TITLE: Fourier transform of $(\vec{r}- \vec{r}_0)/|\vec{r}- \vec{r}_0|^3$
QUESTION [0 upvotes]: Could anyone find the Fourier transform of below expression in 3D, where $r_0$ is a constant vector?
$$\frac{\vec{r}- \vec{r}_0}{|\vec{r}- \vec{r}_0|^3}$$

REPLY [0 votes]: Let $\mathbf F_0 = {\mathbf r}/{|\mathbf r|^3}.$ This has divergence $\nabla \cdot \mathbf F_0 = 4\pi \delta,$ where $\delta$ is the Dirac distribution on $\mathbb R^3$.
Taking the Fourier transform of both sides gives $-i{\mathbf k}\cdot\hat{\mathbf F}_0 = 4\pi.$ Since $\mathbf{F}_0$ is symmetric, so should $\hat{\mathbf F}_0$. Therefore we assert $\hat{\mathbf F}_0 = f(|\mathbf k|) \mathbf{k}$ for some function $f : [0, \infty) \to \mathbb R$.
This gives
$-i(\mathbf k\cdot\mathbf k)f(|\mathbf k|) = -i|\mathbf k|^2 f(|\mathbf k|)= 4\pi$
so $f(|\mathbf k|) = i\,4\pi/|\mathbf k|^2,$ i.e.
$$\hat{\mathbf F}_0 = i\,4\pi \frac{\mathbf k}{|\mathbf k|^2}.$$
The transform of the translated field $\mathbf F = (\mathbf r - \mathbf r_0)/|\mathbf r - \mathbf r_0|^3$ is given by multiplication by a factor:
$$\hat{\mathbf F} = i\,4\pi \frac{\mathbf k}{|\mathbf k|^2} e^{-i \mathbf k \cdot \mathbf r_0}.$$


TITLE: Proving $\prod_{t=1}^{T}\sqrt{1-4x_t^{2}}\leq \exp\left(-2\sum_{t=1}^{T}x_t^{2}\right)$, given $1-x\leq \exp(-x)$
QUESTION [2 upvotes]: Given that $1-x\leq \exp{(-x)}$, prove that
$$\prod_{t=1}^{T}\sqrt{1-4x_t^{2}}\leq \exp\left(-2\sum_{t=1}^{T}x_t^{2}\right)$$

How would one approach this problem. Without the square root I know that
$$1-2\sum_{t=1}^{T}x_t^{2}\leq \exp(-2\sum_{t=1}^{T}x_t^{2})$$
But I'm not sure what to do after that.

REPLY [1 votes]: For all $t$, using the hint, we have
$$1-(2x_t)^2\leq \exp\bigg(-(2x_t)^2\bigg)$$
Thus
$$1-4x_t^2\leq \exp(-4x^2_t)$$
Take the square root on each side
$$\sqrt{1-4x_t^2}\leq \exp(-2x^2_t)$$
Multiplying all these inequalities for $1\leq t\leq T$ yields
$$\prod_{t=1}^T\sqrt{1-4x_t^2}\leq \prod_{t=1}^T\exp(-2x^2_t)$$
In other words
$$\prod_{t=1}^T\sqrt{1-4x_t^2}\leq \exp(-2  \sum_{t=1}^T x^2_t)$$


TITLE: Does $\sum_{n \in \mathbb{Z}} n^2 e^{-n^2 t}$ converge uniformly for $t>0$?
QUESTION [0 upvotes]: Consider the heat equation
$$\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}$$
for $x \in \mathbb{T}:=\mathbb{R}/2\pi\mathbb{Z}, t \geq 0$, with initial condition $u(x,0)=f(x)$ for some $f \in L^1(\mathbb{T})$.
I'm asked to verify that $\sum_{n \in \mathbb{Z}} \hat{f}(n) e^{-n^2 t} e^{inx}$ satisfies this equation for all $x \in \mathbb{T}, t>0$ (where the hat denotes Fourier coefficients).
I'm not sure how to justify differentiating term-by-term with respect to $t$, because I don't know how to prove (or if) the series $\sum_{n\in\mathbb{Z}} n^2 e^{-n^2 t}$ converges uniformly for $t>0$ so that I can apply the Weierstrass M-Test (the Fourier coefficients are bounded because they tend to zero). Perhaps the M-Test is not sufficient. I'd appreciate it if someone could tell me how to justify term-by-term differentiation here.

REPLY [2 votes]: Certainly not uniformly on $(0,\infty)$, but uniformly on $[\varepsilon, \infty)$ for each $\varepsilon > 0$ should be enough.


TITLE: How does a Faraday Cage work?
QUESTION [5 upvotes]: I was reading some articles about lightning and i came across the term "Faraday cage". I know that it refers to an enclosure made out of a conductor that can block electric fields. But what i want to know is the physics behind it in terms of charges and electric fields.

REPLY [4 votes]: There are related questions such as Why does an Faraday Cage protect you, but I don't think they answer fully the question the OP has in the comments.
If you are inside the cage, and it is hit by a large spark, such as lighning or a deliberate electrostatic discharge, the resulting current will stay on the outside of the cage and no current will pass inside, provided the cage mesh size is small enough and thick  enough, that is the gaps between the wires are smaller than the wavelength of the external radiation.

The part that i don't get is that it says that it can cancel the electric field inside the cage when an external electric field is applied.

From Wikipedia Faraday Cages

Externally or internally applied electromagnetic fields produce forces on the charge carriers (usually electrons) within the conductor; the charges are redistributed accordingly (that is, electric currents are generated). Once the charges have rearranged so as to cancel the applied field inside, the currents stop.

Animation showing how a Faraday cage (box) works. When an external electrical field (arrows) is applied, the electrons (little balls) in the metal move to the left side of the cage, giving it a negative charge, while the remaining unbalanced charge of the nuclei give the right side a positive charge. These induced charges create an opposing electric field that cancels the external electric field throughout the box.

Take an internally applied field as it acts the same way as an external field, but might be easier to follow intutively.
Think of the cage as a real version of the the conduction shell,  that you learn about when initially studying electromagnetism. External (or internal) electric fields act to redistribute the electrons,  because of mutual repulsion, (the electrons get as far apart from each other as surface and charge density allows. This causes a temporary flow of current which ceases when the electrons have stopped moving.
Now you bring a charge with you inside the ungrounded cage, this will produce an electric field on the internal side of the cage, as it acts to neutralise the field within the cage. The direct effect of this inner side electron rearrangement is to produce a charge on the ouside of the cage that is equal in magnitude (and with the same  polarity) as the charge you brought in.
In other words, if you bring in a positively charged object, this attracts all the cage's free  electrons to the inner surface of the cage, leaving a net positive charge on the outer surface.  Now it does not make any difference where you place the charge within the shell, the cage generates the same DC electric field that it would generate if it were simply affected by the charge placed inside. The inner and outer charges act to cancel each other out.
Now you ground the cage, providing a route for excess charge to leave the outer side of the shell, so you are left  with the inner side of the shell neutralising the charge you have brought in.

REPLY [1 votes]: A Faraday cage can be considered as an approximation of an ideal hollow conductor. If there is an electric field outside the cage, then it will create forces on the charges on the conductor surface and this will cause the surface charge to be redistributed accordingly so as to cancel the effect of the external field. This we say as achieving  electrostatic equilibrium.

How it works?

A conductor, has free charges (usually electrons) on it's surface. These free charges, in the absence of an electric field are so distributed such that the net electric field created by the whole charges vanish. This state, which we normally call as the neutral state, is actually electrostatic equilibrium. Hence no net current flows through a conductor in the absence of an external field (or when it is in electrostatic equilibrium).
Now, if you place such a conductor in the presence of an electric field, then there is an unbalanced force (given by $F=qE$) now acting on the charges which causes them to make a new rearrangement so as to nullify the effect of the external field and thereby reducing their free energy and to attain equilibrium . This redistribution we see as a current flowing through the conductor surface. The current flows (or the charges redistribute) until electrostatic equilibrium is reached.
So, if there is an external field outside our hollow conductor, it will cause a redistribution of the electrons on the conductor surface cancelling the external field. This prevents any electric field to be present beneath the conductor surface. This phenomenon of shielding is made used in Faraday's cage.
If there is an electric field inside a conductor, which is not grounded, it will again cause a redistribution of electrons on the outer surface due to induction in the inner surface, the inside electric field. If it is grounded, the excess current on the surface flows to the ground, keeping it neutral.
However, it could not block magnetic fields as magnetic fields have no effect on static charges (more precisely, charges under electrostatic equilibrium).


text:
\begin{document}

\title[noncompactness and noncompleteness]{Noncompactness and noncompleteness in isometries of Lipschitz  spaces}

\author{Jes\'us Araujo and Luis Dubarbie}

\address{Departamento de Matem\'aticas, Estad\'istica y
Computaci\'on, Facultad de Ciencias, Universidad de Cantabria,
Avenida de los Castros s/n, E-39071, Santander, Spain.}

\email{araujoj@unican.es} \email{luis.dubarbie@gmail.com}

\thanks{Research partially supported by the Spanish Ministry of Science and Education
(MTM2006-14786). The second author was partially supported by a predoctoral
grant from the University of Cantabria and the Government of Cantabria.}

\keywords{Linear isometry, Banach-Stone theorem, vector-valued Lipschitz function, biseparating map}

\subjclass[2010]{Primary 47B33; Secondary 46B04, 46E15, 46E40, 47B38}

\begin{abstract}
We solve the following  three  questions concerning surjective linear isometries between spaces
of Lipschitz functions $\mathrm{Lip}(X,E)$ and $\mathrm{Lip}(Y,F)$, for strictly convex normed spaces $E$ and $F$ and
metric spaces $X$ and $Y$:
\begin{enumerate}
\item Characterize those base spaces $X$ and $Y$ for which all isometries are weighted composition maps.
\item Give a condition independent of base spaces under which all isometries are weighted composition maps.
\item Provide the general form of an isometry, both when it is a weighted composition map and when it is not.
\end{enumerate}
In particular, we prove that requirements of completeness on $X$ and $Y$ are   not
necessary when $E$ and $F$ are not complete, which is in sharp contrast with results known in the scalar context.
\end{abstract}

\maketitle

\section{Introduction}

It is well known that not all surjective (linear) isometries between spaces of Lipschitz functions
on general metric spaces $X$ and $Y$ can be written as weighted composition maps (see for instance  \cite[p. 61]{W}).
Attempts to identify the isometries which can be described in that way have been  done in three ways,
each  trying to provide an answer to one of the following  questions:
\begin{enumerate}
\item\label{rabanito} Characterize those base spaces $X$ and $Y$ for which all isometries are weighted composition maps.
\item\label{carolo} Give a condition independent of base spaces under which all isometries are weighted composition maps.
\item\label{ocumier} Provide the general form of an isometry, both when it is a weighted composition map and when it is not.
\end{enumerate}

The first question was
studied by Weaver  for a general metric  in the {\em scalar-valued} setting
(see \cite{W1} or \cite[Section 2.6]{W}),
and the second one has been
recently  treated  by Jim\'enez-Vargas and  Villegas-Vallecillos in the more general setting of
{\em vector-valued} functions (see \cite{JV1}).
In the latter,  the Banach spaces   where the functions take values are assumed to be strictly convex.
This is certainly not a heavy restriction, as this type of results is known not to hold
for general Banach spaces. Strict convexity is actually a very common and reasonable assumption,
even if, at least in other contexts, it is not the unique possible (see for instance \cite{A5, B, FJ1, Je}).
 As for the third question, an answer was given by Mayer-Wolf for compact base spaces in the scalar context, not for a general metric $d$, but for powers  $d^{\alpha}$
with $0< \alpha <1$.

 Weaver
 proved that completeness and $1$-connectedness of
 $X$ and $Y$ are
sufficient conditions, and that the weighted composition isometries must have  a very special form.
More concretely, given complete $1$-connected metric spaces $X$ and $Y$ with diameter at most $2$, a linear bijection
$T :\mathrm{Lip}(X) \ra \mathrm{Lip}(Y)$ is an isometry if and only if $T f = \alpha \cdot f \circ h$
 for every $f$, where
$\alpha \in \mathbb{K}$, $\va \alpha \vb =1$, and $h:Y \ra X$ is an isometry. Requirements of $1$-connectedness on both $X$ and $Y$  (that is, they cannot be decomposed into two nonempty
disjoint sets whose distance is greater than or equal to $1$) cannot be dropped  in general.
And, obviously, $\mathrm{Lip}(X)$ and $\mathrm{Lip}(\widehat{X})$ are linearly
isometric when   $X$ is not complete (where $\widehat{X}$ denotes the completion of $X$), so requirements of completeness cannot
be dropped either.

On the other hand, Jim\'enez-Vargas and  Villegas-Vallecillos
gave
a general representation  in the spirit of
the classical Banach-Stone Theorem (along with related results for isometries not necessarily surjective).
Assumptions  include compactness of base metric spaces
and the fact that the isometry fixes
a (nonzero) constant function. The conclusion in the surjective case is that the isometry $T: \mathrm{Lip}(X,E) \ra \mathrm{Lip}(Y,E)$ is of the form  $ T f( y) = J y (f ( h (y)))$  for all
$ f\in {\rm Lip}(X,E)  $ and $y\in Y$,
where $h$ is a bi-Lipschitz homeomorphism  (that is, $h$ and $h^{-1}$ are Lipschitz) from $Y$ onto $X$
and $J$ is a
Lipschitz map   from $Y$ into the set  $I(E,E)$
 of all surjective linear isometries on  the strictly convex Banach space  $E$.
They also proved that this result can be sharpened under stronger hypotheses,
 but the above
assumptions remain basically the same, so that the results do not
provide an "if and only if" description.

Finally, Mayer-Wolf not only characterized the family of compact spaces for which the associated
Lipschitz spaces admitted isometries that were not composition operators, but also gave their general form. In principle, it is not clear  whether or not his results   can be extended  to spaces endowed with a metric not
 of the form $d^{\alpha}$. In fact, the answer, as we will see here, is not completely positive.

The aim of this paper is to give, in the {\em vector-valued} setting, a complete answer to
questions (\ref{rabanito}),(\ref{carolo})  and (\ref{ocumier})  (just assuming strict convexity of $E$ and $F$).
The general answer is not known even in the scalar setting, which can be included here as a special case.
We also prove, on the one hand,  that conditions of compactness can be replaced with just
completeness on base spaces
and, on the other hand, that
even  completeness can be dropped when the
normed spaces $E$ and $F$ are  not complete
(which is in sharp contrast
with the behaviour in the scalar case).

To solve (\ref{carolo}), we show that the condition on the preservation of a constant function (as given in \cite{JV1})
can be replaced with a milder one (see Theorem~\ref{marthahoyo}).
We use it to solve (\ref{ocumier}) (see Theorem~\ref{nonelk}, and more in general
Theorem~\ref{marthahoyo}, Corollary~\ref{bocadillo} and Remark~\ref{sinpase}). Our answer also applies to
results on metrics $d^{\alpha}$ in \cite{M}, and a key to understand the generalization is
Proposition~\ref{pesafrank}.
An answer to (\ref{rabanito}) is given as a direct consequence of the results concerning
(\ref{carolo})  and (\ref{ocumier})
(see Corollaries~\ref{deluxion} and~\ref{rentreeflot}).
As a special case we provide   the natural
 counterpart of the description given in \cite{W1}
 (see Corollary~\ref{siegediesrefe} and Remark~\ref{guelgafunc}).
 We finally mention that we do not use the same techniques
as in  \cite{W1} nor as  in \cite{JV1};
instead we study {\em surjective linear} isometries through biseparating maps, which has proven successful in various contexts (see for instance
\cite{A5, HBN} for recent references).

Other papers where related operators have been recently studied in similar contexts are \cite{AEV, GJ, GJ2, JV, Le} (see also  \cite{FJ, JP, L, M, R, V}).

\section{Preliminaries and notation}

Recall that, given metric spaces $(X, d_1)$ and $(Y, d_2)$, a map $f:X \rightarrow Y$ is said to
be \emph{Lipschitz} if there
exists a constant $k\geq 0$ such that
$
d_2 ( f(x) , f(y) ) \leq k \ d_1(x,y)
$
for each $x,y\in X$, and that
 the \emph{Lipschitz number} of $f$ is
$$L(f):=\mathrm{sup}\left\{\frac{ d_2 (f(x) , f(y) )}{d_1 (x,y)}:x,y\in
X, \ x\neq y\right\}.$$

Given a normed space $E$  (over $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$), we
denote by  $\mathrm{Lip}(X,E)$ the space of all \emph{bounded} $E$-valued
Lipschitz functions  on $X$.
We endow   $\mathrm{Lip}(X,E)$
 with the norm
$
\vc \cdot \vd_{L} :=\mathrm{max}\left\{\vc \cdot \vd_{\infty},L(\cdot)\right\}
$
(where $\vc  \cdot \vd_{\infty}$ denotes
the usual supremum norm).

As a particular case, we can consider in $X$  a power ${d_1}^{\alpha}$     of   the metric $d_1$,
  $0< \alpha <1$.  The corresponding space of all  bounded  $E$-valued
Lipschitz functions  on $X$ with respect to ${d_1}^{\alpha}$ is then denoted by $\mathrm{Lip}^{\alpha} (X,E)$.

Recall also that  a  normed space $E$ is said to be \emph{strictly convex} if
$\vc \mathrm{e}_{1} + \mathrm{e}_{2} \vd <2$ whenever $\mathrm{e}_{1},\mathrm{e}_{2}$ are different vectors of
norm $1$ in $E$ or, equivalently, that
$\vc \mathrm{e}_{1} + \mathrm{e}_{2} \vd = \vc \mathrm{e}_{1} \vd + \vc \mathrm{e}_{2} \vd $
($\mathrm{e}_{1}, \mathrm{e}_{2} \neq 0$)  implies $\mathrm{e}_{1} = \alpha \mathrm{e}_{2}$
for some $\alpha >0$
(see \cite[pp. 332--336]{K}).
From this, it follows that,
given $\mathrm{e}_{1},\mathrm{e}_{2}\in E\setminus \{0\}$,
\begin{equation}\label{strictly}
\vc \mathrm{e}_{1} \vd , \vc \mathrm{e}_{2} \vd<\max\{\vc \mathrm{e}_{1}+\mathrm{e}_{2} \vd,\vc \mathrm{e}_{1}-\mathrm{e}_{2} \vd\}  ,
\end{equation}
which is an inequality we will often use. The fact that a normed space is strictly convex does not imply
that its completion is.  Indeed every infinite-dimensional separable Banach space can be renormed to be not
stricly convex and to contain a strictly convex dense subspace of codimension one (see \cite{GZ}).

From now on, unless otherwise stated, we assume that
$E$ and $F$ are strictly convex normed spaces (including the cases $E=\mathbb{K}$, $F = \mathbb{K}$).

As we mentioned above, on our way to Theorem~\ref{marthahoyo} we will deal with biseparating maps.    Recall that   separating maps are those preserving disjointness of {\em cozero sets} (where the cozero set of
a function
 $f: X \ra E$ is defined as $c(f):=\{x\in X:f(x)\neq 0\}$). More concretely,
we will say that
a   linear map $T:\mathrm{Lip}(X,E)\rightarrow \mathrm{Lip}(Y,F)$ is \emph{separating} if
  $c(Tf)\cap c(Tg)=\emptyset$ whenever $f,g\in \mathrm{Lip}(X,E)$ satisfy $c(f)\cap c(g)=\emptyset$.
Moreover, $T$ is said to be \emph{biseparating} if it is bijective and both $T$ and its inverse
are separating maps.

Obviously, if $f:X \ra E$ is Lipschitz and bounded,
then so  is the map $\vc f \vd : X \ra \mathbb{R}$ defined by $\vc f \vd (x) := \vc f(x) \vd$ for every $x \in X$.
 It is also clear that $\vc f \vd$ can be continuously extended to a Lipschitz function
$\widehat{\vc f \vd} : \widehat{X} \ra \mathbb{R}$
defined on  the completion $\widehat{X}$ of $X$.
More in general, if $x \in \widehat{X} \setminus X$,
we say that $f$ {\em admits an extension to $x$} if it can be continuously extended to a map
$\widehat{f} : X \cup \tl x \tr \ra E$. Clearly, when $E$ is complete and $X$ is not, $f$ admits
a continuous extension to the whole $\widehat{X}$, and the extension $\widehat{f}: \widehat{X} \ra E$
is a Lipschitz function with $\vc \widehat{f} \vd_{\infty} = \vc f \vd_{\infty}$
and $L \pl \widehat{f} \pr = L \pl f \pr$.
For this reason, when $E$ and $F$
are complete, every surjective linear isometry
$T:\mathrm{Lip}(X,E)\rightarrow \mathrm{Lip}(Y,F)$ can be associated in a canonical way to
another one $\widehat{T} :\mathrm{Lip} \pl \widehat{X},E \pr \rightarrow \mathrm{Lip} \pl \widehat{Y},F \pr $
(which coincides with $T$ only if $X$ and $Y$ are complete).

Given $R>0$, we  define in $X$ the following equivalence relation: we put $x \sim_R y$ if there exist
$x_1, \ldots , x_n \in X$ with $x=x_1$, $y=x_n$, and $d(x_i, x_{i+1} ) <R$ for $i=1, \ldots, n-1$.
We call $R$-component each of the equivalence classes of $X$ by $\sim_R$. The set of all $R$-components in $X$
is denoted by $\mathtt{Comp}_R (X)$.

We say that a bijective map $h: Y \ra X$ preserves distances less than $2$
 if $d_1 (h(y),h(y')) = d_2(y,y')$
 whenever $d_2(y,y') <2 $. We denote by $\iso$ the set of all maps $h: Y \ra X$ such that both $h$ and $h^{-1} $
preserve distances less than $2$. Notice that every $h \in \iso$ is  a homeomorphism  and that, when $X$ is bounded,  then  it is also a Lipschitz map (see also Remark~\ref{subidahoyo}).

\begin{definition}\label{martha}
Let $I(E,F)$ be the set of  all linear isometries from $E$ onto $F$.
 We say that a map $T:\mathrm{Lip}(X,E)\rightarrow \mathrm{Lip}(Y,F)$
is a {\em standard} isometry if there exist  $h \in \iso$
  and a map $J: Y \ra I(E,F)$ constant on each $2$-component of $Y$ such that
$$
Tf(y)= Jy (f(h(y)))
$$
for all $f\in \mathrm{Lip}(X,E)$ and $y\in Y$.
\end{definition}

\begin{remark}\label{scientiaedies}
Notice that a standard isometry is indeed a surjective linear isometry. Theorem~\ref{marthahoyo}
 gives a condition under which both classes of operators coincide.
Also, when $Y$ is $2$-connected, the map $J$ is constant, so there exists
a surjective linear isometry  $\mathbf{J} : E \ra F$   such that
$$
Tf(y)= \mathbf{J} (f(h(y)))
$$
for all $f\in \mathrm{Lip}(X,E)$ and $y\in Y$.
In particular, Corollary~\ref{siegediesrefe} roughly says  that this is the only way to obtain  an isometry when  one of the base spaces is
$1$-connected (see also Remark~\ref{guelgafunc}).
\end{remark}

In the definition of standard isometry, we see that $X$ and $Y$ are very much related. In particular, one is complete if and only if the other is. There  are interesting cases which are {\em almost} standard in some sense. For instance, when $E$ is complete, the natural inclusion
$\mathbf{i}_X : \mathrm{Lip}(X,E) \rightarrow \mathrm{Lip} \pl \widehat{X},E \pr $ is not standard if $X$ is not complete, but we immediately obtain a standard isometry from it in a natural way.

On the other hand, when $E$ and $F$ are   complete,  every $T$ can be
written as $T = {\mathbf{i}_Y}^{-1} \circ  \widehat{T} \circ \mathbf{i}_X$. In the following definition, we
distinguish between this kind of isometries
and
{\em nonstandard} isometries.

\begin{definition}\label{desk}
We say that a surjective linear isometry $T:\mathrm{Lip}(X,E)\rightarrow \mathrm{Lip}(Y,F)$ is
{\em nonstandard} if $T$ and $\widehat{T}$ (if it can be defined, that is, if $E$ and $F$ are complete)
are not standard.
\end{definition}

 Just a special family of spaces allows   defining properly nonstandard isometries. We call them spaces of
 {\em type} $\mathbf{A}$.

\begin{definition}\label{nikolabit}
We say that a metric space $X$ is of   type $\mathbf{A}$ if there are a partition of $X$ into two subsets
$\mathfrak{A}$, $\mathfrak{B}$, and a   map $\varphi : \mathfrak{A} \ra \mathfrak{B}$ with the following properties:
\begin{enumerate}
\item\label{cand} $d (x, z) = 1 + d(\varphi(x), z)$ whenever $x \in \mathfrak{A}$ and  $z \in \mathfrak{B}$ satisfy $d(\varphi(x),z) < 1$,
\item\label{ela} $d (x, z) \ge 2$ whenever $x \in \mathfrak{A}$ and  $z \in \mathfrak{B}$  satisfy $d(\varphi(x),z) \ge 1$,
\item $d (x_1, x_2 ) \ge 2 $ whenever $x_1, x_2 \in \mathfrak{A}$ and $\varphi(x_1) \neq \varphi (x_2)$.
\end{enumerate}
For each $E$, the operator  $S_{\varphi} : \cxe \ra \cxe$
defined, for each $f \in \cxe$,  by
\begin{displaymath}
S_{\varphi} f (x) := \left\{ \begin{array}{ll} f (x) & \mbox{if }  x \in \mathfrak{B}
\\
f(\varphi (x)) - f(x) & \mbox{if } x \in \mathfrak{A}
\end{array} \right.
\end{displaymath}
is said to be the {\em purely nonstandard} map associated to $\varphi$.
\end{definition}

\begin{remark}\label{gemacurruscos}
It is easy to check that $S_{\varphi}$ is   linear and bijective. Also
$\vc S_{\varphi} (f) \vd \le 1$ whenever $\vc f \vd \le 1$. Taking into account that
${S_{\varphi}}^{-1} = S_{\varphi}$, this implies that $S_{\varphi}$ is indeed a nonstandard isometry.
Theorem~\ref{nonelk} and Remark~\ref{sinpase} basically say that every nonstandard isometry
is the composition of a standard and a purely nonstandard one.
\end{remark}

Throughout, for each $\mathrm{e}\in E$, the constant function from $X$ into $E$ taking the value $\mathrm{e}$
will be denoted by $\tilde{\mathrm{e}}$.  Also, given a set $A$, $\chi_A$ stands
for the characteristic function on $A$.

As usual, if there is no  confusion both the metric of $X$ and that of $Y$ will
be denoted by $d$.

Given a surjective linear isometry $T:\mathrm{Lip}(X,E)\rightarrow \mathrm{Lip}(Y,F)$,
we denote
$$\mathfrak{A} (T) :=\{y\in Y: T  \tilde{\mathrm{e}} (y)=0 \ \forall  \mathrm{e}\in E\}$$
and
$$\mathfrak{B} (T) := \mathfrak{A} (T)^c.$$
The partition of $Y$ into these two subsets will be very much used in Sections~\ref{purusa} and~\ref{asucar},
and the fact that $\at$ is empty will turn out to be basically equivalent to $T$ being standard.
This property will receive a special name.
 We define Property {\bf P} as follows:
\begin{description}\label{tak}
\item[P] For each $y \in Y$, there exists $\mathrm{e} \in E$ with $T  \tilde{\mathrm{e}} (y) \neq 0$.
\end{description}

\section{Main results}

 We first give some results ensuring that an isometry is standard, and then characterize spaces and describe
the isometries when this is not the case.   Theorem~\ref{marthahoyo} and Corollary~\ref{bocadillo} are proved in
 Section~\ref{rrollo}, and Theorem~\ref{nonelk} in Section~\ref{asucar}.

It is obvious that, by definition, if $T$ is {\em not} nonstandard, then it satisfies Property {\bf P}.
 The converse is given by Theorem~\ref{marthahoyo} and Corollary~\ref{bocadillo}.

\begin{theorem}\label{marthahoyo}
Let  $T:\mathrm{Lip}(X,E)\rightarrow \mathrm{Lip}(Y,F)$ be a surjective linear isometry satisfying Property {\bf P}.
Then $E$ and $F$ are linearly isometric.

Furthermore, if we are in any of the  following two cases:
\begin{enumerate}
\item\label{zala} $X$ and $Y$ are complete,
\item\label{mea} $E$ (or $F$) is not complete,
\end{enumerate}
then $T$ is standard.
\end{theorem}

\begin{remark}\label{subidahoyo}
In Theorem~\ref{marthahoyo}, we cannot in general ensure that the map $h$
is an isometry or that it preserves distances equal to $2$.
Indeed,  following the same ideas as in \cite[Proposition 1.7.1]{W},
if $(Z, d)$ is a metric space with diameter  $\mathrm{diam}(Z, d) >2 $, then there is a new metric
$d' (\cdot, \cdot ) := \min \tl 2, d (\cdot, \cdot ) \tr$ on $Z$ with $\mathrm{diam}(Z, d') = 2$ such that $\mathrm{Lip}(Z, E)$ with respect to $d$
and $\mathrm{Lip}(Z, E)$ with respect to $d'$ are linearly isometric.
On the other hand, notice also that, if $d_1'$ and $d_2'$ are defined in a similar way, then the map
$h: (Y, d_2) \ra  (X, d_1) $ belongs to $\iso$  if and only if
 $h: (Y, d_2') \ra    (X, d_1')$ is an isometry.
\end{remark}

In Theorem~\ref{marthahoyo}, when (\ref{zala}) and (\ref{mea})  do not hold, $E$ and $F$
are complete and $X$ (or $Y$) is not.
In this case, it is easy to see that in general $T$ is not standard.
Nevertheless,
 we have the following result.

\begin{corollary}\label{bocadillo}
Suppose that $E$ and $F$
are complete and $X$ or $Y$ is not. If $T:\mathrm{Lip}(X,E)\rightarrow \mathrm{Lip}(Y,F)$ is a surjective linear isometry satisfying Property {\bf P}, then $\widehat{T} :\mathrm{Lip} \pl \widehat{X},E \pr \rightarrow \mathrm{Lip} \pl \widehat{Y},F \pr $ is standard.
\end{corollary}

 We next give the general form that a nonstandard isometry (or, equivalently, an isometry not satisfying Property {\bf P})
must take.

\begin{theorem}\label{nonelk}
Assume that we are in any of the  following two cases:
\begin{enumerate}
\item\label{kuris} $X$ and $Y$ are complete,
\item\label{masu} $E$ (or $F$) is not complete,
\end{enumerate}
Then there exists a nonstandard isometry $T:\mathrm{Lip}(X,E)\rightarrow \mathrm{Lip}(Y,F)$
if and only if
the following three conditions hold simultaneously:
\begin{enumerate}
\item $X$ and $Y$ are of type $\mathbf{A}$,
\item there exists $h \in \iso$,
\item $E$ and $F$ are linearly isometric.
\end{enumerate}

In this case, $T = S_{\varphi} \circ T'$, where $T' : \cxe \ra \cyf$ is a standard isometry   and
$S_{\varphi} : \cyf \ra \cyf$ is a purely nonstandard isometry.
\end{theorem}

\begin{remark}\label{sinpase}
In the case when $E$ and $F$
are complete and $X$ or $Y$ is not,  if $T:\mathrm{Lip}(X,E)\rightarrow \mathrm{Lip}(Y,F)$ is a nonstandard isometry, then so is $\widehat{T}$, and the description given in Theorem~\ref{nonelk} applies to $\widehat{T}$.
 In particular, $T = {\mathbf{i}_Y}^{-1} \circ  S_{\varphi} \circ T' \circ \mathbf{i}_X$
 where $T' : \cxeh \ra \cyfh$ is a standard isometry   and
$S_{\varphi} : \cyfh \ra \cyfh$ is purely nonstandard.
\end{remark}

A direct consequence (and easy to check) of Theorem~\ref{nonelk} and Remark~\ref{sinpase} is the following.

\begin{corollary}\label{deluxion}
If $E$ is not complete,  then there exists a nonstandard isometry from $\mathrm{Lip}(X,E)$ onto itself if and only if $X$ is of type $\mathbf{A}$. If $E$ is   complete,  then there exists a nonstandard isometry from $\mathrm{Lip}(X,E)$ onto itself if and only if $\widehat{X}$ is of type $\mathbf{A}$.
\end{corollary}

Theorem~\ref{nonelk} says that, under some assumptions, when two spaces of Lipschitz functions are linearly isometric, there
exists in fact a standard isometry between them. The following result is a simple consequence of Theorems~\ref{marthahoyo} and~\ref{nonelk}, Corollary~\ref{bocadillo} and Remark~\ref{sinpase}.

\begin{corollary}\label{rentreeflot}
$\mathrm{Lip}(X,E)$ and $\mathrm{Lip}(Y,F)$ are linearly isometric if and only if  $E$ and $F$ are linearly isometric
and
\begin{itemize}
\item $\iso$ is nonempty (when $E$ and $F$ are not complete).
\item $\isol$ is nonempty (when $E$ and $F$ are   complete).
\end{itemize}
\end{corollary}

We finally adapt the above results to the special case of metrics $d^{\alpha}$, $0 <\alpha <1$. Even if in this
case we just deal with metrics and, consequently, the general form of the isometries between spaces $\cxea$ is
completely given by Theorems~\ref{marthahoyo} and~\ref{nonelk},
Corollary~\ref{bocadillo} and Remark~\ref{sinpase}, it is interesting to see how  the condition of being of type $\mathbf{A}$
can be translated to metrics $d^{\alpha}$. This
 turns out to be more restrictive,
and constitutes a generalization of
the scalar case on compact spaces given in \cite[Theorem 3.3]{M}.

\begin{definition}\label{rebotapel}
Let $0< \alpha <1$. We say that a metric space $(X,d)$ is of {\em type $\mathbf{A}_{\alpha}$} if there are a partition of $X$ into two subsets
$\mathfrak{A}$, $\mathfrak{B}$, and a   map $\varphi : \mathfrak{A} \ra \mathfrak{B}$ with the following properties:
\begin{enumerate}
\item $d(x, \varphi (x)) =1$ for every $x \in \ah$,
\item $d^{\alpha} (x, z) \ge 2$ whenever $x \in \mathfrak{A}$ and  $z \in \mathfrak{B}$, $z \neq \varphi (x)$,
\item $d^{\alpha} (x_1, x_2 ) \ge 2 $ whenever $x_1, x_2 \in \mathfrak{A}$ and $\varphi(x_1) \neq \varphi (x_2)$.
\end{enumerate}
\end{definition}

\begin{proposition}\label{pesafrank}
Let $0< \alpha <1$ and let $(X, d)$ be a metric space.
The following two statements are equivalent:
\begin{enumerate}
\item $\pl X, d^{\alpha} \pr$ is of type $\mathbf{A}$,
\item  $\pl X, d \pr$ is of type $\mathbf{A}_{\alpha}$.
\end{enumerate}
\end{proposition}

Proposition~\ref{pesafrank} will be proved in Section~\ref{asucar}.

It is clear that, since
$X$ is of type $\mathbf{A}_{\alpha}$ if and only if its completion is, the statement of
Corollary~\ref{deluxion} is even simpler when dealing with $\cxea$.

Also, it is immediate to see that, if $(X, d)$ is of type $\mathbf{A}_{\alpha}$, then it is of type
$\mathbf{A}_{\beta}$ for $\alpha < \beta \le 1$. Consequently, by Theorem~\ref{nonelk} and Remark~\ref{sinpase},
we conclude the following.

\begin{corollary}\label{subyecta}
Let $0< \alpha <1$. If there exists a nonstandard isometry between $\cxea$ and $\cyfa$, then there exists
a nonstandard isometry between $\mathrm{Lip}^{\beta} (X, E)$ and $\mathrm{Lip}^{\beta} (Y, F)$ whenever
$\alpha < \beta \le 1$.
\end{corollary}

Obviously, the converse of Corollary~\ref{subyecta} is not true in general. The following example shows somehow the differences between  cases.
\begin{example}\label{retipnol}
Let $X := \{-1\} \cup (0,1) \subset \mathbb{R}$. $X$ is not of type $\mathbf{A}$,
but its completion $\widehat{X} = \{-1\} \cup [0,1]$ is. Neither $X$ nor $\widehat{X}$ are of type
$A_{\alpha}$ for $0 < \alpha <1$. Consequently, we have
\begin{itemize}
\item If $E$ is not complete, then all linear isometries from $\cxe$ onto itself are standard, but there are
nonstandard isometries from $\mathrm{Lip} \pl \widehat{X},E \pr$ onto itself.
\item If $E$ is  complete, then  there are nonstandard isometries from $\cxe$ onto itself. Obviously, by definition of nonstandard isometry, the same holds for
$\mathrm{Lip} \pl \widehat{X},E \pr$.
\item For every $E$ and $\alpha \in (0,1)$, all linear isometries from $\cxea$ onto itself are standard.
The same holds for $\mathrm{Lip}^{\alpha} \pl \widehat{X},E \pr$. In the case when $E$ is complete and $X$ is not, this is due to the special form of $X$.
\end{itemize}
\end{example}

\section{The case when $T$ satisfies Property {\bf P}}\label{rrollo}

In this section, unless otherwise stated, we assume that
$T$ is a linear isometry from
$\mathrm{Lip}(X,E)$ onto $\mathrm{Lip}(Y,F)$ satisfying Property {\bf P}.

Our first goal consists of showing that $T$ is indeed an isometry with respect to the norm $\vc \cdot \vd_{\infty}$. The following two lemmas will be the key tools used to prove it.

\begin{lemma}\label{paco}
Let $f \in \mathrm{Lip}(X,E)$ and $x_0 \in X$ be such that $f(x_0) \neq 0$.
Then there exists $g \in \mathrm{Lip}(X,E)$ with
$\vc g(x_0) \vd = \vc g \vd_{\infty} > L(g)$
such that
$$\vc g (x_0) \vd + \vc f (x_0) \vd = \vc (g + f) (x_0) \vd = \vc g + f \vd_{\infty} > L (g + f).$$
\end{lemma}

\begin{proof}
We put $\mathrm{e}:=f(x_{0})$ and assume without loss of generality that $\vc \mathrm{e} \vd =1$.
We then consider $l \in \mathrm{Lip}(X,E)$ defined by
$
l (x):= \max \{0,2-d(x,x_{0})\} \cdot \mathrm{e}
$
for each $x\in X$.
Clearly $l$ satisfies
$\vc l \vd_{L}=\vc l \vd_{\infty}=\vc l(x_{0}) \vd=2$ and $L(l)\leq 1$,
and also $\vc l(x) \vd< 2$ for all $x\in X$, $x\neq x_{0}$.

Take $n \in \mathbb{N}$ with $n > \vc f \vd_L$. Firstly, it is easy to check that
$\vc (nl +f)(x_{0}) \vd=2n+ 1$. On the other hand, we also have that
$L(nl+f)\leq nL(l)+L(f)< 2n$ and, for $x\in X \setminus \tl x_{0} \tr$ with $d(x, x_0) <2$,
\begin{eqnarray*}
\vc (nl+f)(x) \vd &\leq& n\vc l(x) \vd + \vc f(x) \vd \\
&\leq& 2n-nd(x,x_{0})+\vc f(x_{0}) \vd +  L(f) d(x,x_{0}) \\
&<& 2n+1,
\end{eqnarray*}
whereas if $d(x, x_0 ) \ge 2$, then $l(x)=0$ and
$\vc (nl+f)(x) \vd \leq  \vc f \vd_L < n$.

Consequently, if we define $g:= nl$, the lemma is proved.
\end{proof}

\begin{remark}\label{guelga}
It is easy to see that if $f_1 , f_2 \in \mathrm{Lip}(X,E)$ satisfy $\vc f_1 (x_0) \vd , \vc f_2 (x_0) \vd <  \vc f (x_0) \vd$, then the proof of Lemma~\ref{paco} can be slightly modified (by taking
$n >  \vc f \vd_L, \vc f_1 \vd_L , \vc f_2 \vd_L $) so   that
$\vc g + f_i \vd_{\infty} < \vc g + f \vd_{\infty}$ for $i=1, 2$.
\end{remark}

\begin{lemma}\label{1750}
If $f \in \mathrm{Lip}(X,E)$ satisfies $\vc Tf (y_0) \vd = \vc Tf \vd_{\infty} > L( Tf )$
for some $y_0 \in Y$, then $L(f) \le \vc f \vd_{\infty}$.
\end{lemma}

\begin{proof}
Suppose that $\vc  f \vd_{\infty} < L(f)$. Then,
for each $\mathrm{e} \in E$, there exists $M>0$ such that
$\vc f \vd_{\infty} + M \vc \mathrm{e} \vd < L(f) $,
so $\vc f \pm M \tilde{\mathrm{e}} \vd_{\infty}
< L(f) = L(f \pm M \tilde{\mathrm{e}})$.
 Therefore,
$$
\vc f \pm M \tilde{\mathrm{e}} \vd_{L}=
L(f \pm M \tilde{\mathrm{e}})=L(f)=\vc f \vd_{L}.
$$
Since $T$ is an isometry,
$\vc Tf \pm M T\tilde{\mathrm{e}} \vd_{L}=\vc Tf \vd_{L} = \vc Tf (y_0)  \vd$,
which implies in particular that
$\vc Tf (y_0)  \pm M T\tilde{\mathrm{e}}(y_{0}) \vd\leq\vc Tf (y_{0}) \vd$ and, by Inequality~(\ref{strictly}),
 that $T \tilde{\mathrm{e}} (y_0) =0$ for every $\mathrm{e} \in E$,
which  goes against our hypotheses.
\end{proof}

\begin{remark}\label{hirche}
Notice that, in the proof of Lemma~\ref{1750}, we just use the fact that there exists $\mathrm{e} \in E$ with $T  \tilde{\mathrm{e}} (y_0) \neq 0$, and not the general assumption that Property {\bf P} holds.
\end{remark}

\begin{corollary}\label{mariusz}
$T$ is an isometry with respect to the supremum norm.
\end{corollary}

\begin{proof}
Assume that $\vc f \vd_{\infty} < \vc Tf \vd_{\infty}$, and pick $\epsilon >0$ and $y_0 \in Y$
such that $\vc f \vd_{\infty} + \epsilon < \vc Tf (y_0) \vd$. Next, by Lemma~\ref{paco}, we can take
$g \in \mathrm{Lip}(Y,F)$ with $\vc g(y_0) \vd = \vc g \vd_{\infty} > L(g)$ and such that
$$\vc g (y_0) \vd + \vc Tf (y_0) \vd = \vc (g + Tf) (y_0) \vd = \vc g + Tf \vd_{\infty} > L (g + Tf).$$
Applying Lemma \ref{1750}, we conclude both that
$$L(T^{-1} g) \le \vc  T^{-1} g \vd_{\infty} = \vc g (y_0) \vd$$ and
 that $$L(T^{-1} g+f) \le \vc T^{-1} g+f  \vd_{\infty} = \vc g + Tf \vd_L = \vc (g + Tf) (y_0) \vd .$$
  But this is impossible because
$$
\vc T^{-1} g+ f  \vd_{\infty} < \vc g (y_0) \vd+\vc Tf (y_0) \vd - \epsilon < \vc (g + Tf) (y_0) \vd.
$$
We conclude that $\vc Tf \vd_{\infty} \le \vc f \vd_{\infty}$ for every $f$.
We next see that $T^{-1}$ also satisfies Property {\bf P}, which is enough to prove the equality. By the above, given a nonzero $\mathrm{f} \in F$
we have  $\vc T^{-1} \tilde{\mathrm{f}} \vd_{\infty} = \vc \mathrm{f} \vd$. Also,
 if $\pl  T^{-1} \tilde{\mathrm{f}} \pr (x_0) =0$ for some $x_0 \in X$, then there exists
$k \in \mathrm{Lip}(X,E)$, $k \neq 0$, with $\vc k (x) \vd + \vc  \pl T^{-1} \tilde{\mathrm{f}} \pr (x) \vd \le \vc \mathrm{f} \vd$
for every $x \in X$. By Inequality~(\ref{strictly}), $\vc \mathrm{f} \vd < \vc \tilde{\mathrm{f}} + T k \vd_{\infty}$
or $\vc \mathrm{f} \vd < \vc \tilde{\mathrm{f}} - T k \vd_{\infty}$, which contradicts the paragraph above.
\end{proof}

\begin{remark}\label{hadela}
Notice that, in the proof of Corollary~\ref{mariusz}, we have  seen that  $T^{-1}$ also satisfies Property {\bf P}.
\end{remark}

We are   now ready to see that, under the assumptions we make in this section, every surjective
linear isometry
 is biseparating.

\begin{proposition}\label{bis}
$T $ is biseparating.
\end{proposition}

\begin{proof}
We prove that $T$ is separating. Suppose that it is not, so
there exist $f,g\in \mathrm{Lip}(X,E)$
such that $c(f)\cap c(g)=\emptyset$ but $Tf(y_{0})=\mathrm{f}_1\neq 0$ and $Tg(y_{0})=\mathrm{f}_2\neq 0$
for some $y_{0}\in Y$. Taking into account Inequality~(\ref{strictly}), we can assume without loss of
generality that $\vc \mathrm{f}_2 \vd \leq  \vc \mathrm{f}_1 \vd < \vc \mathrm{f}_1+ \mathrm{f}_2 \vd$.
Now,
by Lemma~\ref{paco} and Remark~\ref{guelga}, there exists $k \in \mathrm{Lip}(Y,F)$ such that
$\vc k + Tf \vd_{\infty} , \vc k + Tg \vd_{\infty} < \vc k + Tf + Tg \vd_{\infty}$.

On the other hand,
since $f$ and $g$
have disjoint cozeros,
$$
\vc T^{-1} k+ f+ g \vd_{\infty}=\max \tl \vc T^{-1} k+ f \vd_{\infty},\vc T^{-1} k+  g \vd_{\infty} \tr,
$$
and consequently
$
\vc k+Tf+Tg \vd_{\infty}=\max\tl \vc k+Tf \vd_{\infty},\vc k+Tg \vd_{\infty} \tr
$,
which is a contradiction.

By Remark~\ref{hadela}, $T^{-1}$ is also separating.
\end{proof}

\begin{remark}\label{pinbo}
In \cite[Theorem 3.1]{AD} (see also comments after it)  a description of biseparating maps
$S: \mathrm{Lip}(X,E)\rightarrow \mathrm{Lip}(Y,F)$ is given, but we cannot use it here because
assumptions of completeness on $X$ and $Y$ are made in \cite{AD}. Under some circumstances automatic continuity
of such $S$ can be achieved and, in that case, the description goes  as follows
 (where $\mathcal{L}(E,F)$ denotes
 the normed space of all
linear and continuous operators from $E$ to $F$):
There exist a  homeomorphism $k: Y \rightarrow X$
and  a   map   $K : Y \rightarrow \mathcal{L}(E, F)$
(which is easily seen to be  also Lipschitz with
$L(K) \le \vc S \vd$) such that
$
Sf(y)=Ky (f(k(y)))
$
for all $f\in \mathrm{Lip}(X,E)$ and $y\in Y$. Also, if both $X$ and $Y$ are bounded, then the map $k$
is bi-Lipschitz.
\end{remark}

\begin{proposition}\label{nuevaescocia}
Given $ \mathrm{e}\in E$, $T \tilde{\mathrm{e}}$ is constant on each $1$-component of $Y$ and
$\vc T \tilde{\mathrm{e}} (y) \vd = \vc \mathrm{e} \vd$ for all $y \in Y$.
\end{proposition}

\begin{proof}
Suppose that this is not the case, but there exist $\mathrm{e}  \in E$, $\vc \mathrm{e}  \vd =1$,
and $y_1, y_2 \in Y$ with $y_1 \sim_1 y_2$, such that $\mathrm{f}_1 := T \tilde{\mathrm{e}} (y_1)$
and $\mathrm{f}_2 := T \tilde{\mathrm{e}} (y_2)$ are different. Of course,
 we may assume without loss of generality that $D:= d(y_1, y_2) <1$
 and that
$ \mathrm{f}_1  \neq 0$. Now, if we  consider $g \in \mathrm{Lip}(Y,F)$ defined by
$$
g ( y):= \max\tl 0,1 - \frac{d(y, y_1 )}{D} \tr \cdot \mathrm{f}_1
$$
for all $y\in Y$, then obviously
$\vc g \vd_L = L (g) =  \vc \mathrm{f}_1 \vd / D > \vc g \vd_{\infty}$.
As a consequence, using Corollary~\ref{mariusz}, $\vc T^{-1} g \vd_L > \vc T^{-1} g \vd_{\infty}$,
and we can take $M>0$ such that
$\vc   T^{-1} g \pm M \tilde{\mathrm{e}} \vd_L = \vc T^{-1} g \vd_L = \vc \mathrm{f}_1 \vd /D$.

Notice also that, as $F$ is strictly convex, either
$\vc \mathrm{f}_1 + M \pl \mathrm{f}_1 - \mathrm{f}_2 \pr  \vd  > \vc \mathrm{f}_1 \vd $
or
$\vc \mathrm{f}_1 - M \pl \mathrm{f}_1 - \mathrm{f}_2 \pr  \vd  > \vc \mathrm{f}_1 \vd $,
that is, either
$$\frac{\vc \pl g + M T \tilde{\mathrm{e}} \pr   (y_1) -
\pl g + M T \tilde{\mathrm{e}} \pr   (y_2)\vd }{d(y_1, y_2)} > \frac{\vc \mathrm{f}_1 \vd}{D}$$
or
$$\frac{\vc   \pl g - M T \tilde{\mathrm{e}} \pr   (y_1) -
\pl g - M T \tilde{\mathrm{e}} \pr   (y_2)\vd }{d(y_1, y_2)} > \frac{\vc \mathrm{f}_1 \vd}{D},$$
which implies that either
$\vc g + M T \tilde{\mathrm{e}}  \vd_L > \vc \mathrm{f}_1 \vd /D$
or
$\vc g - M T \tilde{\mathrm{e}} \vd_L > \vc \mathrm{f}_1 \vd /D$,
yielding a contradiction.

Finally suppose that $T \tilde{\mathrm{e}} (y) = \mathrm{f}$ for all $y $ in  $B \in \mathtt{Comp}_1 (Y)$.
Since  by Proposition~\ref{bis} $T^{-1}$ is separating,
$c \pl T^{-1} \pl \chi_B \cdot T \tilde{\mathrm{e}} \pr \pr \cap c \pl T^{-1} \pl \chi_{Y \setminus B} \cdot T \tilde{\mathrm{e}} \pr \pr = \emptyset$. This implies that $\mathrm{e}$ is
the only nonzero  value taken by $T^{-1} \pl \chi_B \cdot T \tilde{\mathrm{e}} \pr$ and, since $T^{-1}$ is an isometry with respect to $\vc \cdot \vd_{\infty}$, we have that $\vc \mathrm{e} \vd = \vc \mathrm{f} \vd$.
\end{proof}

\begin{lemma}\label{lavaderomartha}
There exists a bijection $H : \mathtt{Comp}_1 (X) \ra \mathtt{Comp}_1 (Y)$ and, for each $A \in \mathtt{Comp}_1 (X)$,
a surjective linear isometry $\mathbf{J}_A : E \ra F$ with the property
that
$T \pl \chi_A \cdot \tilde{\mathrm{e}} \pr = \chi_{H(A)} \cdot \widetilde{\mathbf{J}_A (\mathrm{e})} = \chi_{H(A)} \cdot T    \tilde{\mathrm{e}}   $ for every
$\mathrm{e} \in E$.
\end{lemma}

\begin{proof}
Fix $A \in \mathtt{Comp}_1 (X)$ and $\mathrm{e} \in E$ with $\vc \mathrm{e} \vd =1$, and take
 $f := \chi_A \cdot \tilde{\mathrm{e}}$, $g := \chi_{X \setminus A} \cdot \tilde{\mathrm{e}}$ in $\mathrm{Lip}(X,E)$.
We have that $c(f) \cap c(g) = \emptyset$,  so by Proposition~\ref{bis} $Tf$ and $Tg$ have disjoint cozeros.
Then, by Proposition~\ref{nuevaescocia}, $Tf(y), Tg(y) \in \tl 0, T \tilde{\mathrm{e}} (y) \tr$
for all $y  \in Y$.
Now, suppose that $y \sim_1 y'$,
and that $Tf(y) \neq 0$ and
$Tf (y') =0$. We
 can assume that $d(y, y') <1$.
Since $\vc Tf (y) \vd =1$, we deduce $ L(Tf) \ge 1/ d(y,y') >1$, which
is impossible because $\vc f \vd_L =1$.

Reasoning similarly with $T^{-1}$,
$Tf = \chi_B \cdot \tilde{\mathrm{f}}$
for some $1$-component $B$ in $Y$ and some norm-one vector $\mathrm{f} \in F$. The conclusion is now easy.
\end{proof}

\begin{lemma}\label{stomas}
Given $A, B \in \mathtt{Comp}_1 (X)$, if $\min \tl d(A, B) , d(H(A), H(B)) \tr <2$, then
$d(A, B) = d(H(A), H(B))$ and $\mathbf{J}_A   = \mathbf{J}_B$.
\end{lemma}

\begin{proof}
Put $D_1 := d  (A, B)$, $ D_2 := d  (H(A), H(B))$.
 Due to the symmetric r\^{o}les
of $H$ and $H^{-1}$  with respect to $T$ and $T^{-1}$, we can assume without loss of generality that
$D_1 \le D_2$.
Pick $\mathrm{e} \in E$ with $\vc \mathrm{e} \vd =1$, and define
 $ f := \pl \chi_A  - \chi_B \pr \cdot \tilde{\mathrm{e}} \in \mathrm{Lip}(X,E)$. We easily see that
 $\vc f \vd_L = L(f) = 2/D_1$  and, since
$L (Tf) = \vc \mathbf{J}_A (\mathrm{e} ) + \mathbf{J}_B (\mathrm{e} ) \vd  / D_2
\le 2/D_2$,
we necessarily have    $D_1 = D_2$ and
$\vc \mathbf{J}_A (\mathrm{e} ) + \mathbf{J}_B (\mathrm{e} ) \vd =2$, so $\mathbf{J}_A (\mathrm{e}  ) = \mathbf{J}_B (\mathrm{e} )$ because $F$ is strictly convex.
\end{proof}

\begin{corollary}\label{borjas}
There exists a map $J: Y \ra I(E, F)$ which is constant on each $2$-component of $Y$ and such that
$T \tilde{\mathrm{e}} (y) = J y (\mathrm{e})$ for all $\mathrm{e} \in E$ and $y \in Y$.
\end{corollary}

\begin{proof}
We define $Jy:= \mathbf{J}_A $ if $y \in H(A)$ and $A \in \mathtt{Comp}_1 (X)$.
Applying Lemma~\ref{stomas},
 the result follows.
\end{proof}

\begin{lemma}\label{gatosinchicle}
Let $(y_n)$ be a  sequence in $Y$ which is not a Cauchy sequence and such that
all $y_n$ are pairwise different.
Then there exist infinite   subsets $A_1$ and $A_2$
of $\tl y_n : n \in \mathbb{N} \tr$ with  $d \pl A_1, A_2 \pr >0$.
\end{lemma}

\begin{proof}
Taking a subsequence if
necessary,
we have that there exists $\epsilon>0$ such that
 $d (y_{2n} , y_{2n-1} ) \ge 3 \epsilon $ for all
$n  \in \mathbb{N}$. Let $  A:= \tl y_n : n \in \mathbb{N} \tr$. Now we have two possibilities: either there exists $n_0$ such that
$B \pl y_{n_0} ,  \epsilon \pr$ contains infinitely many $y_n$ or $A \cap B \pl y_k ,  \epsilon \pr$ is
 finite for every $k$. In the first case, it is clear that $A_1 := A \cap B \pl y_{n_0} ,  \epsilon \pr$
 and $A_2 := \tl y_{2n} : y_{2n -1} \in A_1 \tr \cup \tl y_{2n-1} : y_{2n} \in A_1 \tr$ satisfy
$d(A_1, A_2) \ge \epsilon$. In the second case,   we can find a subsequence $\pl y_{n_k} \pr$
with $d \pl y_{n_k}, y_{n_l} \pr > \epsilon$ when $k \neq l$, and the result follows easily.
\end{proof}

 In Lemma~\ref{uberrimo} and Corollary~\ref{estela} we do not necessarily assume that base spaces are not complete, so it could be the case that $\widehat{X} =X$ and $\widehat{Y} =Y$.

\begin{lemma}\label{uberrimo}
Given
$x_0 \in \widehat{X}$, there exists
$y_0 \in \widehat{Y}$ such that $\widehat{ Tf } (y_0) = 0$ whenever $f \in \mathrm{Lip}(X,E)$ satisfies
$\widehat{ f } (x_0) =0$.
\end{lemma}

\begin{proof}
Fix $\mathrm{e} \in E$ with $\vc \mathrm{e} \vd =1$, and let
$$A := \tl f \cdot \tilde{\mathrm{e}} : f \in \mathrm{Lip}(\widehat{X}) , \ f (x_0) =1 , \ \forall \epsilon > 0 \sup_{d(x , x_0) \ge \epsilon} \va f(x) \vb <1   \tr .$$
We will see that there exists a unique point $y_0 \in \widehat{Y}$ such that
 $\widehat{\vc Tf \vd} (y_0) =1$ for every $f \in A$.

Fix $f_0 \in A$. By Corollary~\ref{mariusz}, taking into account that $\vc f_0 \vd_{\infty} =1$,
there exists a sequence $(y_n)$ in $Y$ such that $\vc Tf_0 (y_n) \vd \ge 1 - 1/n$ for each $n \in \mathbb{N}$.
Let us see that it is a Cauchy sequence. Suppose that this is not the case. Either if all $y_n$ are pairwise different (by using Lemma~\ref{gatosinchicle})
 or not, we see that  there exist   subsets $A_1, A_2$ of $\tl y_n : n \in \mathbb{N} \tr$ such that
$d(A_1, A_2) >0$ and  $\sup_{y_n \in A_i} \vc Tf_0 (y_n) \vd =1$, $i=1,2$. Then we take
$g_1, g_2 \in \mathrm{Lip}(\widehat{Y})$ with  $0 \le g_1, g_2 \le 1$ such that
$g_1 (A_1) = 1$, $g_2  (A_2) = 1$, and $g_1 g_2 \equiv 0$. It is immediate that $\vc Tf_0 + g_i Tf_0 \vd_{\infty}  =2$
for $i=1,2$.
 Since, again by Corollary~\ref{mariusz},
 $\vc T^{-1} \pl g_i Tf_0 \pr \vd_{\infty} =1$,
 we
deduce that $ \widehat{\vc  T^{-1} \pl g_i Tf_0 \pr  \vd} (x_0) =1$ for $i=1,2$, which goes against the fact that
$T^{-1}$ is separating. Consequently $(y_n)$ is a Cauchy sequence and converges to a point $y_0 \in \widehat{Y}$,
which obviously satisfies $\widehat{\vc Tf_0 \vd} (y_0) =1$. Now it is straightforward to see that
$\widehat{\vc Tf \vd} (y_0) =1$ for every $f \in A$.

Next suppose that $f \in \mathrm{Lip}(X, E)$ satisfies
 $\widehat{f} (x_0)=0$. Then, given $\epsilon >0$, there exists
$f_{\epsilon} \in \mathrm{Lip}(X, E)$ such that $\widehat{f_{\epsilon}} \equiv 0$
 on a neighborhood of $x_0$ and
$\vc f - f_{\epsilon} \vd_{\infty} \le \epsilon$. We can  take $f_{\epsilon}' \in A$ with
 $c(f_{\epsilon}') \cap c(f_{\epsilon}) = \emptyset$, and we
deduce from the paragraph above that  $\widehat{\vc Tf_{\epsilon} \vd} \equiv 0$ on a neighborhood of $y_0$ in $\widehat{Y}$; in
particular $\widehat{\vc Tf_{\epsilon}  \vd }  (y_0) =0$. Since $\vc T f -  T f_{\epsilon} \vd_{\infty} \le \epsilon$ (by Corollary~\ref{mariusz}),
 we conclude that $ \widehat{ \vc Tf   \vd} (y_0) \le \epsilon$, and we are done.
\end{proof}

\begin{corollary}\label{estela}
There exists a bijective map $h : \widehat{Y} \ra \widehat{X}$ such that
 $ Tf   (y) = J y (\widehat{f } (h (y)))$
whenever  $y \in Y$
and $f \in \mathrm{Lip}(X,E)$ admits a continuous extension to $h(y)$.
\end{corollary}

\begin{proof}
Let $x_0$ and $y_0$ be as in Lemma~\ref{uberrimo}. Since $T^{-1}$ is also biseparating,
there exists $x_1 \in \widehat{X}$ such that $ \widehat{ f } (x_1) =0$ whenever $\widehat{ Tf } (y_0) = 0$ and,
in particular, whenever $ \widehat{ f } (x_0) =0$. Now, as $\mathrm{Lip}(\widehat{X}, E)$   separates points
in $\widehat{X}$, we deduce that $x_1 = x_0$.
As a consequence, it is straightforward to see that Lemma~\ref{uberrimo} gives us a bijective map between
 $\widehat{X}$ and $\widehat{Y}$, which we denote by $h: \widehat{Y} \ra \widehat{X}$, satisfying
 $\widehat{ Tf  } (y) =0$  if and only if  $\widehat{ f  } (h(y)) =0$.
Finally,
if $f \in \mathrm{Lip}(X,E)$ can be continuously extended to $h(y)$, say
$\widehat{f } (h(y)) = \mathrm{e} \in E$, then
$\widehat{\pl f - \tilde{\mathrm{e}}  \pr } (h(y)) =0$,
and the representation follows from
Corollary~\ref{borjas}.
\end{proof}

\begin{remark}\label{pliniuz}
As in the proof of Corollary~\ref{estela}, the bijection $k: \widehat{X} \ra \widehat{Y}$ associated to $T^{-1}$ satisfies $\widehat{ T^{-1} g  } (x) =0$  if and only if  $\widehat{ g  } (k(x)) =0$,
$g \in \mathrm{Lip}(Y,F)$. This implies that $k = h^{-1}$.

\end{remark}

\begin{lemma}\label{ester}
If $E$ is not complete, then there exists a sequence $(\mathrm{e}_n)$ in $E$
with $\vc \mathrm{e}_n \vd \le 1/4^n$ such that $\sum_{n=1}^{\infty} \mathrm{e}_n$ does not converge in $E$.
\end{lemma}

\begin{proof}
Clearly, there exists a nonconvergent sequence $(\mathrm{u}_n)$ in $E$ satisfying
$\vc \mathrm{u}_n - \mathrm{u}_{n+1} \vd \le 1/4^n$ for every
 $n \in \mathbb{N}$. It is then easy to check that it is enough to define
$ \mathrm{e}_n := \mathrm{u}_n - \mathrm{u}_{n+1}$ for each $n$.
\end{proof}

\begin{corollary}\label{sara}
If $E$  is not complete,
then the map $h$ given in
Corollary~\ref{estela} is a bijection from $Y$ onto $X$.
\end{corollary}

\begin{proof}
We will prove  first that $h(y) \in X$ whenever $y \in Y$. If this is not the case,
then take $y \in Y$ with $h(y) \in \widehat{X} \setminus X$. For each $n \in \mathbb{N}$, let
$$
f_n (x) := \max \tl 0,1 - 2^n d \pl x,   h(y) \pr   \tr
$$
for all $x \in   X$. It is clear that each $f_n$ belongs to
$\mathrm{Lip}(X)$
and that $L(f_n) \le  2^n$. It is easy to see that, since $\mathrm{Lip}(X,\widehat{E})$ is complete, if we take $(\mathrm{e}_n)$ in $E$ as in Lemma~\ref{ester}, then
$f:= \sum_{n=1}^{\infty} f_n \cdot \widetilde{\mathrm{e}_n}$ belongs to
$\mathrm{Lip}(X,\widehat{E})$, and since all values are taken in $E$, to  $\mathrm{Lip}(X,E)$.
Thus, by Corollary~\ref{mariusz},
$$
\lim_{k \ra \infty} \vc Tf - \sum_{n=1}^k T \pl f_n \cdot  \widetilde{\mathrm{e}_n} \pr \vd_{\infty}=0.
$$
Finally, by Corollary~\ref{estela}, this implies that
$Tf (y) = \sum_{n=1}^{\infty} J y (\mathrm{e}_n)$, which belongs to $\widehat{F} \setminus  F$,
and $Tf$ takes values outside $F$, which is absurd.

We  deduce  from Remark~\ref{pliniuz} that $h(Y) =X$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{marthahoyo}]
Taking into account Corollaries~\ref{borjas}, \ref{estela} and \ref{sara}, it is enough  to show that $h \in \iso$. Let $y_1, y_2 \in Y $ be such that $d(y_1, y_2) < 2$.
We are going to see that $D:= d(h(y_1), h(y_2)) \le d(y_1, y_2)$.

Pick $\mathrm{e} \in E$ with $\vc \mathrm{e} \vd=1$ and define $g \in \mathrm{Lip}(X,E)$ by $$
g ( x):= \max\tl -1 ,1 - \frac{ 2 \ d(x, h(y_1) )}{D} \tr \cdot \mathrm{e}
$$
for every $x \in X$. We have that $\vc g \vd_{\infty}=1$, $L(g) = 2/D$, $ g (h(y_1)) = \mathrm{e} $,
and $ g (h(y_2)) = - \mathrm{e}$. Obviously, by Corollary~\ref{borjas}, $J y_1 = J y_2$, and
$$
1 < \frac{2}{d(y_1, y_2)} =  \frac{\vc J y_1 (\mathrm{e})  - J y_2 (- \mathrm{e}) \vd}{d(y_1, y_2)}
 =   \frac{\vc T g (y_1) - T g (y_2) \vd}{d(y_1, y_2)} \le \vc Tg \vd_L,
$$
which implies that $\vc g \vd_L >1$, and then  $ \vc g \vd_L = L(g) = 2/D $. This means that
$\vc Tg \vd_L =2/D$, and consequently
$
2/d(y_1, y_2)  \le  2/D
$.
The other inequality can be seen in a similar way working with $T^{-1}$ (see Remark~\ref{pliniuz}).
\end{proof}

\begin{proof}[Proof of Corollary~\ref{bocadillo}]
The fact that $\widehat{T}$ satisfies Property {\bf P} follows easily from Proposition~\ref{nuevaescocia}. The conclusion is then immediate by Theorem~\ref{marthahoyo}.
\end{proof}

\section{The distance between $\at$ and $\bt$}\label{purusa}

Propositions~\ref{siegediesref} and~\ref{floater} will be used in Section~\ref{asucar}.

\begin{proposition}\label{siegediesref}
Let $T:\mathrm{Lip}(X,E)\rightarrow \mathrm{Lip}(Y,F)$ be a surjective linear isometry. If $ \mathfrak{A} (T) \neq \emptyset$, then $d \pl \mathfrak{A} (T) , \mathfrak{B} (T) \pr \ge 1$.
\end{proposition}

\begin{proof}
Obviously $ \mathfrak{B} (T) \neq \emptyset$. Suppose first that $d  ( \mathfrak{A} (T),  \mathfrak{B} (T))<1$, and  take
$y_{0}\in  \mathfrak{B} (T)$ and $\epsilon >0$ with
$d  (y_{0},  \mathfrak{A} (T))<1-2\epsilon$.
We then  select $\mathrm{f}\in F$, $\vc \mathrm{f} \vd=1$, and define  $ l \in \mathrm{Lip}(Y,F)$ by
$
l(y):= \max\{0,2-d(y,y_{0})\}\cdot \mathrm{f}
$
for every $y\in Y$. We have that
$\vc l \vd_{L}=\vc l \vd_{\infty}=\vc l(y_{0}) \vd=2$, $L(l)\leq 1$,
and  $\vc l(y) \vd< 2$ for all $y\in Y \setminus \tl y_0 \tr$.

Now, by Lemma~\ref{1750} (see also Remark~\ref{hirche}), we have that $L(T^{-1}l) \le \vc T^{-1}l \vd_{\infty}$.
Consequently $\vc T^{-1}l \vd_{L}=\vc T^{-1}l \vd_{\infty}$, and then
$\vc T^{-1}l \vd_{\infty}=2$. Therefore,  there is a point $x_{0}$ in $X$ such that
$
\vc T^{-1}l(x_{0}) \vd> 2-\epsilon
$,
that is, $T^{-1}l(x_{0})=\alpha \mathrm{e}$ for some  $\mathrm{e}\in E$, $\vc \mathrm{e} \vd=1$,
and $\alpha\in \mathbb{R}$, $\alpha>2-\epsilon$. Next, obviously
$$
\vc \tilde{\mathrm{e}}+T^{-1}l \vd_{L}
\geq \vc \mathrm{e}+T^{-1}l(x_{0}) \vd
= \vc \pl 1 + \alpha  \pr \mathrm{e} \vd
 > 3- \epsilon,
$$
so $\vc T\tilde{\mathrm{e}}+ l \vd_{L} > 3-\epsilon$.
Since $L(T\tilde{\mathrm{e}}+ l)\leq L(T\tilde{\mathrm{e}})+L(l)\leq 2$,
this implies that $\vc T\tilde{\mathrm{e}}+ l \vd_{\infty} > 3-\epsilon$, and hence
  the set
$B := \tl y \in Y :  \vc \pl T \tilde{\mathrm{e}}+ l \pr (y) \vd > 3-\epsilon \tr$ is nonempty.

Notice that, since
$\vc T \tilde{\mathrm{e}} \vd_{\infty}  \le 1$,
all points $y \in B$ must satisty $\vc l(y) \vd >2-\epsilon$, which is equivalent to $d(y,y_{0})<\epsilon$.
Thus, for some $y_{1}$ with $d(y_{1}, y_{0})  < \epsilon$,
we have $\vc T\tilde{\mathrm{e}}(y_{1})+ l(y_{1}) \vd > 3-\epsilon$, which implies that
$\vc T\tilde{\mathrm{e}}(y_{1})\vd > 1-\epsilon$.
On the other hand, taking into account that $d  (y_{0}, \mathfrak{A} (T) )<1-2\epsilon$, there exists $y_{2}\in \mathfrak{A} (T) $ with
$d(y_{0},y_{2})\leq 1-2\epsilon$. Finally, observe that
$$
\frac{\vc T\tilde{\mathrm{e}}(y_{1})-T\tilde{\mathrm{e}}(y_{2})\vd}{d(y_{1},y_{2})}
=\frac{\vc T\tilde{\mathrm{e}}(y_{1})\vd}{d(y_{1},y_{2})}
>\frac{1-\epsilon}{1-2\epsilon+\epsilon}=1,
$$
which allows us to conclude that $L(T\tilde{\mathrm{e}})>1$, in contradiction with
the fact that $\vc \mathrm{e} \vd=1$ and $T$ is an isometry.
\end{proof}

\begin{proposition}\label{floater}
Let $T:\mathrm{Lip}(X,E)\rightarrow \mathrm{Lip}(Y,F)$ be a surjective linear isometry. If $ y_0 \in \mathfrak{A} (T) $, then $d \pl y_0 , \mathfrak{B} (T) \pr = 1$.
\end{proposition}

\begin{proof}
Suppose on the contrary  that  there exists
$s \in (0, 1)$ such that $d \pl B (y_0, s) , \mathfrak{B} (T) \pr > 1+s$. Take
 $f \in \mathrm{Lip}(Y)$ with $c(f) \subset B(y_0, s)$ and such that $0 \le f \le s$, $f(y_0) =s$, and
$L (f) \le 1$. Let $\mathrm{e} \in E$ and $\mathrm{f} \in F$ have norm $1$.
It is easy to check that
$\vc f \cdot  \mathrm{f} \pm   T\tilde{\mathrm{e}} \vd \le 1$, whereas, since
$T^{-1} \pl f \cdot  \mathrm{f} \pr \neq 0$, Inequality~(\ref{strictly}) implies that
$$\vc T^{-1} \pl f \cdot  \mathrm{f} \pr + \tilde{\mathrm{e}} \vd_{\infty} >1$$ or
$$\vc T^{-1} \pl f \cdot  \mathrm{f} \pr - \tilde{\mathrm{e}} \vd_{\infty} >1 ,$$
contradicting the fact that $T$ is an isometry.
\end{proof}

We next see that Property {\bf P}
holds when $Y$
is $1$-connected.  Obviously, the same result holds if $X$ is $1$-connected (see Remark~\ref{hadela}).

\begin{corollary}\label{siegediesrefe}
 Let $Y$  be $1$-connected and suppose that $\mathrm{Lip}(X,E) $ and
$ \mathrm{Lip}(Y,F)$ are linearly isometric. Then $X$ is also $1$-connected and
every surjective linear isometry $T:\mathrm{Lip}(X,E)\rightarrow \mathrm{Lip}(Y,F)$
satisfies Property {\bf P}.
\end{corollary}

\begin{proof}
By Proposition~\ref{siegediesref}, Property {\bf P}
holds when $Y$
is $1$-connected.

The fact that $X$ is $1$-connected can be easily deduced from the representation of $T$ in
Theorem~\ref{marthahoyo} or that of $\widehat{T}$ in Corollary~\ref{bocadillo} (taking into account that
a metric space is $1$-connected if and only if so is its completion).
\end{proof}

\begin{remark}\label{guelgafunc}
An immediate consequence of Corollary~\ref{siegediesrefe} is that, when $X$ (or $Y$) is $1$-connected, every surjective linear isometry $T:\mathrm{Lip}(X,E)\rightarrow \mathrm{Lip}(Y,F)$
is standard in any of the cases (\ref{zala}), (\ref{mea}),  given in Theorem~\ref{marthahoyo}.
\end{remark}

\section{The case when $T$ does not satisfy Property {\bf P}}\label{asucar}

In this section, unless otherwise stated, we assume that
$T$ is a linear isometry from
$\mathrm{Lip}(X,E)$ onto $\mathrm{Lip}(Y,F)$ that {\em does not} satisfy Property {\bf P} (that is, $\mathfrak{A} (T)  \neq \emptyset$). We will make use of Theorem~\ref{marthahoyo}, so we  also assume  that
 we are in any of the  following two cases:
\begin{enumerate}
\item $X$ and $Y$ are complete,
\item $E$ (or $F$) is not complete.
\end{enumerate}
It is then clear by Proposition~\ref{siegediesref} that $X$ is complete
if and only if both $\as$ and $\bs$ are complete.

We will introduce two isometries on spaces of Lipschitz functions
defined on $\as$ and $\bs$. The fact that these new isometries turn out to be standard
  will allow us to obtain a description of $T$.

\begin{lemma}\label{1750ref}
Suppose  that  $f \in \mathrm{Lip}(X,E)$ satisfies $f \equiv 0$ on
$\mathfrak{B} (T^{-1})$. Then $Tf \equiv 0$ on $\mathfrak{B} (T)$.
\end{lemma}

\begin{proof}
Suppose on the contrary that there exists $y_0 \in \mathfrak{B} (T)$ with $T f (y_0)  \neq 0$. By
Lemma~\ref{paco}, we can find $g \in \mathrm{Lip}(Y,F) $ with
$\vc g(y_0) \vd = \vc g \vd_{\infty} > L(g)$
such that
$$\vc g (y_0) \vd + \vc Tf (y_0) \vd = \vc (g + Tf) (y_0) \vd = \vc g + Tf \vd_{\infty} > L (g + Tf).$$

We see that
\begin{eqnarray*}
\sup_{x \in \mathfrak{B} (T^{-1}) } \vc T^{-1} g (x) + f(x) \vd
&=&\sup_{x \in \mathfrak{B} (T^{-1}) } \vc T^{-1} g (x) \vd \\
&\le&  \vc g \vd \\
&<& \vc (g + Tf) (y_0) \vd.
\end{eqnarray*}
On the other hand, if we put $ \mathrm{f} := (g + Tf) (y_0) $, since  $T^{-1}  \tilde{\mathrm{f}} \equiv 0$
on $\mathfrak{A} (T^{-1})$, there exists $n \in \mathbb{N}$ such that
\begin{eqnarray*}
\sup_{x \in \mathfrak{A} (T^{-1}) }  \vc T^{-1} g (x) + f(x) +  n T^{-1}  \tilde{\mathrm{f}} (x) \vd
&<& \sup_{x \in \mathfrak{B} (T^{-1}) }  \vc T^{-1} g (x) + f(x) + n T^{-1}  \tilde{\mathrm{f}} (x) \vd \\
&<& (n+1) \vc \mathrm{f} \vd,
\end{eqnarray*}
so if we denote $k:=  T^{-1} \pl  g  + n \tilde{\mathrm{f}} \pr + f $, then we see that
$\vc k \vd_{\infty} < \vc Tk \vd$. Consequently,  $\vc k \vd_{\infty} < L (k)$ and there exists
$\mathrm{e} \in E$ with $T \tilde{\mathrm{e}} (y_0) \neq 0$  such that
 $\vc k \pm  \tilde{\mathrm{e}} \vd_{\infty}
< L(k) = L(k \pm \tilde{\mathrm{e}})$.

Also $L (Tk) = L (g + Tf) <\vc \mathrm{f} \vd$, so if we assume $n$ big enough, then $\vc Tk \vd = \vc Tk \vd_{\infty}$.
 Therefore, $$
\vc  T \pl k \pm \tilde{\mathrm{e}} \pr \vd_{L} = \vc k \pm \tilde{\mathrm{e}} \vd_{L}=
L(k) = \vc T k \vd_{\infty} = \pl n+1 \pr \vc \mathrm{f} \vd .$$
This implies that
$$ \vc \pl n+1 \pr  \mathrm{f} \pm  T \tilde{\mathrm{e}} (y_0)  \vd  = \vc  T \pl k \pm \tilde{\mathrm{e}} \pr (y_0)  \vd
\le \pl n+1 \pr \vc \mathrm{f} \vd ,$$
which goes against Inequality~(\ref{strictly}).
\end{proof}

 Using Proposition~\ref{siegediesref},
we see that  the subspace $$\bxe := \tl f \in  \cxe : f \pl \as \pr \equiv 0 \tr$$ is isometrically isomorphic to $\bie$,
 via the restriction map.
In the same way,  $$\axe := \tl f \in  \cxe : f \pl \bs \pr \equiv 0 \tr$$
and  $\aie$ are isometrically isomorphic.  Let denote by $I_{\bs} : \bie \ra \bxe$ and $I_{\as} : \aie \ra \axe$,
respectively, the corresponding natural isometries.
In particular we can write in a natural way $$\cxe = \axe \oplus \bxe = \aie \oplus \bie ,$$ where this equality
has to be seen as a direct sum just  in the {\em linear} sense.

Next, let $R_{\bt} : \cyf \ra \bif$ be the operator sending each function to its restriction.

\begin{lemma}\label{parados}
The map  $$T_{\bh} := R_{\bt} \circ T \circ I_{\bs} : \bie \ra \bif $$ is a surjective linear isometry.
\end{lemma}

\begin{proof}
Notice first that if $f \in \bxe$ and $g \in \cxe$ satisfy $f \equiv g $ on $\bs$, then $\vc f \vd \le \vc g \vd$.

$T_{\bh}$ is linear and, by Lemma~\ref{1750ref}, it is easy to check that  it is surjective. We next see that it is
an isometry.  Of course this is equivalent to show that $\vc R_{\bt} \circ T (f) \vd = \vc T (f) \vd$
for every $f \in \bxe$, and it is clear that $\vc R_{\bt} (T (f) ) \vd \le  \vc T (f) \vd$.
Since $\vc R_{\bt} (T (f) ) \vd = \vc I_{\bt} \pl R_{\bt} (T (f) ) \pr \vd$, the fact that
$\vc R_{\bt} (T (f) ) \vd <  \vc T (f) \vd$ is equivalent to that
$$\vc T^{-1} \pl I_{\bt} \pl R_{\bt} \pl T \pl f \pr \pr \pr \pr \vd < \vc f \vd ,$$
which goes against the first comment in this proof.
\end{proof}

\begin{lemma}\label{altziturria}
$T_{\ah} := {I_{\at}}^{-1} \circ T \circ I_{\as} : \aie \ra \aif $ is standard.
\end{lemma}

\begin{proof}
Suppose that this is not the case. Since $\at = \ath \cup \bth$, we are in fact saying that $\ath \neq \emptyset$.

For $\mathrm{e} \in E$ with $\vc \mathrm{e} \vd =1$, we have
$$ \vc T \pl   \tilde{\mathrm{e}} +   \chi_{\ash} \cdot \mathrm{e} \pr \vd =2 .$$
Notice  that both $T \tilde{\mathrm{e}}   \equiv 0$ and
$T \pl \chi_{\as} \cdot \mathrm{e} \pr \equiv 0$ on $\ath$,
so $T \pl \chi_{\bs} \cdot \mathrm{e} \pr \equiv 0$  on $\ath$. On the other hand, by Lemma~\ref{1750ref},
$c \pl T \pl \chi_{\ash} \cdot \mathrm{e} \pr \pr \subset \ath$, and consequently, since
 $$ \vc T    \tilde{\mathrm{e}} \vd =  \vc T \pl    \chi_{\ash} \cdot \mathrm{e} \pr \vd =1 =  \vc T \pl   \tilde{\mathrm{e}} +   \chi_{\ash} \cdot \mathrm{e} \pr \vd_{\infty} ,$$
 there are sequences  $(y_n)$ in $\ath$ and $(z_n) $ in $\bt$ with
\begin{eqnarray*}
2 &=& \lim_{n \ra \infty} \frac{\vc  T \pl   \tilde{\mathrm{e}} +   \chi_{\ash} \cdot \mathrm{e} \pr   (y_n) -  T \pl   \tilde{\mathrm{e}} +   \chi_{\ash} \cdot \mathrm{e} \pr   (z_n)  \vd}
{d \pl y_n, z_n \pr}
\\ &=& \lim_{n \ra \infty} \frac{\vc T \pl \chi_{\ash} \cdot \mathrm{e} \pr (y_n) -  T \pl \chi_{\bs} \cdot \mathrm{e} \pr  (z_n)  \vd}
{d \pl y_n, z_n \pr} \\
&=& \lim_{n \ra \infty} \frac{\vc  T \pl   \chi_{\bs} \cdot \mathrm{e} +   \chi_{\ash} \cdot \mathrm{e} \pr   (y_n) -  T \pl   \chi_{\bs} \cdot \mathrm{e} +   \chi_{\ash} \cdot \mathrm{e} \pr   (z_n)  \vd}
{d \pl y_n, z_n \pr} \\
&\le& \vc T \pl  \pl  \chi_{\bs} +   \chi_{\ash}  \pr \cdot \mathrm{e} \pr \vd \\
&\le& 1.
\end{eqnarray*}

We conclude that $\ath$ is empty.
\end{proof}

 It is easy to check that  $T_{\bh}$ satisfies Property {\bf P}, so by Theorem~\ref{marthahoyo}, it is standard.
 We deduce the following result, which allows us to give the values on $\bt$ and on $\at$ of the images of all functions in $\cxe$ and $\axe$, respectively.

 \begin{corollary}\label{josecerecedo}
 There exist
\begin{enumerate}
\item  $\hb \in \isob$   and  $\ha \in \isoa $,
\item and maps $\jb : \bt \ra I(E,F)$ and $\ja: \at \ra I(E,F)$
constant on each $2$-component of $\bt$ and $\at$, respectively,
\end{enumerate}
 such that
\begin{enumerate}
\item $
Tf(y)= \jb y (f(\hb (y)))
$
for all $f\in \cxe$ and $y\in \bt$, and
\item\label{facundo} $
Tf(y)= \ja y (f(\ha (y)))
$
for all $f\in \axe$ and $y\in \at$.
\end{enumerate}
\end{corollary}

\begin{lemma}\label{silvieta}
Let $y_0 \in \at$ and $A \subset \bs$ be such that $d(\ha (y_0), A) =1$. If $f \in \bxe$ satisfies
$f (A) \equiv \mathrm{e} \in E$, then $$T f (y_0) =  - \ja y_0 \pl \mathrm{e} \pr .$$
\end{lemma}
\begin{proof}
Notice first that, since $y_0 \in \at$, $ T \tilde{\mathrm{e}} (y_0) = 0$, and consequently, by Corollary~\ref{josecerecedo},  $  T \pl \chi_{\bs} \cdot \mathrm{e} \pr \pl y_0 \pr =  - T \pl \chi_{\as} \cdot \mathrm{e} \pr \pl y_0 \pr = - \ja y_0 \pl \mathrm{e} \pr  $.

Next we  prove the result through several steps. We denote $\mathrm{a} :=  \ja y_0 \pl \mathrm{e} \pr$ for short.

\begin{step}\label{zelada}
Assume that $\vc \mathrm{e} \vd = 1 =\vc f \vd$.
\end{step}

Consider $k' \in \cx$ defined by $k ' (x) := \max \tl 0, 1 - d(x, \ha (y_0))  \tr $
for every $x \in X$, and $k \in \axe$ defined by $k := - k' \cdot \mathrm{e}$. It is easy to see that $(k + f ) (\ha (y_0)) = - \mathrm{e}$ and that $(k + f ) (x) =  \mathrm{e}$ for every $x \in A$. As a consequence,
$\vc k + f \vd =2$.

Suppose now that $T f (y_0) = \mathrm{f} \neq -   \mathrm{a}  $. By  Corollary~\ref{josecerecedo}, $Tk (y_0) = - \mathrm{a}$
and, since $\vc  f  \vd_{\infty}  =1$, we can take $M <2$ such that
$$
\vc T (k + f) (y_0) \vd =  \vc  \mathrm{-a} + \mathrm{f}  \vd
< M.
$$
Consequently there exists $0<r<1$ such that $\vc T \pl k + f     \pr (y) \vd < M$ for every
$y \in B(y_0, r)$. On the other hand, for $y \in \at$ with $d(y, y_0 ) \ge r$,
\begin{eqnarray*}
\vc T k (y) \vd &=& \vc T (- k' \cdot \mathrm{e} )   (y) \vd \\
&=& \vc    \max \tl 0, 1 - d(\ha (y), \ha (y_0))  \tr  \cdot  \ja y (\mathrm{e})   \vd \\
&\le& 1- r,
\end{eqnarray*}
so $\vc T (k + f) (y) \vd \le 2-r$. Since $\vc T (k + f) (y) \vd = \vc T f (y) \vd \le 1$ for every $y \in \bt$,
 we deduce that $\vc T (k + f)  \vd_{\infty} < 2 = \vc T (k + f)  \vd$. Let $M>0$ with $M + \vc T (k + f)  \vd_{\infty} <2$, and
$y \in \bt$ such that $\hb (y) \in A$ and $d(\hb (y), \ha (y_0)) < 1+M/2$. Define $\mathrm{b} := M T \tilde{\mathrm{e}} (y) \in F$.
 By Corollary~\ref{josecerecedo},   $T^{-1} \tilde{\mathrm{b}} (\hb (y)) = M \mathrm{e}$, and consequently
 \begin{eqnarray*}
2 d (\hb (y), \ha (y_0))    &<& 2 + M  \\
&=& \vc \pl k + f + T^{-1} \tilde{\mathrm{b}} \pr \pl \hb \pl y \pr \pr - \pl k + f + T^{-1} \tilde{\mathrm{b}} \pr \pl \ha \pl y_0 \pr \pr \vd ,
\end{eqnarray*}
 against the fact that $\vc   k + f + T^{-1} \tilde{\mathrm{b}} \vd =2$.

\begin{step}\label{sinapsis}
Assume that $\vc \mathrm{e} \vd = 1 =\vc f \vd_{\infty}$.
\end{step}

It is easy to check that  if $n \ge L (  f)$,
then $\vc    n \chi_{\bs} \cdot \mathrm{e}  \vd_{\infty} = \vc   n \chi_{\bs} \cdot \mathrm{e}  \vd = n$  and that
$$\vc  f +  n \chi_{\bs} \cdot \mathrm{e}  \vd_{\infty} = \vc  f +  n \chi_{\bs} \cdot \mathrm{e}  \vd = n+1 .$$
Using Step~\ref{zelada}, $T (   n \chi_{\bs} \cdot \mathrm{e} ) (y_0) = - n \mathrm{a}$ and
$T ( f +  n \chi_{\bs} \cdot \mathrm{e} ) (y_0) = - (n +1)  \mathrm{a}$. The conclusion is easy.

\begin{step}\label{aurillaque}
Assume that $\mathrm{e} =0$.
\end{step}

Of course we must prove that $Tf (y_0) =0$. Fix $\mathrm{d} \in E$ with norm $1$. Consider $m \in \bxe$ defined by
$
m (x):= \max \{0,1-d(x,A)\} \cdot \mathrm{d}
$
for each $x\in X$.
We easily check that $\vc   m \vd_{\infty}= 1 = \vc \mathrm{d} \vd$,
and  if we assume that $\vc   f \vd \le 1$, then $\vc f(x) \vd \le d(x, A)$ for every $x$.
As in  the proof of
Lemma~\ref{paco}, we see that
$\vc   m +  f  \vd_{\infty}= 1 = \vc \mathrm{d} \vd$.
The conclusion follows immediately from Step~\ref{sinapsis}.

\medskip

The rest of the proof  is easy.
\end{proof}

\begin{corollary}\label{indoxina}
Suppose that $A_1, A_2 \subset \bs$ satisfy $d(\ha (y_0), A_i ) =1$ for $i=1,2$. Then $d(A_1, A_2) = 0$.
\end{corollary}

\begin{proof}
Just assume that $d(A_1, A_2) > 0$ and apply  Lemma~\ref{silvieta}
to any  $  f \in \bxe$
such that $  f   (A_i) \equiv (-1)^i \mathrm{e} \neq 0$ for $i=1, 2$. This leads to two different values for
$Tf (y_0)$.
\end{proof}

\begin{corollary}\label{quena}
Let $y_0 \in  \at$. Then there exists exactly one point $\varphi (y_0) $ in $ \bt$
such that $d \pl \hb \pl \varphi (y_0) \pr, \ha \pl y_0 \pr  \pr = 1$. Also,
$$T f (y_0) =  -   \ja y_0 (f \pl \hb \pl \varphi \pl  y_0 \pr \pr  \pr ) $$
for every $f \in \bxe$.
\end{corollary}

\begin{proof}
By Lemma~\ref{gatosinchicle} and Corollary~\ref{indoxina}, we deduce that if $(x_n)$ is a sequence in $X$ such that $d (\ha (y_0), x_n ) \le 1 + 1/n$
for each $n \in \mathbb{N}$, then it is a Cauchy sequence, so there is a limit $x_0$ in $ \widehat{X}$, which necessarily belongs to $\widehat{\bs}$. Obviously the point $x_0$ does not depend on the sequence we take.

We next assume that $\widehat{X}$ is not complete
 and prove that
$x_0 \in \bs$.
 If this is not the case,
for each $n \in \mathbb{N}$, let
$$
f_n (x) := \max \tl 0,1 -   d \pl x,  B \pl x_0 , 1/n \pr \pr   \tr
$$
for all $x \in   X$. It is clear that each $f_n$ belongs to
$\cx$. Since $\mathrm{Lip}(X,\widehat{E})$ is complete, if we take $(\mathrm{e}_n)$ in $E$ as in Lemma~\ref{ester}, then
$f:= \sum_{n=1}^{\infty} f_n \cdot \widetilde{\mathrm{e}_n}$ belongs to
$\mathrm{Lip}(X,\widehat{E})$, and since all values are taken in $E$, to  $\mathrm{Lip}(X,E)$, and indeed to $\bxe$.
Thus, since $f = \lim_{k \ra \infty } \sum_{n=1}^k f_n \cdot \widetilde{\mathrm{e}_n}$, we deduce
from Lemma~\ref{silvieta}
that
\begin{eqnarray*}
 Tf (y_0) &=& \lim_{k \ra \infty}  \sum_{n=1}^k T \pl f_n \cdot  \widetilde{\mathrm{e}_n} (y_0) \pr
\\  &=&  -  \lim_{k \ra \infty} \sum_{n=1}^{k}   \ja y_0  \pl \mathrm{e}_n  \pr \\
 &=& -     \sum_{n=1}^{\infty} \ja y_0 \pl \mathrm{e}_n  \pr ,
\end{eqnarray*}
  which belongs to $\widehat{F} \setminus  F$. This is absurd.

  If we define $\varphi (y_0) := {h_{\bh}}^{-1} (x_0) \in \bt$, then we are done.
\end{proof}

\begin{proposition}\label{nohayna}
For every $y \in \at$, $\ja y = - \jb \varphi (y)$.
\end{proposition}

\begin{proof}
Fix $y \in \at$ and $\mathrm{e} \in E$, and let $\mathrm{f} := T \tilde{\mathrm{e}} (\varphi (y))$. Then
$T^{-1} \tilde{\mathrm{f}} (\hb (\varphi (y))) = \mathrm{e}$ and $T^{-1} \tilde{\mathrm{f}} \equiv 0$ on $\as$.
We conclude from Corollary~\ref{josecerecedo} that $\jb \varphi (y) (\mathrm{e}) = \mathrm{f}$,
 and from Corollary~\ref{quena} that $- \ja y (\mathrm{e}) = \mathrm{f}$.
\end{proof}

Next result follows now easily from Corollaries~\ref{josecerecedo} and~\ref{quena}, and Proposition~\ref{nohayna}.

\begin{corollary}\label{errabia}
For $y \in \at$ and $f \in \cxe$,
\begin{eqnarray*}
Tf (y) &=& -\ja y (f (\hb (\varphi (y)))) + \ja y (f (\ha (y))) \\
&=& \jb \varphi (y)  (f (\hb (\varphi (y)))) - \jb \varphi (y) (f (\ha (y))) .
\end{eqnarray*}
\end{corollary}

\begin{corollary}\label{kamisabieja}
Let $y_0 \in \at$. If $y \in \bt$ is such that   $  d(y, \varphi (y_0 )) \ge 2$, then
$$d (y, y_0) \ge 2 .$$
\end{corollary}

\begin{proof}
Let $\mathrm{e}_1, \mathrm{e}_2 \in E$ be vectors with norm $1$   and such that $\jb \varphi(y_0) (\mathrm{e}_1) = \jb y (\mathrm{e}_2)  $. Define $f:= f_1 - f_2 \in \bxe$, where
 $$f_1 (x) := \max \tl 0, 1- d(x, \hb \pl \varphi (y_0) \pr ) \tr \cdot \mathrm{e}_1$$
and $$ f_2 (x) := \max \tl 0, 1- d(x, \hb \pl y \pr ) \tr \cdot \mathrm{e}_2 $$
 for every $x \in X$. Obviously, $\vc f_1 \vd = 1 = \vc f_2 \vd$, so to show that $\vc f \vd =1$,
 it is enough to see that, if $d(x, \hb \pl \varphi (y_0) \pr ) , d(z, \hb \pl y \pr ) <1$, then
 $\vc f_1 (x) \vd + \vc f_2 (z) \vd \le d(x, z)$. Taking into account that
 $$2 \le  d(\hb \pl y \pr , \hb \pl \varphi (y_0 ) \pr ) \le d(z, \hb \pl y \pr ) + d(x,z ) + d(x, \hb \pl \varphi (y_0) \pr ) , $$
 it follows that
 \begin{eqnarray*}
 \vc f_1 (x) \vd + \vc f_2 (z) \vd &=& \pl 1- d(x, \hb \pl \varphi (y_0) \pr ) \pr + \pl 1 - d(z, \hb \pl y \pr ) \pr \\
 &\le& d(x, z).
 \end{eqnarray*}

On the other hand,  by Corollary~\ref{errabia}, $ Tf (y_0) =  \jb \varphi (y_0) (\mathrm{e}_1) = Tf (\varphi (y_0))$,
and by the way we have taken $\mathrm{e}_1$ and $ \mathrm{e}_2 $, we have
 $ Tf (y) = - \jb y  (\mathrm{e}_2) = - \jb \varphi(y_0) (\mathrm{e}_1) $.
 We conclude that, since $\vc Tf \vd = 1$, $$2 = \vc Tf (y) - Tf (y_0) \vd \le d(y, y_0). $$
\end{proof}

\begin{corollary}\label{nitanmal}
Let $y_0 \in \at$. Given $y \in \bt$, if  $0 \le d(y, \varphi (y_0 )) <1$, then
$$d (y, y_0) = 1 + d (y, \varphi (y_0 )) ,$$
and if $1 \le d(y, \varphi (y_0 )) <2$, then
$$d (y, y_0) \ge 2 .$$
\end{corollary}

\begin{proof}
Fix $\mathrm{e} \in E$ with norm $1$ and let
$f (x) :=  \min \tl 1,  d(x , \hb (\varphi (y_0)) ) \tr \cdot \mathrm{e} $ for every $x \in X$.
Let $y \in \bt$ with $ d(y, \varphi (y_0 )) <2$. Taking into account that
 $\vc f \vd =1$,  Corollaries~\ref{josecerecedo} and~\ref{errabia} give
\begin{eqnarray*}
d(y, y_0)  &\ge& \vc Tf (y) - T f (y_0) \vd\\
 &=& \vc \min \tl 1,  d(\hb (y) , \hb (\varphi (y_0)) ) \tr \cdot \jb y (\mathrm{e}) + \jb y (\mathrm{e}) \vd \\
  &\ge& \min \tl 2, d( \hb (y) , \hb (\varphi (y_0)) ) + 1 \tr.
\end{eqnarray*}
The conclusion is immediate.
\end{proof}

\begin{corollary}\label{tillohablador}
If $y_1, y_2 \in \at$ satisfy $\varphi (y_1) \neq \varphi (y_2)$, then
$d (y_1, y_2 ) \ge 2  $.
\end{corollary}

\begin{proof}
Suppose that $M:= d(y_1, y_2) /2 <1$, so by Corollary~\ref{josecerecedo} $\ja y_1 = \ja y_2$.
Put $ N:=  d(\hb (\varphi (y_1)) , \hb (\varphi (y_2))) $ and,
 for a fixed $\mathrm{e} \in E$ with norm $1$, let
$$f (x ) := \max \tl - M, M - d( x , \ha (y_1) ) \tr \cdot \mathrm{e} $$
and
$$g (x) := \max \tl 0 , N - d(x , \hb (\varphi (y_1)) ) \tr \cdot \mathrm{e} $$
for every $x \in X$.
If we take
 $A >0$ such that $M + AN <1$, then
$k := \chi_{\as} f  - A \chi_{\bs} g $ has norm $1$. Also, by Corollary~\ref{errabia}, $Tk (y_1) = M \ja y_1 (\mathrm{e}) +  A N \ja y_1 (\mathrm{e})$ and
$Tk (y_2) = - M \ja y_2 (\mathrm{e}) = - M \ja y_1 (\mathrm{e})$. Consequently
$$ \vc \frac{Tk (y_1) - Tk(y_2)}{d(y_1, y_2)} \vd >1,$$ which is impossible.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{nonelk}]
 Corollaries~\ref{kamisabieja}, ~\ref{nitanmal} and \ref{tillohablador} show that $Y$ is of type $\mathbf{A}$.
We consider the associated
purely nonstandard map $S_{\varphi} : \cyf \ra \cyf$ and see that, given $\mathrm{e} \in E$, $\mathrm{e} \neq 0$,
 the composition $S_{\varphi} \circ T : \cxe \ra \cyf$
satisfies $S_{\varphi} \circ T \pl \tilde{\mathrm{e}} \pr = \jb \varphi (y) (\mathrm{e}) \neq 0$ if $y \in \at$ and
$S_{\varphi} \circ T \pl \tilde{\mathrm{e}} \pr = \jb   y (\mathrm{e}) \neq 0$ if $y \in \bt$, that is, the composition satisfies Property {\bf P}.

This implies that $S_{\varphi} \circ T$ is standard.  Since $S_{\varphi} = {S_{\varphi}}^{-1}$, we have that
$T = S_{\varphi} \circ \pl S_{\varphi} \circ T \pr$, and we are done.
\end{proof}

\begin{remark}
It is easy to check that, if  $h: Y \ra X$
and $J : Y \ra I(E, F)$ are the
associated maps to $S_{\varphi} \circ T$ in the proof of Theorem~\ref{nonelk}, then
 $h \equiv \ha$ on $\at$ and $h \equiv \hb$ on $\bt$. In the same way,
$J \equiv - \ja$ on $\at$ and $J \equiv  \jb$ on $\bt$. Finally, it is also apparent that,
given $A \in \mathtt{Comp}_2 (\bs)$, $A = A' \cap \bs$, where $A' \in \mathtt{Comp}_2 (X)$, and that
a similar fact does not necessarily hold for the elements in
 $ \mathtt{Comp}_2 (\as)$.
\end{remark}

\begin{proof}[Proof of Proposition~\ref{pesafrank}]
Suppose that $\pl X, d^{\alpha}\pr$ is of type $\mathbf{A}$. Then
$$d^{\alpha} (z, y) = 1 + d^{\alpha} (z, \varphi (y)) $$ whenever
$y \in \at$ and $z \in \bt$ satisfy  $0 < d^{\alpha} (z, \varphi (y)) <1$.
In such case,  $d^{\alpha} (z, y) < d  (z, y) $ and $d (z, \varphi (y )) < d^{\alpha} (z, \varphi (y))  $,
and this implies
$$d (z, y) > 1 + d (z, \varphi (y)) = d(y , \varphi (y) )  + d (z, \varphi (y)) , $$
which is impossible. We deduce that, if $d  (z, \varphi (y )) <1$, then $z = \varphi (y)$.
The rest of the proof is easy.
\end{proof}

\section{Acknowledgements}
The authors wish  to thank the referee for his/her  valuable remarks, which helped them to improve
the first version of the paper.

config: arxiv


TITLE: Finding a specific series sum $\sum_{k=1}^{i-1}(\frac{((ak^3+a(1-2i+2n)k^2+2b)}{2ak})^{-1}$
QUESTION [0 upvotes]: I am stuck in this specific sum equation as below.
Sum it $k$ is from $1$ to $i-1$ and others are static variables.
Please help to find $\sum_{k=1}^{i-1}(\frac{((ak^3+a(1-2i+2n)k^2+2b)}{2ak})^{-1}$
I appreciated your helps.
Thank you.

REPLY [0 votes]: You are considering $$S_i=\sum_{k=1}^{i-1} \frac{2ak}{ak^3+a(1-2i+2n)k^2+2b}$$ I think that what I should do is to write
$$ak^3+a(1-2i+2n)k^2+2b=a(k-r)(k-s)(k-t)$$ Now using partial fraction decomposition
$$ \frac{2ak}{ak^3+a(1-2i+2n)k^2+2b}$$ write $$\frac{2 r}{(r-s) (r-t) (k-r)}-\frac{2 s}{(r-s) (s-t) (k-s)}-\frac{2 t}{(r-t) (t-s)   (k-t)}$$ to face three summations looking like
$$I_x=\sum_{k=1}^{i-1} \frac 1 {k-x}=H_{i-x-1}-H_{-x}$$
The problem will be to find the expressions of the roots $(r,s,t)$ but from a formal point of view we have the result.


TITLE: Finding a nonzero polynomial involving the floor function
QUESTION [2 upvotes]: Find a nonzero polynomial $P(x,y)$ where the coefficients are integers such that $P(\lfloor a \rfloor, \lfloor 2a \rfloor) = 0$  $\forall a \in$ $\mathbb{R}$

REPLY [0 votes]: Hint: $\,2 \lfloor a \rfloor -1 \le 2a-1 \lt \lfloor 2a \rfloor \le 2a \lt 2 \left(\lfloor a \rfloor +1\right) = 2 \lfloor a \rfloor + 2\,$, so $2 \lfloor a \rfloor\le \lfloor 2a \rfloor \lt 2 \lfloor a \rfloor + 2\,$. It follows that $\,\lfloor 2a \rfloor\,$ can only take one of the two integer values $\,2 \lfloor a \rfloor\,$ or $\,2 \lfloor a \rfloor+1\,$, and therefore:
$$\big(\lfloor 2a \rfloor  -
 2 \lfloor a \rfloor \big)\big(\lfloor 2a \rfloor  -
 2 \lfloor a \rfloor - 1\big) = 0 \quad\quad\forall a \in \mathbb{R}$$


TITLE: How do I compute the length of the following curve?
QUESTION [0 upvotes]: I have the following problem.
I need to compute the length of the curve given by

$z^2=2x,3y=xz$ between $(0,0,0)$ and $(\frac{1}{2},\frac{1}{6},1)$

I first want to find my curve, so geometrically I could draw it with geogebra but I can't see how to find my curve. Could maybe someone give me a hint?
Thanks a lot

REPLY [1 votes]: Hint: Use the fact that $x=\frac{z^2}2$ and that $y=\frac{xz}3=\frac{z^3}6$.

REPLY [1 votes]: Parametrize your curve $\vec{r}(t) = (x(t), y(t), z(t))$, and then compute the arc-length integral of the curve:
$$
\int_{t_{1}}^{t_{2}} |\vec{r}'(t)| dt
$$
Hint: Define $y$, $z$ in terms of $x \in [0, 1/2]$


TITLE: What does the "\" mean in this statement about regular languages?
QUESTION [1 upvotes]: I have having difficulty finding the meaning of the backslash in the statement. I don't know if it could be a division, an OR operator or something else. If someone could provide information about it, that would be very helpful!

REPLY [0 votes]: The backslash like symbol is used to indicate elements present in set A and not in set B, in the case of A\B. In your question {aa,ab,ba,bb} \ {aa,bb}, the elements aa and bb are present in both sets, so they are excluded. Thus the answer is the remaining elements in the first set which are ab and ba.


TITLE: Is there a Horn formula which is equivalent to $(p \lor q)$?
QUESTION [1 upvotes]: Is there a Horn formula which is equivalent to $(p \lor q)$?
Hi I have to answer the following question:
Given any formula $\phi$, is it possible to find a Horn formula equivalent to $\phi$?
I know that a formula is Horn's if in its conjunctive normal form all clauses are Horn's.
But in this case $\phi$ is already in it's conjunctive normal form, but $\phi$ has 2 positive literal.
So, this would be a example that proves that what I have to answer is not possible, right?
Thanks in advance
UPDATE
I just thought of something; let me know if I'm right.
So a Horn Clause has the form:
a) $(p_1 ∧...∧ p_n) → q$
b) $ \lnot p_1 \lor \lnot p_2, ... , \lnot p_n $
Since $(p \lor q)$ is not equivalent to $(p \land q) → s$, and
$(p \lor q)$ is not equivalent to $(\lnot p \lor \lnot q)$
There $(p \lor q)$ can't be equivalent to any Horn formula.
Is this reasoning right?

REPLY [0 votes]: Any formula $\varphi$ which is equivalent to a Horn formula has the following property: if $v_1$ and $v_2$ are two valuations that make $\varphi$ true, then the valuation $v_1\land v_2$ also makes $\varphi$ true.
The formula $\varphi=(p\lor q)$ does not have this property, since it is true under the valuations $(T,F)$ and $(F,T)$ but false under the valuation $(T\land F,F\land T)=(F,F)$. Therefore $(p\lor q)$ is not equivalent to a Horn formula.
(More generally, if a first-order sentence is equivalent to a Horn sentence, then it has the property: if it's true in two models $\mathfrak A$ and $\mathfrak B$, then it's true in their direct product $\mathfrak A\times\mathfrak B$. However, not every first-order sentence with this property is equivalent to a Horn sentence.)


TITLE: Can the map sending a presentation to its group be considered as a functor?
QUESTION [20 upvotes]: It is well-known that the functor $Grp \to Set$ sending a group $G$ to its underlying set $UG$ has a left adjoint, the functor $Set\to Grp$ sending a set $X$ to the free group $FX$.
I wonder whether one can consider the following modification of the functor $F$: to each presentation $\langle S\mid R\rangle$ we assign the group $FS/N$, where $N$ is the normal subgroup induced by $R$.

Is this a functor?
For this to be true, there need to be a category of presentations. Is there such a category?
Does the functor $\langle S\mid R\rangle\mapsto FS/N$ have a right adjoint (maybe similar to $U$)?

Of course, my questions are not precise, because I haven't specified what the morphisms between presentations are. The question is if one can do all this in a sensible way.

REPLY [19 votes]: Yes, yes, yes, to all three questions.
And it can be done very generally and very nicely for universal algebra $\ $ [or even for first order structures].
Let's fix an algebraic signature consisting of named finitary operation symbols, and a variety $\mathcal{Alg}$ of algebras of that signature that we're working on. (In your question, the groups.)
As in your question, we use $F:\mathcal{Set}\to\mathcal{Alg}$ for the free functor and $U:\mathcal{Alg}\to\mathcal{Set}$ for the underlying set functor.
A presentation $\langle A,E\rangle$ is a pair where $A$ is a set and $E$ is a subset of equations, i.e. pairs of terms (in other words $E\subseteq F(A)^2$) $\ $ [or, $E$ is a set of atomic formulas].
In the case of groups, an equation can be arranged on one side, leaving the identity on the other side, so then $E\subseteq F(A)$ is enough, as learned.
We can define a morphism of presentations $\langle A,E\rangle\to\langle B,\Sigma\rangle$ to simply be a function $f:A\to B$ such that
$$(\tau,\sigma)\in E\ \implies\  (Ff(\tau),\,Ff(\sigma))\in\Sigma\,.$$
By an interpretation of the presentation $\langle A,E\rangle$ in an algebraic structure $\mathfrak B$ we mean a function $h:A\to U\mathfrak B$ such that all elements of $E$ are evaluated equal $\,$ [all atomic formulas of $E$ are true], i.e. for every $(\tau,\sigma)\in E$ we have
$$\varepsilon_{\mathfrak B}\left( Fh(\tau)\right) \ =\ \varepsilon_{\mathfrak B}\left( Fh(\sigma)\right)\ \text{ that is, }\
(\tau,\sigma)\in\ker(\varepsilon_{\mathfrak B}\circ Fh)$$
where $\varepsilon:FU\to 1_{\mathcal{Alg}}$ is the counit of the free-forgetful adjunction.
Observe that we can compose morphisms of presentations with interpretations as well as with homomorphisms, in an associative way, since all are just functions.
So this actually connects up the category of presentations $\mathcal{Pres}$ with the category of algebras $\mathcal{Alg}$ by (so called hetero-)morphisms between them, yielding to a bigger category, containing both $\mathcal{Pres}$ and $\mathcal{Alg}$ as a full subcategory.
Such an animal is called a profunctor, and it is determined by the restriction of its hom functor to the heteromorphisms, i.e. of a functor of the form $\mathcal{Pres}^{op}\times\mathcal{Alg}\to\mathcal{Set}$.

Now, every presentation $\langle A,E\rangle$ has a reflection in $\mathcal{Alg}$, namely $FA/(E)$ where $(E)$ is the congruence relation generated by $E$.
This defines the free functor on presentations.
Conversely, every algebraic structure $\mathfrak B$ has a coreflection in $\mathcal{Pres}$, namely $\langle U\mathfrak B,\Delta_{\mathfrak B}\rangle$ where $\Delta_{\mathfrak B}$ is the set of all equations that can be written on letters of $U\mathfrak B$ and hold in $\mathfrak B$ $\,$ [the positive diagram of $\mathfrak B$].
This defines the right adjoint.
Note that this is a fully faithful functor.


TITLE: Is this the correct curvature of the space curve?
QUESTION [2 upvotes]: So lets say we have a space curve $r=\langle\cos t,\sin t,3t\rangle$ and we need to find the curvature at the point $P(1,0,0)$
$r'(t)=\langle-\sin t,\cos t,3\rangle$
$r''(t)=\langle-\cos t,-\sin t,0\rangle$
$||r'(t)||=\sqrt{(-\sin t)^2+(\cos t)^2+(3)^2} = \sqrt{1+9} = \sqrt{10}$
$||r'(t)||^3=(\sqrt{10})^3=10\sqrt{10}$
$r'(t) \times r''(t)$ = $\langle3\sin t,-3\cos t,1\rangle$
$||r'(t) \times r''(t)|| = \sqrt{(3\sin t)^2+(-3\cos t)^2+1^2}=\sqrt{10}$
$k=\frac{||r(t) \times r''(t)||}{||r'(t)||^3}=\frac{\sqrt{10}}{10\sqrt{10}}=\frac{1}{10}$
Is this correct? I think that's all you need to do for the problem, right?

REPLY [1 votes]: Looks good to me. Another way is to reparametrize by arclength and use the simpler curvature formula.
Let $r(t) = \langle \cos(t),\sin(t),3t \rangle$. Then the arclength is written:
$$ s(t) = \int_0^t ||r'(\tau)||_2d\tau=\sqrt{10}t \;\;\;\implies\;\;\;t = \frac{s}{\sqrt{10}} $$
So, using this parametrization, we get:
\begin{align} r(s) &= \langle \cos(s/\sqrt{10}),\sin(s/\sqrt{10}),3t/\sqrt{10} \rangle\\
r'(s) &= \langle -\sin(s/\sqrt{10})/\sqrt{10},\cos(s/\sqrt{10})/\sqrt{10},3/\sqrt{10} \rangle\\
r''(s) &= \langle -\cos(s/\sqrt{10})/{10},-\sin(s/\sqrt{10})/{10},0 \rangle
\end{align}
The curvature is then computable by a simpler formula, i.e:
$$
\kappa(s) = ||r''(s)||_2=\sqrt{(-\cos(s/\sqrt{10})/{10})^2 + (-\sin(s/\sqrt{10})/{10}})^2 = \frac{1}{10}
$$
as you computed as well.