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Let $X$ be a set. We say that $\psi:X\times X\to[0,\infty)$ is a CND (conditionally negative definite) kernel if there is a Hilbert space $\mathcal{H}$ and a map $f:X\to\mathcal{H}$ such that \begin{align*} \psi(x,y)=\|f(x)-f(y)\|^2,\quad\forall x,y\in X. \end{align*}

Let us consider the following property for a countable group $\Gamma$.

(P) There is a CND kernel $\psi:\Gamma\times\Gamma\to[0,\infty)$ and a constant $C>0$ such that \begin{align*} \psi(sx,sy)\leq\psi(x,y)+C,\quad\forall s,x,y\in\Gamma, \end{align*} and $\psi(x,1)\to\infty$ as $x\to\infty$.

This can be thought of as an "almost equivariant" coarse embedding into a Hilbert space.

Question: Besides the class of groups that are not coarsely embeddable, are there groups that do not satisfy (P)?

The motivation behind this question comes from the fact that this property characterises groups admitting a proper uniformly Lipschitz affine action on a subspace of an $L^1$ space; see 1.

1 I. Vergara. Almost invariant CND kernels and proper uniformly Lipschitz actions on subspaces of $L^1$. arXiv:2109.12949

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    $\begingroup$ Just for context, a group for which this exists with $C=0$ is called a group with the Haagerup property. Do you know an example of a non-Haagerup group with such a kernel? $\endgroup$
    – YCor
    Commented May 12, 2023 at 19:37
  • $\begingroup$ @YCor: OP's paper provides such examples, e.g. all residually finite hyperbolic groups satisfying property (T) also satisfy (P). $\endgroup$ Commented May 13, 2023 at 14:47

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