I asked a similar question for the case of compact groups not long ago in math.stackexchange. Now I understand that the answer was "yes", and I want to modify that question. This is also related to my recent question here.

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 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)?