I don't expect to find an explicit counterexample to my question, because ~~any example which was known to have the Haagerup property yet not have AP would have given an exact group without AP, and the existence of such groups~~
 the existence of exact groups without AP was open until the recent(ish) work of Lafforgue--de la Salle and de Laat--Haagerup. My understanding is that all these examples of groups without AP all have Kazhdan's property (T), so can't have the Haagerup property.


On the other hand, it is not obvious to me why the Haagerup property should imply AP. But perhaps some MO readers will know of an argument that proves such an implication? If not, what is an example of a group which is known to have the Haagerup property yet which is not known to have AP?


(One characterization of the Haagerup property for discrete groups is that there should exist a net of normalized, positive-definite functions on the group, each of which vanishes at infinity, such that the net converges pointwise to the constant function 1. The definition of the AP is given [here](https://mathoverflow.net/questions/199122/).) 


(**Correction 2015-04-15:** the original version mistakenly claimed, tacitly, that the Haagerup property implies exactness. My thanks to N. Ozawa for pointing out that this is not known either.) 



For discrete groups, does the Haagerup property imply the AP of Haagerup-Kraus?
The answer is no. It's solved in the following paper by Osajda.
<http://arxiv.org/abs/1406.5015>


