is in $\mathcal{U}_w$ and the map $$\Phi \mapsto \Phi \ast g$$ is continuous from $\mathcal{U}_w$ into $\mathcal{U}_w$.

Take now a (strongly continuous) unitary representation $\pi$ of $G$ acting on a separable complex Hilbert space $\mathcal{H}$ with scalar product $\langle{\cdot},{\cdot}\rangle_\mathcal{H}$ linear in the first entry. We assume that $\pi$ is reproducing, namely there exists a vector $u \in \mathcal{H}$ such that the corresponding voice transform $$Vv(x) = \langle{v},{\pi(x)u}\rangle_\mathcal{H},
  \qquad
  v \in \mathcal{H},\, x \in G,$$ is an isometry from $\mathcal{H}$ into $L_2(G)$. We observe that this implies that $V$ is injective, whence $\operatorname{span} \left\{\pi(x)u\right\}_{x \in G}$ is dense in $\mathcal{H}$.

We denote by $K$ the reproducing kernel $$K(x)
  =Vu(x)
  =\langle{u},{\pi(x)u}\rangle_\mathcal{H},
  \qquad
  x \in G, v \in \mathcal{H},
\label{eq:KernelDefinition}$$ which is a bounded continuous function and enjoys the following basic properties

$$\begin{aligned}
    & \overline{K} = \widecheck{K}, \label{eq:32a} \\
    & \sum_{i,j = 1}^n c_i \overline{c_j} K(x_i^{-1} x_j) \geq 0,
      \qquad
      c_1, \ldots, c_n \in \mathbb{C},\
      x_1,\ldots,x_n\in G, \label{eq:32b} \\
    & K \ast K = K \in L_2(G). \label{eq:32c}
  \end{aligned}$$

In general, $\pi$ is not assumed to be irreducible, but the reproducing assumption implies that $u$ is a cyclic vector. Properties  and  uniquely define the representation $\pi$ up to a unitary equivalence, see Theorem 