%~Mouliné par MaN_auto v.0.29.1 2023-11-10 19:16:59 \documentclass[CRMATH,Unicode,XML]{cedram} \TopicFR{Analyse fonctionnelle} \TopicEN{Functional analysis} \usepackage{amssymb} \DeclareMathOperator{\supp}{supp} \DeclareMathOperator{\cotype}{cotype} \DeclareMathOperator{\type}{type} \newcommand{\norm}[1]{\lVert#1\rVert} \newcommand{\Norm}[1]{\big\lVert#1\big\rVert} \newcommand{\NORM}[1]{\left\lVert#1\right\rVert} \newcommand{\triple}[1]{|\!|\!|#1|\!|\!|} \newcommand{\TRIPLE}[1]{\Big|\!\Big|\!\Big|#1\Big|\!\Big|\!\Big|} \newcommand{\maps}[1]{\mathop{\overset{#1}\longrightarrow}} \newcommand{\N}{\mathbb N} \newcommand{\M}{\mathbb M} \newcommand{\om}{\omega} \newcommand{\eps}{\epsilon} \newcommand{\iso}{\cong} \newcommand{\sym}{\vartriangle} \newcommand{\saa}{\Rightarrow} \newcommand{\equi}{\Leftrightarrow} \newcommand{\Del}{ \; \Big| \;} \newcommand{\del}{ \; \big| \;} \newcommand{\ku} {\mathcal} \newcommand{\ov}{\overline} \newcommand{\inv}{^{-1}} \graphicspath{{./figures/}} \newcommand*{\mk}{\mkern -1mu} \newcommand*{\Mk}{\mkern -2mu} \newcommand*{\mK}{\mkern 1mu} \newcommand*{\MK}{\mkern 2mu} \hypersetup{urlcolor=purple, linkcolor=blue, citecolor=red} \newcommand*{\relabel}{\renewcommand{\labelenumi}{(\theenumi)}} \newcommand*{\romanenumi}{\renewcommand*{\theenumi}{\roman{enumi}}\relabel} \newcommand*{\Romanenumi}{\renewcommand*{\theenumi}{\Roman{enumi}}\relabel} \newcommand*{\alphenumi}{\renewcommand*{\theenumi}{\alph{enumi}}\relabel} \newcommand*{\Alphenumi}{\renewcommand*{\theenumi}{\Alph{enumi}}\relabel} \let\oldexists\exists \renewcommand*{\exists}{\mathrel{\oldexists}} \let\oldforall\forall \renewcommand*{\forall}{\mathrel{\oldforall}} \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 $00$, 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 $i4$. 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}. \bibliographystyle{crplain} \bibliography{CRMATH_Rosendal_20230430} \end{document}