text:
\begin{document}

\maketitle

\begin{abstract}

In this paper we consider the inverse problem of determining a
rigid inclusion inside a thin plate by applying a couple field at
the boundary and by measuring the induced transversal displacement
and its normal derivative at the boundary of the plate. The plate
is made by non-homogeneous, linearly elastic and isotropic
material. Under suitable a priori regularity assumptions on the
boundary of the inclusion, we prove a constructive stability
estimate of log type. Key mathematical tool is a recently proved optimal three
spheres inequality at the boundary for solutions to the
Kirchhoff-Love plate's equation.

\medskip

\noindent\textbf{Mathematics Subject Classification (2010)}:
Primary 35B60. Secondary 35B30, 35Q74, 35R30.

\medskip

\noindent \textbf{Keywords}: Inverse problems, elastic plates,
stability estimates, unique continuation, rigid inclusion.
\end{abstract}

\centerline{}

\section{Introduction}

In this paper we consider the inverse problem of the stable
determination of a rigid inclusion embedded in a thin elastic
plate by measuring the transverse displacement and its normal
derivative at the boundary induced by a couple field applied at
the boundary of the plate. We prove that the stability estimate of
log-log type found in \cite{M-R-V2} can be improved to a single
logarithm in the case in which the plate is made of isotropic
linear elastic material. {}From the point of view of applications,
modern requirements of structural condition assessment demand the
identification of defects using non-destructive methods, and,
therefore, the present results can be useful in quality control of
plates. We refer, among other contributions, to Bonnet and
Constantinescu \cite{B-C} for a general overview of inverse
problems arising in diagnostic analysis applied to linear
elasticity and, in particular, to plate theory (\cite[Section $5.3$]{B-C}),
and to \cite{K} for the identification of a stiff
inclusion in a composite thin plate based on wavelet analysis of
the eigenfunctions.

In order to describe our stability result, let us introduce the
Kirchhoff-Love model of thin, elastic isotropic plate under
infinitesimal deformation; see, for example, \cite{Gu}. Let the
middle plane of the plate $\Omega$ be a bounded domain of $\R^2$
with regular boundary. The rigid inclusion $D$ is modelled as a
simply connected domain compactly contained in $\Omega$. Under the
assumptions of vanishing transversal forces in $\Omega$ and
assigned couple field $\widehat{M}$ acting on $\partial \Omega$,
the transversal displacement $w \in H^2(\Omega)$ of the plate
satisfies the following mixed boundary value problem

\begin{center}
\( {\displaystyle \left\{
\begin{array}{lr}
     {\rm div}({\rm div} (
      {\mathbb P}\nabla^2 w))=0,
      & \mathrm{in}\ \Omega \setminus \overline {D},
        \vspace{0.25em}\\
      ({\mathbb P} \nabla^2 w)n\cdot n=-\widehat M_n, & \mathrm{on}\ \partial \Omega,
          \vspace{0.25em}\\
      {\rm div}({\mathbb P} \nabla^2 w)\cdot n+(({\mathbb P} \nabla^2
      w)n\cdot \tau),_s
      =(\widehat M_\tau),_s, & \mathrm{on}\ \partial \Omega,
        \vspace{0.25em}\\
      w|_{\overline{D}} \in \mathcal{A}, &\mathrm{in}\ \overline{D},
          \vspace{0.25em}\\
        \frac{\partial w^e}{\partial n} = \frac{\partial w^i}{\partial n}, &\mathrm{on}\ \partial {D},
          \vspace{0.25em}\\
\end{array}
\right. } \) \vskip -8.9em
\begin{eqnarray}
& & \label{eq:1.dir-pbm-incl-rig-1}\\
& & \label{eq:1.dir-pbm-incl-rig-2}\\
& & \label{eq:1.dir-pbm-incl-rig-3}\\
& & \label{eq:1.dir-pbm-incl-rig-4}\\
& & \label{eq:1.dir-pbm-incl-rig-5}
\end{eqnarray}
\end{center}

coupled with the \emph{equilibrium conditions} for the rigid
inclusion $D$

\begin{multline}
  \label{eq:1.equil-rigid-incl}
    \int_{\partial D} \left ( {\rm div}({\mathbb P} \nabla^2 w)\cdot n+(({\mathbb P} \nabla^2
  w)n\cdot \tau),_s \right )g - (({\mathbb P} \nabla^2 w)n\cdot n)
  g_{,n} =0, \\   \quad \hbox{for every } g\in \mathcal{A},
\end{multline}

where $\mathcal{A}$ denotes the space of affine functions. We
recall that, {}from the physical point of view, the boundary
conditions
\eqref{eq:1.dir-pbm-incl-rig-4}-\eqref{eq:1.dir-pbm-incl-rig-5}
correspond to ideal connection between the boundary of the rigid
inclusion and the surrounding elastic material, see, for example,
\cite[Section $10.10$]{O-R}. The unit vectors $n$ and $\tau$ are
the outer normal to $\Omega \setminus \overline{D}$ and the
tangent vector to $\partial D$, respectively. Moreover, we have
defined $w^e \equiv w|_{ \Omega \setminus \overline{D}}$ and
$w^i\equiv w|_{\overline{D} }$. The functions $\widehat M_{\tau}$,
$\widehat M_n$ are the twisting and bending component of the
assigned couple field $\widehat M$, respectively. The plate tensor
$\mathbb P$ is given by $\mathbb P = \frac{h^3}{12}\mathbb C$,
where $h$ is the constant thickness of the plate and $\mathbb C$
is the non-homogeneous Lam\'{e} elasticity tensor describing the
response of the material.

The existence of a solution $w \in H^2(\Omega)$ of the problem
\eqref{eq:1.dir-pbm-incl-rig-1}--\eqref{eq:1.equil-rigid-incl} is
ensured by general results, provided that $\widehat{M}\in
H^{-\frac{1}{2}}(\partial\Omega, \R^2)$, with $\int_{\partial
\Omega} \widehat{M}_i = 0$, for $i=1,2$, and $\mathbb P$ is
bounded and strongly convex. Let us notice that $w$ is uniquely
determined up to addition of an affine function.

Let us denote by $w_i$ a solution to
\eqref{eq:1.dir-pbm-incl-rig-1}--\eqref{eq:1.equil-rigid-incl} for
$D=D_i$, $i=1,2$. In order to deal with the stability issue, we
found it convenient to replace each solution $w_i$ with
$v_i=w_i-g_i$, where $g_i$ is the affine function which coincides
with $w_i$ on $\partial D_i$, $i=1,2$. By this approach,
maintaining the same letter to denote the solution, the
equilibrium problem
\eqref{eq:1.dir-pbm-incl-rig-1}--\eqref{eq:1.dir-pbm-incl-rig-5} can
be rephrased in terms of the following mixed boundary value
problem with homogeneous Dirichlet conditions on the boundary of
the rigid inclusion

\begin{center}
\( {\displaystyle \left\{
\begin{array}{lr}
     {\rm div}({\rm div} (
      {\mathbb P}\nabla^2 w))=0,
      & \mathrm{in}\ \Omega \setminus \overline {D},
        \vspace{0.25em}\\
      ({\mathbb P} \nabla^2 w)n\cdot n=-\widehat M_n, & \mathrm{on}\ \partial \Omega,
          \vspace{0.25em}\\
      {\rm div}({\mathbb P} \nabla^2 w)\cdot n+(({\mathbb P} \nabla^2
      w)n\cdot \tau),_s
      =(\widehat M_\tau),_s, & \mathrm{on}\ \partial \Omega,
        \vspace{0.25em}\\
      w=0, &\mathrm{on}\ \partial D,
          \vspace{0.25em}\\
        \frac{\partial w}{\partial n} = 0, &\mathrm{on}\ \partial {D},
          \vspace{0.25em}\\
\end{array}
\right. } \) \vskip -8.9em
\begin{eqnarray}
& & \label{eq:1.dir-pbm-incl-rig-1bis}\\
& & \label{eq:1.dir-pbm-incl-rig-2bis}\\
& & \label{eq:1.dir-pbm-incl-rig-3bis}\\
& & \label{eq:1.dir-pbm-incl-rig-4bis}\\
& & \label{eq:1-dir-pbm-incl-rig-5bis}
\end{eqnarray}

\end{center}

for which there exists a unique solution $w\in H^2(\Omega\setminus
\overline{D})$. The arbitrariness of this normalization, related
to the fact that $g_i$ is unknown, $i=1,2$, leads to the following
formulation of the stability issue.

\textit{Given an open portion $\Sigma$ of $\partial \Omega$,
satisfying suitable regularity assumptions, and given two solutions $w_i$ to
\eqref{eq:1.dir-pbm-incl-rig-1bis}--\eqref{eq:1-dir-pbm-incl-rig-5bis} when $D=D_i$, $i=1,2$, satisfying,
for some $\epsilon>0$,

\begin{equation}
  \label{eq:cond_Dir_stab}
  \min_{g\in \mathcal{A}}\left\{ \| w_1 - w_2 - g \|_{L^2(\Sigma)}
  + \left\| \frac{\partial}{\partial n}(w_1 - w_2 - g) \right\|_{L^2(\Sigma)}\right\}\leq \epsilon,
\end{equation}

to evaluate the rate at which the Hausdorff distance $d_{\cal H}(
\overline{D_1},\overline{D_2} )$ between $D_1$ and $D_2$ tends to
zero as $\epsilon$ tends to zero.}

In this paper we prove the following quantitative stability
estimate of log type for inclusions $D$ of $C^{6,\alpha}$ class:

\begin{equation}
    \label{eq:log-estimate}
    d_{\cal H}( \overline{D_1},\overline{D_2} ) \leq C | \log \epsilon |^{-\eta},
\end{equation}

where $C$, $\eta$, $C>0$ and $\eta>0$, are constants only
depending on the a priori data, see Theorem \ref{theo:Main} for a
precise statement.

The above estimate is an improvement of the log-log type stability
estimate found in \cite{M-R-V2}, although it must be said that the
latter is not restricted to isotropic materials and also applies
to less regular inclusions (e.g., $D$ of $C^{3,1}$ class). It is
worth to notice that a single logarithmic rate of convergence for
the fourth order elliptic equation modelling the deflection of a
Kirchhoff-Love plate is expected to be optimal, as it is in fact
for the analogous inverse problem in the scalar elliptic case,
which models the detection of perfectly conducting inclusions in
an electric conductor in terms of measurements of potential and
current taken on an accessible portion of the boundary of the
body, as shown by the counterexamples due to Alessandrini
(\cite{Al}), Alessandrini and Rondi (\cite{Al-R}), see also \cite{Dc-R}.

The methods used to prove \eqref{eq:log-estimate} are inspired to
the approach presented in the seminal paper \cite{A-B-R-V} where,
for the first time, it was shown how logarithmic stability
estimates for the inverse problem of determining unknown
boundaries can be derived by using quantitative estimates of
Strong Unique Continuation at the Boundary (SUCB) which ensure a
polynomial vanishing rate of the solutions satisfying homogeneous
Dirichlet or Neumann conditions at the boundary. Precisely, in
\cite{A-B-R-V} the key tool was a Doubling Inequality at the
boundary established by Adolfsson and Escauriaza in \cite{l:ae}.

Following the direction traced in \cite{A-B-R-V}, other kinds of
quantitative estimates of the SUCB turned out to be crucial
properties to prove optimal stability estimates for inverse
boundary value problems with unknown boundaries in different
frameworks, see for instance \cite{S} where the case of Robin
boundary condition is investigated. Let us recall, in the context
of the case of thermic conductors involving parabolic equations,
the three cylinders inequality and the one-sphere two-cylinders
inequality at the boundary (\cite{C-R-V1},
\cite{C-R-V2}, \cite{E-F-V}, \cite{E-V}, \cite{Ve1}), and similar
estimate at the boundary for the case of wave equation with time
independent coefficients \cite{S-Ve}, \cite{Ve2}, \cite{Ve3}.

In the present paper, the SUCB property used to improve the double
logarithmic estimate found in \cite{M-R-V2} takes the form of an
optimal three spheres inequality at the boundary. This latter
result was recently proved in \cite{A-R-V} for isotropic elastic
plates under homogeneous Dirichlet boundary conditions, and leads
to a Finite Vanishing Rate at the Boundary (Proposition
\ref{prop:SUCP_boundary}).

Other main mathematical tools are quantitative estimates of Strong
Unique Continuation at the Interior, essentially based on a three
spheres inequality at the interior obtained in \cite{M-R-V1} which
allows to derive quantitative estimates of unique continuation
{}from Cauchy data (Proposition \ref{prop:Cauchy1}), a Lipschitz
estimate of Propagation of Smallness (Proposition \ref{prop:LPS})
and the Finite Vanishing Rate at the Interior (Proposition
\ref{prop:SUCP_interior}) for the solutions to the plate equation.

Let us observe that estimate \eqref{eq:log-estimate} is the first
stability estimate with optimal rate of convergence in the
framework of linear elasticity. Indeed, up to now, the analogous
estimate for the determination, within isotropic elastic bodies,
of rigid inclusions (\cite{Morassi-Rosset2}), cavities
(\cite{Morassi-Rosset1}) or pressurized cavities (\cite{A-B-R})
show a double logarithmic character, and the same convergence rate
has been established by Lin, Nakamura and Wang for star-shaped
cavities inside anisotropic elastic bodies (\cite{L-N-W}).

The plan of the paper is as follows. Main notation and a priori
information are presented in section \ref{Notation}. In section
\ref{Main}, we first state some auxiliary propositions regarding
the estimate of continuation {}from Cauchy data (Proposition
\ref{prop:Cauchy1}) and {}from the interior (Proposition
\ref{prop:LPS}), and the determination of the finite vanishing
rate of the solutions to the plate equation at the interior
(Proposition \ref{prop:SUCP_interior}) and at the Dirichlet
boundary (Proposition \ref{prop:SUCP_boundary}). Finally, in the
second part of section \ref{Main} we give a proof of the main
theorem (Theorem \ref{theo:Main}).

\section{Notation}
\label{Notation}

Let $P=(x_1(P), x_2(P))$ be a point of $\R^2$.
We shall denote by $B_r(P)$ the disk in $\R^2$ of radius $r$ and
center $P$ and by $R_{a,b}(P)$ the rectangle of center $P$ and sides parallel
to the coordinate axes, of length $2a$ and $2b$, namely
$R_{a,b}(P)=\{x=(x_1,x_2)\ |\ |x_1-x_1(P)|<a,\ |x_2-x_2(P)|<b \}$. To simplify the notation,
we shall denote $B_r=B_r(O)$, $R_{a,b}=R_{a,b}(O)$.

Given a bounded domain $\Omega$ in $\R^2$ we shall denote
\begin{equation}
  \label{eq:Omega_rho}
   \Omega_\rho=\{x\in \Omega\ |\ \hbox{dist}(x,\partial\Omega)>\rho\}.
\end{equation}

\noindent When representing locally a boundary as a graph, we use
the following definition.

\begin{definition}
  \label{def:reg_bordo} (${C}^{k,\alpha}$ regularity)
Let $\Omega$ be a bounded domain in ${\R}^{2}$. Given $k,\alpha$,
with $k\in\N$, $0<\alpha\leq 1$, we say that a portion $S$ of
$\partial \Omega$ is of \textit{class ${C}^{k,\alpha}$ with
constants $r_{0}$, $M_{0}>0$}, if, for any $P \in S$, there
exists a rigid transformation of coordinates under which we have
$P=0$ and
\begin{equation*}
  \Omega \cap R_{r_0,2M_0r_0}=\{x \in R_{r_0,2M_0r_0} \quad | \quad
x_{2}>g(x_1)
  \},
\end{equation*}
where $g$ is a ${C}^{k,\alpha}$ function on
$[-r_0,r_0]$
satisfying

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

\begin{equation*}
\|g\|_{{C}^{k,\alpha}([-r_0,r_0])} \leq M_0r_0,
\end{equation*}
where
\begin{equation*}
\|g\|_{{C}^{k,\alpha}([-r_0,r_0])} = \sum_{i=0}^k  r_0^i\sup_{[-r_0,r_0]}|g^{(i)}|+r_0^{k+\alpha}|g|_{k,\alpha},
\end{equation*}

\begin{equation*}
|g|_{k,\alpha}= \sup_ {\overset{\scriptstyle t,s\in [-r_0,r_0]}{\scriptstyle
t\neq s}}\left\{\frac{|g^{(k)}(t) - g^{(k)}(s)|}{|t-s|^\alpha}\right\}.
\end{equation*}

\end{definition}

Given a bounded domain $\Omega$ in $\R^2$ such that $\partial
\Omega$ is of class $C^{k,\alpha}$, with $k\geq 1$, we consider as
positive the orientation of the boundary induced by the outer unit
normal $n$ in the following sense. Given a point
$P\in\partial\Omega$, let us denote by $\tau=\tau(P)$ the unit
tangent at the boundary in $P$ obtained by applying to $n$ a
counterclockwise rotation of angle $\frac{\pi}{2}$, that is

\begin{equation}
    \label{eq:2.tangent}
        \tau=e_3 \times n,
\end{equation}

where $\times$ denotes the vector product in $\R^3$ and $\{e_1,
e_2, e_3\}$ is the canonical basis in $\R^3$.

In the sequel we shall denote by $C$ constants which may change
{}from line to line.

\subsection{A priori information}
\medskip
\noindent {\it i) A priori information on the domain.}

Let us consider a thin plate
$\Omega\times[-\frac{h}{2},\frac{h}{2}]$ with middle surface
represented by a bounded domain $\Omega$ in $\R^2$ and having
uniform thickness $h$, $h<<\hbox{diam}(\Omega)$.

We shall assume
that, given $r_0$, $M_1>0$,

\begin{equation}
   \label{eq:bound_area}
\hbox{diam}(\Omega) \leq M_1 r_0.
\end{equation}

We shall also assume that $\Omega$ contains an open simply
connected rigid inclusion $D$ such that

\begin{equation}
   \label{eq:compactness}
    \hbox{dist}(D, \partial \Omega) \geq r_0.
\end{equation}

Moreover, we denote by $\Sigma$ an open portion
within $\partial \Omega$ representing the part of the boundary
where measurements are taken.

Concerning the regularity of the boundaries, given
$M_0\geq\frac{1}{2}$ and $\alpha$, $0<\alpha\leq 1$, we assume that
\begin{equation}
   \label{eq:reg_Omega}
\partial\Omega \hbox{ is of } class\  C^{2,1}
\ with\ constants\ r_0, M_0,
\end{equation}

\begin{equation}
   \label{eq:reg_Sigma}
\Sigma \hbox{ is of } class\  C^{3,1} \ with\ constants\
r_0, M_0.
\end{equation}

\begin{equation}
   \label{eq:reg_D}
\partial D \hbox{ is of } class\  C^{6,\alpha}
\ with\ constants\ r_0, M_0.
\end{equation}

Let us notice that, without loss of generality, we have chosen
$M_0\geq\frac{1}{2}$ to ensure that $B_{r_0}(P) \subset
R_{r_0,2M_0r_0}(P)$ for every $P \in
\partial \Omega$.

Moreover, we shall assume that for some $P_0\in\Sigma$ and some $\delta_0$, $0<\delta_0<1$,

\begin{equation}
   \label{eq:large_enough}
   \partial\Omega\cap
R_{r_0,2M_0r_0}(P_0)\subset\Sigma,
\end{equation}

and that
\begin{equation}
   \label{eq:small_enough}
   |\Sigma|\leq(1-\delta_0)|\partial\Omega|.
\end{equation}

\medskip
\noindent {\it ii) Assumptions about the boundary data.}

On the Neumann data $\widehat{M}$ we assume that

\begin{equation}
   \label{eq:reg_M}
\widehat{M}\in L^2(\partial \Omega,\R^2),\quad
(\widehat{M}_n,(\widehat{M}_\tau),_s)\not\equiv 0,
\end{equation}

\begin{equation}
   \label{eq:supp_M}
\hbox{supp}(\widehat{M})\subset\subset\Sigma,
\end{equation}

the (obvious) compatibility condition
\begin{equation}
   \label{eq:M_comp}
    \int_{\partial \Omega} \widehat{M}_i = 0, \quad i=1,2,
\end{equation}

and that, for a given constant $F>0$,

\begin{equation}
\label{eq:M_frequency}
   \frac{\|\widehat{M}\|_{L^2(\partial
   \Omega ,\R^2)}}{ \|\widehat{M}\|_{H^{-\frac{1}{2}}(\partial \Omega,\R^2)}}\leq
   F.
\end{equation}

{\it iii) Assumptions about the elasticity tensor.}

Let us assume that the plate is made by elastic isotropic material, the plate
tensor $\mathbb P$ is defined by

\begin{equation}
  \label{eq:3.emme_compact}
   \mathbb P A = B \left [ (1-\nu)A^{sym}+\nu (tr A)I_2 \right ],
\end{equation}

for every $2 \times 2$ matrix $A$, where $I_2$ is the $2\times 2$
identity matrix and $\hbox{tr}(A)$ denotes the trace of the matrix
$A$. The
  \emph{bending stiffness} (per unit length) of the plate is given by the
  function

\begin{equation}
  \label{eq:3.stiffness}
  B(x)=\frac{h^3}{12}\left(\frac{E(x)}{1-\nu^2(x)}\right),
\end{equation}

where the \emph{Young's modulus} $E$ and the \emph{Poisson's coefficient} $\nu$
can be written in terms of the Lam\'{e} moduli as follows

\begin{equation}
  \label{eq:3.E_nu}
  E(x)=\frac{\mu(x)(2\mu(x)+3\lambda(x))}{\mu(x)+\lambda(x)},\qquad\nu(x)=\frac{\lambda(x)}{2(\mu(x)+\lambda(x))}.
\end{equation}

Hence, in this case, the displacement equation of equilibrium
\eqref{eq:1.dir-pbm-incl-rig-1} is

\begin{equation}
    \label{eq:equil-eq-isotropic}
    {\rm div}\left ({\rm div} \left ( B((1-\nu)\nabla^2w + \nu \Delta w I_2) \right ) \right )=0, \qquad\hbox{in
    } \Omega.
\end{equation}

We make the following strong convexity assumptions on the Lam\'e moduli

\begin{equation}
  \label{eq:3.Lame_convex}
  \mu(x)\geq \alpha_0>0,\qquad 2\mu(x)+3\lambda(x)\geq\gamma_0>0, \qquad \hbox{a.e. in } \Omega,
\end{equation}

where $\alpha_0$, $\gamma_0$ are positive constants.

We assume that the Lam\'{e} moduli $\lambda,\mu$ satisfy the following regularity assumptions

\begin{equation}
  \label{eq:C4Lame}
  \|\lambda\|_{C^4(\overline{\Omega})}, \quad \|\mu\|_{C^4(\overline{\Omega})}\leq \Lambda_0.
\end{equation}

Under the above assumptions, by standard variational arguments
(see, for example, \cite{l:agmon}), problem
\eqref{eq:1.dir-pbm-incl-rig-1bis}--\eqref{eq:1-dir-pbm-incl-rig-5bis}
has a unique solution $w\in H^2(\Omega\setminus \overline{D})$
satisfying

\begin{equation}
    \label{eq:w_stima_H2}
    \|w\|_{H^2(\Omega\setminus \overline{D})}\leq C r_0^2\|\widehat{M}\|_{H^{-\frac{1}{2}}(\partial\Omega,\R^2)},
\end{equation}

where $C>0$ only depends on $\alpha_0$, $\gamma_0$, $M_0$, and $M_1$.

In the sequel, we shall refer to the set of constants $\alpha_0$,
$\gamma_0$, $\Lambda_0$, $\alpha$, $M_0$, $M_1$, $\delta_0$ and
$F$ as to the \textit{a priori} data.

\section{Statement and proof of the main result}
\label{Main}

Here and in the sequel we shall denote by $G$ the connected
component of $\Omega \setminus   \overline{(D_1 \cup D_2)}$ such
that $\Sigma \subset \partial{G}$.

\begin{theo}[Stability result]
  \label{theo:Main}

Let $\Omega$ be a bounded domain in $\R^2$ satisfying
\eqref{eq:bound_area} and \eqref{eq:reg_Omega}. Let $D_i$,
$i=1,2$, be two simply connected open subsets of $\Omega$
satisfying \eqref{eq:compactness} and \eqref{eq:reg_D}. Moreover,
let $\Sigma$ be an open portion of $\partial\Omega$ satisfying
\eqref{eq:reg_Sigma}, \eqref{eq:large_enough} and
\eqref{eq:small_enough}. Let $\widehat{M}\in
L^2(\partial\Omega,\R^2)$ satisfy
\eqref{eq:reg_M}--\eqref{eq:M_frequency} and let the plate tensor
$\mathbb{P}$ given by \eqref{eq:3.emme_compact} with Lam\'e moduli
satisfying the regularity assumptions \eqref{eq:C4Lame} and the
strong convexity condition \eqref{eq:3.Lame_convex}. Let $w_i\in
H^2(\Omega \setminus \overline{D_i})$ be the solution to
\eqref{eq:1.dir-pbm-incl-rig-1bis}--\eqref{eq:1-dir-pbm-incl-rig-5bis} when $D=D_i$, $i=1,2$. If, given
$\epsilon>0$, we have

\begin{equation}
    \label{eq:small_L2}
\min_{g \in
\cal{A}} \left\{\|w_1 - w_2 -g \|_{L^2(\Sigma)}+r_0
\left\|\frac{\partial}{\partial n}
(w_1 - w_2 -g) \right\|_{L^2(\Sigma)}\right\}\leq
\epsilon,
\end{equation}

then we have

\begin{equation}
    \label{eq:small_Haus_inclusion}
d_{\cal H}( \overline{D_1},\overline{D_2} ) \leq
r_0\omega\left(\frac{\epsilon}
{r_0^2\|\widehat{M}\|_{H^{-\frac{1}{2}}(\partial
\Omega,\R^2)}}\right),
\end{equation}

where $\omega$ is an increasing continuous function on
$[0,\infty)$ which satisfies
\begin{equation}
    \label{eq:omega_log}
\omega(t)\leq C(|\log t|)^{-\eta}, \quad\hbox{for every }t, \
0<t<1,
\end{equation}
and $C$, $\eta$, $C>0$, $\eta>0$, are constants only
depending on the a priori data.
\end{theo}

The proof of Theorem \ref{theo:Main} is obtained {from} a sequence
of propositions. The following proposition can be derived by
merging Proposition $3.4$ of \cite{M-R-V2} and geometrical
arguments contained in Proposition $3.6$ of \cite{A-B-R-V}.

\begin{prop}[Stability Estimate of Continuation
{from} Cauchy Data {\cite[Proposition 3.4]{M-R-V2}}]
  \label{prop:Cauchy1}
Let the hypotheses of Theorem \ref{theo:Main} be satisfied. We
have

\begin{equation}
   \label{eq:Cauchy1}
\int_{(\Omega\setminus \overline{G})\setminus \overline{D_1}}|\nabla^2 w_1|^2\leq
r_0^2\|\widehat{M}\|_{H^{-\frac{1}{2}}(\partial \Omega
,\R^2)}^2\omega\left(\frac{\epsilon}
{r_0^2\|\widehat{M}\|_{H^{-\frac{1}{2}}(\partial
\Omega ,\R^2)}}\right),
\end{equation}

\begin{equation}
   \label{eq:Cauchy1bis}
\int_{(\Omega\setminus \overline{G})\setminus \overline{D_2}}|\nabla^2 w_2|^2\leq
r_0^2\|\widehat{M}\|_{H^{-\frac{1}{2}}(\partial \Omega
,\R^2)}^2\omega\left(\frac{\epsilon}
{r_0^2\|\widehat{M}\|_{H^{-\frac{1}{2}}(\partial
\Omega ,\R^2)}}\right),
\end{equation}

where $\omega$ is an increasing continuous function on
$[0,\infty)$ which satisfies
\begin{equation}
   \label{eq:omega_cauchy_loglog}
\omega(t)\leq C(\log|\log t|)^{-\frac{1}{2}},\qquad \hbox{for every }
t<e^{-1},
\end{equation}
with $C>0$ only depending on
$\alpha_0$, $\gamma_0$, $\Lambda_0$, $M_0$, $M_1$ and $\delta_0$.

Moreover, there exists $d_0>0$, with $\frac{d_0}{r_0}$ only depending on $M_0$,
such that if
$d_{\cal H}( \overline{\Omega\setminus D_1},\overline{\Omega\setminus D_2} )\leq d_0$ then
\eqref{eq:Cauchy1}--\eqref{eq:Cauchy1bis} hold with $\omega$
given by

\begin{equation}
   \label{eq:omega_cauchy_log}
\omega(t)\leq C|\log t|^{-\sigma},\qquad \hbox{for  every }
t<1,
\end{equation}

where $\sigma>0$ and $C>0$ only depend on $\alpha_0$, $\gamma_0$, $\Lambda_0$,
$M_0$, $M_1$, $\delta_0$, $L$ and $\frac{\tilde
r_0}{r_0}$.
\end{prop}

Next two propositions are quantitative versions of the SUCP
property at the interior for solutions to the plate
equilibrium problem. Precisely, Proposition \ref{prop:LPS} has
global character and gives a lower bound of the strain energy
density over any small disc compactly contained in
$\Omega\setminus \overline{D}$ in terms of the Neumann boundary
data. Proposition \ref{prop:SUCP_interior} establishes a
polynomial order of vanishing for solutions to the plate problem
at interior points of $\Omega \setminus \overline{D}$.

\begin{prop}[Lipschitz Propagation of Smallness {\cite[Proposition 3.3]{M-R-V2}}]
  \label{prop:LPS}
Let $\Omega$ be a bounded domain in $\R^2$ satisfying
\eqref{eq:bound_area} and \eqref{eq:reg_Omega}. Let $D$ be an open
simply connected subset of $\Omega$ satisfying
\eqref{eq:compactness}, \eqref{eq:reg_D}. Let $w\in H^2(\Omega
\setminus \overline{D})$ be the solution to
\eqref{eq:1.dir-pbm-incl-rig-1bis}--\eqref{eq:1-dir-pbm-incl-rig-5bis},
coupled with the equilibrium condition
\eqref{eq:1.equil-rigid-incl}, where the plate tensor $\mathbb{P}$
is given by \eqref{eq:3.emme_compact} with Lam\'e moduli
satisfying the regularity assumptions \eqref{eq:C4Lame} and the
strong convexity condition \eqref{eq:3.Lame_convex} and with
$\widehat{M}\in L^2(\partial\Omega,\R^2)$ satisfying
\eqref{eq:reg_M}--\eqref{eq:M_frequency}.

There exists $s>1$, only depending on $\alpha_0$, $\gamma_0$,
$\Lambda_0$, $M_0$ and $\delta_0$, such that for every $\rho>0$
and every $\bar x\in (\Omega\setminus \overline{D})_{s\rho}$, we
have

\begin{equation}
   \label{eq:LPS}
\int_{B_\rho(\bar x)}|\nabla^2 w|^2\geq
\frac{Cr_0^2}{\exp\left[A\left(\frac{r_0}{\rho}\right)^B\right]}
\| \widehat{M} \|_{H^{-\frac{1}{2}}(\partial\Omega,\R^2)}^2,
\end{equation}
where $A>0$, $B>0$ and $C>0$ only depend on $\alpha_0$,
$\gamma_0$, $\Lambda_0$, $M_0$, $M_1$, $\delta_0$ and $F$.
\end{prop}

\begin{rem}
\label{rem:exp_order}
The exponential character of the dependence on $\rho$ in \eqref{eq:LPS} comes {}from the fact that this global estimate follows {}from the trace-type inequality
\begin{equation*}
\| \widehat{M} \|_{H^{-\frac{1}{2}}(\partial\Omega,\R^2)}\leq C\| \nabla ^2 w\|_{L^2(\Omega\setminus \overline{D})},
\end{equation*}
with $C$ only depending on $M_0$, $M_1$, $\delta_0$ and $\Lambda_0$ (see \cite[Lemma 4.6]{M-R-V2}) and {}from a lower bound of the strain energy density over the disc $B_\rho(\overline{x})$ in terms of the strain energy density over all the domain $\Omega\setminus \overline{D}$. The latter estimate requires a geometrical construction involving a number of iterated applications of the three spheres inequality depending in the radius $\rho$ which leads to an exponential dependence. The local polynomial vanishing rate at the interior is given in the following proposition.

\end{rem}

\begin{prop}[Finite Vanishing Rate at the Interior]
\label{prop:SUCP_interior} Under the hypotheses of Proposition
\ref{prop:LPS}, there exist $\overline{c}_0<\frac{1}{2}$ and
$C>1$, only depending on $\alpha_0$, $\gamma_0$, $\Lambda_0$, such
that, for every $\overline{r}>0$ and for every $x \in \Omega
\setminus \overline{D}$ such that $B_{ \overline{r}}(x) \subset
\Omega \setminus \overline{D}$, and for every $r_1<r_2=
\overline{c}_0 \overline{r}$, we have

\begin{equation}
  \label{eq:SUCP_interior}
   \int_{B_{r_{1}}(x)} |\nabla^2 w|^2 \geq
   C\left(\frac{r_1}{\overline{r}}\right)^{\tau_0}
   \int_{B_{\overline{r}}(x)} |\nabla^2 w|^2,
\end{equation}

where $\tau_0\geq 1$ only depends on $\alpha_0$, $\gamma_0$,
$\Lambda_0$, $\delta_0$ and $F$.
\end{prop}

\begin{proof}

The above estimate is based on the following three spheres
inequality at the interior, which was obtained in \cite[Theorem
6.6]{M-R-V1}: there exist $c_0$, $0<c_0<1$, and $C>1$ only
depending on $\alpha_0$, $\gamma_0$, $\Lambda_0$, such that for
every $\overline{r}>0$, for every $x\in (\Omega \setminus
\overline{D})$ such that $B_{\overline{r}}(x) \subset \Omega
\setminus \overline{D}$, and for every $r_1 < r_2 < r_3 <
c_0\overline{r}$, we have

\begin{equation}
  \label{eq:3sph}
   \int_{B_{r_{2}}(x)}|\nabla ^2w|^{2} \leq C
   \left (
   \frac{r_3}{r_1}
   \right )^C
   \left(  \int_{B_{r_{1}}(x)}|\nabla ^2w|^{2}
   \right)^{\vartheta_0}\left(  \int_{B_{r_{3}}(x)}|\nabla ^2w|^{2}
   \right)
   ^{1-\vartheta_0},
\end{equation}

where

\begin{equation}
  \label{eq:3sph-esponente}
   \vartheta_0 =
   \frac
   {\log \left (
   \frac{c_0 r_3}{r_2}
   \right )    }
   { 2  \log \left (
   \frac{r_3}{r_1}
   \right )   }.
\end{equation}

Inequality \eqref{eq:SUCP_interior} can be derived by exploiting
the optimality of the exponent $\vartheta_0$ and by reassembling
the terms in \eqref{eq:3sph}.

\end{proof}

As noticed in the introduction, our key SUCB property is stated in
the following proposition, which is the counterpart at the
boundary $\partial D$ of Proposition \ref{prop:SUCP_interior}.

\begin{prop}[Finite Vanishing Rate at the Boundary]
\label{prop:SUCP_boundary} Under the hypotheses of Proposition
\ref{prop:LPS}, there exist $\overline{c}<\frac{1}{2}$ and $C>1$,
only depending on $\alpha_0$, $\gamma_0$, $\Lambda_0$, $M_0$,
$\alpha$, such that, for every $x\in \partial D$ and for every
$r_1<r_2= \overline{c} r_0$,

\begin{equation}
  \label{eq:SUCP_boundary}
   \int_{B_{r_{1}}(x)\cap (\Omega\setminus \overline{D})}w^2 \geq
   C\left(\frac{r_1}{r_0}\right)^\tau
   \int_{B_{r_{0}}(x)\cap (\Omega\setminus \overline{D})}w^2.
\end{equation}

where $\tau\geq 1$ only depends on $\alpha_0$, $\gamma_0$, $\Lambda_0$,
$M_0$, $\alpha$, $M_1$, $\delta_0$ and $F$.
\end{prop}

\begin{proof}
By Corollary $2.3$ in \cite{A-R-V}, there exist $c<1$, only
depending on $M_0$, $\alpha$, and $C>1$ only depending on
$\alpha_0$, $\gamma_0$, $\Lambda_0$, $M_0$, $\alpha$, such that,
for every $x\in \partial D$ and for every $r_1<r_2<cr_0$,

\begin{equation}
    \label{eq:suc1}
\int_{B_{r_1}(x)\cap(\Omega\setminus \overline{D})}w^2 \geq C\left(\frac{r_1}{r_0}\right)^{\frac{\log B}{\log \frac{cr_0}{r_2}}}
\int_{B_{r_0}(x)\cap(\Omega\setminus \overline{D})}w^2,
\end{equation}

where $B>1$ is given by
\begin{equation}
    \label{eq:suc2}
B= C\left(\frac{r_0}{r_2}\right)^C\frac{\int_{B_{r_0}(x)\cap(\Omega\setminus \overline{D})}w^2}{\int_{B_{r_2}(x)\cap(\Omega\setminus \overline{D})}w^2},
\end{equation}
Let us choose in the above inequalities $r_2 = \overline{c}r_0$, with $\overline{c}=\frac{c}{2}$.

By \eqref{eq:w_stima_H2} we have

\begin{equation}
   \label{eq:suc3}
\int_{B_{r_0}(x)\cap(\Omega\setminus \overline{D})}w^2\leq  C r_0^6\|\widehat{M}\|_{H^{-\frac{1}{2}}(\partial \Omega,\R^2)}^2,
\end{equation}

with $C$ depending on $\alpha_0$, $\gamma_0$, $M_0$, $\alpha$,
$M_1$. By interpolation estimates for solutions to elliptic
equations (see, for instance, \cite[Lemma 4.7]{A-R-V}, stated for
the case of hemidiscs, but which holds also in the present
context), we have that

\begin{equation*}
\int_{B_{r_2}(x)\cap(\Omega\setminus \overline{D})}w^2\geq Cr_2^4
\int_{B_{\frac{r_2}{2}}(x)\cap(\Omega\setminus \overline{D})}|\nabla^2w|^2,
\end{equation*}
with $C$ depending on $\alpha_0$, $\gamma_0$ and $\Lambda_0$.
By Proposition \ref{prop:LPS} and recalling the definition of $r_2$, we derive
\begin{equation}
    \label{eq:suc4}
\int_{B_{r_2}(x)\cap(\Omega\setminus \overline{D})}w^2\geq C r_0^6\|\widehat{M}\|_{H^{-\frac{1}{2}}(\partial \Omega,\R^2)}^2,
\end{equation}
with $C$ depending on $\alpha_0$, $\gamma_0$, $\Lambda_0$, $M_0$, $\alpha$, $M_1$, $\delta_0$ and $F$.

By \eqref{eq:suc3}--\eqref{eq:suc4}, we can estimate $B$ {}from above, obtaining the thesis.
\end{proof}

\begin{proof}[Proof of Theorem \ref{theo:Main}]

In order to estimate the Hausdorff distance between the
inclusions,

\begin{equation}
    \label{eq:1.0}
    \delta = d_{\cal H}( \overline{D_1}, \overline{D_2}),
\end{equation}

it is convenient to introduce the following auxiliary distances:

\begin{equation}
    \label{eq:1.0bis}
   d = d_{\cal H}( \overline{\Omega \setminus D_1}, \overline{\Omega \setminus
    D_2}),
\end{equation}

\begin{equation}
    \label{eq:1.0ter}
    d_m = \max \left \{ \max_{x \in \partial D_1} \hbox{dist}(x, \overline{\Omega \setminus
    D_2}),  \max_{x \in \partial D_2} \hbox{dist}(x, \overline{\Omega \setminus
    D_1}) \right \}.
\end{equation}

Let $\eta >0$ such that

\begin{equation}
    \label{eq:1.1}
    \max_{i=1,2} \int_{ (\Omega \setminus \overline{G}) \setminus
    \overline{D_i}} |\nabla^2 w_i|^2 \leq \eta.
\end{equation}

Following the arguments presented in \cite{A-B-R-V}, we first
control $d_m$ in terms of $\eta$, and then we use this estimate to control $d$ in
terms of $\eta$. Finally, by improving the results in \cite[proof
of Theorem $1.1$, step 2]{C-R-V2}, we estimate $\delta$ in terms of
$d$.

Let us start by proving the inequality

\begin{equation}
    \label{eq:1.2}
    d_m
    \leq
    C r_0
    \left (
    \frac{ \eta   }{ r_0^2 \| \widehat{M}  \|_{H^{-1/2}(\partial \Omega)}^2  }
    \right )^{\frac{1}{\tau}},
\end{equation}

where $\tau$ has been introduced in Proposition \ref{prop:SUCP_boundary} and $C$ is a positive constant only
depending on the a priori data.

Let us assume, without loss of generality, that there exists $x_0
\in \partial D_1$ such that

\begin{equation}
    \label{eq:2.1}
    \hbox{dist}(x_0, \overline{\Omega \setminus D_2}) =d_m>0.
\end{equation}

Since $B_{d_m}(x_0) \subset D_2 \subset \Omega \setminus
\overline{G}$, we have

\begin{equation}
    \label{eq:2.2}
    B_{d_m}(x_0) \cap (\Omega \setminus \overline{D_1}) \subset
    ( \Omega \setminus
\overline{G} ) \setminus \overline{D_1}
\end{equation}

and then, by \eqref{eq:1.1},

\begin{equation}
    \label{eq:2.3}
    \int_{  B_{d_m}(x_0) \cap (\Omega \setminus \overline{D_1}) } |\nabla^2 w_1|^2 \leq \eta.
\end{equation}

Let us assume that

\begin{equation}
    \label{eq:3.1}
    d_m < \overline{c} r_0,
\end{equation}

where $\overline{c}$ is the positive constant appearing in
Proposition \ref{prop:SUCP_boundary}. Since $w_1=0$, $\nabla w_1
=0$ on $\partial D_1$, by Poincar\'{e} inequality (see, for
instance, \cite[Example 4.4]{A-M-R-08})   and noticing that $d_m
\leq \hbox{diam} (\Omega)\leq M_1 r_0$, we have

\begin{equation}
    \label{eq:3.2}
    \eta \geq \frac{C}{r_0^4}  \int_{ B_{d_m}(x_0) \cap (\Omega \setminus \overline{D_1}) } w_1^2
    ,
\end{equation}

where $C>0$ is a positive constant only depending on $\alpha$,
$M_0$, $M_1$.

By Proposition \ref{prop:SUCP_boundary}, we have

\begin{equation}
    \label{eq:4.1}
    \eta \geq \frac{C}{r_0^4} \left(\frac{d_m}{r_0}\right)^\tau \int_{ B_{r_0}(x_0) \cap (\Omega \setminus \overline{D_1}) } w_1^2,
\end{equation}

where $C>0$ is a positive constant only depending on $\alpha_0$, $\gamma_0$, $\Lambda_0$,$\alpha$,
$M_0$, $M_1$ and $F$.

By Lemma 4.7 in \cite{A-R-V}, we have

\begin{equation}
    \label{eq:4.2}
    \eta \geq C\left(\frac{d_m}{r_0}\right)^\tau \int_{ B_{\frac{r_0}{2}}(x_0) \cap (\Omega \setminus \overline{D_1}) } | \nabla^2w_1|^2,
\end{equation}

where $C>0$ is a positive constant only depending on $\alpha_0$, $\gamma_0$, $\Lambda_0$,$\alpha$,
$M_0$, $M_1$.

By Proposition \ref{prop:LPS}, we have

\begin{equation}
    \label{eq:4.3}
    \eta \geq C\left(\frac{d_m}{r_0}\right)^\tau r_0^2\|\widehat{M}\|_{H^{-1/2}(\partial\Omega)}^2,
\end{equation}

where $C>0$ is a positive constant only depending on $\alpha_0$, $\gamma_0$, $\Lambda_0$,$\alpha$,
$M_0$, $M_1$, $\delta_0$, $F$, {}from which we can estimate $d_m$

\begin{equation}
    \label{eq:5.1}
    d_m
    \leq
    C r_0
    \left (
    \frac{ \eta   }{ r_0^2 \| \widehat{M}  \|_{H^{-1/2}(\partial \Omega)}^2  }
    \right )^{\frac{1}{\tau}},
\end{equation}
where $C>0$ is a positive constant only depending on $\alpha_0$, $\gamma_0$, $\Lambda_0$,$\alpha$,
$M_0$, $M_1$, $\delta_0$, $F$.

Now, let us assume that
\begin{equation}
    \label{eq:5.2}
    d_m
    \geq
    \overline{c} r_0.
\end{equation}
By starting again {}from \eqref{eq:2.3}, and applying Proposition \ref{prop:LPS} and recalling $d_m\leq M_1r_0$, we easily have
\begin{equation}
    \label{eq:5.4}
    d_m
    \leq
    C r_0
    \left (
    \frac{ \eta   }{ r_0^2 \| \widehat{M}  \|_{H^{-1/2}(\partial \Omega)}^2  }
    \right ),
\end{equation}
where $C>0$ is a positive constant only depending on $\alpha_0$, $\gamma_0$, $\Lambda_0$,
$M_0$, $M_1$, $\delta_0$, $F$. Assuming $\eta\leq r_0^2\| \widehat{M}  \|_{H^{-1/2}(\partial \Omega)}^2$, we obtain \eqref{eq:1.2}.

Without loss of generality, let $y_0\in \overline{\Omega\setminus D_1}$ such that
\begin{equation}
    \label{eq:7.1}
\hbox{dist}(y_0,\overline{\Omega\setminus D_2})=d.
\end{equation}

It is significant to assume $d>0$, so that $y_0\in D_2\setminus D_1$. Let us define
\begin{equation}
    \label{eq:7.2}
    h=\hbox{dist}(y_0,\partial D_1),
\end{equation}
possibly $h=0$.

There are three cases to consider:

i) $h\leq \frac{d}{2}$;

ii) $h> \frac{d}{2}$, $h\leq \frac{d_0}{2}$;

iii) $h> \frac{d}{2}$, $h> \frac{d_0}{2}$.

\medskip
\noindent Here the number $d_0$, $0<d_0<r_0$, is such that
$\frac{d_0}{r_0}$ only depends on $M_0$, and it is the same
constant appearing in Proposition \ref{eq:Cauchy1}. In particular,
Proposition $3.6$ in \cite{A-B-R-V} shows that there exists an
absolute constant $C>0$ such that if $d \leq d_0$, then $d\leq
Cd_m$.

\medskip
\noindent
\emph{Case i).}

By definition, there exists $z_0\in \partial D_1$ such that
$|z_0-y_0|=h$. By applying the triangle inequality, we get
$\hbox{dist}\left(z_0, \overline{\Omega\setminus D_2}\right)\geq
\frac{d}{2}$. Since, by definition, $\hbox{dist}\left(z_0,
\overline{\Omega\setminus D_2}\right)\leq d_m$, we obtain $d\leq
2d_m$.

\medskip
\noindent
\emph{Case ii).}

It turns out that $d<d_0$ and then, by the above recalled property, again we have that
$d\leq Cd_m$, for an absolute constant $C$.

\medskip
\noindent
\emph{Case iii).}

Let $\widetilde{h}=\min\{h,r_0\}$. We obviously have that $B_{\widetilde{h}}(y_0)\subset \Omega\setminus \overline{D_1}$ and
$B_d(y_0)\subset D_2$. Let us set

$$
d_1=\min\left\{\frac{d}{2},\frac{\overline{c_0}d_0}{4}\right\}.
$$
Since $d_1<d$ and $d_1<\widetilde{h}$, we have that
$B_{d_1}(y_0)\subset D_2\setminus \overline{D_1}$ and therefore
$\eta\geq \int_{B_{d_1}(y_0)}|\nabla^2 w_1|^2$.

Since $\frac{d_0}{2}<\widetilde{h}$,
$B_{\frac{d_0}{2}}(y_0)\subset \Omega\setminus \overline{D_1}$ so
that we can apply Proposition \ref{prop:SUCP_interior} with
$r_1=d_1$, $\overline{r}=\frac{d_0}{2}$, $r_2=\overline{c_0}\
\overline{r}$, obtaining $\eta\geq
C\left(\frac{2d_1}{d_0}\right)^{\tau_0}\int_{B_{\frac{d_0}{2}}(y_0)}|\nabla^2
w_1|^2$, with $C>0$ only depending on $\alpha_0$, $\gamma_0$,
$\Lambda_0$, $\delta_0$ and $F$. Next, by Proposition
\ref{prop:LPS}, recalling that $\frac{d_0}{r_0}$ only depends on
$M_0$, we derive that
$$
d_1
    \leq
    C r_0
    \left (
    \frac{ \eta   }{ r_0^2 \| \widehat{M}  \|_{H^{-1/2}(\partial \Omega)}^2  }
    \right )^{\frac{1}{\tau_0}},
$$
where $C>0$ only depends on $\alpha_0$, $\gamma_0$, $\Lambda_0$, $M_0$, $M_1$, $\delta_0$ and $F$. For $\eta$ small enough, $d_1<\frac{\overline{c_0}d_0}{4}$, so that
$d_1=\frac{d}{2}$ and
$$
d
    \leq
    C r_0
    \left (
    \frac{ \eta   }{ r_0^2 \| \widehat{M}  \|_{H^{-1/2}(\partial \Omega)}^2  }
    \right )^{\frac{1}{\tau_0}}.
$$
Collecting the three cases, we have

\begin{equation}
    \label{eq:9bis.1}
    d
    \leq
    C r_0
    \left (
    \frac{ \eta   }{ r_0^2 \| \widehat{M}  \|_{H^{-1/2}(\partial \Omega)}^2  }
    \right )^{\frac{1}{\tau_1}},
\end{equation}
with $\tau_1=\max\{\tau,\tau_0\}$ and $C>0$ only depends on $\alpha_0$, $\gamma_0$, $\Lambda_0$, $\alpha$, $M_0$, $M_1$, $\delta_0$ and $F$.

Finally, let us estimate $\delta$ in terms of $d$. By \eqref{eq:9bis.1}, for $\eta$ small enough, we have that
$$
d<\frac{r_0}{4\sqrt{1+M_0^2}}.
$$
Let us notice that if a point $y$ belongs to $\overline{D_1} \setminus D_2$, then $\hbox{dist}(y,\partial D_1)\leq d$.

Without loss of generality let $\overline{x}\in \overline{D_1}$
such that $\hbox{dist}(\overline{x},\overline{D_2})=\delta>0$.
Then $\overline{x} \in \overline{D_1} \setminus D_2$ and therefore
$\hbox{dist}(\overline{x},\partial D_1)\leq d$.

Let $w\in \partial D_1$ such that
$|w-\overline{x}|=\hbox{dist}(\overline{x},\partial D_1)\leq d$.

Letting $n$ the outer unit normal to $D_1$ at $w$, we may write
$\overline{x} =w-|w-\overline{x}|n$. By our regularity assumptions
on $D_1$, the truncated cone $C(\overline{x},
-n,2(\frac{\pi}{2}-\arctan M_0))\cap
B_{\frac{r_0}{4}}(\overline{x})$ having vertex $\overline{x}$,
axis $-n$ and width $2(\frac{\pi}{2}-\arctan M_0)$, in contained
in $D_1$.

On the other hand, by definition of $\delta$,
$B_\delta(\overline{x})\subset \Omega\setminus \overline{D_2}$, so
that the truncated cone $C(\overline{x},
-n,2(\frac{\pi}{2}-\arctan M_0))\cap
B_{\min\{\delta,r_0/4\}}(\overline{x})$ is contained in
$D_1\setminus \overline{D_2}$.

Let us see that $\delta<\frac{r_0}{4}$. In fact if, by
contradiction, $\delta\geq\frac{r_0}{4}$, we can consider the
point $z=\overline{x}-\frac{r_0}{4}n$. Since $z\in D_1\setminus
D_2$, as noticed above, $\hbox{dist}(z,\partial D_1)\leq d$. On
the other hand, by using the fact that $|z-w|\leq \frac{r_0}{2}$
and by the regularity of $D_1$, it is easy to compute that
$\hbox{dist}(z,\partial D_1)\geq \frac{r_0}{4\sqrt{1+M_0^2}}$,
obtaining a contradiction.

Hence $\min\{\delta, \frac{r_0}{4}\}=\delta$ and, by defining
$\overline{z}= \overline{x}-\delta n$ and by analogous
calculations, we can conclude that $\delta\leq (\sqrt{1+M_0^2})d$,
which is the desired estimate of $\delta$ in terms of $d$.

By Proposition \ref{prop:Cauchy1},
\begin{equation}
    \label{eq:ultima}
    d
    \leq
    C r_0
    \left (\log\left|\log\left(\frac{ \epsilon   }{ r_0^2 \| \widehat{M}  \|_{H^{-1/2}(\partial \Omega)}^2  }\right)\right|
    \right )^{-\frac{1}{2\tau_1}},
\end{equation}

with $\tau_1\geq 1$ and $C>0$ only depends on $\alpha_0$, $\gamma_0$,
$\Lambda_0$, $\alpha$, $M_0$, $M_1$, $\delta_0$ and $F$. By this
first rough estimate, there exists $\epsilon_0>0$, only depending
on on $\alpha_0$, $\gamma_0$, $\Lambda_0$, $\alpha$, $M_0$, $M_1$,
$\delta_0$ and $F$, such that, if $\epsilon\leq \epsilon_0$, then
$d\leq d_0$. Therefore the second part of Proposition
\ref{prop:Cauchy1} applies and the thesis follows.

\end{proof}

\section*{Acknowledgements}
Antonino Morassi is supported by PRIN 2015TTJN95 ``Identificazione
e diagnostica di sistemi strutturali complessi''. Edi Rosset and
Sergio Vessella are supported by Progetto GNAMPA 2017 ``Analisi di
problemi inversi: stabilit\`a e ricostruzione'', Istituto
Nazionale di Alta Matematica (INdAM). Edi Rosset is supported by
FRA2016 ``Problemi inversi, dalla stabilit\`a alla ricostruzione",
Universit\`a degli Studi di Trieste.

config: arxiv


text:
\begin{document}

\title{A quantitative notion of redundancy for infinite frames}

\author[J. Cahill]{Jameson Cahill}
\address{Department of Mathematics, University of Missouri, Columbia, MO 65211-4100, USA}
\email{jameson.cahill@gmail.com}

\author[P. G. Casazza]{Peter G. Casazza}
\address{Department of Mathematics, University of Missouri, Columbia, MO 65211-4100, USA}
\email{pete@math.missouri.edu}

\author[A. Heinecke]{Andreas Heinecke}
\address{Department of Mathematics, University of Missouri, Columbia, MO 65211-4100, USA}
\email{ah343@mail.missouri.edu}

\thanks{The authors were supported by NSF DMS 0704216 and 1008183.}

\begin{abstract}
Bodmann, Casazza and Kutyniok introduced a quantitative notion
of redundancy for finite frames - which they called
{\em upper and lower redundancies} - that match better with
an intuitive understanding of redundancy for finite frames \gk{ in a Hilbert
space}.
 The objective of this paper is to see how much of this
 theory generalizes to infinite frames.
  \end{abstract}

\keywords{Frames, Linearly Independent Sets, Redundancy,
Spanning Sets.}

\subjclass{Primary: 94A12; Secondary: 42C15, 15A04, 68P30}

\maketitle

\section{Introduction}

The customary notion of redundancy for a finite frame $\{\phi_i\}_{i=1}^N$ in $\cH_n$ is
to use $\frac{N}{n}$.  Many people have felt for a long time that this was not really
satisfactory since it assigns {\em redundancy 2} to each of the following frames (where
$\{e_i\}_{i=1}^n$ is an orthonormal basis for $\cH_n$):

\[ \Phi_{1} = \{e_1,e_1,e_2,e_2,\ldots,e_n,e_n\};\]
\[ \Phi_2 = \{e_1,\ldots,e_1,e_2,e_3,\dots,e_n\},\ \ \mbox{where $e_1$ occurs $(n+1)$-times,}\]
\[ \Phi_3 = \{e_1,0,e_2,0,\ldots,e_n,0\}.\]

The frame $\Phi_1$ has redundancy 2 and is a disjoint union of two spanning sets and
a disjoint union of two linearly independent sets.  This description of redundancy is
informative.  But for $\Phi_2$, the frame is heavily concentrated in one dimension of the
space.  In particular, this frame is made up of just one spanning set and it requires
(n+1)-linearly independent sets to represent it.  Finally, the frame $\Phi_3$ is made
up of one orthonormal basis and a collection of zero vectors.  Assigning this frame
redundancy 2 is quite misleading.  Although zero vectors are important in some
areas of frame theory, such as filter bank theory, counting them in redundancy gives
no useful information.  What is important, is to keep track of the number of zero vectors
while not letting them artificially increase redundancy.
\medskip

In this paper, we generalize the results of \cite{BCK} by applying their quantitative
notion of redundancy for finite frames \gk{in a Hilbert space $\cH$},  to infinite frames.
Most of the results carry over easily, but a few fail in this setting .

Concerning infinite-dimensional Hilbert spaces,
much work has been done on the idea of {\em deficits, excesses
and redundancy} \cite{BCHL03, BCHL03a,BCHL06,BCHL06a,BCHL06b, BCL09}.
In \cite{BL07}, the authors provide
a meaningful quantitative notion of redundancy which applies to general
infinite frames. In their work, redundancy
is defined as the reciprocal of a so-called frame measure function, which is a
function of certain averages of inner products of frame elements with their
corresponding dual frame elements.  More recently, in \cite{BCL09}, it is shown that
$\ell_1$-localized frames satisfy
several properties intuitively linked to redundancy such as that any frame with
redundancy \gk{greater} than one should contain in it a frame with redundancy arbitrarily
close to one, the redundancy of any frame for the whole space should be greater than
or equal to one, and that the redundancy of a Riesz basis should be exactly one,
were proven for this notion.  Our approach is slightly different in that we are interested
in how many spanning sets or linearly independent sets are in the frame.  However,
our notion does not capture much information about infinite frames whose frame
vectors are not bounded.  We will give examples to show the problems with
our notion of redundancy for unbounded frames.

\subsection{Review of Frames}

We start by fixing our terminology while briefly reviewing the basic definitions
related to frames. \gk{Let} $\cHn$ denote an $n$-dimensional
real or complex Hilbert space and $\cH$ denotes
a finite or infinite dimensional Hilbert space.  A {\em frame} for a Hilbert
space $\cH$ is a family of vectors $\{\phi_i\}_{i\in I}$
(with $|I|$ finite or infinite)  for which there exists constants $0 < A \le B < \infty$
such that
\[
A\norm{x}^2 \leq \sum_{i\in I} |\langle x, \varphi_i \rangle |^2 \leq B\norm{x}^2
\quad \mbox{for all } x \in \cHn.
\]
When $A$ is chosen as the largest possible value and $B$ as the smallest
for these inequalities to hold,
then we call them the {\em (optimal) frame constants}.
If $A$ and $B$ can be chosen as $A=B$, then the frame is called {\em $A$-tight}, and if
$A=B=1$ is possible, $\Phi$ is a {\em Parseval frame}. A frame is called
{\em equal-norm}, if there exists some $c>0$ such that $\|\varphi_i\|=c$ for all
$i=1,\ldots,N$, and it is {\em unit-norm} if $c=1$.

Apart from providing redundant expansions, frames can also serve as an analysis tool.
In fact, they allow the analysis of data by studying the associated {\em frame coefficients}
$(\langle x, \varphi_i \rangle)_{i\in I}$, where the operator $T_\Phi$
defined by
$T_\Phi: \cHn \to \ell_2I)$, $x \mapsto (\langle x,\varphi_i\rangle)_{i\in I}$
is called the \emph{analysis operator}. The adjoint $T^*_\Phi$ of the analysis operator is typically
referred to as the {\em synthesis operator} and satisfies $T^*_\Phi((c_i)_{i\in I}) = \sum_{i\in I}
c_i\varphi_i$.
The main operator associated with a frame, which provides a stable reconstruction process, is the
{\em frame operator}
\[
S_\Phi=T^*_\Phi T_\Phi : \cHn \to \cHn, \quad x \mapsto \sum_{i\in I} \langle x,\varphi_i\rangle \varphi_i,
\]
a positive, self-adjoint invertible operator on $\cHn$. In the case of a Parseval frame, we have $S_\Phi=\Id_{\cHn}$.
In general, $S_\Phi$ allows reconstruction of a signal $x \in \cHn$ through the
reconstruction formula
\beq \label{eq:expansion}
x = \sum_{i\in I}\langle x,S_\Phi^{-1} \varphi_i\rangle \varphi_i.
\eeq
The sequence $(S_\Phi^{-1} \varphi_i)_{i=1}^N$ which can be shown to form a frame itself, is often
referred to as the {\em canonical dual frame}.

We note that the choice of coefficients in the expansion \eqref{eq:expansion}
is generally not the only possible one. If the frame is linearly dependent --
which is typical in applications --
then there exist infinitely many choices of coefficients $(c_i)_{i=1}^N$ leading to expansions of
$x \in \cH$ by
\beq \label{eq:sparseexpansion}
x = \sum_{i\in I} c_i \varphi_i.
\eeq
This fact, for instance, ensures resilience to erasures and noise. The particular choice of coefficients
displayed in \eqref{eq:expansion} is the smallest in $\ell_2$ norm \cite{Chr03}, hence contains the least energy.
A different paradigm has recently received rapidly increasing attention \cite{BDE09}, namely to choose
the coefficient sequence to be sparse in the sense of having only few non-zero entries, thereby allowing
data compression while preserving perfect recoverability.

For a more extensive introduction to frame theory, we refer the interested reader to the books
\cite{Dau92,Mal98,Chr03} as well as to the survey papers \cite{KC07a,KC07b}.

\subsection{\gk{An Intuition-Driven Approach to Redundancy}}
\label{subsec:intuition}

 In order to properly define a quantitative notion
of redundancy, the authors of \cite{BCK} first gave a list of desiderata that a notion is required
to satisfy.

\subsection{Desiderata}
\label{subsec:desiderata}

Summarizing and analyzing the requirements we have discussed, we state the following list of
desired properties for an upper redundancy  $\cR^+_\Phi$ and a lower redundancy $\cR^-_\Phi$
of a frame $\Phi = (\varphi_i)_{i\in I}$ for a finite or infinite dimensional
real or complex Hilbert space $\cH$.

\renewcommand{\labelenumi}{{\rm [D\arabic{enumi}]}}

\begin{enumerate}\item\label{D0}{\em Zero Vectors.}  Redundancy should not count
zero vectors.
\item\label{DEE} {\em Generalization.} If $\Phi$ is an equal-norm Parseval frame, then
in this special case the customary notion of redundancy shall be attained, i.e., $\cR^-_\Phi = \cR^+_\Phi $.
\item\label{DNyquist} {\em Nyquist Property.} The condition $\cR^-_\Phi = \cR^+_\Phi$ shall characterize
tightness of a normalized version of $\Phi$, thereby supporting the intuition that upper and lower
redundancy being different implies `non-uniformity' of the frame. In particular, $\cR^-_\Phi =
\cR^+_\Phi = 1$ shall be equivalent to orthogonality as the `limit-case'.
\item\label{Duplow} {\em Upper and Lower Redundancy.} Upper and lower redundancy shall be
`naturally' related by $0 < \cR^-_\Phi \le \cR^+_\Phi < \infty$.
\item\label{Dadditiv} {\em Additivity.} Upper and lower redundancy shall be subadditive and superadditive,
respectively, with respect to unions of frames. They shall be additive provided that the redundancy
is uniform, i.e., $\cR^-_\Phi=\cR^+_\Phi$.
\item\label{DInvariance} {\em Invariance.} Redundancy shall be invariant under the action of a unitary
operator on the frame vectors, under scaling of single frame vectors, as well as under permutation,
since intuitively all these actions should have no effect on, for instance, robustness against erasures,
\gk{which is one property redundancy shall intuitively measure.}
\item\label{Daverob} {\em Spanning Sets.} The lower redundancy shall measure the maximal number of
spanning sets of which the frame consists. This immediately implies that the lower
redundancy is a measure for robustness of the frame against erasures in the sense that any set of
a particular number of vectors can be deleted yet leave a frame.
\item\label{Dmaxrob} {\em Linearly Independent Sets.} The upper redundancy shall measure the minimal
number of linearly independent sets of which the frame consists.

\end{enumerate}

It is straightforward to verify that for the special type of frames consisting of orthonormal basis vectors, each
repeated a certain number of times, the upper and lower redundancies given by the maximal or minimal
number of repetitions satisfy these conditions. The challenge is now to extend this definition to all
frames in such a way that as many of these properties as possible are preserved.

\renewcommand{\labelenumi}{{\rm (\roman{enumi})}}

\section{Defining Redundancy}

\subsection{Definitions}

As explained before, we first introduce a local redundancy given in \cite{BCK}, which
encodes the concentration of frame vectors around one point. Since the
norms of the frame vectors do not matter for concentration, we normalize
the given frame and also consider only points on the unit sphere $\SSn=\{x \in \cHn: \|x\|=1\}$
in $\cH$. Hence another way to view local redundancy is by considering
it as some sort of density function on the sphere.  A consequence of normalizing the
frame vectors, is that our new set may no longer be Bessel.

We now define a notion of local redundancy. \gk{For this, we remark that} throughout the paper, we
let \gk{$\langle y \rangle$ denote the span of some $y \in \cHn$ and $P_{\langle y \rangle}$
the orthogonal projection onto $\langle y \rangle$.}

\begin{definition}
Let $\Phi = (\varphi_i)_{i\in I}$ be a frame for $\cH$. For
each $x \in \SSn$, the {\em redundancy function}
$\cR_\Phi : \SSn \to \RR^+$ is defined by
\[
\cR_\Phi(x) = \sum_{i\in I} \|P_{\langle \varphi_i \rangle} (x)\|^2

.
\]
\end{definition}

We might think about the function $\mathcal R_\Phi$ as a redundancy pattern on the sphere,
which measures redundancy at each single point. Also notice that this notion is reminiscent
of the fusion frame condition \cite{CKL08}, here for rank-one projections.

In contrast to \cite{BCK}, this redundancy function may not assume its maximum or
minimum on the unit sphere and in general both the max and min of this function
could be infinite.

We now define:

\begin{definition} \label{def:upplowred}
Let $\Phi = (\varphi_i)_{i\in I}$ be a frame for  $\cH$.
Then the {\em upper redundancy of $\Phi$} is defined by
\[
\cR^+_\Phi = \sup_{x \in \SSn} \cR_\Phi(x)
\]
and the {\em lower redundancy of $\Phi$} by
\[
\cR^-_\Phi = \inf_{x \in \SSn} \cR_\Phi(x).
\]
Moreover, $\Phi$ has a {\em uniform redundancy}, if
\[
\cR^-_\Phi = \cR^+_\Phi.
\]
\end{definition}

This notion of redundancy hence equals the upper and lower frame bound of the
normalized version of the frame - which could now be infinite.

\section{The Case of Infinite Redundancy}

\subsection{Main Result}

With the previously defined quantitative notion of upper and lower redundancy, we can
now verify the  properties from Subsection \ref{subsec:desiderata} which hold (and those
which do not hold) in the infinite dimensional setting.
\begin{theorem}
\label{T1}
Let $\Phi = (\varphi_i)_{i\in I}$ be a frame for an $\infty$-dimensional real or complex Hilbert space   $\cHn$ and assume that $\cR^+< \infty$.
\begin{enumerate}
\item[{\rm [D1]}] {\em Generalization.} If $\Phi$ is an equal-norm Parseval frame, then
\[
\cR^-_\Phi = \cR^+_\Phi.
\]
\item[{\rm [D2]}] {\em Nyquist Property.} The following conditions are equivalent:
\bitem
\item[{\rm (i)}] We have $\cR^-_\Phi = \cR^+_\Phi$.
\item[{\rm (ii)}] The normalized version of $\Phi$ is tight.
\eitem
Also the following conditions are equivalent.
\bitem
\item[{\rm (i')}] We have $\cR^-_\Phi = \cR^+_\Phi = 1$.
\item[{\rm (ii')}] $\Phi$ is orthogonal.
\eitem
\item[{\rm [D3]}] {\em Upper and Lower Redundancy.} We have
\[
0 < \cR^-_\Phi \le \cR^+_\Phi < \infty.
\]
\item[{\rm [D4]}] {\em Additivity.} For each orthonormal basis $(e_i)_{i=1}^n$,
\[
\cR^\pm_{\Phi\cup(e_i)_{i=1}^n} = \cR^\pm_\Phi + 1.
\]
Moreover, for each frame $\Phi'$ in $\cHn$,
\[
\cR^-_{\Phi\cup\Phi'} \ge \cR^-_\Phi + \cR^-_{\Phi'}
\quad \mbox{and} \quad
\cR^+_{\Phi\cup\Phi'} \le \cR^+_\Phi + \cR^+_{\Phi'}.
\]
In particular, if $\Phi$ and $\Phi'$ have uniform redundancy, then
\[
\cR^-_{\Phi\cup\Phi'} = \cR_\Phi + \cR_{\Phi'} = \cR^+_{\Phi\cup\Phi'}.
\]
\item[{\rm [D5]}] {\em Invariance.}
Redundancy is invariant under application of a unitary operator $U$ on $\cHn$, i.e.,
\[
\cR^\pm_{U(\Phi)} = \cR^\pm_{\Phi},
\]
under scaling of the frame vectors, i.e.,
\[
\cR^\pm_{(c_i \varphi_i)_{i=1}^N} = \cR^\pm_{\Phi}, \quad c_i \mbox{ scalars},
\]
and under permutations, i.e.,
\[
\cR^\pm_{(\varphi_{\pi(i)})_{i=1}^N} = \cR^\pm_{\Phi}, \quad \pi \in S_{\{1,\ldots,N\}},
\]
\item[{\rm [D6]}] {\em Spanning Sets.} In the finite setting, $\Phi$ contains
$\lfloor \cR^-_\Phi\rfloor$ disjoint spanning sets.
In the infinite dimensional setting, this property fails as we will show with
an example.
\item[{\rm [D7]}] {\em Linearly Independent Sets.}
$\Phi$
can be partitioned into $\lceil \cR^+_\Phi\rceil$ linearly independent sets.

\end{enumerate}
\end{theorem}

\begin{proof}
The properties [D1] is true because redundancy is
the upper and lower frame bounds of the normalized version
of the frame.

The first part of [D2] is true by definition and for the
second part of [D3], it is well known that a unit norm Parseval frame must
be an orthonormal basis.

Property [D4] follows easily from the argument in \cite{BCK}.

Property [D5] is ovbious.

Property [D6] fails as we will see in the next section.

[D7] follows from Theorem 4.2 of \cite{CKLV} which states that:  Every Bessel sequence
$\{\varphi_i\}_{i\in I}$ with Bessel bound $B$ and $\|\varphi_i\|\ge c$ for all $i\in I$
(in our case $c=1$), can be decomposed into $\lceil B/c^2\rceil$ linearly independent
sets.
\end{proof}

The redundancy function gives little information near the extreme
cases - as was true in the finite dimensional case, as the following example shows.

\begin{example}
\label{example:F3}
{\rm We add a example in which the frame is not merely composed of vectors from the unit basis
$\{e_1, \ldots, e_n\}$. Letting $0 < \varepsilon < 1$, we choose $\Phi_4 = (\varphi_i)_{i\in I}$
as
\[
\varphi_i = \left\{\begin{array}{rcl}
e_1 & : & i=1,\\
\sqrt{1-\varepsilon^2} e_1 + \varepsilon e_i & : & i \neq 1\\
e_i&:& i>N.
\end{array} \right.
\]
This frame is strongly concentrated around the vector $e_1$. We first observe that
\[
\cR_{\Phi_3}(e_1) = \sum_{i=1}^N \|P_{\langle \varphi_i \rangle}(e_1)\|^2
= 1 + \sum_{i=2}^N \absip{e_1}{\sqrt{1-\varepsilon^2} e_1 + \varepsilon e_i}^2
= 1 + (N-1)(1-\varepsilon^2).
\]
However, this is not the maximum, which is in fact attained at the average point of the
frame vectors. But in order to avoid clouding the intuition by technical details,
we omit this analysis, and observe that
\[
1 + (N-1)(1-\varepsilon^2) \le \cR_{\Phi_4}^+ < N.
\]
Since
\[
\cR_{\Phi_4}(e_2) = \sum_{i=1}^N \|P_{\langle \varphi_i \rangle}(e_2)\|^2
= \sum_{i=2}^N \absip{e_2}{\sqrt{1-\varepsilon^2} e_1 + \varepsilon e_i}^2
= \varepsilon^2,
\]
we can conclude similarly, that
\[
0 < \cR_{\Phi_4}^- \le \varepsilon^2.
\]}
\end{example}
The frame $\Phi_3$ shows that the new redundancy notion gives little information
near the extreme cases:  $\cR^- \approx 0$ and $\cR^+$ large,
but becomes increasingly more accurate as $\cR^-$ and $\cR^+$ become closer to one another.
By [D2], the frame $\Phi_4$ is not orthogonal, nor is it tight. [D6] is not applicable
for this frame, since $\lfloor \cR_{\Phi_4}^- \rfloor = 0$ although there does exist a
partition into one spanning set.  Now, [D7] implies that this frame can be partitioned into $N-1$ linearly
independent sets. Again, we see that we can do better than this by merely taking the
whole frame which happens to be linearly independent. As before, we observe that [D7]
is not sharp for large values of $\cR^+$.  However, these become increasingly accurate
as $\cR^-$ and $\cR^+$ approach each other.

\section{Infinite Equal Norm Parseval Frames}

It is easy to construct infinite equal norm
Parseval frames for which the norms of the vectors are aribitrarily
close to one.
\vskip12pt

\begin{theorem}
For any $r\le 1$, there is an equal norm Parseval frame $\{\varphi_i\}_{i=1}^{\infty}$
for $\ell_2$ with $\|\varphi_i\|^2 = r$, for all $i=1,2,\ldots$.
\end{theorem}

\begin{proof}
Given the orthonormal basis $\{e^{2\pi int}\}_{n\in \ZZ}$, let $E\subset [0,1]$ be
a measurable set for which $|E|=r$.  Then
\[ \left \{ e^{2\pi int}\chi_E \right \}_{n\in \ZZ}\]
is a Parseval frame of norm $r$ vectors.
\end{proof}

\section{Infinite Parseval Frames}

Since {\em linear independence} is so weak in the infinite dimensional setting,
we will now see that (equal norm Parseval) frames can have some surprising
properties.  This will affect our work in this area.

\begin{example}\label{E1}
For every natural number $k\in \NN$
there is an equal norm Parseval frame for $\ell_2$ which can be written as
$j$-linearly independent and disjoint spanning sets, for all $j=1,2,\ldots,k$.
\end{example}

\begin{proof}
It is straightforward to choose families of vectors $\{f_{ij}\}_{i,j=1}^{\infty}$ satisfying:
\vskip12pt
(1)  The vectors $\{f_{ij}\}_{i,j=1}^{\infty}$ are linearly independent.
\vskip12pt
(2)   For each $j=1,2,\ldots$, we have that span $\{f_{ij}\}_{i=1}^{\infty}$ is dense in $\ell_2$.
\vskip12pt

It follows that if we apply Grahm-Schmidt to $\{f_{ij}\}_{i=1}^{\infty}$
for each $j=1,2,\ldots$, we get a sequence of orthonormal basis
$\{g_{ij}\}_{i=1}^{\infty}$ for $\ell_2$, with the property that $\{g_{ij}\}_{i,j=1}^{\infty}$
is a linearly independent set.
Fix $k\in \NN$ and consider the family:
\[ \left \{ \frac{1}{\sqrt{k}}f_{ij}\right \}_{i=1,j=1}^{\  \infty,\ k}.\]
This family clearly has the desired properties.
\end{proof}

\begin{example}
There is a Parseval frame for $\ell_2$ which can be written as
$j$-linearly independent and disjoint spanning sets, for all $j=1,2,\ldots,\infty$.
(Note that $j=\infty$ is included here).
\end{example}

\begin{proof}
We use the family $\{g_{ij}\}_{i,j=1}^{\infty}$ from Example \ref{E1} and form
the Parseval frame
\[ \left \{ \frac{1}{2^j}g_{ij} \right \}_{i,j=1}^{\infty}.\]

This is the required family.
\end{proof}

\section{More on the Infinite Version of Property [D6]}

In this section we will further examine property [D6].  Unfortunately, it is
dangerously
close to Kadison-Singer.  First, we give an alternative proof of Corollary 2.4
of \cite{BCPS09}.

\begin{definition}
A family of vectors $\{\varphi_i\}_{i=1}^{\infty}$ is $\omega$-{\bf independent}
if whenever
\[ \sum_{i=1}^{\infty}a_i\varphi_i =0,\]
it follows that $a_i=0$, for all $i=1,2,\ldots$.  If we have this property only for
all $\{a_i\}_{i=1}^{\infty}\in \ell_2$, we say the family of vectors is $\ell_2$-{\bf independent}.
\end{definition}

\begin{theorem}
Let $\{Pe_i\}_{i=1}^{\infty}$ be a Parseval frame in $\cH$.  If $I\subset \NN$, the following
are equivalent:

1.    The family $\{Pe_i\}_{i\in I}$ spans $P(\cH)$.

2.  The family $\{(I-P)e_i\}_{i\in I^c}$ is $\ell_2$-independent.
\end{theorem}

\begin{proof}
\noindent $(1)\Rightarrow (2)$:
Assume that $\{(I-P)e_i\}_{i\in I^c}$ is not $\ell_2$-independent.  Then there exists
scalars $\{b_i\}_{i\in I^c}\in \ell_2$ so that
\[ \sum_{i\in I^c}b_i(I-P)e_i =0.\]
It follows that
\[ f = \sum_{i\in I^c}b_ie_i = \sum_{i\in I^c}b_iPe_i \in P(\cH).\]
Thus,
\[ \langle f,Pe_j\rangle = \langle Pf,e_j\rangle = \sum_{i\in I^c}b_i\langle e_i,e_j\rangle
=0,\ \ \mbox{if $j\not= i$. i.e. if $j\in I$}.\]
So $f \perp span\ \{Pe_i\}_{i\in I}$, and this family is not spanning for $P(\cH)$.

\vskip12pt
\noindent $(2)\Rightarrow (1)$:
First assume there is an $f\in P(\cH)$ so that $f \perp span\ \{Pe_i\}_{i\in I}$.
Then, $f=\sum_{i\in I}a_iPe_i$.  Also,
\[ \langle f,Pe_i\rangle = \langle Pf,e_i\rangle = \langle f,e_i\rangle =0,
\ \ \mbox{for all $i\in I$}.\]
Hence, $f=\sum_{i\in I^c}b_ie_i$, with not all $b_i=0$ and $\{b_i\}_{i\in I^c}\in \ell_2$.  Thus,
\[ \sum_{i\in I^c}b_ie_i = f = Pf = \sum_{i\in I^c}b_iPe_i.\]
i.e.
\[ \sum_{i\in I^c}b_i(I-P)e_i =0.\]
That is, $(I-P)e_i\}_{i\in I^c}$ is not $\ell_2$-independent.
\end{proof}

\begin{corollary}
The property [D6] is true for unit norm 2-tight frames
 if and only if whenever $\{\varphi_i\}_{i=1}^{\infty}$ is a
unit norm 2-tight frame then there is a partition $\{I_1,I_2\}$ of $\NN$ so that
$\{\varphi_i\}_{i\in I_j}$ is $\ell_2$-independent for $j=1,2$.
\end{corollary}

If we examine the proof above, we see that what is proved is really
the following:

\begin{corollary}
Let $\{e_i\}_{i\in I}$ be an orthonormal basis for
$\cH$ and $\{Pe_i\}_{i\in I}$ be a Parseval frame for $P(\cH)$.
If
\[ \varphi \in span\ \{e_i\}_{i\in J}\cap P(\cH),\]
then $\varphi \perp span\ \{Pe_i\}_{i\in J^c}$.

In particular,
the following are equivalent for a subset $J\subset I$:

(1) We have
\[ span\ \{e_i\}_{i\in J}\cap P(\cH) \not= \{0\},\]

(2)  We have
\[ span\ \{Pe_i\}_{i\in J^c} \not= \cH.\]

(3)  The family $\{(I-P)e_i\}|_{i\in J}$ is not $\ell_2$-independent.
\end{corollary}

\begin{remark}
Note that the above corollary unifies the finite linearly
independent result with the infinite one.  i.e.  The same theorem above
holds with $|I|$ finite and {\em linearly independent} for part (3).
\end{remark}

\begin{remark}
The above also raises the question if there is an infinite dimensional
Rado-Horn Theorem.  But we are not sure what it should say at this time.
\end{remark}

We note that property [D6] is an infinite dimensional version of a result
from \cite{CFMT}.  In this paper, using variations of the discrete
Fourier transform matrices, the authors construct families of unit norm
2-tight frames for $\cH_n$, so that whenever you partition the frame vectors into two subsets,
the lower Riesz bound of one of the subsets is on the order of $1/n$.
For a counter-example to  [D6], we are
looking for unit norm 2-tight frames for $\ell_2$ so that whenever you divide the
frame vectors into two sets, one of them is not $\ell_2$-independent.

Finally, let us observe that we can find an equal norm Parseval frame with
the above properties.  This example is due to Bodmann, Casazza, Paulsen,
and Speegle.

\begin{example}
For any $E\subset [0,1]$ measurable, the family
\[ \left \{ e^{2\pi int}\chi_E \right \}_{n\in \ZZ},\]
can be written as $k$, $\ell_2$-independent spanning sets for all
$k=1,2,\ldots,\infty$.
\end{example}

\section{Some Notes}

It is possible that there is a better notion of redundancy than that given in
\cite{BCK}.  The problem with that notion is that if we apply
an invertible operator
to a frame, we get different redundancy.  Intuitively,
this should not give a different
value.  A possible alternative definition is:

\begin{definition}
Given a frame $\Phi=\{\varphi_i\}_{i=1}^N$ in $\cH_n$ with frame operator $S$,
let
or
each $x \in \SSn$, the {\em redundancy function}
$\cR_\Phi : \SSn \to \RR^+$ is defined by
\[
\cR_\Phi(x) = \sum_{i=1}^N \|P_{\langle S^{-1/2}(\varphi_i) \rangle} (x)\|^2

.
\]
\end{definition}

\begin{definition} \label{def:upplowred}
Let $\Phi = (\varphi_i)_{i\in I}$ be a frame for  $\cH$.
Then the {\em upper redundancy of $\Phi$} is defined by
\[
\cR^+_\Phi = \sup_{x \in \SSn} \cR_\Phi(x)
\]
and the {\em lower redundancy of $\Phi$} by
\[
\cR^-_\Phi = \inf_{x \in \SSn} \cR_\Phi(x).
\]
Moreover, $\Phi$ has a {\em uniform redundancy}, if
\[
\cR^-_\Phi = \cR^+_\Phi.
\]
\end{definition}

This notion of redundancy equals the upper and lower frame bounds of the
normalized version of the canonical Parseval frame to $\Phi$.

This definition seems to lose some of the properties of the original definition - which
needs to be checked - such as

[D4]  Do these hold?  Especially, if we add a Parseval frame to a frame,
do these
redundancies increase by 1?

[D5]  Does \[
\cR^\pm_{(c_i \varphi_i)_{i=1}^N} = \cR^\pm_{\Phi}, \quad c_i
\mbox{ scalars},
\]

Everything else seems to hold.  But we do pick up a new result that
redundancy is
invariant under application of an invertible operator.

\begin{theorem}
If $\Phi = \{\varphi_i\}_{i=1}^N$ is a frame for $\cH_n$ and $T$ is an
invertible
operator on $\cH_n$, then for all $x\in \cH_n$ we have
\[ \cR_{T(\Phi)}^+(x)= \cR_{\Phi}^+(x),\  \ \mbox{and}\ \
\cR_{T(\Phi)}^-(x) =
\cR_{\Phi}^-(x).\]
\end{theorem}

\begin{proof}
Let $S$ (resp. $S_T$) be the frame operator for $\Phi$ (resp. $T(\Phi)$).
Then
$S^{-1/2}\Phi$ is equivalent to $\Phi$ which is equivalent to $T(\Phi)$
which is
equivalent to $S_1^{-1/2}(T\Phi)$.  Since both of these frames are
Parseval, it follows
from \cite{CKov} that there is a unitary operator $U$ satisfying:
\[ U[(S^{-1/2}(\Phi)] = S_1^{-1/2}(T\Phi).\]
The result is obvious from here.
\end{proof}

\noindent {\bf Alert}:  Unfortunately, this new idea for a definition
does not work.  We will now give an example to show that the upper
frame bound of the normalized version of a Parseval frame is not
related to the number of linearly independent sets we can partition
the family into.

\vskip12pt

\begin{example}  Fix $N$ and let $\{e_i\}_{i=1}^N$ be an orthonormal
basis for $\cH_N$.  We will build a Parseval frame for $\cH_N$ in
pieces.  First, we build the sets
\[ \frac{1}{\sqrt{2N}}e_1+\frac{1}{\sqrt{2N}}e_i,\ \
\frac{1}{\sqrt{2N}}e_1-\frac{1}{\sqrt{2N}}e_i\ \ \mbox{for all
$i=2,3,\ldots,N$}.\]
This family is a frame for $\cH_N$ with frame operator having
eigenvectors $\{e_i\}_{i=1}^N$ and respective eigenvalues
\[ \{\frac{N-1}{N},\frac{1}{N},\frac{1}{N} ,\cdots, \frac{1}{N}\}\]
Hence, if we add to this family the vectors
\[ \{\frac{1}{\sqrt{N}}e_1\} \cup \left \{ \sqrt{1-\frac{1}{N}}e_i \right \}_{i=2}^N,\]
then we will have a Parseval frame which can clearly be devided into
3 linearly independent sets.  Namely, divide the first set into
\[ \left \{\frac{1}{\sqrt{2N}}e_1+\frac{1}{\sqrt{2N}}e_i \right \}_{i=2}^N\]
and
\[ \left \{\frac{1}{\sqrt{2N}}e_1-\frac{1}{\sqrt{2N}}e_i \right \}_{i=2}^N\]
and the third set is already linearly independent.  However, if we
normalize the vectors, we get a frame:
\[ \left \{\frac{1}{\sqrt{2}}e_1 \pm \frac{1}{\sqrt{2}}e_i\right \}_{i=2}^N,\]
plus $\{e_i\}_{i=2}^N$.  For this family, if we check the
frame bound at say $e_1$ we get:
\[ \sum_{i=2}^N |\langle e_1,\frac{1}{\sqrt{2}}e_1\pm \frac{1}{\sqrt{2}}
e_i \rangle |^2 = \frac{N-1}{2}.\]
That is, the upper frame bound of the normalized version of this
Parseval frame is unrelated to the number of linearly independent
sets we can divide it into.
\end{example}

\section{Concluding Remarks}

In the case of infinite redundancy, it is possible that our upper frame bound
is infinity.  It is not clear at this time if anything can be concluded from this
case.


TITLE: Covariance of Newton's equation of motion under Galilean transformations
QUESTION [0 upvotes]: I'm reading Arnold's mathematical methods of classical mechanics and in the section he talks about newton's equation ($\ddot{\boldsymbol{x}}=\boldsymbol{F}(\boldsymbol{x}, \boldsymbol{\dot{x}},t)$) he says that this is invariant under Galilean transformations.
Right, I understand this, but now i wanna show mathematically that this happens
-For covariance under rotations assuming that i already now that the left hand side of newton's equation does not depend explicitly on time: let $\boldsymbol{x}:{\phi}(t)$ be a solution for $\ddot{\boldsymbol{x}}=\boldsymbol{F}(\boldsymbol{x}, \boldsymbol{\dot{x}},t)$. Then if it's invariant under rotations, a rotation $\boldsymbol{\chi}=G\boldsymbol{x}$ is also a solution. So it satifies
$$\boldsymbol{\ddot{\chi}}=\boldsymbol{F}(\boldsymbol{\chi},\boldsymbol{\dot{\chi}})=G\boldsymbol{\ddot{x}}=G\boldsymbol{F}(\boldsymbol{x},\boldsymbol{\dot{x}})$$
Therefore,
$$\boldsymbol{F}(\boldsymbol{\chi},\boldsymbol{\dot{\chi}})=G\boldsymbol{F}(\boldsymbol{x},\boldsymbol{\dot{x}})\\
\implies \boldsymbol{F}(G\boldsymbol{x},G\boldsymbol{\dot{x}})=G\boldsymbol{F}(\boldsymbol{x},\boldsymbol{\dot{x}})$$
and that's the condition that Arnold showed. As i said, i assumed that it was already shown that the function $\boldsymbol{F}$ has no explicit dependence on time.

HERE IS MY QUESTION: how do i show, in a similar way as i did to rotations , that in fact $\boldsymbol{F}$ does not depend on time in a time translation $t \rightarrow t+S$? if i do it, i think i'll be able to do the same for space translation. Otherwise, if i'm not right in my approach please tell what i did wrong.

REPLY [0 votes]: Let $T=t+S$.  Then
\begin{align}
\frac{d}{dT}=\frac{d}{dt}\left(\frac{dt}{dT}\right)=\frac{d}{dt}
\end{align}
since $S$ is constant.
Thus: $\frac{dG}{dt}=\frac{dG}{dT}$ etc.


TITLE: Maple 2019 does not evaluate trigonometric functions as functions!
QUESTION [0 upvotes]: I need to calculate a function of $\tanh(x)$ in Maple.
The problem is that when I enter my function (or any other trigonometric functions) in Maple, the answer looks like this:
tanh(x/sqrt(6));
                         -116                      -116
      2.73949338633639 10     + 2.73949338633639 10     I
sin(x);
                         -116                      -116
      2.73949338633639 10     + 2.73949338633639 10     I
There is no problem with polynomial functions like x^2+x^5!

REPLY [2 votes]: This can happen when the Maple kernel thinks that its license authentication has been cracked.
There have been some known cases where this detection is a false positive. The latest point-release Maple 2019.2.1 should be better, with respect to that.
Point-release upgrades are downloadable (free) from www.maplesoft.com
If you have a valid license, and are using the latest point-release for your version, and still encounter the problem then contact support@maplesoft.com
Your question is not about mathematics, per se, and so is off-topic in this forum.


TITLE: Constructing a tangent line to a point on the curve $y = x^3$ using a compass and straight edge?
QUESTION [0 upvotes]: There are number of ways of constructing a tangent line to the curve $y = x^2$ using a compass and straight edge. Does anyone know of a way of constructing a tangent line to the curve $y = x^3$ using a compass and straight edge?

REPLY [1 votes]: My understanding is that you are given the graph of $y = x^3$ plotted in a Cartesian coordinate system and an arbitrary point on the graph, and you are to construct a line tangent to the graph at that point without using any knowledge of the location of the origin of the coordinates or the orientation of the axes.
You have already given a valid construction, so this answer will focus on how we can prove it is valid.
The construction uses the following fact:

If $A$, $B$, and $C$ are three distinct points on the curve $y = x^3$ with $x$-coordinates $x_A$, $x_B$, and $x_C$ respectively, then $A$, $B$, and $C$ are collinear if and only if $x_A + x_B + x_C = 0.$

Proof of "only if": suppose $A$, $B$, and $C$ are collinear. Let $y = ax + b$ be the equation of the line on which the three points lie.
Then the cubic polynomial $x^3 - ax - b$ has distinct roots at $x_A$, $x_B$, and $x_C$,
and therefore $x^3 - ax - b = (x - x_A)(x - x_B)(x - x_C)$.
Multiplying out the expression on the right and setting coefficients of the polynomials on both sides equal, we find that the coefficient of $x^2$ on the right is
$-(x_A + x_B + x_C)$
whereas the coefficient on the left is $0$, so $x_A + x_B + x_C = 0.$
Proof of "if": Suppose $x_A + x_B + x_C = 0.$
Let $y = ax + b$ be the equation of the line on which the points $A$ and $B$ lie.
Then the cubic polynomial $x^3 - ax - b$ has roots at $x_A$, $x_B$, and some value $x_0$, and therefore $x^3 - ax - b = (x - x_A)(x - x_B)(x - x_0)$
and $x_A + x_B + x_0 = 0.$ Therefore $x_0 = -x_A - x_B = x_C.$
This implies that the line $AB$ intersects the curve $y = x^3$ at a point $P$ with $x$-coordinate $x_C$.
But since $y = x^3$ is the graph of a function of $x$, there is only one point on the curve with $x$-coordinate $x_C$, namely $C$.
Hence the line $AB$ intersects the curve at $C$ and the three points are collinear.
We also use the following fact:

If $A$ and $B$ are distinct points on the curve $y=x^3$ with $x$-coordinates $x_A$ and $x_B$ respectively, then the line $AB$ is tangent to the curve at $B$ if and only if $x_A + 2x_B = 0.$

The proof is similar to the previous proof, except that in the "only if" direction we assume the polynomial $x^3 - ax - b$ has a double root at $x_B$, whereas in the "if" direction we have to prove $x_B$ is a double root.

For ease of exposition, in the following I will write $x_P$ to denote the $x$ coordinate of point $P$, where the letter $P$ can be replaced by the name of any point.
Then the construction works as follows.
Let an arbitrary point $A$ on the curve $y = x^3$ be given,
with the condition only that $A$ is not the point $(0,0).$
Find a line $ABC$ that intersects the curve in two other points $B$ and $C$.
Therefore $x_A + x_B + x_C = 0$.
Let $D$ be the midpoint of $B$ and $C$; then
$x_D = \frac12(x_B + x_C) = -\frac12 x_A$.
Similarly, find a different line $AEF$ intersecting the curve at two other points $E$ and $F$. Therefore $x_A + x_E + x_F = 0$.
Let $G$ be the midpoint of $E$ and $F$; then
$x_G = \frac12(x_E + x_F) = -\frac12 x_A$.
Therefore $x_G = -\frac12 x_A = x_D$ and the line $DG$ is the line
$x = -\frac12 x_A$.
(Note that the $x$-coordinates of $D$ and $G$ are not only equal,
but they are derived by essentially the same procedure.)
Construct the line $AH$ through $A$ parallel to $DG$. Therefore $x_H = x_A$.
Take a point $I$ on the curve distinct from $A$.
Reflect the point $I$ through $H$ to produce the point $J$.
Therefore $x_I + x_J = 2x_H = 2x_A.$
Construct a line through $J$ parallel to $AH$, intersecting the curve at $K$.
Then $x_K = x_J$ and $x_I + x_K = 2x_A.$
If the line $IK$ intersects the curve at a third point $L$,
then $x_L = -2x_A.$
(If there is not a third intersection, you have chosen $I$ so that $IK$ is tangent to the curve; try a different point $I$.)
Now construct the line $AL$. Since $x_L + 2x_A = 0,$ this line is tangent to the curve $y = x^3$ at $A.$

Note that I specified that $A$ is not the point $(0,0)$ in the proof of the construction. If $A$ is $(0,0)$ then all lines through $A$ that intersect the curve in two other points intersect the curve in points that are symmetric about $A$.
Therefore if we attempt to follow the construction as described, we will find that
$D = G = A$.
Therefore the points $D$ and $G$ do not determine a line.
This is an easily resolved difficulty, because we can simply start the construction at some point other than $A$ in order to find a line parallel to the $y$ axis.
The real difficulty is that the tangent at $A$ does not pass through any other point on the curve, so we cannot construct a distinct point $L$ on the curve that determines the tangent line.
If we suppose that we are able to detect that $A$ is the midpoint of $B$ and $C$,
then we could construct a line through $A$ parallel to the $y$ axis, then a line through $A$ perpendicular to the first line.
But I'm not aware that the standard toolbox of compass-and-straightedge constructions includes the ability to recognize that you have arrived exactly at a previously given point.

The construction for the curve $y = x^3$ generalizes to a construction for
$y = p(x)$ for any cubic polynomial $p(x)$,
provided that $A$ is not at the inflection point of the curve,
because the graph of $y = p(x)$ is simply a translation of a linear transformation of the graph of $y = x^3$, and the essential steps of the construction
(the pairs of parallel lines and the proportional positions of the named points along the named lines)
are all preserved by linear transformations and by translations.


TITLE: Sphere volume. Calculation formula from school textbook.
QUESTION [0 upvotes]: I had a question from the course of school stereometry: why is the volume of a ball calculated exactly according to the formula $V=\frac{4}{3} \pi R^3$, why is $4/3$ used?
I did not find the necessary information in the textbook and on the Internet.
I will be glad to have a good explanation.

REPLY [4 votes]: Below, see a half ball, a cone, and a cyclinder (all sideways and therefore looking like a half disk, a triangle, a rectangle).
The all have radius $r$ and height $r$.
Look at the plane at heihgt $h$ above the ground (dotted line).
The intersection with the cylinder is of course a circle of radius $r$ and area $\pi r^2$;
the intersection with th econe is a circle of radius $h$ and area $\pi h^2$;
the intersection with the half ball is a circle of radius $\sqrt{r^2-h^2}$ (use Pythagoras to find the radius) and area $\pi(r^2-h^2)$.
Hence - independently of $h$ - the coross section area of the half ball plus the cross section area of the cone equals the cross section area of the cylinder:
$$\pi(r^2-h^2)+\pi h^2=\pi r^2. $$

Convince yourself that this means that the same equation holds for the volumes:
$$ V_{\text{half sphere}} + V_{\text{cone}} = V_{\text{cylinder}}.$$
You certainly know that the volume of the cylinder is $\pi r^3$ and that the volume of the cone is $\frac 13\pi r^3$ (where the factor $\frac13$ comes in exactly the same way as it does for pyramids).
Therefore, the volume of the half sphere is $\frac23\pi r^3$ and the volume of the full sphere is
$$ \frac43\pi r^3.$$


TITLE: General Topology: Quotient Map
QUESTION [0 upvotes]: Take $X,Y$ to be two topological spaces.
$p: X \to Y$ be continuous
Suppose $\ f: Y \to X$ such that  $\ p \circ f = id_Y$.
Claim: $p$ is a quotient map.
It suffices to show that $p$ is surjective and sets are open in X $\iff$ open in Y.
$\bf{Surjectivity}:$ We know $p(f(x))=id_Y$ does this mean that $f(x)=y$? If so, we're done.
What does this notion of "right inverse" mean?

REPLY [4 votes]: If $p:X\to Y$, $f:Y\to X$ are continuous function between topological space such that $p\circ f=\mathrm{id}_y$, then $p$ is a quotient map.
For $y\in Y$ we have $p(f(y))=y$, hence $p$ is surjective.
Let $V$ be a subset of $Y$.
If $V$ is open, then $p^{-1}[V]$ is open because $p$ is continuous.
Conversely, if $p^{-1}[V]$ is open, then $V=f^{-1}[p^{-1}[V]]$ thus $V$ is open in $Y$ because $f$ is continuous.


TITLE: Is the Euler product formula always divergent for 0<Re(s)<1?
QUESTION [21 upvotes]: It is known that the Euler product formula converges for $\Re(s)>1$
(and there it represents the Riemann zeta function).
My question: Is the Euler product always divergent for
$0 < \Re(s) < 1$ ?
I thought that the absolute value of the Euler product formula is positively divergent under the above condition. Is it apparent?

REPLY [3 votes]: The simplest argument is that the domain of convergence of generalized Dirichlet series are always half planes.   Since there is a pole at s=1,  it can only converge for $Re(s) >1$.
But the question is still very interesting for two reasons.   If you consider instead Dirichlet L-functions based on non-principal characters, then there is no pole at $s=1$ and the above argument does not apply.    In recent work with Franca,  we argued that the Euler product in fact converges for $Re(s)> 1/2$.
The other reason is that for Riemann zeta and other principal characters,  a truncated Euler product can still be made sense of for $Re(s) > 1/2$.
Furthermore,  the Euler product formula is valid in the limit $Im(s) \to \infty$ without truncation in a manner that can be made precise.    This work is on  math.NT


TITLE: Dijkstra's Algorithm Proof Explanation
QUESTION [1 upvotes]: I read over this link from a previous question regarding Dijkstra's algorithm but I'm confused as to some of the conclusions drawn.
In the graphic within the question, there are a bunch of inequalities set up.
I'd like to explain how I currently see/understand it, but I know I'm misssing something.
$l(p)$ in the algorithm is the total length of the path $P$, which is some s-v path.
$l(p')$ is some sub-path of $P$ that  does not leave $P$;
$l(x, y)$ is the first edge that leaves $P$;
and $d(x)$ is the length of the shortest s-x path.
So I understand that that $l(p)$ would be the largest value since it's just SOME s-v path not necessarily the minimal s-v path (which would be $\pi(v)$).
I understand that $l(p') \geq d(x)$ since $d(x)$ is the minimal $l(p')$, but what is the intuition behind $l(p') + l(x, y) \geq d(x) + l(x, y)$?
And furthermore, I don't understand the intuition behind $\pi(y) \geq \pi(v)$ in the example.
If anybody could explain it, that would be great. Thank you.

REPLY [1 votes]: I understand that $l(p') \geq d(x)$ since $d(x)$ is the minimal $l(p')$, but what is the intuition behind $l(p') + l(x, y) \geq d(x) + l(x, y)$?

Well, this is just a basic algebra rule. If you agree that $a \geq b$, then it follows that $a + c \geq b + c$. Here's a reference if you need it.

And furthermore, I don't understand the intuition behind $\pi(y) \geq \pi(v)$ in the example.

This is exactly what the question that you linked is asking, it seems. To paraphrase the answer, at this point, we are assuming that $v$ is the next vertex that the algorithm adds, which means precisely that $\pi(v) \leq \pi(y)$ for all other vertices $y$ that are not yet added. If this weren't true, then this wouldn't be the Djikstra algorithm.
To see why this isn't circular reasoning, it helps to step back and look at what the algorithm is doing, and what we're trying to prove about it. I don't know what reference you have for how the algorithm runs, but the one on Wikipedia is fine. That's how it runs, and we want to prove that this ultimately gives the shortest distance between the starting vertex and the destination. $\pi(y) \geq \pi(v)$ is a consequence of step 6 in Wikipedia's description of the algorithm. We picked $v$ because its distance is shorter than anything else at that point.


TITLE: Euclidean Algorithm - find $\gcd(172, 20)$ and solve $172a + 20b = 1000$.
QUESTION [2 upvotes]: I am revising for an exam and have just realised that the euclidean algorithm questions in past exams are much harder than in homeworks! So i need some help please.
I have a question here, i already have the solution to it but i dont understand it that well, so i would like someone to please explain what is going on ect, and dont worry about avoiding giving me the full solution as i already have it so a full solution would be good!
Here is the question:
Use euclids Algorithm to find $gcd(172,20)$.
Hence solve $172a + 20b = 1000$ giving all solutions in terms of parameter t. List any solutions for which $a,b>0$.
So it wants the answers in terms of parameter t, which is something i am not familiar with and cannot find in my books.
I have the solution in the solutions paper as
$$172= 8\times20+12 $$
$$20=1\times 12 + 8$$
$$12=1 \times 8 +4$$
$$8=2\times 4+0$$
thus $gcd(172,20)=4$
then
$$4= 12-8=12-(20-12)=2\times 12 - 20$$
$$=2\times(172-8\times20)-20$$
$$=2 \times 172 -17 \times20$$
multiplying by 250 we get
$$1000=500 \times 172 -4250 \times 20$$
so one solution is $a=500, b=-4250$
So, I understand everything up until here!!
Now it says,
Generally, $$a=500-t \frac{20}{4}=500-t$$
           $$b=-4250+t \frac{172}{4}=-4250+43t$$
To have $a,b>0$, we need $$500-5t>0, (100>t)$$
to have $a>b$ we need $$-4250+43t>0,(t>98.8)$$$
Need $t=99$, $a=5$, $b=7$
Could someone please help explain what is going on in this last bit of working out? I am pretty good at euclidean algorithm usually but havent seen this type before an am panicking  bit as so close to exam! Any help appreciated how to find t and the other positive solutions would be great!
Please could you show me exactly how it is done here as my lecturer is quite strict with us using his exact methods. Many thanks

REPLY [1 votes]: You have
$$
172a + 20b = 1000
$$
and you found one of the solutions, $a = 500, b = -4250$. I will try to explain parametrization and then finding specific solutions.
Parametrisation
You have two variables $a,b$ and one equation so it can by parametrised by one parameter. Let's find $p_1$ and $p_2$, s.t.
$$
172*(500 + p_1*t) + 20*(-4250 + p_2*t) = 1000.
$$
We have
$$
86000 + 172 * p_1 * t - 85000 + 20*p_2*t = 1000.
$$
so
$$
172 * p_1 * t + 20*p_2*t = 0.
$$
We can divide by $gcd(20,172)=4$ and get
$$
(43 * p_1 + 5*p_2)*t = 0.
$$
Now, either $t$ is $0$ and we have the same solution or $(43 * p_1 + 5*p_2) = 0$ which is true for $p_1 = 5$ and $p_2 = -43$.
So we have parametrization:
a_t = 500 + 5*t
b_t = -4250 - 43*t
Solution
Also, another condition is that : We are looking for such $t$ that $a_t,b_t >0$ so we simply require
$$
a_t = 500 + 5*t > 0 \text{ and } b_t = -4250 - 43*t > 0
$$
Now
$$
-100 < t \text{ and } -98.8  > t,
$$
so $t = -99$ works. We get $a_t = 5$ and $b_t = 7$.
I hope this clarifies the idea in this particular example. In order to get general formulas just try with symbols.


TITLE: Can there exist a `natural' finitely generated group with an undecidable word problem?
QUESTION [18 upvotes]: There are naturally occurring groups that have undecidable algorithmic problems.  For instance, $F_2\times F_2$ has undecidable generalized word problem (membership problem for subgroups) and there is a semidirect product $\Bbb Z^4\rtimes F_4$ with undecidable conjugacy problem.  But to the best of my knowledge every finitely generated group with an undecidable word problem is directly constructed from a Turing machine variant or a non-recursive, recursively enumerable set.

Can there exist a naturally occurring finitely generated group with an undecidable word problem?

Of course this problem is not well-defined because `natural' is not well defined in this setting.  It should somehow mean things built up from well studied classes of groups (hyperbolic groups, polycyclic groups, solvable groups, linear groups) via natural operations, e.g., extensions, amalgamations, maybe direct limit (I am not sure I like allowing this last option since some strange groups are direct limits of hyperbolic groups).

REPLY [15 votes]: If you take a subgroup $H$ of $F_2\times F_2$ with undecidable membership problem and take the HNN extension of $F_2\times F_2$ where the stable letter commutes with $H$, you get a (finitely presented) group with undecidable word problem. I do not know how natural it is since $H$ encodes a Turing machine. The only known "natural" example of universal algebras with undecidable word problem are free modular lattices (Freese, 1980).
As a more natural examples you can take McKenzie-Thompson group that simulates all recursive functions or Valiev's (Boone-Collins) universal group that simulates all Turing machines, see comments below.


TITLE: Probability of the maximum (Levy Stable) random variable in a list being greater than the sum of the rest?
QUESTION [6 upvotes]: Given a list of identical and independently distributed Levy Stable random variables, $(X_0, X_1, \dots, X_{n-1})$, what is the is the probability that the maximum exceeds the sum of the rest?  i.e.:
$$ M = \text{Max}(X_0, X_1, \dots, X_{n-1}) $$
$$ \text{Pr}( M > \sum_{j=0}^{n-1} X_j - M ) $$
Where, in Nolan's notation, $X_j \in S(\alpha, \beta=1, \gamma, \delta=0 ; 0)$, where $\alpha$ is the critical exponent, $\beta$ is the skew, $\gamma$ is the scale parameter and $\delta$ is the shift.  For simplicity, I have taken the skew parameter, $\beta$, to be 1 (maximally skewed to the right) and $\delta=0$ so everything has its mode centered in an interval near 0.
From numerical simulations, it appears that for the region of $0 < \alpha < 1$, the probability converges to a constant, irregardless of $n$ or $\gamma$.  For $1 < \alpha < 2$ it appears to go as $O(1/n^{\alpha - 1})$ (maybe?) irregardless of $n$ or $\gamma$.  For $\alpha=1$ it's not clear (to me) but appears to be a decreasing function dependent on $n$ and $\gamma$.
I have tried making a heuristic argument to the in the form of:
$$\text{Pr}( M > \sum_{j=0}^{n-1} X_j - M) \le n \text{Pr}( X_0 - \sum_{j=1}^{n-1} X_j > 0 )$$
Then using formula's provided by Nolan (pg. 27) for the parameters of the implicit r.v. $ U = X_0 - \sum_{j=1}^{n-1} X_j$ combined with the tail approximation:
$$ \text{Pr}( X > x ) \sim \gamma^{\alpha} c_{\alpha} ( 1 + \beta ) x^{-\alpha} $$
$$ c_{\alpha} = \sin( \pi \alpha / 2) \Gamma(\alpha) / \pi $$
but this leaves me nervous and a bit unsatisfied.
Just for comparison, if $X_j$ were taken to be uniform r.v.'s on the unit interval, this function would decrease exponentially quickly.  I imagine similar results hold were the $X_j$'s Gaussian, though any clarification on that point would be appreciated.
Getting closed form solutions for this is probably out of the question, as there isn't even a closed form solution for the pdf of Levy-Stable random variables, but getting bounds on what the probability is would be helpful.  I would appreciate any help with regards to how to analyze these types of questions in general such as general methods or references to other work in this area.
Note:  There didn't seem to be much interest from this site so I've cross-posted to math.stackexchange.com here.

REPLY [0 votes]: When $\alpha<1$ your heuristic argument above is correct I think, at least for the totally asymmetric case that you considered ($\beta=1$). Totally asymmetric stable random variables with $\alpha<1$ are non-negative. Thus one of the random variables being larger than the sum of the rest implies that it must be the maximum. Thus,
$$
P\left(M > \sum_{j=0}^{n-1} X_j - M \right) = \sum_{i=0}^{n-1}P\left( X_i > \sum_{j\neq i} X_j \right) = n P\left(X_0 > \sum_{j=1}^{n-1} X_j \right)
$$
Now, if the tail asymptotics of $X$ are $P(X>x) \sim C_0 x^{-\alpha}$ then by conditioning on the sum $\sum_{j=1}^{n-1}$ we obtain that
\begin{align*}
\lim_{n\rightarrow\infty}  n P\left(X_0 > \sum_{j=1}^{n-1} X_j \right)
&= \lim_{n\rightarrow\infty} E\left[ n C_0 \left( \sum_{j=1}^{n-1} X_j \right)^{-\alpha} \right] \\
&= \lim_{n\rightarrow\infty} \frac{C_0 n}{n-1} E\left[ \left( \frac{1}{(n-1)^{1/\alpha}} \sum_{j=1}^{n-1} X_j \right)^{-\alpha} \right] \\
&= \lim_{n\rightarrow\infty}\frac{ C_0 n}{n-1} E\left[ X^{-\alpha} \right]\\
&= C_0 E\left[ X^{-\alpha} \right].
\end{align*}
Note that the first equality can be justified by the dominated convergence theorem by using the fact that instead of using the tail asymptotics we use that there is an upper bound $P(X>x) \leq C' x^{-\alpha}$ for some constant $C'<\infty$.


TITLE: Mean value theorem with trigonometric functions
QUESTION [2 upvotes]: Let $f(x) = 2\arctan(x) + \arcsin\left(\frac{2x}{1+x^2}\right)$

Show that $f(x)$ is defined for every $ x\ge 1$
Calculate $f'(x)$ within this range
Conclude that $f(x) = \pi$ for every $ x\ge 1$

Can I get some hints how to start? I don't know how to start proving that  $f(x)$ is defined for every $ x\ge 1$ and I even had problems calculating $f'(x)$
Thanks everyone!

REPLY [3 votes]: The expression for $f(x)$ is defined whenever
$$
-1\le\frac{2x}{1+x^2}\le1
$$
that is
\begin{cases}
2x\le 1+x^2\\
2x\ge -(1+x^2)
\end{cases}
or
\begin{cases}
x^2-2x+1\ge0\\
x^2+2x+1\ge0
\end{cases}
which is satisfied for every $x$.
Computing the derivative of
$$
g(x)=\arcsin\frac{2x}{1+x^2}
$$
just requires some patience; consider $h(x)=2x/(1+x^2)$; then
$$
g'(x)=\frac{1}{\sqrt{1-h(x)^2}}h'(x)
$$
by the chain rule. Now
$$
1-h(x)^2=1-\frac{4x^2}{(1+x^2)^2}=\frac{1+2x^2+x^4-4x^2}{(1+x^2)^2}=
\frac{(1-x^2)^2}{(1+x^2)^2}
$$
so
$$
\frac{1}{\sqrt{1-h(x)^2}}=\frac{1+x^2}{|1-x^2|}
$$
Note the absolute value! Always remember that $\sqrt{a^2}=|a|$. On the other hand, we can write $\sqrt{(1+x^2)^2}=1+x^2$, because this expression is always positive.
Now we tackle $h'(x)$:
$$
h'(x)=2\frac{1\cdot(1+x^2)-x\cdot 2x}{(1+x^2)^2}=\frac{2(1-x^2)}{(1+x^2)^2}
$$
and so, putting all together, we have
$$
g'(x)=\frac{1+x^2}{|1-x^2|}\frac{2(1-x^2)}{(1+x^2)^2}=
\frac{2}{1+x^2}\frac{1-x^2}{|1-x^2|}
$$
Note that this is not defined for $x=1$ or $x=-1$.
Thus you have
$$
f'(x)=\frac{2}{1+x^2}+\frac{2}{1+x^2}\frac{1-x^2}{|1-x^2|}
$$
For $x>1$ we have $|1-x^2|=x^2-1$, so $\frac{1-x^2}{|1-x^2|}=-1$
and therefore
$$
f'(x)=0
$$
for $x>1$ and the function is constant in the interval $(1,\infty)$. Since it is continuous at $1$ it is constant also on $[1,\infty)$. The constant value is
$$
f(1)=2\arctan1+\arcsin1=2\frac{\pi}{4}+\frac{\pi}{2}=\pi
$$
Note also that, for $x<-1$, we can draw the same conclusion and
$$
f(x)=-\pi \qquad(x\le -1)
$$
On the other hand, for $-1<x<1$ we have
$$
f'(x)=\frac{4}{1+x^2}
$$
so
$$
f(x)=k+4\arctan x
$$
(because in $(-1,1)$ the two functions have the same derivative), where $k$ is constant; this can be evaluated by seeing that $f(0)=0$, so $k=0$.
Thus your function $f$ can also be represented by
$$
f(x)=\begin{cases}
-\pi & \text{for $x\le-1$}\\
4\arctan x & \text{for $-1<x<1$}\\
\pi & \text{for $x\ge 1$}
\end{cases}
$$
Note. I showed the full exercise because it uses just basic facts. I wanted to underline how splitting the computation into pieces can lead to the result in an easier way than by having gigantic expressions to deal with.
Here's a picture of the graph.


TITLE: Find the minimum values of $a^3$ and $b^3$ from the following given cubic function.
QUESTION [2 upvotes]: Today after teaching AM,GM,HM relations my teacher proposed this question.
It isn't a HW though, since we don't have to tell that we solved the question and submit a solution.
Question

Given that $a,b$ are two positive real numbers.
  If $$f(x)=2x^3+ax^2+bx+4=0$$
  Find the minimum values of $a^3$ and $b^3$.

Well I have progressed a bit.
Applying $AM\ge GM${taking all the terms in $f(x)$ positive}
We get $8x^6ab\le0$ which is impossible since $x$ can't be $0$ and $a,b$ are positive.
So we have a conclusion that roots of $f(x)$ are negative.
Now putting $x=-X$ where $X$ is positive real number, we get
$2X^3+bX=aX^2+4$
This is all useful I got. Using AM,GM on the both the sides one by one yields two inequalities of no use.
Thanks for any solution or hint in advance.

REPLY [1 votes]: The minimum value of a positive number does not exist, which says that a minimal value of $a^3$ does not exist and a minimum value of $b^3$ does not exist.