L^{\infty}(\om)}\right\}.\]

Our boundary measurements are represented by the
Dirichlet-to-Neumann map. As a matter of fact, since we will
restrict our measurements to boundary data that have support on
some subset of the boundary, we will make use of a local
Dirichlet-to-Neumann map.

\begin{definition}[The Local Dirichlet-to-Neumann map]
\label{DNmap}
Let $\om$ be a bounded domain of Lipschitz class and let $\Sigma$ be an open portion of $\der\om$.
We denote by $H^{ \frac{1}{2} }_{co}(\Sigma)$ the function space
\[
H^{ \frac{1}{2} }_{co}(\Sigma):=\left\{\phi\in H^{ \frac{1}{2} }(\der\om)\,\ |\ \,\mbox{supp }\phi\subset\Sigma\right\}
\]
and by $H^{- \frac{1}{2} }_{co}(\Sigma)$ the topological dual of $H^{ \frac{1}{2} }_{co}(\Sigma)$. We denote by $<\cdot,\cdot>$ the dual pairing between $H^{ \frac{1}{2} }_{co}(\Sigma)$ and
$H^{- \frac{1}{2} }_{co}(\Sigma)$ based on the $L^2(\Sigma)$ scalar product, that is
$<f,g> = r_0^{-2}\int_{\partial\Omega}fg$, for every $f,g\in L^2(\partial\Omega)$.
 Then, given $\psi\in H^{ \frac{1}{2} }_{co}(\Sigma)$,  there exists a unique vector valued function $u\in H^1(\om)$ weak solution to the Dirichlet
 problem

\begin{equation}\label{1}
    \left\{\begin{array}{rcl}
             \mbox{div}(\CC  \nabhat u)& = & 0, \quad \mbox{ in }\om, \\
             u & = & \psi, \quad \mbox{ on }\der\om.
           \end{array}\right.
\end{equation}

We define the local Dirichlet-to-Neumann linear map
$\Lambda_{\CC}^{\Sigma}$ as follows:
\[
\Lambda_{\CC}^{\Sigma}:\psi\in  H^{ \frac{1}{2} }_{co}(\Sigma)\rightarrow
(\CC\nabhat u)n|_{\Sigma}\in H^{- \frac{1}{2} }_{co}(\Sigma),
\]
\end{definition}

where $n$ is the exterior unit vector to $\Omega$.

Note that for $\Sigma=\der\om$ we get the usual
Dirichlet-to-Neumann map. For this reason we will set
$\Lambda_{\CC}:=\Lambda_{\CC}^{\der\om}$.

The map $\Lambda_{\CC}^{\Sigma}$ can be identified with the bilinear form on $H^{ \frac{1}{2} }_{co}(\Sigma)\times H^{ \frac{1}{2} }_{co}(\Sigma)$ by
\begin{equation}\label{DNbilineare}
\widetilde{\Lambda}_{\CC}^{\Sigma}(\psi, \phi):=<\Lambda_{\CC}^{\Sigma}\psi, \phi>=r_0^{-2}\int_{\om}\CC \nabhat u:\nabhat v,
\end{equation}
for all $\psi,\phi \in H^{ \frac{1}{2} }_{co}(\Sigma)$ and where $u$ solves (\ref{1})
and $v$ is any $H^1(\om)$ function such that $v=\phi$ on $\der\om$.

We shall denote by $\|\cdot\|_{\star}$ the usual norm in the linear space $\mathcal{L}\left(H^{ \frac{1}{2} }_{co}(\Sigma),H^{- \frac{1}{2} }_{co}(\Sigma)\right)$.
Let us observe that, {}from our convention on the homogeneity of the norms,
we have in particular that

\[
\left\|\Lambda_{\CC}^{\Sigma}\right\|_{\star}=\sup
\left|r_0^{-2}\int_{\Omega}\CC\nabhat u:\nabhat
v\right|,
\]
where the sup is taken for $\phi$, $\psi\in H^{ \frac{1}{2} }_{co}(\Sigma)$,
$\|\phi\|_{ H^{ \frac{1}{2} }_{co}}(\Sigma) = \|\psi\|_{ H^{ \frac{1}{2}}_{co}}(\Sigma) =1$, being
$u$ the solution to (\ref{I1.1}) and $v\in H^1(\Omega)$ any extension of $\phi$.

\subsection{A priori assumptions and statement of the main result}\label{subsec2.3}

Our main assumptions are:

\bigskip

\noindent{\bf (A1)} $\om\subset\RR^3$ is an open bounded domain such that
\[
\Omega \ \hbox{is of class } C^{0, 1}, \ \hbox{with constants } \,
r_0, L,
\]
and we assume that
\[\ov{\om}=\cup_{j=1}^N \ov{D}_j,\]
where  $D_j$, $j=1,\ldots,N$, are connected and pairwise disjoint
domains of class $C^{0,1}$ with constants $r_{0}$, $L$, such that
there exists a constant $A>0$ such that
\[
|D_j|\leq Ar_0^3,\quad j=1,\cdots, N.
\]

We also assume that there exists one region, say $D_1$, such that
$\der D_1\cap\der\om$ contains the open  portion $\Sigma$ where
the measurements are taken. Moreover, for every
$j\in\{2,\ldots,N\}$ there exist $j_1,\ldots,j_M\in\{1,\ldots,N\}$
such that
\[
D_{j_1}=D_1,\quad D_{j_M}=D_j,
\]
and, for every $k=2,\ldots,M$, the set
\[
\der D_{j_{k-1}}\cap \der D_{j_k}
\]
contains a portion $\Sigma_k\subset\om$.

\noindent
Furthermore, for $k=1,\ldots,M$, we assume  there exists $P_k\in\Sigma_k$ and a rigid transformation of coordinates such that $P_k=0$ and for all $k=1,\cdots, M$
\label{walkway}\begin{eqnarray*}
\Sigma_k\cap Q_{r_0,L} &=& \{x\in Q_{r_0,L}\ |\  x_3=\varphi_k(x')\}, \\
D_{j_k}\cap Q_{r_0,L} &=& \{x\in  Q_{r_0,L}\ |\  x_3<\varphi_k(x')\}, \\
 D_{j_{k-1}}\cap  Q_{r_0,L} &=& \{x\in  Q_{r_0,L}\ |\
 x_3>\varphi_k(x')\},
\end{eqnarray*}
where $\Sigma_1\subset\Sigma$, with $\varphi_k\in C^{1,\alpha}(\mathbb{R}^2)$ such that
\[
\varphi_k(0)=|\nabla\varphi_k(0)|=0,\quad \|\varphi_k\|_{C^{1,\alpha}(\mathbb{R}^2)}\leq Lr_0,
\]
and where  we set  $D_{j_{0}}:=\mathbb{R}^3\backslash
\overline{\Omega}$.  Finally, let
\[
D_0:= \{x\in Q_{ \frac{2}{3}r_0,L} |
\varphi_1(x')<x_3<\frac{2}{3}r_0L\}
\]
and let $\Omega_0=\textit{Int}(\overline{\Omega\cup D_0})$.
Observe that $D_0$ and $\Omega_0$ are of class $C^{0,1}$, with
constants $r_1,L_1$ such that $L_1= \tan \left ( \frac{\pi}{4} +
\frac{1}{2} \arctan L \right )$ and $r_1 = \frac{2}{3} \frac{
\sqrt{  1+L^2  }}  { \sqrt{1+L_1^2}  }r_0$.

For simplicity we will call $D_{j_1},\ldots,D_{j_M}$ \textit{a
chain of domains connecting $D_1$ to $D_j$}. For any
$k\in\{1,\ldots,M\}$ we will denote by $n_k$ the exterior unit
vector to $\partial D_k$ in $P_k$.

\bigskip

\noindent{\bf (A2)} We assume that the tensor $\CC$ is piecewise
constant

\begin{equation}\label{isotrelisa}
\CC=\sum_{j=1}^N\CC_j \chi_{D_j}(x),
\end{equation}

where
\[\CC_j=\lambda_j I_3\otimes I_3 +2\mu_j{\II}_{Sym},\]

with constant Lam\'{e} coefficients $\lambda_j$ and $\mu_j$
satisfying \eqref{strong-convexity}.

In what follows we shall refer to the constants $L$, $\alpha$,
$A$, $N$, $\alpha_0$, $\beta_0$ as to the \emph{a priori data}.

In the sequel we will introduce a number of constants that we will
always denote by $C$. The values of these constants might differ
{}from one line to the other.

The main result of this paper is the following stability result.

\begin{theo}\label{teo2.1}

Let $\om$ and $\Sigma$ satisfy {\bf (A1)} and let the tensors $\CC$
and $\overline{\CC}$ satisfy {\bf (A2)}. Then there exists a positive
constant $C$ depending only on the a priori data such that

\begin{equation}\label{stabil}
\|\CC-\overline{\CC}\|_{\infty}\leq C r_0^{}\left\|\Lambda^{\Sigma}_{\CC}-\Lambda^{\Sigma}_{\overline{\CC}}\right\|_{\star}.
\end{equation}

\end{theo}

\subsection{Some basic properties of the Lam\'{e} system}\label{sec4}

\subsubsection{Alessandrini's identity}\label{subsec-aless}

Alessandrini's identity is a key relation connecting the
Dirichlet-to-Neumann maps and a volume integral. It was originally
derived in \cite{Al1} within the conductivity framework. Its
extension to our context is as follows. Given $u_1$ and $u_2$
solutions to
\[
 \mbox{div}(\CC^{k}  \nabhat u_k) =  0, \quad \mbox{ in }\om,
 \quad k=1,2,
\]

we have

\begin{equation}
    \label{AlessId}
    \int_{\om}(\CC^{1}-\CC^{2})\nabhat u_1:\nabhat
    u_2=r_0^2<(\Lambda_{\CC^{1}}-\Lambda_{\CC^{2}})u_2,u_1>,
\end{equation}

where $\Lambda_{\CC^{1}}$, $\Lambda_{\CC^{2}}$ denotes the
Dirichlet-to-Neumann map corresponding to $\CC^{1}$, $\CC^{2}$
respectively.

\subsubsection{Singular solutions}\label{subsec-sing}

In a suitable coordinate system, let us consider the set
\[
D=\{(x',x_3) \ |\ x_3<\varphi(x')\},
\]
where $\varphi\in C^{1,\alpha}(\mathbb{R}^2)$ is such that
\[
\varphi(0)=|\nabla\varphi(0)|=0,\quad \|\varphi\|_{C^{1,\alpha}(\mathbb{R}^2)}\leq Lr_0.
\]
Let
\[\CC_b=\CC+(\CC^D-\CC)\chi_{D},\]
where $\CC$ and $\CC^D$ are constant isotropic elasticity tensors
satisfying \eqref{strong-convexity}. Given $y\in\mathbb{R}^3$,
let us consider the \emph{normalized} fundamental solution
$\Gamma^D(\cdot,y)$ defined by

\begin{equation}
    \label{norm_fund_sol}
    \left\{\begin{array}{rl}
             \mbox{div}(\CC_b\nabhat \Gamma^D(\cdot,y)) & = -\delta_yI_3, \\
             \lim_{|x|\rightarrow\infty}\Gamma^D(x,y) & = 0.
           \end{array}\right.
\end{equation}

The following result, derived in \cite{AdCMR}, holds.

\begin{prop}\label{fundamental}

There exists a unique normalized fundamental solution
$\Gamma^D(\cdot, y) \in C^0(\mathbb{R}^3\setminus \{y\})$.
Moreover, for every $x \in \RR^3$, $x \neq y$, we have

\begin{equation}
  \label{bound1}
   \Gamma^D(x,y) = (\Gamma^D(y,x))^T,
\end{equation}

\begin{equation}
  \label{bound2}
   |\Gamma^D(x,y)| \leq C |x-y|^{-1},
\end{equation}

\begin{equation}
  \label{bound3}
   |\nabla_x \Gamma^D(x,y)| \leq C |x-y|^{-2},
\end{equation}

where the constant $C>0$ only depends on $L$, $\alpha$, $\alpha_0$
and $\beta_0$.

\end{prop}

In particular, for $\varphi=0$, we have $D=\R_-^3$ and we
will denote the fundamental solution by $\Gamma$. An explicit
expression of $\Gamma$ has been obtained by Rongved in \cite{R}.

A crucial result in our analysis is the following asymptotic
estimate of $\Gamma^D$ in terms of $\Gamma$ derived in
\cite[Theorem 8.1]{AdCMR}.

\begin{prop}\label{asymptotic}
Let $y=(0,0,h)$, where $0<h< \frac{L r_0}{8 \sqrt{1+L^2} }$. Then
\begin{equation}\label{asymp1}
|(\Gamma^D-\Gamma)(x,y)|\leq
\frac{C}{r_0}\left(\frac{|x-y|}{r_0}\right)^{-1+\alpha},\quad
\forall x\in Q_{\frac{r_0}{8 \sqrt{1+L^2} },L}\cap D,
\end{equation}

\begin{equation}\label{asymp2}
|(\nabla_x\Gamma^D-\nabla_x\Gamma)(x,y)|\leq
\frac{C}{r^2_0}\left(\frac{|x-y|}{r_0}\right)^{-2+\frac{\alpha^2}{3\alpha+2}},\quad
\forall x\in Q^-_{\frac{r_0}{12 \sqrt{1+L^2} }, L}\cap D,
\end{equation}
where $C$ only depends on $L,\alpha, \alpha_0$ and $\beta_0$.
\end{prop}

Let $\CC$ be an isotropic elasticity tensor satisfying {\bf
(A2)}. We still denote by $\CC$ its  extension to $\Omega_0$ such
that $\CC|_{D_0}=\CC_0$ is the isotropic tensor with Lam\'{e}
parameters $\lambda_0=0$ and $\mu_0=1$. This extended tensor is
still an isotropic elasticity tensor of the form
\begin{equation}\label{isotr1}
\CC=\sum_{j=0}^N\CC_j \chi_{D_j}(x),
\end{equation}
where each $\CC_j$, $j=0,\ldots,N$, has Lam\'{e} parameters
satisfying \eqref{strong-convexity}.

For all possible interfaces $\Sigma_k$ introduced in \textbf{(A1)}
let
\[
\Sigma_k'=\Sigma_k\cap Q_{\frac{r_0}{3}, L},
\]
and denote by $\mathcal{F}:=\cup_{k}\Sigma_k'$. Let
\[y\in \cup_{j=0}^N D_j\cup \mathcal{F}, \quad r^*=r^*(y)=\min \left \{\frac{r_0}{3}, \textit{dist}(y, \cup_{j=0}^N \der D_j\backslash\mathcal{F}) \right
\}.
\]
Let us consider the sphere  $B_{r^*}(y)$. Then, either $B_{r^*}(y)\cap
\mathcal{F}=\emptyset$, so that  $B_{r^*}(y)\subset D_j$  for some
$j\in \{1,\cdots,N\}$ and  we define $\CC_y=\CC_j$, or
$B_{r^*}(y)\cap  \mathcal{F}\neq\emptyset$ and under our
regularity assumptions there exist exactly two domains, say,
$D_{j-1}$ and $D_j$, intersecting $B_{r^*}(y)$ and, in this case,
we define
$\CC_y=\CC_{j-1}+(\CC_{j}-\CC_{j-1})\chi_{\{x_3<\varphi_j(x')\}}$.
In the latter expression, $\varphi_j$ is the function whose graph
contains $\Sigma_j$, according to \textbf{(A1)}.

Let $\Gamma'(\cdot,y)$ denote the normalized fundamental solution
to

\begin{equation}\label{fundamental_solution}
\mbox{div}( \CC_y\nabhat \Gamma'(\cdot, y))=-\delta_y I_3, \quad
\text{ in }\RR^3.
\end{equation}

\begin{prop}\label{Green} Let $\om_0$ and  $\CC$ satisfy
\textbf{(A1)} and \textbf{(A2)}. Then, for any $y\in \cup_{j=0}^N
D_j\cup\mathcal{F}$, there exists a unique matrix-valued
function $G(\cdot,y)\in C^0(\Omega_0 \setminus \{y\},
\mathcal{L}(\RR^3,\RR^3))$ such that

\begin{equation}
    \label{green-2}
    \int_{\om_0}\CC\nabhat G(\cdot,y):\nabhat \phi=\phi(y), \quad \forall\phi\in C^\infty_0(\om_0, \mathcal{L}(\RR^3,\RR^3)),
\end{equation}

\begin{equation}
    \label{green-1}
    G(\cdot,y)=0, \quad \text{ on }\der\om_0,
\end{equation}

and

\begin{equation}\label{green1}
    \|G(\cdot,y)\|_{H^1(\om_0\setminus B_r(y))}\leq \frac{C}{\sqrt{rr_0}}, \quad\forall r\leq
    r^*,
\end{equation}

where $C>0$ depends only on $\alpha_0$, $\beta_0$, $A$, $N$, $\alpha$ and $L$.\\
Furthermore, for every $c_1>1$, if  $\textit{dist}(y, \cup_{j=0}^N
\der D_j\backslash\mathcal{F})\}\geq \frac{r_0}{c_1}$, then
\begin{equation}\label{energy}
\|G(\cdot,y)-\Gamma'(\cdot, y)\|_{H^1(\om_0)}\leq \frac{C}{r_0},
\end{equation}

where $C>0$ depends only on $\alpha_0$, $\beta_0$, $A$, $N$,
$\alpha$, $L$ and $c_1$. Finally
\begin{equation}\label{symgreen}
G(x,y)=G(y,x)^T, \quad \mbox{for every }x,y \in \left (
\cup_{j=0}^N D_j \right ) \cup \mathcal{F}, \  \ x \neq y.
\end{equation}

\end{prop}

The proof follows the lines of the proof of Proposition 3.1 in
\cite{BFV}.

\subsubsection{Three spheres inequality}\label{subsec-3spheres}

A mathematical tool which plays an important role in the proof of
Theorem \ref{teo2.1} is the following three spheres inequality for
solutions to the Lam\'{e} system.

\begin{prop}\label{3spheres}

Let $u \in H^1(B_R)$ be a solution to the Lam\'{e} system

\begin{equation}
    \label{3s-1}
    \mbox{div}( \CC \widehat{\nabla} u) = 0, \quad \hbox{in } B_R,
\end{equation}

where $\CC$ is a constant isotropic elasticity tensor satisfying
\eqref{strong-convexity}. For every $r_1$, $r_2$, $r_3$, with $0 <
r_1 < r_2 < r_3 \leq R$, we have

\begin{equation}
    \label{3s-2}
    \|u\|_{L^\infty(B_{r_2})} \leq C
    \|u\|_{L^\infty(B_{r_1})}^\delta
    \|u\|_{L^\infty(B_{r_3})}^{1-\delta},
\end{equation}

where $C>0$ and $\delta \in (0,1)$ only depend on $\alpha_0$,
$\beta_0$, $ \frac{r_2}{r_3}$, $ \frac{r_1}{r_3}$.

\end{prop}

For a proof, see \cite{Al-M}.

\section{Proof of the main result}\label{sec5}

Let $j\in\{1,\ldots,N\}$ be such that
\[
\|\CC-\overline{\CC}\|_{L^{\infty}(D_j)}=\|\CC-\overline{\CC}\|_{L^{\infty}(\Omega_0)}
\]
and let $D_{j_1},\ldots,D_{j_M}$ be a chain of domains, defined
according to \textbf{(A1)}, connecting $D_1$ to $D_j$. For the
sake of brevity, set $D_k=D_{j_k}$, $k=1,\ldots, M$. Let

\begin{equation}
    \label{path-and-complem}
    {\cal W}_k=\textit{Int}(\cup_{j=0}^k \overline{D}_j), \quad {\cal
    U}_k=\Omega_0\backslash \overline{{\cal W}_k}, \quad k=0,\dots, M-1.
\end{equation}

Note that here the tensors $\CC$ and $\overline{\CC}$ are extended as in
(\ref{isotr1}) in all of $\Omega_0$. Finally, for
$y,z\in(\cup_{j=0}^k D_j)\cup (\cup_{j=1}^k \Sigma'_j)$, let us
define the matrix-valued function
\[
{\cal S}_k(y,z):=\int_{{\cal U}_k}(\CC-\overline{\CC})(x)\nabhat
G(x,y):\nabhat  \overline{G}(x,z)\, dx ,
\]
whose entries are given by
\[
{\cal S}^{(p,q)}_k(y,z):=\int_{{\cal U}_k}(\CC-\overline{\CC})(x)\nabhat
G(x,y)e_p:\nabhat \overline{G}(x,z)e_q\,dx,\quad p,q=1,2,3,
\]

and where $G(\cdot,y)$ and $\overline{G}(\cdot,z) $ denotes
respectively the
singular solution of Proposition
 \ref{Green} corresponding to the tensors $\CC$ and $\overline{\CC}$,
 respectively. Let us denote ${\cal S}_k^{(\cdot,q)}= \sum_{i=1}^3{\cal
 S}_k^{(i,q)}e_i$, ${\cal S}_k^{(p, \cdot)}=  \sum_{i=1}^3{\cal
 S}_k^{(p,i)}e_i$, where $e_i$, $i=1,2,3$, are the fundamental
 unit vectors of $\R^3$.

Proceeding similarly to in \cite[Proposition 4.4]{BFV}, one can
see that the functions ${\cal S}_k^{(\cdot,q)}$, ${\cal S}_k^{(p, \cdot)}$ are solutions
to the Lam\'e system with elasticity tensor $\CC$ defined in \eqref{isotr1} in the weak sense
clarified below.

\begin{prop}\label{sol}

Let $\mathcal{R}_k:=(\cup_{j=0}^k \der D_j)\backslash
(\cup_{j=1}^k \Sigma'_j)$ and let $y,z\in(\cup_{j=0}^k D_j)\cup
(\cup_{j=1}^k \Sigma'_j)$.  Then, ${\cal S}_k^{(\cdot,q)}(\cdot,
z)$ and ${\cal
S}_k^{(p,\cdot)}(y, \cdot)$ belong to
$H^1_{loc}(\mathcal{W}_k\backslash\mathcal{R}_k, \RR^3)$ and

\begin{equation}\label{soly}
    \int_{\mathcal{W}_k}\CC  \nabhat_y {\cal S}_k^{(\cdot,q)}(y,z):\nabla\phi(y)\, dy=0,\quad\forall \phi\in C^{\infty}_0(\mathcal{W}_k\backslash\mathcal{R}_k,
    \RR^3), \quad \forall q\in\{1,2,3\},
\end{equation}

\begin{equation}\label{solz}
\int_{\mathcal{W}_k}\CC  \nabhat_z {\cal
S}_k^{(p,\cdot)}(y,z):\nabla\phi(z)\, dz=0,\quad\forall \phi\in
C^{\infty}_0(\mathcal{W}_k\backslash\mathcal{R}_k, \RR^3), \quad
\forall p\in\{1,2,3\}.
\end{equation}

\end{prop}

\begin{proof}[Proof of Theorem \ref{teo2.1}]

Let
 \[
 E:=\|\CC-\overline{\CC}\|_{L^{\infty}(\Omega_0)}
 \]
 \[
 \epsilon:=r_0\left\|\Lambda_{\CC}^\Sigma-\Lambda_{\overline{\CC}}^\Sigma\right\|_\star
 \]
and let
\[
K_0:=\left \{ x\in D_0 | \textit{dist}(x,\der\Omega)\geq
\frac{r_0}{24} \right \}.
\]
By H\"older inequality and (\ref{green1}) we have

\begin{equation}\label{aprioriest}
    |\mathcal{S}_k^{(p,q)}(y,z)|\leq
    \frac{CE}{\sqrt{d(y)d(z)}},\quad\forall \ p,q=1,2,3,
\end{equation}

where $d(y)=d(y,\mathcal{U}_k)$, $d(z)=d(z,\mathcal{U}_k)$ and
$C>0$ only depends on $\alpha_0$,$\beta_0$, $A$, $N$, $\alpha$ and
$L$.

By Alessandrini's identity \eqref{AlessId} applied to
$u_1(\cdot)=G(\cdot,y)l$ and $u_2(\cdot)=\overline{G}(\cdot,z)m$, for
$y,z\in K_0$ and for $l,m\in\mathbb{R}^3$ with $|l|=|m|=1$, we get

\begin{equation}
    \label{small}
    \left|\int_{\om_0}(\CC-\overline{\CC})(x)\nabhat G(x,y)\,l:\nabhat\overline{G}(x,z)\,m\, dx\right| \leq \frac{C}{r_0}\varepsilon,
\end{equation}

where $C>0$ depends on the a priori data only.\\
We now proceed iteratively with respect to the index $k$. \\
{\bf First step: $k=0$.}\\
For $y,z\in K_0$, let us consider
\[
{\cal S}_0^{(p,q)}(y,z):=\int_{{\cal U}_0}(\CC-\overline{\CC})(x)\nabhat
G(x,y)e_p:\nabhat \overline{G}(x,z)e_q\, dx, \quad p,q \in \{1,2,3\}.
\]

{}From (\ref{small}) we get
\[
|{\cal S}_0^{(p,q)}(y,z)|\leq \frac{C}{r_0}\epsilon,\quad \forall
y,z\in K_0,
\]

where $C>0$ depends only on the a priori data. Let us fix $z\in K_0$ and $q\in\{1,2,3\}$.
Recalling that, for fixed $q\in\{1,2,3\}$, ${\cal S}^{(\cdot,q)}_0(\cdot ,z)$ is solution to
(\ref{soly}), we shall propagate the smallness with respect to the
first variable {}from the point $Q_1=P_1 + \frac{r_0}{6}L n_1$ to
$y_r=P_1+rn_1$, for every $r \in \left(0,
\frac{r_0}{24\sqrt{1+L^2}}\right]$, by iterating the three spheres
inequality \eqref{3s-2} over a chain of balls of decreasing
radius and contained in a suitable cone with vertex at $P_1$ and
axis in the direction $n_1$, obtaining

\begin{equation}
    \label{holderest}
    |{\cal S}^{(\cdot,q)}_0(y_r,z)|\leq \frac{CE}{\sqrt{rr_0}}\left(\frac{\epsilon}{E}\right)^{\delta^{\eta}},
\end{equation}

where
\begin{equation}
   \label{eta_def}
y_r=P_1+rn_1, \ \eta=b\left|\ln \left(a\frac{r}{r_0}\right)\right|+1, \quad  r
\in \left(0, \frac{r_0}{24\sqrt{1+L^2}}\right],
\end{equation}
$\delta\in (0,1)$ only depends on $\alpha_0$, $\beta_0$, and $a>0,\ b>0$ only
depend on $L$.

Now, for fixed $p\in\{1,2,3\}$, let us consider a solution ${\cal
S}^{(p,\cdot)}_0(y_r,\cdot)$ to (\ref{solz}). Then, a similar
procedure leads to

\begin{equation}\label{holderest2}
    |{\cal S}^{(p, \cdot)}_0(y_r,z_{ \overline{r}})|\leq
    \frac{CE}{\sqrt{\overline{r}r}}\left(\frac{\epsilon}{E}\right)^{\delta^{\eta+\overline{\eta}}},
\end{equation}

where
\[
z_{\overline{r}}=P_1+\overline{r}n_1,
\quad\overline\eta=b\left|\ln a\frac{\overline{r}}{r_0}\right|+1, \quad \overline{r} \in \left(0, \frac{r_0}{24\sqrt{1+L^2}}\right].
\]

Hence, for every $r, \overline{r} \in \left(0, \frac{r_0}{24\sqrt{1+L^2}}\right]$

\begin{equation}\label{Small1}
|{\cal S}^{(p, q)}_0(y_r,z_{\overline{r}})|\leq
\frac{CE}{\sqrt{\overline{r}r}}\left(\frac{\epsilon}{E}\right)^{\delta^{\eta+\overline{\eta}}},
\quad \forall \  p,q=1,2,3.
\end{equation}

Let $\overline{r}=cr$, with $c\in \left[\frac{2}{3}, \frac{4}{5}\right]$, $p=q=3$
and let $r_1=\frac{r_0}{12\sqrt{1+L^2}}$. Let us write

\begin{equation}\label{split}
{\cal S}_0(y_r,z_{\overline{r}})\,e_3\cdot e_3=I_1+I_2,
\end{equation}

where

\begin{equation}\label{I1}
    I_1= \int_{B_{r_1}\cap D_{1}}\!\!\!\!\!\!(\CC-\overline{\CC})(x)\nabhat G(x,y_r) \,e_3:\nabhat\overline{G}(x,z_{\overline{r}})\,e_3\,dx
\end{equation}

\begin{equation}\label{I2}
I_2= \int_{\Omega\backslash (B_{r_1}\cap
D_{1})}\!\!\!\!\!\!(\CC-\overline{\CC})(x)\nabhat G(x,y_r)
\,e_3:\nabhat\overline{G}(x,z_{\overline{r}})\,e_3\,dx \ .
\end{equation}

Here and in the sequel, $B_{r_1}$ denotes $B_{r_1}(P_1)$. Then,
{}from (\ref{green1}) we have

\begin{equation}\label{Small4}
|I_2|\leq \frac{CE}{r_0}.
\end{equation}

{}From (\ref{Small1}), (\ref{split}) and (\ref{Small4}) we derive

\begin{equation}\label{Bound1}
    |I_1|\leq CE\left( \frac{1}{r}\left(\frac{\epsilon}{E}\right)^{\delta^{\eta+\overline{\eta}}}+ \frac{1}{r_0}\right).
\end{equation}

We rewrite $I_1$ as follows

\begin{equation}\label{decomp}
I_1=\int_{B_{r_1}\cap D_{1}}\!\!\!\!\!\!(\CC_1-\overline{\CC}_1)\nabhat
\Gamma'(x,y_r) \,e_3:\nabhat \overline{\Gamma}'(x,z_{\overline{r}})\,e_3\,dx+A_1+A_2+A_3,
\end{equation}

where $\Gamma'$, $\overline{\Gamma'}$ is the normalized
fundamental solution to (\ref{norm_fund_sol}) corresponding to the pair of elasticity tensors
$(\CC=\CC_0,\CC^D=\CC_1)$, $(\CC=\overline{\CC}_0,\CC^D=\overline{\CC}_1)$ respectively, $D=D_1$ and
\[
A_1=\int_{B_{r_1}\cap D_{1}}\!\!\!\!\!\!(\CC_1-\overline{\CC}_1)(\nabhat
G-\nabhat\Gamma')(x,y_r) \,e_3:(\nabhat\overline{G}-\nabhat\overline{\Gamma}')(x,z_{\overline{r}})\,e_3\,dx \ ,
\]
\[
A_2=\int_{B_{r_1}\cap D_{1}}\!\!\!\!\!\!(\CC_1-\overline{\CC}_1)(\nabhat
G-\nabhat\Gamma')(x,y_r) \,e_3:\nabhat\overline{\Gamma}'(x,z_{\overline{r}
})\,e_3\,dx \ ,
\]
\[
A_3=\int_{B_{r_1}\cap
D_{1}}\!\!\!\!\!\!(\CC_1-\overline{\CC}_1)\nabhat\Gamma'(x,y_r)
\,e_3:(\nabhat\overline{G}-\nabhat\overline{\Gamma}')(x,z_{\overline{r}})\,e_3\,dx \
.
\]

By \eqref{energy} and (\ref{bound3}) we obtain

\begin{equation}\label{Bound2}
|A_1|\leq \frac{CE}{r_0} ,
\end{equation}

\begin{equation}\label{Bound3}
|A_2|, |A_3|\leq \frac{CE}{\sqrt{rr_0}} ,
\end{equation}

where $C>0$ only depends on the a priori data. {}From (\ref{decomp})-(\ref{Bound3}) we get

\begin{equation}\label{Bound4}
|I_1|\geq\left| \int_{B_{r_1}\cap
D_{1}}\!\!\!\!\!\!(\CC_1-\overline{\CC}_1)\nabhat \Gamma'(x,y_r)
\,e_3:\nabhat\overline{\Gamma}'(x,z_{\overline{r}
})\,e_3\,dx\right|-\frac{CE}{\sqrt{rr_0}} ,
\end{equation}
where $C>0$ only depends on the a priori data. {}From (\ref{Bound1}) and (\ref{Bound4}) we
obtain

\begin{equation}
    \label{Bound3bis}
    \left| \int_{B_{r_1}\cap D_{1}}\!\!\!\!\!\!(\CC_1-\overline{\CC}_1)\nabhat \Gamma'(x,y_r) \,e_3:\nabhat\overline{\Gamma}'(x,z_{\overline{r}})\,e_3\,dx\right|\leq
    \frac{CE}{r}\left( \left(\frac{\epsilon}{E}\right)^{\delta^{\eta+\overline{\eta}}}+ \left ( \frac{r}{r_0}  \right )^{\frac{1}{2}}
    \right)\ ,
\end{equation}

where $C>0$ only depends on the a priori data.

Let us denote by $\Gamma$ and $\overline{\Gamma}$ the Rongved fundamental
solutions corresponding to the tensors
$\CC_0\chi_{\mathbb{R}^3_+}+\CC_1\chi_{\mathbb{R}^3_-}$ and
$\overline{\CC}_0\chi_{\mathbb{R}^3_+}+\overline{\CC}_1\chi_{\mathbb{R}^3_-}$,
respectively. Let

\begin{equation}
    \label{decomp2}
    \int_{B_{r_1}\cap D_{1}}\!\!\!\!\!\!(\CC_1-\overline{\CC}_1)\nabhat \Gamma'(x,y_r) \,e_3:\nabhat\overline{\Gamma}'(x,z_{\overline{r}
    })\,e_3\,dx=:B_1+B_2+B_3\ ,
\end{equation}

where
\[
B_1=\int_{B_{r_1}\cap D_{1}}\!\!\!\!\!\!(\CC_1-\overline{\CC}_1)\nabhat
\Gamma(x,y_r) \,e_3:\nabhat\overline{\Gamma}(x,z_{\overline{r}})\,e_3\,dx \ ,
\]
\[
B_2=\int_{B_{r_1}\cap D_{1}}\!\!\!\!\!\!(\CC_1-\overline{\CC}_1)\nabhat
(\Gamma'-\Gamma)(x,y_r) \,e_3:\nabhat\overline{\Gamma}(x,z_{\overline{r}
})\,e_3\,dx \ ,
\]
\[
B_3=\int_{B_{r_1}\cap
D_{1}}\!\!\!\!\!\!(\CC_1-\overline{\CC}_1)\nabhat\Gamma'(x,y_{
r})\,e_3:(\nabhat\overline{\Gamma}'-\nabhat\overline{\Gamma})(x,z_{\overline{r}})
\,e_3\,dx \ .
\]
To estimate $B_2$ and $B_3$, we observe that
\[
B_{r_1}\cap D_{1}\subset \left(D_1\cap
Q^-_{\frac{r_0}{12\sqrt{1+L^2}},L}\right)\cup\left\{(x',x_3): 0\leq
x_3\leq \frac{L}{r_0^{\alpha}}|x'|^{1+\alpha}\right\} \ .
\]

In $D_1 \cap Q^-_{ \frac{r_0}{ 12 \sqrt{1+L^2}  }}$ we can apply
the asymptotic estimate \eqref{asymp2} so that, recalling also
\eqref{bound3}, we have

\[
|B_2| \leq \frac{CE}{r_0^\gamma} \int_{\RR_{-}^3} |x-y_r|^{\gamma
-2} |x -z_{\overline{r}}|^{-2} dx +  CE \int_{0\leq x_3 \leq
\frac{L}{r_0^\alpha}|x'|^{1+\alpha}}
|x-y_r|^{-2}|x-z_{\overline{r}}|^{-2} dx \ ,
\]

where $\gamma = \frac{\alpha^2}{3\alpha+2} < \frac{1}{2}$ and
$C>0$ depend only on $\alpha_0$, $\beta_0$, $L$ and $\alpha$. The
first integral can be easily estimated by passing to cylindrical
coordinates and by applying H\"{o}lder inequality, obtaining

\[
    \frac{CE}{r_0^\gamma} \int_{\RR_{-}^3} |x-y_r|^{\gamma -2} |x
    -z_{\overline{r}}|^{-2} dx  \leq \frac{CE}{r_0} \left (
    \frac{r}{r_0} \right )^{\gamma -1} \ .
\]

The estimate of the second integral is not straightforward. First,
by performing the change of variables $y= \frac{x}{r}$, we have

\begin{multline*}
    CE \int_{0\leq x_3 \leq \frac{L}{r_0^\alpha}|x'|^{1+\alpha}}
    |x-y_r|^{-2}|x-z_{\overline{r}}|^{-2} dx \leq
    \\
    \leq \frac{CE}{r} \int_{\RR^2} \left ( \int_0^{
    \frac{L}{r_0^\alpha}r^\alpha |y'|^{1+\alpha}} \frac{1}{ (|y'|^2
    +(y_3-1)^2)(|y'|^2 +(y_3 -c)^2)} dy_3 \right ) dy_1 dy_2 .
\end{multline*}

By splitting $\RR^2$ as the union of $A= \left\{ y' \in \RR^2 | \ |y'|
\geq \left ( \frac{2L}{r_0^\alpha} \right )^{-
\frac{1}{1+\alpha}}r^{ - \frac{\alpha}{1+\alpha}} \right\}$ and $B= \RR^2\setminus A $, we have

\begin{equation*}
    CE \int_{0\leq x_3 \leq \frac{L}{r_0^\alpha}|x'|^{1+\alpha}}
    |x-y_r|^{-2}|x-z_{\overline{r}}|^{-2} dx \leq
    \frac{CE}{r_0} \left(\left ( \frac{r}{r_0} \right )^{ \frac{\alpha -1 }{\alpha
    +1}}+ \left ( \frac{r}{r_0} \right )^{ \alpha -1}\right),
\end{equation*}

with $C>0$ only depending on $\alpha_0$, $\beta_0$, $L$ and
$\alpha$, where we have used the fact that $|y_3| < \frac{1}{2}$
in $B$, so that $|y_3 - 1| > \frac{1}{2}$ and $|y_3 -c| \geq c-
\frac{1}{2} \geq \frac{1}{6}$.

The term $B_3$ in \eqref{decomp2} can be estimated similarly,
obtaining

\begin{equation}\label{Bound5}
    |B_2|,|B_3|\leq
    \frac{CE}{r_0}\left(\left(\frac{r}{r_0}\right)^{\gamma-1}+\left(\frac{r}{r_0}\right)^{\alpha-1}\right),
\end{equation}

where $C>0$ only depends on $\alpha_0$, $\beta_0$, $L$, $\alpha$,
and $\gamma= \frac{\alpha^2}{3\alpha + 2 } <\frac{1}{2}$.

We split $B_1$ as follows
\[
B_1=C_1+C_2+C_3 ,
\]
where
\[
C_1=\int_{B^-_{r_1}}\!\!\!\!\!\!(\CC_1-\overline{\CC}_1)\nabhat
\Gamma(x,y_r) \,e_3:\nabhat \overline{\Gamma}(x,z_{\overline{r}})\,e_3\,dx \ ,
\]
\[
C_2=\int_{B^+_{r_1}\cap D_1}\!\!\!\!\!\!(\CC_1-\overline{\CC}_1)\nabhat
\Gamma(x,y_r) \,e_3:\nabhat\overline{\Gamma}(x,z_{\overline{r}})\,e_3\,dx \ ,
\]
\[
C_3=\int_{B^-_{r_1}\backslash
D_1}\!\!\!\!\!\!(\CC_1-\overline{\CC}_1)\nabhat \Gamma(x,y_r)
\,e_3:\nabhat\overline{\Gamma}(x,z_{\overline{r}})\,e_3\,dx \ .
\]

Since
\[
B_{r_1}^+\cap D_{1}\subset \left\{(x',x_3)| 0\leq x_3\leq
\frac{L}{r_0^{\alpha}}|x'|^{1+\alpha}\right\}
\]
and
\[
B^-_{r_1}\backslash D_1\subset \left\{(x',x_3)|
-\frac{L}{r_0^{\alpha}}|x'|^{1+\alpha}\leq x_3\leq 0\right\},
\]
we can estimate the terms $C_2$ and $C_3$ similarly to the second
addend of $B_2$, getting

\begin{equation}
    \label{Bound6}
    |C_2|,|C_3|\leq
    \frac{CE}{r_0}\left(\frac{r}{r_0}\right)^{\alpha-1},
\end{equation}

where $C>0$ only depends on $\alpha_0$, $\beta_0$, $L$, $\alpha$.

Finally, to estimate $C_1$, we use the following property of the
Rongved fundamental solution
\[
\Gamma(\xi,y_0)=h\Gamma(h\xi,hy_0),
\quad\overline{\Gamma}(\xi,y_0)=h\overline{\Gamma}(h\xi,hy_0), \quad \forall
\xi\neq y_0, \ \forall h>0.
\]

Then

\begin{equation}\label{changevariable}
C_1=\frac{1}{r}\int_{B^-_{\frac{r_1}{r}}}\!\!\!\!\!\!(\CC_1-\overline{\CC}_1)(x)\nabhat
\Gamma(x,e_3) \,e_3:\nabhat\overline{\Gamma}(x,ce_3)\,e_3\,dx
\end{equation}

and by (\ref{Bound3bis})--(\ref{Bound6}) we obtain

\begin{equation}\label{Bound7}
\left|\int_{B^-_{\frac{r_1}{r}}}\!\!\!\!\!\!(\CC_1-\overline{\CC}_1)(x)\nabhat \Gamma(x,e_3) \,e_3:\nabhat\overline{\Gamma}(x,ce_3)\,e_3\,dx\right|
\leq
CE \left ( \left(\frac{r}{r_0}\right)^{\gamma}+\left(\frac{\epsilon}{E}\right)^{\delta^{\eta+\overline{\eta}}}\right)
\ ,
\end{equation}

where $C>0$ only depends on the a priori data.
{}From \eqref{bound3} and since $c\in \left[\frac{2}{3}, \frac{4}{5}\right]$
and $r_1=\frac{r_0}{12\sqrt{1+L^2}}$, we derive

\begin{equation}\label{Bound8}
\left|\int_{\mathbb{R}_-^3\backslash {B^-_{\frac{r_1}{r}}}}\!\!\!\!\!\!(\CC_1-\overline{\CC}_1)(x)\nabhat \Gamma(x,e_3) \,e_3:\nabhat\overline{\Gamma}(x,ce_3)\,e_3\,dx\right|\\
\leq CE\frac{r}{r_0} \ ,
\end{equation}

where $C>0$ only depends on $\alpha_0$, $\beta_0$, $L$, $\alpha$.
{}From (\ref{Bound7}), (\ref{Bound8}), since $\gamma<\alpha$,
$\gamma<\frac{1}{2}$ and $c\geq \frac{2}{3}$, by (\ref{holderest})
we have

\begin{equation}\label{Bound9}
\left|\int_{\mathbb{R}_-^3}\!\!\!\!\!\!(\CC_1-\overline{\CC}_1)(x)\nabhat \Gamma(x,e_3) \,e_3:\nabhat\overline{\Gamma}(x,ce_3)\,e_3\,dx\right|\\
\leq CEf\left(\frac{r}{r_0}\right), \quad \forall r\leq
\frac{r_0}{24\sqrt{1+L^2}},
\end{equation}

where

\[
f(\rho)=\rho^{\gamma}+\left(\frac{\epsilon}{E}\right)^{\delta^{2+A|\ln
B\rho|}} \ ,
\]
where $0 < \rho\leq \frac{1}{24\sqrt{1+L^2}}$, and $A$, $B>0$ only depend on $L$.

By an appropriate choice of $\rho=\rho(\epsilon)$, we get

\begin{equation}\label{Bound9bis}
    \left|\int_{\mathbb{R}_-^3}\!\!\!\!\!\!(\CC_1-\overline{\CC}_1)(x)\nabhat \Gamma(x,e_3) \,e_3:\nabhat\overline{\Gamma}(x,ce_3)\,e_3\,dx\right|\\
    \leq
    CE\left|\ln\frac{\epsilon}{E}\right|^{\frac{-\gamma}{2A|\ln\delta|}}\
    ,
\end{equation}
where $C>0$ only depends on $\alpha_0$, $\beta_0$, $L$, $\alpha$.
Applying Proposition 3.2 of \cite{BFV} we have

\begin{equation*}
    \int_{\mathbb{R}_-^3}\!\!\!\!\!\!(\CC_1-\overline{\CC}_1)(x)\nabhat \Gamma(x,e_3) \,e_3:\nabhat\overline{\Gamma}(x,ce_3)\,e_3\,dx
    =
    (\Gamma(e_3,ce_3)-\overline{\Gamma}(e_3,ce_3))e_3 \cdot e_3
\end{equation*}

and then

\begin{equation}
    \label{Bound10}
    \left|\Gamma(e_3,ce_3)-\overline{\Gamma}(e_3,ce_3)\right|\leq
    CE\left|\ln\frac{\epsilon}{E}\right|^{\frac{-\gamma}{2A|\ln\delta|}}\
    ,
\end{equation}

where $C>0$ only depends on the a priori data.
Now, by using the explicit form of the Rongved fundamental
solution and proceeding as in \cite{BFV} (Section 4.2), it can be
shown that \eqref{Bound10} implies

\begin{equation}
    \label{Bound11}
    \|\CC_1-\overline{\CC}_1\|_\infty \leq CE\omega_1\left(\frac{\epsilon}{E}\right),
\end{equation}

where
\[
\omega_1(t)=|\ln t|^{\frac{-\gamma}{2A|\ln\delta|}}
\]
and $C>0$ only depends on the a priori data.

If  $\|\CC-\overline{\CC}\|_{\infty}=\|\CC_1-\overline{\CC}_1\|_\infty$, then we
get

\[
\|\CC-\overline{\CC}\|_{\infty}=E\leq
\frac{\epsilon}{\omega_1^{-1}(\frac{1}{C})} \ ,
\]

and the claim follows. Otherwise, we proceed with the next step. \\
{\bf Second step: $k=1$}

In this case, let us consider the function
\begin{multline*}
{\cal S}_1^{(p,q)}(y,z)=\int_{\Omega}(\CC-\overline{\CC})(x)\nabhat G(x,y)e_p:\nabhat \overline{G}(x,z)e_q\, dx-\\
-\int_{D_1}(\CC-\overline{\CC})(x)\nabhat G(x,y)e_p:\nabhat \overline{G}(x,z)e_q\, dx,\quad p,q=1,2 ,3.
\end{multline*}
{}From (\ref{green1}), (\ref{small}) and (\ref{Bound11}) we get
\begin{equation}\label{Bound12}
|{\cal S}_1^{(p,q)}(y,z)|\leq
\frac{CE}{r_0}\omega_1\left(\frac{\epsilon}{E}\right),\quad \forall\ y,z\in K_0,
\end{equation}
where $C>0$ only depends on the a priori data.
Proceeding similarly to what previously described for ${\cal S}_0^{(\cdot,q)}(\cdot,z)$, and by regularity estimates  for elliptic systems, we derive that for every $P\in\Sigma'_1$ and every $r$, $0<r\leq \frac{r_0}{24\sqrt{1+L^2}}$,
\begin{equation}\label{holderest3}
|{\cal S}_1^{(p,q)}(y'_r,z)|+r_0 |\nabla{\cal S}_1^{(p,q)}(y'_r,z)|\leq \frac{CE}{r_0}\left(\omega_1\left(\frac{\epsilon}{E}\right)\right)^{\delta^{\eta}},\quad\forall z\in K_0,
\end{equation}

where $y'_r=P+rn_1$ and $\eta$ is the same constant defined in (\ref{eta_def}).
Let us recall the following regularity estimates due to Li and Nirenberg \cite{LN}
\begin{equation}\label{Bound13}
\|\nabla{\cal S}_1^{(p,q)}(\cdot,z)\|_{L^{\infty}( \overline{D}_0\cap Q_{\frac{r_0}{3},L} )}+r_0^{\beta}\left|\nabla{\cal S}_1^{(p,q)}(\cdot,z)\right|_{\beta, \overline{\Gamma}_0\cap Q_{\frac{r_0}{3},L} }\leq \frac{CE}{r^2_0},
\end{equation}
where $\beta=\frac{\alpha}{2(1+\alpha)}$ and $C>0$ only depends on the a priori data.

{}From (\ref{holderest3}) and (\ref{Bound13}) we have that, for every $P\in \Sigma'_1$, $z\in K_0$ and $r\in \left(0, \frac{r_0}{24\sqrt{1+L^2}}\right]$,

\begin{equation}\label{Bound14}
|{\cal S}_1^{(p,q)}(P,z)|+r_0|\nabla{\cal S}_1^{(p,q)}(P,z)|\leq \frac{CE}{r_0}\left(\left(\frac{r}{r_0}\right)^{\beta}+\left(\omega_1\left(\frac{\epsilon}{E}\right)\right)^{\delta^{\eta}}\right),
\end{equation}
where $C$ only depends on the a priori data.
By a suitable choice of $r=r(\epsilon)$, we get
\begin{equation}\label{Bound16}
|{\cal S}_1^{(p,q)}(P,z)|+r_0|\nabla{\cal S}_1^{(p,q)}(P,z)|\leq  \frac{CE}{r_0}\widetilde{\omega}_2\left(\frac{\epsilon}{E}\right),
\end{equation}
where
\[
\widetilde{\omega}_2(t) =\left|\ln |\ln\left(t\right)|\right|^{\frac{-\gamma}{2A|\ln\delta|}}.
\]
By the transmission conditions
\[
{\cal S}_1^{(p,\cdot)}(\cdot,z)|_{D_0} = {\cal S}_1^{(p,\cdot)}(\cdot,z)|_{D_1}, \quad
\CC^0\nabla {\cal S}_1^{(p,\cdot)}(\cdot,z)n_1|_{D_0} =
\CC^1\nabla {\cal S}_1^{(p,\cdot)}(\cdot,z)n_1|_{D_1}, \quad \mbox{on }\Sigma_1,
\]
and by (\ref{Bound16}) we have, for every $p\in\{1,2,3\}$,
\begin{equation}\label{Bound17}
\|{\cal S}_1^{(p,\cdot)}(\cdot,z)|_{D_1}\|_{H^{\frac{1}{2}}(\Sigma_1')}+r_0\|\nabla{\cal S}_1^{(p,\cdot)}(\cdot,z)|_{D_1}\|_{H^{-\frac{1}{2}}(\Sigma_1')}\leq \frac{CE}{r_0}\tilde{\omega}_2\left(\frac{\epsilon}{E}\right),
\end{equation}
where $C$ only depends on the a priori data.

We can adapt the arguments in \cite{ARRV}  (see in particular Lemma 6.1 and Theorem 6.2) to obtain the following stability estimate for the Cauchy problem for ${\cal S}_1^{(p,\cdot)}(\cdot,z)$ in
$Q_{\frac{r_0}{3}L}(P_1)\cap D_1$

\begin{equation}\label{Bound18}
\|{\cal S}_1^{(p,q)}(\cdot,z)|\|_{L^{\infty}(B_{\overline{\rho}}(R_1))}\leq \frac{CE}{r_0}\left(\tilde{\omega}_2\left(\frac{\epsilon}{E}\right)\right)^{\xi},
\end{equation}
where $R_1=P_1-dn_1$, $d=
\frac{r_0}{3}\frac{L\sqrt{1+L^2}}{1+\sqrt{1+L^2}}$, $\overline{\rho} =\frac{r_0}{12}\frac{L}{1+\sqrt{1+L^2}}$, and the constants $\xi\in (0,1)$ and $C>0$ only depend on the a priori data.
Observe that by Proposition 5.5 in \cite{ARRV} there exists $h_1>0$, with $\frac{h_1}{r_0}$ depending only on $L$, such that
\[
(D_1)_h =\{x\in D_1\ |\ d(x,\partial D_1)>h\}\textrm{ is connected },\forall h\leq h_1,
\]
Let $ \overline{h}=\min\left\{h_1,\frac{r_0}{3}\frac{L\sqrt{1+L^2}}{1+\sqrt{1+L^2}}\right\}$. Then, $\frac{\overline{h}}{r_0}$ depends only on $L$, $\overline{(D_1)}_{\overline{h}}$ is connected and contains the points $R_1$ and $Q_2=P_2+\frac{8}{5}\overline{h}\sqrt{1+L^2}n_2$, with $P_2\in \Sigma_2$ as in \textbf{($A_1$)}. Let $\gamma$ be an arc contained in $\overline{(D_1)}_{\overline{h} }$ connecting $R_1$ with $Q_2$.
By iterating the three spheres
inequality \eqref{3s-2} first over a chain of balls with centers on $\gamma$ and then over a chain of balls of decreasing radius and contained in a suitable cone with vertex at $P_2$ and
axis in the direction $n_2$, we obtain
\[
|{\cal S}_1^{(p,q)}(y_r,z)|\leq \frac{CE}{\sqrt{rr_0}}\left(\tilde{\omega}_2\left(\frac{\epsilon}{E}\right)\right)^{\tau\delta^{\eta}},\quad\forall z\in K_0, \forall\ p,q\in\{1,2,3\},
\]
where $y_r=P_2+rn_2$, $\tau\in(0,1)$ and $C>0$ only depend on the a priori data. Repeating the estimating procedure for the function ${\cal S}_1^{(\cdot,q)}(y_r,\cdot)$  we obtain
\[
|{\cal S}_1^{(p,q)}(y_r,z_{\overline{r}})|\leq \frac{CE}{r}\left(\omega^*_2\left(\frac{\epsilon}{E}\right)\right)^{\tau\delta^{\eta +\overline{\eta}}},
\quad \forall\ p,q\in\{1,2,3\},
\]
where $C$ only depend on the a priori data and
\[
\omega^*_2(t)=\left|\ln\left|\ln\left|\ln\left(\frac{\epsilon}{E}\right)\right|\right|\right|^{\frac{-\gamma}{2A|\ln\delta|}}.
\]
Now, choosing a coordinate system centered at $P_2$, with $e_3=n_2$, and denoting by $\Gamma$ and $\overline{\Gamma}$ the Rongved solutions corresponding to the tensors $\CC_1\chi_{\mathbb{R}^3_+}+\CC_2\chi_{\mathbb{R}^3_-}$ and $\overline{\CC}_1\chi_{\mathbb{R}^3_+}+\overline{\CC}_2\chi_{\mathbb{R}^3_-}$, we get that
\[
|(\Gamma(e_3,ce_3)-\overline{\Gamma}(e_3,ce_3))e_3\cdot e_3|\leq CE \omega_2\left(\frac{\epsilon}{E}\right),
\]
where $C$ only depend on the a priori data and
\[
\omega_2(t)=\left|\ln\left|\ln\left|\ln\left|\ln\left(\frac{\epsilon}{E}\right)\right|\right|\right|\right|^{\frac{-\gamma}{2A|\ln\delta|}},
\]
so that, proceeding as in \cite{BFV}, we have
\[
\|\CC_2-\overline{\CC}_2\|\leq CE \omega_2\left(\frac{\epsilon}{E}\right).
\]
If $E=\|\CC_2-\overline{\CC}_2\|_\infty$, then
\[
\|\CC-\overline{\CC}\|_{\infty}=E\leq \frac{\epsilon}{\omega_2^{-1}(\frac{1}{C})}
\]
and the claim follows.
Otherwise, we proceed similarly iterating the procedure up to $k=j$ obtaining the desired result.
\end{proof}

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TITLE: Let $A$ be an abelian finitely generated free group and $A/B$ be a torsion group. Show that $rank(A)=rank(B)$.
QUESTION [2 upvotes]: Let $A$ be an abelian free group that is finitely generated, and let $B\subset A$ be a subgroup of $A$ such that $A/B$ is a torsion group. Show that $rank(A)=rank(B)$.
From the hypothesis, I know that $A=<x_1,..., x_n>$, so $rank(A)=n$. Since the set $\{x_1, ..., x_n\}$ clearly generates $B$, then I have that $rank(B)\leq n$. Now, the idea that I have is to show that if $S\subset B$ generates $B$, then it also generates $A$, but I don't know if this is true, and I don't know how to prove it. I was trying to do this, and here's what I got:
Since $A/B$ is a torsion group, for $a\in A$, there exists a natural number $m$ such that $ma\in B$, so $ma$ can be written as a linear combination of elements in $S$. (I don't know if this is useful).
Any hint would be very appreciated! Thank you!

REPLY [0 votes]: You can show $B$ has finite index in $A$, say $[A:B]=n$. Then the map $a\rightarrow a^n$ is a monomorphism from $A$ to $B$, so rank ($A$)$\le$rank($B$). The reverse inequality is also true and that gives you what you want.