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\title{Separation ratios of maps between Banach spaces}

\author{\firstname{Christian} \lastname{Rosendal}}
\address{Department of Mathematics\\University of Maryland, 4176 Campus Drive - William E. Kirwan Hall, College Park, MD 20742-4015, USA}
\email{rosendal@umd.edu}
%\urladdr{sites.google.com/view/christian-rosendal/}
\subjclass{46B80}

\keywords{\kwd{Coarse embeddings}
\kwd{Banach spaces}}
\thanks{The research for this article was supported by the NSF through award DMS 2204849.}
\CDRGrant[NSF]{DMS 2204849}

\begin{abstract}
Under the weak assumption on a Banach space $E$ that $E\oplus E$ embeds isomorphically into $E$, we provide a characterisation of when a Banach space $X$ coarsely embeds into $E$ via a single numerical invariant.
\end{abstract}

\begin{document}
\maketitle

\section{Introduction}

The concept of coarse embeddability between metric spaces can be viewed as a large scale analogue of uniform embeddability and may most easily be understood in terms of the moduli associated with a map. However, as we are exclusively concerned with Banach spaces, these moduli can further be reduced to a couple of numerical invariants.

\begin{defi}
For a (generally discontinuous and nonlinear) map $X\maps \phi E$ between two Banach spaces we define the \emph{exact compression coefficient} $\ov \kappa(\phi)$, the \emph{compression coefficient} $\kappa(\phi)$ and the \emph{expansion coefficient} $\om(\phi)$ by
\[
\ov \kappa(\phi)=\sup_{t<\infty}\;\;\inf_{\norm{x-y}= t}\;\;\Norm{\phi(x)-\phi(y)},
\]
\[
\kappa(\phi)=\sup_{t<\infty}\;\;\inf_{\norm{x-y}\geqslant t}\;\;\Norm{\phi(x)-\phi(y)},
\]
and
\[
\om(\phi)=\inf_{t>0}\;\;\sup_{\norm{x-y}\leqslant t}\;\;\Norm{\phi(x)-\phi(y)}.
\]
\end{defi}

To avoid certain trivialities, we shall tacitly assume that all Banach spaces have dimension at least $2$ and hence, in particular, that the infima and suprema above are taken over non-empty sets. Let us first note the obvious fact that $\om(\phi)=0$ if and only if $\phi$ is uniformly continuous. On the other hand, $\om(\phi)<\infty$ if and only if $\phi$ is \emph{Lipschitz for large distances}, that is,
\[
\Norm{\phi(x)-\phi(y)}\leqslant K\norm{x-y}+K
\]
for some constant $K$ and all $x,y\in X$. Similarly, assumptions on $\kappa(\phi)$ correspond to known conditions on the map $\phi$. We summarise these as follows.
\begin{enumerate}
\item $\om(\phi)=0$, that is, $\phi$ is uniformly continuous,
\item $\om(\phi)<\infty$, that is, $\phi$ is Lipschitz for large distances,
\item $\kappa(\phi)=\infty$, that is, $\phi$ is \emph{expanding},
\item $\kappa(\phi)>0$, that is, $\phi$ is \emph{uncollapsed}.
\end{enumerate}
Note that the three coefficients above are all positive homogenous, in the sense that
\[
\kappa(\lambda\phi)=\lambda\cdot\kappa(\phi)
\]
for all $\lambda>0$ and similarly for $\ov \kappa(\phi)$ and $\om(\phi)$. In particular, this means that the following quantity is invariant under rescaling $\phi$.

\begin{defi}\label{sep rat}
The \emph{separation ratio} of a map $X\maps \phi E$ is the quantity
\[
\ku R(\phi)=\frac{\kappa(\phi)}{\om(\phi)},
\]
where we set $\frac a\infty=\frac 0a=0$ for all $a\in [0,\infty]$ and $\frac a0=\frac \infty a=\frac \infty 0=\infty$ for all $0<a<\infty$.
\end{defi}

Whereas $\phi$ being a uniform embedding cannot be directly expressed via the coefficients above, we note that $\phi$ is a \emph{coarse embedding} provided that $\om(\phi)<\infty$ and $\kappa(\phi)=\infty$, that is, if $\phi$ is Lipschitz for large distances and is expanding. We thus see that
\[
\ku R(\phi)=\infty
\]
if and only if $\phi$ is either uniformly continuous and uncollapsed (e.g., a uniform embedding) or if $\phi$ is a coarse embedding.

Motivated in part by the still open problem of deciding whether a Banach space $X$ coarsely embeds into a Banach space $E$ if and only it uniformly embeds, the papers~\cite{Braga1, Braga2, naor-nets, equivariant, almosttransitive, geomgroups} contain various constructions for producing uniform and coarse embeddings or obstructions to the same. In particular, in~\cite{geomgroups} (see Theorem~1.16 therein) we showed that, provided that $E\oplus E$ isomorphically embeds into $E$, then a uniformly continuous and uncollapsed map $X\maps\phi E$ gives rise to a simultaneously uniform and coarse embedding of $X$ into $E$. However, as shown by A.~Naor~\cite{naor-nets}, there are Lipschitz for large distance maps that are not even close to any uniformly continuous map. For the exclusive purpose of coarse embeddability, our main result, Theorem~\ref{main}, removes the problematic assumption of uniform continuity of $\phi$.

\begin{theo}\label{main}
Suppose $X$ and $E$ are Banach spaces so that $E\oplus E$ isomorphically embeds into $E$. Then $X$ coarsely embeds into $E$ if and only if
\[
\sup_{\smash{\phi}}\;\ku R(\phi)=\infty,
\]
where the supremum is taken over all maps $X\maps \phi E$.
\end{theo}

The proof of Theorem~\ref{main} also allows us to address another issue, namely, the preservation of cotype under different forms of embeddability. For this, consider the following conditions on a map $X\maps\phi E$.
\begin{enumerate}\setcounter{enumi}{4}
\item $\ov\kappa(\phi)=\infty$, that is, $\phi$ is \emph{almost expanding},
\item $\ov\kappa(\phi)>0$, that is, $\phi$ is \emph{almost uncollapsed}.
\end{enumerate}
Also, the map $\phi$ is said to be \emph{solvent} provided that there are constants $R_1,R_2,\ldots$ so that
\[
R_n\leqslant \norm{x-y}\leqslant R_n+n \;\;\;\saa\;\;\; \Norm{\phi(x)-\phi(y)}\geqslant n.
\]
Provided that $\phi$ is Lipschitz for large distances, $\phi$ is solvent if and only if it is almost expanding (see ~\cite[Lemma~8]{equivariant}). In analogy with Definition~\ref{sep rat}, we then define the \emph{exact separation ratio} of $\phi$ to~be
\[
\ov{\ku R}(\phi)=\frac {\ov\kappa(\phi)} {\om(\phi)}.
\]
As $\kappa(\phi)\leqslant \ov \kappa(\phi)$, we then have $\ku R(\phi)\leqslant \ov{\ku R}(\phi)$. Also, $\ov{\ku R}(\phi)=\infty$ exactly when $\phi$ is either uniformly continuous and almost uncollapsed or is Lipschitz for large distances and solvent.

In connection with this, B.~Braga~\cite{Braga2} strengthened work by M.~Mendel and A.~Naor~\cite{mendel} to show that, if $X$ maps into a Banach space $E$ with non-trivial type by a map that is either uniformly continuous and almost uncollapsed or is Lipschitz for large distances and solvent, then
\[
\cotype(X)\leqslant \cotype(E).
\]
The following statement therefore covers both cases of Braga's result and seemingly provides the ultimate extension in this direction.

\begin{theo}\label{main2}
Suppose $X$ and $E$ are Banach spaces so that
\[
\sup_{\phi} \ov{\ku R}(\phi) =\infty
\]
and that $E$ has non-trivial type. Then
\[
\cotype(X)\leqslant \cotype(E).
\]
\end{theo}

Problem 7.4 in Braga's paper~\cite{Braga2} asks what can be deduced about a space $X$ that admits a map $X\maps \phi E$ that is just Lipschitz for large distances and almost uncollapsed, i.e. so that $\ov{\ku R}(\phi)>0$. That is, will restrictions on the geometry of $E$ also lead to information about the geometry of $X$? In Example~\ref{constant=1}, we show that this is not always so. Indeed, if $X$ is separable and $E$ is infinite-dimensional, one can always find a map $X\maps \phi E$ that is both Lipschitz for large distances and uncollapsed, i.e., so that ${\ku R}(\phi)>0$, and after renorming $E$ one can even obtain ${\ku R}(\phi)\geqslant1$. On the other hand, Theorem~\ref{main2} provides a positive answer to Braga's question under the alternative assumption $\sup_{\phi} \ov{\ku R}(\phi) =\infty$.






\section{Proofs}\label{sec:proofs}
Before proving our main results, let us introduce four functional moduli that lie behind the definitions of the (exact) compression and expansion coefficients.
\begin{defi}[Compression moduli]
For a (generally discontinuous and nonlinear) map $X\maps \phi E$ between two Banach spaces we define the \emph{exact compression modulus} $\ov \kappa_\phi\colon [0,\infty[\,\to [0,\infty[$
\[
\ov\kappa_\phi(t)=\inf\big\{\norm{\phi(x)-\phi(y)}\del \norm{x-y}= t\big\}
\]
and the \emph{compression modulus} by $\ov\kappa_\phi\colon [0,\infty[\,\to [0,\infty[$ by
\[
\kappa_\phi(t)=\inf\big\{\norm{\phi(x)-\phi(y)}\del \norm{x-y}\geqslant t\big\}.
\]
\end{defi}

Thus, $\ov\kappa_\phi$ is the pointwise largest map so that $
\ov\kappa_\phi\big(\norm{x-x}\big)\leqslant \Norm{\phi(x)-\phi(y)} $ for all $x,y\in X$, while $\kappa_\phi(t)=\inf_{r\geqslant t}\ov\kappa_\phi(r)$ is the pointwise largest monotone map satisfying the same inequality.



\begin{defi}[Expansion moduli]
For a map $X\maps \phi E$ between Banach spaces, the \emph{exact expansion modulus} $\ov\om_\phi\colon [0,\infty[\,\to [0,\infty]$ is defined by
\[
\ov\om_\phi(t)=\sup\big\{\norm{\phi(x)-\phi(y)}\del \norm{x-y}= t\big\},
\]
and the \emph{expansion modulus} $\om_\phi\colon [0,\infty[\,\to [0,\infty]$ by
\[
\om_\phi(t)=\sup\big\{\norm{\phi(x)-\phi(y)}\del \norm{x-y}\leqslant t\big\}.
\]
\end{defi}

The following are evident.
\[
\kappa_\phi(t)\leqslant \ov \kappa_\phi(t)\leqslant \ov \om_\phi(t)\leqslant \om_\phi(t).
\]



We recall that, to avoid trivialities, all Banach spaces are assumed to have dimension at least $2$. Thus, suppose $X\maps \phi E$ is a map and that $t>0$ and $x,y\in X$. Let $n\geqslant 1$ be minimal so that $\norm{x-y}\leqslant nt$, whereby $(n-1)t\leqslant \norm{x-y}$ and pick $z_0=x,z_1,z_2,\ldots, z_n=y$ so that $\norm{z_{i-1}-z_i}=t$ for $i=1,\ldots, n$. Then
\[
\Norm{\phi(x)-\phi(y)}\;\leqslant\; \sum_{i=1}^n\Norm{\phi(z_{i-1})-\phi(z_i)}\;\leqslant\; n\cdot\ov\om_\phi(t)\leqslant \frac{\ov\om_\phi(t)}{t}\norm{x-y}+\ov\om_\phi(t).
\]
In turn, this shows that
\[
\om_\phi(s)\leqslant \frac{\ov\om_\phi(t)}{t}s+\ov\om_\phi(t)
\]
for all $s,t>0$ and so $\limsup_{s\to 0_+}\om_\phi(s)\leqslant \inf_{t>0}\ov\om_\phi(t)$. Because $\om_\phi$ is non-decreasing, the limit $\lim_{s\to 0_+}\om_\phi(s)=\inf_{s>0}\om_\phi(s)$ exists, whereby
\[
\inf_{t>0}\ov\om_\phi(t)
\leqslant \liminf_{t\to 0_+}\ov\om_\phi(t)
\leqslant \limsup_{t\to 0_+}\ov\om_\phi(t)
\leqslant\lim_{t\to 0_+}\om_\phi(t)
\leqslant \inf_{t>0}\ov\om_\phi(t).
\]
All in all, we find that
\[
\om(\phi)=\inf_{t>0}\om_\phi(t) =\lim_{t\to 0_+}\om_\phi(t) =\lim_{t\to 0_+}\ov\om_\phi(t)=\inf_{t>0}\ov\om_\phi(t).
\]
In particular, we would obtain nothing new by introducing an \emph{exact expansion coefficient} by $\ov{\om}(\phi)=\inf_{t>0}\ov\om_\phi(t)$, since this is just the expansion coefficient itself. Furthermore, if ${\om}(\phi)<\infty$, then $\phi$ is Lipschitz for large distances, that is,
\[
\Norm{\phi(x)-\phi(y)}\leqslant K\norm{x-y}+K
\]
for some constant $K$ and all $x,y\in X$.


Next, the definition of the separation ratio may initially be difficult to parse, so let us briefly restate it more explicitly.

\begin{lemm}
For a map $X\maps \phi E$ and a constant $K>0$, we have
\[
\ku R(\phi)>K
\]
if and only if there are constants $\Delta,\delta, \Lambda, \lambda>0$ so that
\[
\norm{x-y}\geqslant \Delta\;\saa\; \Norm{\phi(x)-\phi(y)}\geqslant \delta,
\]
\[
\norm{x-y}\leqslant \Lambda\;\saa\; \Norm{\phi(x)-\phi(y)}\leqslant \lambda
\]
and $\frac\delta\lambda>K$.
\end{lemm}


\begin{proof}
Note that, if $\ku R(\phi)>K$, we may find $\Delta,\Lambda>0$ so that $\frac{\kappa_\phi(\Delta)}{\om_\phi(\Lambda)}>K$. Letting $\delta=\kappa_\phi(\Delta)$ and $\lambda=\om_\phi(\Lambda)$, the two implications follow.

Conversely, if the two implications hold for some $\Delta,\delta, \Lambda, \lambda>0$ so that $\frac\delta\lambda>K$, then
\[
\ku R(\phi)=\frac{\sup_{t<\infty}\kappa_\phi(t)}{\inf_{t>0}\om_\phi(t)}\geqslant \frac{\kappa_\phi(\Delta)}{\om_\phi(\Lambda)}\geqslant \frac\delta\lambda>K,
\]
which verifies the lemma.
\end{proof}



\begin{proof}[Proof of Theorem~\ref{main}]As noted, if $X\maps \phi E$ is a coarse embedding between arbitrary Banach spaces, then $\ku R(\phi)=\infty$, which proves one direction of implication. Also, under the stated assumption on $E$, by Theorem~1.16 in~\cite{geomgroups}, we have that $X$ coarsely embeds into $E$ if and only if $\ku R(\phi)=\infty$ for some map $X\maps \phi E$. So suppose instead only that $\sup_{\phi}\ku R(\phi)=\infty$. We then construct a coarse embedding $X\maps \psi E$ as follows.

Because $E\oplus E$ embeds isomorphically into $E$, we may inductively construct two sequences $E_n, Z_n$ of closed linear subspaces of $E$ all isomorphic to $E$ so that
\[
E_{n+1}\oplus Z_{n+1}\subseteq Z_n.
\]
Concretely, we simply begin with an isomorphic copy $E\oplus E$ inside of $E$ and let $E_1$ and $Z_1$ be respectively the first and second summand. Again, pick an isomorphic copy of $E\oplus E$ inside of $Z_1$ with first and second summand denoted respectively $E_2$ and $Z_2$, etc. It thus follows that
\[
E\;\supseteq\; E_1\oplus Z_1\;\supseteq\; E_1\oplus E_2\oplus Z_2\;\supseteq\; E_1\oplus E_2\oplus E_3\oplus Z_3\;\supseteq\; \ldots
\]
is a decreasing sequence of closed linear subspaces of $E$. Let
\[
V_n =E_1\oplus E_2\oplus \cdots\oplus E_n\oplus Z_n
\]
and set $V=\bigcap_{n=1}^\infty V_n$. We note that $V$ is a closed linear subspace of $E$ in which each $E_n$ is a closed subspace complemented by a bounded projection~$V\maps{P_n}E_n$ so that $E_m\subseteq \ker P_n$ whenever $n\neq m$. On the other hand, we have no uniform bound on the norms $\norm{P_n}$.

Fix now a sequence of isomorphisms $E\maps {T_n}E_n$ and find maps $X\maps {\theta_n} E$ with $\ku R(\theta_n)>n\,2^n\norm{P_n}\norm{T_n}\norm{T_n\inv}$. Observe that, for all $t>0$,
\[
\kappa_{T_n\circ\theta_n}(t)\geqslant \frac{\kappa_{\theta_n}(t)}{\norm{T_n\inv}},
\]
whereas
\[
\om_{T_n\circ\theta_n}(t)\leqslant {\norm{T_n}}\cdot \om_{\theta_n}(t),
\]
which shows that
\[
\ku R(T_n\circ\theta_n)\geqslant \frac{\ku R(\theta_n)}{\norm{T_n}\norm{T_n\inv}}\geqslant n\,2^n\norm{P_n}.
\]
Setting $\phi_n=T_n\circ\theta_n$, we find that $\lim_n\frac{\ku R(\phi_n)}{2^n\norm{P_n}}=\infty$. The conclusion of the theorem therefore\linebreak follows directly from Lemma~\ref{unif saa coarse} below.
\end{proof}


\begin{lemm}\label{unif saa coarse}
Suppose $X$ and $E$ are Banach spaces and $E\maps{P_n} E$ is a sequence of bounded linear projections onto subspaces $E_n\subseteq E$ so that $E_m\subseteq \ker P_n$ for all $m\neq n$. Assume also that there is a sequence of maps
\[
X\maps {\phi_n} E_n
\]
so that
\[
\lim_n\frac {\ku R(\phi_n)}{2^n\norm{P_n}}=\infty.
\]
Then $X$ coarsely embeds into $E$.
\end{lemm}



\begin{proof}
By composing with a translation, we may suppose that $\phi_n(0)=0$ for each $n$. Because $\lim_n\frac {\ku R(\phi_n)}{2^n\norm{P_n}}=\infty$, we may also find constants $\Delta_n,\delta_n,\Lambda_n, \lambda_n>0$ so that
\[
\norm{x-y}\geqslant \Delta_n\;\saa\; \Norm{\phi_n(x)-\phi_n(y)}\geqslant \delta_n
\]
and
\[
\norm{x-y}\leqslant \Lambda_n\;\saa\; \Norm{\phi_n(x)-\phi_n(y)}\leqslant \lambda_n,
\]
while
\[
\lim_n\frac{\delta_n}{\lambda_n2^n\norm{P_n}}=\infty.
\]
For every $n$, we let
\[
\psi_n(x)=\frac{1}{\lambda_n2^n}\cdot\phi_n\Big(\tfrac{\Lambda_n}n\cdot x\Big).
\]
Then
\begin{equation}\label{upper}
\begin{split}
\norm{x-y}\leqslant n &\;\saa\; \Norm{\tfrac{\Lambda_n}n\cdot x-\tfrac{\Lambda_n}n\cdot y}\leqslant \Lambda_n\\
&\;\saa\; \NORM{\phi_n\big(\tfrac{\Lambda_n}n\cdot x\big)-\phi_n\big(\tfrac{\Lambda_n}n\cdot y\big)}\leqslant \lambda_n\\
&\;\saa\; \norm{\psi_n(x)-\psi_n(y)}\leqslant 2^{-n}.
\end{split}
\end{equation}
Similarly,
\begin{equation}\label{lower}
\begin{split}
\norm{x-y}\geqslant \frac{n\Delta_n}{\Lambda_n} &\;\saa\; \Norm{\tfrac{\Lambda_n}n\cdot x-\tfrac{\Lambda_n}n\cdot y}\geqslant \Delta_n\\
&\;\saa\;\NORM{\phi_n\big(\tfrac{\Lambda_n}n\cdot x\big)-\phi_n\big(\tfrac{\Lambda_n}n\cdot y\big)}\geqslant \delta_n\\
&\;\saa\; \Norm{\psi_n(x)-\psi_n(y)}\geqslant \frac{\delta_n}{\lambda_n2^n}.
\end{split}
\end{equation}
In particular, if $\norm{x-y}\leqslant m$, then $\norm{x-y}\leqslant n$ for all $n\geqslant m$, whereby
\[
\sum_{n=1}^\infty\Norm{\psi_{n}(x)-\psi_{n}(y)}\leqslant \sum_{n=1}^{m-1}\Norm{\psi_{n}(x)-\psi_{n}(y)}+\sum_{n=m}^\infty2^{-n}<\infty.
\]
Also, $\psi_n(0)=0$ for all $n$, which shows that, for all $x\in X$,
\[
\sum_{n=1}^\infty\Norm{\psi_{n}(x)}<\infty
\]
and so the series $\sum_{n=1}^\infty\psi_{n}(x)$ is absolutely convergent in $E$. We may therefore define a map $X\maps \psi E$ by letting
\[
\psi(x)=\sum_{n=1}^\infty\psi_{n}(x).
\]


We now verify that $\psi$ is a coarse embedding of $X$ into $E$. First, let $m\geqslant 1$ be any given natural number and suppose that $x,y\in X$ satisfy $\norm{x-y}\leqslant m$. Then we may find $z_0=x, z_1, z_2, \ldots, z_m=y$ so that $\norm{z_{i-1}-z_{i}}\leqslant 1$ for all $i$ and so, in particular, $\norm{\psi_n(z_{i-1})-\psi_n(z_i)}\leqslant 2^{-n}$ for all $n$. It thus follows that
\begin{align*}
\Norm{\psi(x)-\psi(y)} &=\NORM{\sum_{n=1}^\infty\psi_{n}(x)-\sum_{n=1}^\infty\psi_{n}(y)}\\
&\leqslant \sum_{n=1}^{m-1} \Norm{\psi_n(x)-\psi_n(y)} + \sum_{n=m}^\infty\Norm{\psi_n(x)-\psi_n(y)}\\
& \stackrel{\eqref{upper}}{\leqslant} \sum_{n=1}^{m-1} \Norm{\psi_n(z_0)-\psi_n(z_m)} + \sum_{n=m}^\infty2^{-n}\\
&\leqslant \sum_{n=1}^{m-1} \NORM{\sum_{i=1}^m\big(\psi_n(z_{i-1})-\psi_n(z_i)\big)} + 2^{-m+1}\\
&\leqslant \sum_{n=1}^{m-1} \sum_{i=1}^m\Norm{\psi_n(z_{i-1})-\psi_n(z_i)} + 2^{-m+1}\\
&\leqslant \sum_{n=1}^{m-1} \sum_{i=1}^m2^{-n} + 2^{-m+1}\\
&= \sum_{n=1}^{m-1} m2^{-n} + 2^{-m+1}\\
&< m+ 2^{-m+1}.
\end{align*}
In other words, for all $m$ and $x,y\in X$, we have
\[
\norm{x-y}\leqslant m\;\saa\; \Norm{\psi(x)-\psi(y)}< m+ 2^{-m+1}.
\]
Conversely, if $m$ is any given number, find $n$ large enough so that $\frac{\delta_n}{\lambda_n2^n\norm{P_n}}\geqslant m$. Then, if $\norm{x-y}\geqslant \frac{n\Delta_n}{\Lambda_n}$, we have
\begin{align*}
\norm{\psi(x)-\psi(y)} &\geqslant\frac1{ \norm{P_n}}\Norm{P_{n}\psi(x)-P_{n}\psi(y)}\\
&=\frac1{ \norm{P_n}} \Norm{\psi_n(x)-\psi_n(y)}\\
& \stackrel{\eqref{lower}}{\geqslant} \frac{\delta_n}{\lambda_n2^n\norm{P_n}}\\
&\geqslant m.
\end{align*}
Taken together, these two conditions show that $\psi$ is a coarse embedding.
\end{proof}

The gluing presented in Lemma~\ref{unif saa coarse} may be contrasted with the so-called barycentric gluing technique discussed in detail in~\cite{baudier}. In our gluing above, the purpose is to improve the metric qualities of maps $X\to E$, whereas the barycentric gluing allows one to paste together a sequence of maps $nB_X\to E$ defined on larger and larger balls of $X$, but where, on the other hand, the metric qualities of the maps are not improved. It is not clear whether the two techniques may be combined.

\begin{proof}[Proof of Theorem~\ref{main2}]
Suppose $X$ and $E$ are Banach spaces so that
\[
\sup_{\phi}\;\ov{\ku R}(\phi)=\infty,
\]
and $E$ have non-trivial type, i.e., $\type(E)>1$. We then note that also $\type\big(\ell_2(E)\big)=\type(E)>1$ and $\cotype\big(\ell_2(E)\big)=\cotype(E)$. Thus, if we can show that $X$ maps into $\ell_2(E)$ by a map that is Lipschitz for large distances and solvent, then, by the previously mentioned result of Braga (\cite[Theorem~1.3]{Braga2}), we will have that
\[
\cotype(X)\leqslant \cotype\big(\ell_2(E)\big)=\cotype(E).
\]
So fix a sequence of maps $X\maps{\phi_n}E$ so that $\ov{\ku R}(\phi_n)> n2^n$ for all $n\geqslant 1$. This means that there are $\Delta_n,\delta_n, \Lambda_n, \lambda_n>0$ so that
\[
\norm{x-y}=\Delta_n\;\saa\; \Norm{\phi_n(x)-\phi_n(y)}\geqslant \delta_n
\]
and
\[
\norm{x-y}\leqslant \Lambda_n\;\saa\; \Norm{\phi_n(x)-\phi_n(y)}\leqslant \lambda_n
\]
and $\tfrac{\delta_n}{\lambda_n}>n2^n$. We then define $\psi_n$ by $\psi_n(x)=\tfrac{1}{2^n\lambda_n}\phi_n\big(\tfrac{\Lambda_n}nx\big)$ and note that, as in~\eqref{upper} and~\eqref{lower},
\[
\norm{x-y}\leqslant n\;\saa\; \Norm{\psi_n(x)-\psi_n(y)}\leqslant 2^{-n},
\]
whereas
\[
\norm{x-y}=\tfrac{n\Delta_n}{\Lambda_n}\;\saa\; \Norm{\psi_n(x)-\psi_n(y)}\geqslant n.
\]
We finally define $X\maps\psi \ell_2(E)$ by $\psi(x)=\big(\psi_1(x),\psi_2(x),\ldots\big)$ and note that $\psi$ is well-defined by the above and satisfies $\om(\psi)\leqslant \om_\psi(1)\leqslant 1$ and $\ov\kappa(\psi)\geqslant \ov\kappa_\psi\big(\tfrac{n\Delta_n}{\Lambda_n}\big)\geqslant n$ for all $n$. So $\psi$ is Lipschitz for large distances and almost expanding, whence, by~\cite[Lemma~8]{equivariant}, $\psi$ is Lipschitz for large distances and solvent.
\end{proof}

Another way to prove Theorem~\ref{main2} is first to establish an analogue to Theorem~\ref{main} for the quantity $\sup_{\phi}\;\ov{\ku R}(\phi)$ in place of $\sup_{\phi}\;{\ku R}(\phi)$. This is done by observing that the proof of Theorem~\ref{main} above can be changed to prove the following statement.

\begin{theo}
Suppose $X$ and $E$ are Banach spaces so that $E\oplus E$ isomorphically embeds into $E$. Assume also that
\[
\sup_{\phi}\;\ov{\ku R}(\phi)=\infty,
\]
then there is a map $X\maps\phi E$ that is Lipschitz for large distances and solvent.
\end{theo}

In order to obtain Theorem~\ref{main2}, one then notes that $\ell_2(E)\oplus \ell_2(E)\iso\ell_2(E)$ and so, if $\sup_{\phi}\;\ov{\ku R}(\phi)=\infty$, where the supremum is taken over all maps $X\maps\phi E$, we have a map $X\maps\psi \ell_2(E)$ that is both Lipschitz for large distances and solvent.


\section{Examples}\label{sec:examples}
For every pair of Banach spaces $X$ and $E$, we define the \emph{coarse embeddability ratio} of $X$ in $E$ to be the numerical invariant
\[
\ku{CR}(X,E)=\sup\big\{\ku R(\phi)\del \phi\colon X\to E \text{ is a map }\!\big\}.
\]
This is simply the quantity appearing in Theorem~\ref{main}, which therefore states that, under very mild assumptions on $E$, we have $\ku{CR}(X,E)=\infty$ if and only if $X$ coarsely embeds into $E$. As the next example shows, the main interest lies in the case when $\ku{CR}(X,E)>1$, whereas $\ku{CR}(X,E)=1$ is easily obtained.

\begin{exem}\label{constant=1}
If $X$ is separable and $E$ is a Banach space that admits an infinite \emph{equilateral} set, that is, an infinite subset $A\subseteq E$ so that, for some $\delta>0$,
\[
\norm{x-y}=\delta
\]
for all distinct $x,y\in A$, then we have $\ku{CR}(X,E)\geqslant 1$. To see this, let $(Y_x)_{x\in A}$ be a partition of $X$ indexed by the set $A$ into subsets $Y_x\subseteq X$ of diameter at most $1$ and let $X\maps \phi E$ be defined by
\[
\phi(y)=x \;\equi \; x\in A\;\&\; y\in Y_x.
\]
Observe that, if $\norm{y-y'}>1$, then $y$ and $y'$ must belong to different pieces of the partition and so $\norm{\phi(y)-\phi(y')}=\delta$. On the other hand, $\norm{\phi(y)-\phi(y')}\leqslant \delta$ for all $y,y'\in X$, so we see that $
\kappa_\phi(t)=\delta$ for all $t>1$, whereas $\om_\phi(t)\leqslant \delta$ for all $t>0$. So $\ku R(\phi)\geqslant 1$.

In particular, this reasoning applies when $E$ is one of the classical Banach spaces $\ell_p$, $c_0$, $L_p$ or even the Tsirelson space $T$. Indeed, in these spaces, the standard unit basis $(e_n)_{n=1}^\infty$ is an infinite equilateral set (or, in the case of Tsirelson's space, $(e_n)_{n=2}^\infty$ is equilateral). Here we remark that $T^*$ is the reflexive space originally constructed and described by B.~S.~Tsirelson~\cite{tsirelson}, while $T$ is its $\ell_1$-asymptotic dual whose explicit construction was given by T.~Figiel and W.~B.~Johnson~\cite{figiel}.

Let us also observe that, if $E$ is infinite-dimensional, then $E$ admits an equivalent renorming with respect to which it has an infinite equilateral set. Indeed, since $E$ is infinite-dimensional, it contains a normalised basic sequence $(e_n)_{n=1}^\infty$. We define a new equivalent norm $\triple\cdot$ on the closed linear space $[e_n]_{n=1}^\infty$ by letting
\[
\Biggl|\!\Biggl|\!\Biggl|\sum_{n=1}^\infty a_ne_n\Biggr|\!\Biggr|\!\Biggr| =\sup\Biggl\{
\Biggl\|\sum_{n\in I} a_ne_n\Biggr\|+\Biggl\|\sum_{n\in J} a_ne_n\Biggr\| \;\Biggm| I, J \text{ are intervals and $i<j$ for all $i\in I$ and $j\in J$}\Biggr\}.
\]
As $\norm{e_n}=1$ for all $n$, we find that $\triple{e_i-e_j}=2$ for all $i<j$ and so $(e_n)_{n=1}^\infty$ is an equilateral set of the norm $\triple\cdot$. It now suffices to notice that $\triple\cdot$ extends to an equivalent norm on all of $E$.
\end{exem}

Example~\ref{constant=1} illustrates that the embeddability ratio $\ku{CR}(X,E)$ is sensitive to the specific norm on $E$, but not to the choice of norm on $X$. On the other hand, the condition $\ku{CR}(X,E)=\infty$ only depends on the isomorphism class of $E$. Note also that, if $X$, $Y$ and $Z$ are Banach spaces so that $\ku{CR}(X,Y)=\infty$, then
\[
\ku{CR}(X,Z)\geqslant \ku{CR}(Y,Z).
\]

An important non-embeddability result was recently established by F.~Baudier, G.~Lancien and T.~Schlumprecht~\cite{bls}, who showed that the separable Hilbert space $\ell_2$ does not coarsely embed into Tsirelson's space $T^*$. It is known that $T^*$ is minimal, that is, $T^*$ embeds isomorphically into all of its infinite-dimensional subspaces (see~\cite[Chapter~VI]{casazza}). Also, $T^*$ has an unconditional basis and can therefore be written as a direct sum of two infinite-dimensional subspaces. It therefore follows that $T^*\oplus T^*$ embeds isomorphically into $T^*$ and thus $E=T^*$ satisfies the assumption of Theorem~\ref{main}. It follows that the coarse embeddability ratio $\ku{CR}(\ell_2,T^*)$ is finite and we now proceed to give an upper bound.

\begin{prop}
If $T^*$ denotes Tsirelson's space, then
\[
\ku{CR}(\ell_2,T^*)\leqslant 4.
\]
\end{prop}

\begin{proof}
We rely on the analysis of~\cite{bls}, which also contains additional details about the construction below. For the proof, assume towards a contradiction that $\ell_2\maps{\phi}E$ satisfies $\ku R(\phi)>4$. Then by pre and post-composing $\phi$ with dilations we can suppose that, for some constants $\Delta>0$ and $\delta>4$, we have
\[
\norm{x-y}\geqslant \Delta\;\saa\; \Norm{\phi(x)-\phi(y)}\geqslant \delta
\]
and
\[
\norm{x-y}\leqslant \sqrt 2\;\saa\; \Norm{\phi(x)-\phi(y)}\leqslant 1.
\]
Let $(e_n)_{n=1}^\infty$ be the standard unit vector basis for $\ell_2$ and set $\eps=\frac{\delta-4}2$. Let also $k$ be large enough so that $\sqrt{2k}\geqslant \Delta$ and let $[\N]^k$ be the collection of all $k$-element subsets of $\N$ equipped with the \emph{Johnson metric},
\[
d_J(A,B)=\frac{|A\sym B|}2.
\]
Observe that $d_J$ is simply the shortest-path metric on the graph whose vertices is $[\N]^k$ and where two vertices $A$ and $B$ are connected by an edge provided that $|A\sym B|=2$. Let then $f\colon [\N]^k\to T^*$ be defined by
\[
f(A)=\phi\Big(\sum_{n\in A}e_n\Big).
\]
Observe that, if $d_J(A,B)=1$, then
\[
\NORM{\sum_{n\in A}e_n-\sum_{n\in B}e_n}=\sqrt{|A\sym B|}=\sqrt 2
\]
and so $\norm{f(A)-f(B)}\leqslant 1$. Thus, $f$ is Lipschitz with constant $1$.

By Proposition~4.1 of~\cite{bls} there is an infinite subset $\M\subseteq \N$ and some $y\in T^*$ so that, for any $A\in [\N]^k$ with $A\subseteq \M$, there are vectors $y_1^A, \ldots, y_k^A\in T^*$ with $\norm{y_i^A}\leqslant 1$ so that $y, y_1^A, \ldots, y_k^A$ form a finite block basis of the standard unit vector basis for $T^*$, $k\leqslant \min\supp(y^A_1)$ and
\[
\Norm{f(A)-(y+y_1^A+\cdots+y_k^A)}<\eps.
\]
In particular, for all $A,B\in [\N]^k$, $A,B\subseteq \M$, we have that
\begin{align*}
\norm{f(A)-f(B)} &< \Norm{y_1^A+\cdots+y_k^A}+\Norm{y_1^B+\cdots+y_k^B}+2\eps\\
&\leqslant 2 +2 +2\eps\\
&\leqslant \delta,
\end{align*}
where the second bound follows from (2.13) in~\cite{bls}. On the other hand, for any two disjoint $A, B\in [\N]^k$, we have
\[
\left\|\sum_{n\in A}e_n-\sum_{n\in B}e_n\right\|=\sqrt{2k}\geqslant \Delta,
\]
which implies that $\norm{f(A)-f(B)}\geqslant \delta$ and thus contradicts the preceding upper bound.
\end{proof}


The following still unsolved problem provides the main theoretical motivation for our investigations here.

\begin{enonce}{Problem}
Suppose $X$ and $E$ are Banach spaces. Is it true that $X$ coarsely embeds into $E$ if and only if it uniformly embeds?
\end{enonce}

\begin{enonce}{Problem}
Suppose $X$ and $E$ are Banach spaces so that $\ku{CR}(X,E) >1$. Does it follow that $\ku{CR}(X,E)=\infty$?
\end{enonce}


\subsection*{Acknowledgements}

I am very grateful for the extensive feedback and criticisms I got from B.~Braga on a first version of this paper and for suggesting a link with cotype that led to Theorem~\ref{main2}.

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