I asked a similar question for the case of compact groups not long ago in [math.stackexchange](https://math.stackexchange.com/questions/1114047/is-a-matrix-element-of-a-norm-continuous-representation-always-a-trigonometric-p). Now I understand that the answer was "yes", and I want to modify that question. This is also related to my recent question [here](https://mathoverflow.net/questions/194823/trigonometric-polynomials-on-non-compact-and-non-abelian-groups).


Let us say that a unitary representation $\pi:G\to {\mathcal B}(H)$ of a locally compact group $G$ is *norm-continuous*, if it is continuous with respect to the usual operator norm in ${\mathcal B}(H)$:
$$
t\_i\to t\quad\Longrightarrow\quad ||\pi(t\_i)-\pi(t)||\to 0.
$$


Let $\pi:G\to {\mathcal B}(H)$ be a *norm-continuous* unitary representation of a locally compact group $G$ in a Hilbert space $H$. Consider a matrix element of $\pi$, i.e. a function of the form 
$$
f(t)=\langle\pi(t)x,y\rangle,\quad t\in G,
$$
where $x,y\in H$. 


From the [paper by A.I.Shtern](http://link.springer.com/article/10.1134%2FS1061920808040134) it follows that if $G$ is compact, then $f$ is automatically a trigonometric polynomial, i.e. a (finite) linear combination of matrix elements of some unitary irreducible representations $\pi\_i:G\to {\mathcal B}(H\_i)$:
$$
f(t)=\sum\_{i=1}^n\lambda\_i\cdot\langle\pi\_i(t)x\_i,y\_i\rangle,\quad t\in G,
$$
($x\_i,y\_i\in H\_i$, and $\lambda\_i\in{\mathbb C}$).


A question:



> 
> Is the same true for wider classes of groups, in particular for Moore groups (i.e. where every unitary irreducible representation is finite-dimensional)?
> 
> 
> 



Is a matrix element of a norm continuous representation always a trigonometric polynomial?
In general, the answer is "no".


Let $G=\Bbb Z$; then all trigonometric polynomials are linear combinations of $f\_\xi\colon t\mapsto\xi^t$, $\xi\in\Bbb C$, $|\xi|=1$. On the other hand, consider the space $L^2(\Bbb Z)$, i.e., the space of sequences $\{a\_i\,|\,i\in\Bbb Z\}$ such that $\sum\_i|a\_i|^2<\infty$. This space has a canonical orthonormal basis $e\_i$, $i\in\Bbb Z$, and $\Bbb Z$ acts on it by the shifts $t\colon e\_i\mapsto e\_{i+t}$. Let, further, $x=y=e\_0$; the resulting "matrix entry" $f(t)$ is $0\mapsto1$, $t\mapsto0$ for $t\ne0$. This function is not a trigonometric polynomial, as for any finite collection $f\_{\xi\_i}$, $i=1,\ldots,N$, the only linear combination taking value $0$ at $t=1,\ldots,N$ is the trivial one (Vandermonde determinant $\ne0$); hence, it cannot take value $1$ at $t=0$.


