An irreducible continuous unitary representation $\pi$ of $G$ is said to be integrable, if the map $\phi(x)=\langle\pi(x)\zeta,\zeta\rangle$ is integrable on $G$, where that $\zeta\in H(\pi)$.


**Question1**: Suppose $G$ is an unimodular locally compact group. Can we say that $G$ always has an (irreducible) integrable representation?


**Question 2**: If the answer of the above question is no, is there an example of a non-compact group for which the answer of question 1 is yes?



Existence of an integrable representation
[Thanks to Loren Spice for fixing the references and pointing out a silly error/mis-statement in the original version of this answer.]




---


The answer to Q1 is no for $G={\mathbb Z}$, since all its irreducible representations are one-dimensional, and hence all the associated coefficient functions ${\mathbb Z} \to {\mathbb C}$ take values in ${\mathbb T}$.


The answer to Q2 is positive, but I am not aware of any easy approach that naturally leads one to an explicit example. However, explicit examples are known, by the following line of thought.


First: it is a non-trivial result that $G$ having an integrable irreducible representation $\pi$ implies that $\{\pi\}$ is an open subset of the unitary dual $\widehat{G}$. (This result is, to my knowledge, due to Duflo and Moore in [Duflo and Moore - On the regular representation of a nonunimodular locally compact group](http://www.sciencedirect.com/science/article/pii/0022123676900793) ([MSN](http://www.ams.org/mathscinet-getitem?mr=393335)); but see also [Valette - Minimal projections, integrable representations and property (T)](https://link.springer.com/article/10.1007%2FBF01193846) ([MSN](http://www.ams.org/mathscinet-getitem?mr=773186)) for an easier proof in the unimodular case).


So for Q2, one should look for unimodular non-compact $G$ for which $\widehat{G}$ has an open point $\{\pi\}$, and then check if $\pi$ really is integrable.
Note that since a function $\phi$ of the form in your question is always continuous and bounded, if it is integrable then it belongs to every $L^p$. In particular integrable (irreducible) representations are always *square-integrable*, and there is a substantial literature on which groups admit square-integrable representations.


Moreover, it turns out that for certain classes of groups -- including all connected nilpotent Lie groups -- an irreducible unitary representation $\pi$ is integrable if and only if it is *square-integrable*, if and only if $\pi$ is an open point in the unitary dual. See Theorem 3 of [Barnes - The role of minimal idempotents in the representation theory of locally compact groups](https://doi.org/10.1017/S0013091500003114) ([MSN](http://www.ams.org/mathscinet-getitem?mr=597128)).


Therefore, to get a positive answer to Q2, it suffices to exhibit a connected nilpotent Lie group $G$ such that $\widehat{G}$ has at least one open point. A concrete example of such a group is the *reduced Heisenberg group*, defined as $G = H\_3({\mathbb R})/Z$ where


$$ H\_3({\mathbb R}) = \left\{ \left( \begin{matrix} 1 & p & \theta \\ 0 & 1 & q \\ 0 & 0 & 1 \end{matrix} \right) \;\colon\; p,q,\theta\in {\mathbb R} \right\}
\quad,\quad
Z = \left\{ \left( \begin{matrix} 1 & 0 & n \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right) \;\colon\; n\in {\mathbb Z} \right\}
 $$
with an explicit integrable irreducible representation $\pi: G \to {\mathcal B}(L^2({\mathbb R}))$ given by
$$
(\pi(p,q,\theta+{\mathbb Z})\;f )(s) = e^{2\pi i \theta}e^{2\pi i qs}f(s+p)
$$


