Continuity requirement for unitary representations Let $G$ be a locally compact group and let $\pi : G \rightarrow U(\mathcal{H}_\pi)$ be a group homomorphism into the group of unitary operators on a separable complex Hilbert space $\mathcal{H}_\pi$. Suppose that for every $v \in \mathcal{H}_\pi$ the map
$$G \ni g \mapsto \langle \pi(g)v,v\rangle \in \mathbb{C}$$
is measurable. I would like to show that this implies that $\pi$ is a unitary representation, that is, $\pi$ is continuous with respect to the strong operator topology on $\mathrm{B}(\mathcal{H}_\pi)$, meaning that for any $v \in \mathcal{H}_\pi$ the map
$$G \ni g \mapsto \pi(g)v \in \mathcal{H}_\pi$$
is continuous.
The separability of $\mathcal{H}_\pi$ is crucial for the problem at hand. However, I don't see how I can use this together with the fact that the map
$$G \ni g \mapsto \langle \pi(g)v,v\rangle \in \mathbb{C}$$
is measurable in order to deduce the continuity requirement above.
I was hoping that you could explain how to prove the statement above.
Thank you in advance.
 A: In what follows, I give a full proof of the statement above. I found particularly useful the arguments given by S. A. Gaal in his book ''Linear Analysis and Representation Theorey'' (see e.g. Theorems V.7.1, V.7.3), which I found thanks to the comment given by Ruy above.
To begin with, we show that for every $v,w \in \mathcal{H}$ the map $$G \ni g  \longmapsto \langle \pi_g v,w \rangle \in \mathbb{C}$$ is also measurable. Indeed, since sums of measurable functions are measurable, we see that
\begin{align*}
     \sum_{k=1}^4 i^k\langle \pi_g (v+i^kw), v+i^kw \rangle &= 4 \langle \pi_g v,w \rangle
 \end{align*}
depends measurable on $g$, too. We now give some preparatory arguments. Denote by $m$ a left Harr measure on $G$. Note that we can associate with $\pi$ a bounded representation $\pi_* : L^1(G) \rightarrow \mathrm{B}(\mathcal{H})$ via the weak integral
\begin{equation*}
     \pi_*(f)u = \int_G \pi_g u f(g) \mathrm{d} m(g),
 \end{equation*}
which means we define $\pi_*(f)u$ by the Fréchet-Riesz representation theorem and the formula
\begin{equation*}
     \langle \pi_*(f)u, w \rangle = \int_G \langle \pi_g u, w \rangle f(g) \mathrm{d} m(g)
 \end{equation*}
for all $w \in \mathcal{H}$. This is possible as $g \mapsto \langle \pi_g u, w \rangle$ is assumed to be measurable. Moreover, one has
\begin{equation*}
     \Big \vert \int_G \langle \pi_g u, w \rangle f(g) \mathrm{d} m(g) \Big \vert \leq \int_G \vert \langle \pi_g u, w \rangle \vert \vert f(g) \vert \mathrm{d} m(g) \leq \Vert u \Vert \Vert w \Vert \Vert f \Vert_1 
 \end{equation*}
and hence $\Vert \pi_*(f) \Vert \leq \Vert f \Vert_1$. Next, we decompose $\mathcal{H}$ into a direct sum of cyclic subspaces with respect to $\pi_*$. Let $v \in \mathcal{H}$ and define the closed $\pi_*$-invariant subspace
\begin{equation*}
     \mathcal{H}(v) := \overline{\langle \pi_*(L^1(G)) v \rangle_\mathbb{C}}.
 \end{equation*}
It is possible that $\mathcal{H}(v) = \{0\}$. Clearly, $\pi_*\vert_{\mathcal{H}(v)}$ is a cyclic representation with cyclic generator $v$. We are going to apply Zorn's lemma: consider families consisting of non-zero subspaces $\mathcal{H}_i$, $(i \in I)$, such that $\mathcal{H}_i \perp \mathcal{H}_j$ if $i \neq j$ and $\pi_*\vert_{\mathcal{H}_i}$ is cyclic. We partially order the collection of such families $\mathcal{H}_i$, $(i \in I)$, by the inclusion relation. Every linearly ordered subset has an upper bound, namely the union of the individual families in the chain. Thus by Zorn's lemma there exists a family $\mathcal{H}_i$, $(i \in I)$, which is maximal with respect to inclusion. Then
\begin{equation*}
     \mathcal{H}_0 = \Big ( \bigoplus_{i \in I} \mathcal{H}_i \Big ) ^\perp
 \end{equation*}
consists only of such vectors $v$ for which $\pi_*(L^1(G))v = 0$: otherwise $\mathcal{H}(v)$ could be joined to our family in contradiction to its maximality. Thus
\begin{equation*}
     \mathcal{H} = \mathcal{H}_0 \oplus \bigoplus_{i \in I} \mathcal{H}_i
 \end{equation*}
is the desired cyclic decomposition. Using the separability of $\mathcal{H}$, we can require $I$ to be countably infinite. We claim that $\mathcal{H}_0 = \{0\}$. Let $(v_1,v_2,\dots)$ be an orthonormal basis in $\mathcal{H}$ and let $v \in \mathcal{H}_0$. Then one has
\begin{equation*}
     \langle \pi_*(f)v,v_n \rangle = \int_G \langle \pi_g(v), v_n \rangle f(g) \mathrm{d} m(g) = 0
 \end{equation*}
for all $f \in L^1(G)$ and $n \geq 1$. Hence $\langle \pi_g(v), v_n \rangle = 0$ for $n \geq 1$ and for $m$-almost every $g \in G$. Since  $(v_1,v_2,\dots)$ is an orthonormal basis, we deduce that $\pi_g(v) = 0$ for $m$-almost every $g \in G$. Since $G$ is of positive or infinite measure there is $g_0 \in G$ such that $\pi_{g_0}(v) = 0$ and so $\pi_{g_0}$ being invertible we obtain $v = 0$. Since $v$ was arbitrary we conclude that $\mathcal{H}_0 = \{0\}$, as claimed. Now we go over to the proof of the main statement. Suppose first that $\pi_*$ is cyclic and define the linear subspace
\begin{equation*}
     \mathcal{H}' := \{\pi_*(f)v \: \vert \: f \in C_0(G)\}
 \end{equation*}
where $v$ is a cyclic vector for $\pi_*$. Then $\mathcal{H}'$ is dense in $\mathcal{H}$ because for every $f' \in L^1(G)$ there is an $f \in C_0(G)$ such that $\Vert f - f' \Vert_1 < \varepsilon$. The main idea to conclude the proof is to define a function $\tau : G \rightarrow \mathrm{B}(\mathcal{H})$ which will turn out to be a unitary representation and which satisfies $\tau = \pi$, showing that $\pi$ is a unitary representation. Given $g_0 \in G$, we first define $\tau_{g_0}$ only on $\mathcal{H}'$ by
\begin{equation*}
     \tau_{g_0} \xi = \tau_{g_0} \pi_*(f)v = \pi_*(\lambda_{g_0}f)v,
 \end{equation*}
where $\xi = \pi_*(f)v$ and $\lambda : G \rightarrow \mathrm{B}(L^1(G))$ is defined by $\lambda_g f(h) = f(g^{-1}h)$. We show that $\tau_{g_0}$ is an isometry on $\mathcal{H}'$:
\begin{align*}
     \Vert \tau_{g_0} \xi \Vert^2 &= \langle \pi_*(\lambda_{g_0}f)v, \pi_*(\lambda_{g_0}f)v \rangle \\
     &= \int_G \lambda_{g_0}f(g) \langle \pi_g v, \pi_*(\lambda_{g_0}f)v \rangle \mathrm{d} m(g) \\
     &= \int_G \int_G \lambda_{g_0}f(g) \overline{\lambda_{g_0}f(h)}\langle \pi_g v, \pi_h v \rangle \mathrm{d} m(g) \mathrm{d} m(h) \\
     &= \int_G \int_G f(g) \overline{f(h)}\langle \pi_g v, \pi_h v \rangle \mathrm{d} m(g) \mathrm{d} m(h) \\
     &= \langle \pi_*(f)v, \pi_*(f)v \rangle \\
     &= \Vert \xi \Vert^2,
 \end{align*}
where we used that $\pi$ is unitary and the fact that $m$ is left-invariant. Since $\mathcal{H}'$ is dense in $\mathcal{H}$ one can extend $\tau_{g_0}$ to a unitary transformation
\begin{equation*}
     \tau_{g_0}: \mathcal{H} \longrightarrow \mathcal{H}.
 \end{equation*}
Moreover, from
\begin{equation*}
     \tau_g (\tau_h \xi) = \tau_g \pi_*(\lambda_h f)v = \pi_*(\lambda_g \lambda_h f)v = \pi_*(\lambda_{gh} f)v = \tau_{gh} \xi,
 \end{equation*}
it follows that $\tau$ is a homomorphism with values in the group of unitary operators on $\mathcal{H}$. We shall now prove that $\tau$ is a unitary representation (satisfying the continuity requirement). Given $\xi = \pi_*(f)v$, one has
\begin{align*}
     \Vert \tau_g \xi - \xi \Vert &= \Vert \pi_*(\lambda_g f)v - \pi_*(f)v \Vert = \Vert \pi_*(\lambda_g f - f)v \Vert \\
     & \leq \Vert v \Vert \vert \pi_*(\lambda_g f - f) \Vert \leq \Vert v \Vert \Vert \lambda_g f - f \Vert_1.
 \end{align*}
Since $f \in C_0(G)$ by the right uniform continuity of $f$ we have $\Vert \lambda_g f - f \Vert_\infty \rightarrow 0$ as $g \rightarrow e$. Thus by the same reasoning as in the proof of the continuity of the left regular representation of $G$ one has $\Vert \lambda_g f - f \Vert_1 \rightarrow 0$ as $g \rightarrow e$. Hence $\tau_g \xi \rightarrow \xi$ as $g \rightarrow e$ for every fixed $\xi \in \mathcal{H}'$. Since $\mathcal{H}'$ is dense in $\mathcal{H}$ for any given $w \in \mathcal{H}$ and $\varepsilon > 0$ we can find a $\xi$ in $\mathcal{H}'$ such that $\Vert w - \xi \Vert < \varepsilon/3$. Since $\tau$ is unitary $\Vert \tau_g (w -\xi) \Vert < \varepsilon/3$ for every $g \in G$. We know the existence of a neighbourhood $U$ of $e$ in $G$ such that $\Vert \tau_g \xi - \xi \Vert < \varepsilon /3$ for every $g \in U$. Thus given $\varepsilon > 0$ we found $U$ such that
\begin{equation*}
     \Vert \tau_g w -w \Vert \leq \Vert \tau_g (w - \xi) \Vert + \Vert \tau_g \xi - \xi \Vert + \Vert \xi - w \Vert \leq \varepsilon
 \end{equation*}
for all $g \in U$. Hence $\tau$ is strongly continuous at $e$. Now we prove that $\tau = \pi$: for all $g_0 \in G$, $\xi \in \mathcal{H}'$, and $w \in \mathcal{H}$ one has
\begin{align*}
     \langle \tau_{g_0} \xi, w \rangle &= \langle \pi_*(\lambda_{g_0} f) v,w \rangle = \int_G f({g_0}^{-1} g) \langle \pi_g v, w \rangle \mathrm{d} m(g) \\
     &= \int_G f(g) \langle \pi_{g_0 g} v,w \rangle \mathrm{d} m(g) = \langle \pi_*(f) v, \pi_{g_0}^* w \rangle \\
     &= \langle \xi, \pi_{g_0}^* w \rangle = \langle \pi_{g_0} \xi, w \rangle.
 \end{align*}
This shows that $\tau_{g_0} = \pi_{g_0}$ on $\mathcal{H}'$ and consequently also on $\mathcal{H}$. Finally, we turn to the case where $\pi_*$ is not cyclic. As shown above we can decompose $\mathcal{H}$ into a direct sum
\begin{equation*}
     \mathcal{H} = \bigoplus_{i = 1}^\infty \mathcal{H}_i,
 \end{equation*}
where $\mathcal{H}_i$ for $i\geq 1$ is such that $\pi_*\vert_{\mathcal{H}_i}$ is cyclic. In order to prove that $\pi$ is strongly continuous at $e$ let $v \in \mathcal{H}$ and $\varepsilon > 0$ be given. We can write
\begin{equation*}
     v = w + v_{i_1} + \cdots + v_{i_n}
 \end{equation*}
where $v_{i_k} \in \mathcal{H}_k$ for $k = 1, \dots, n$ and $\Vert w \Vert < \varepsilon/4$. By the arguments we gave above we can choose neighbourhoods $U_k$ for $k = 1, \dots, n$ such that
\begin{equation*}
     \Vert \pi_g v_{i_k} - v_{i_k} \Vert < \frac{\varepsilon}{2n}
 \end{equation*}
for all $g \in U_k$. Hence for all $g \in \bigcap_{k=1}^n U_k$, one has
\begin{equation*}
     \Vert \pi_g v - v \Vert \leq 2\Vert w \Vert + \Vert \pi_g v_{i_1} - v_{i_1} \Vert + \cdots + \Vert \pi_g v_{i_n} - v_{i_n} \Vert < \varepsilon
 \end{equation*}
which is the desired conclusion and completes the proof.
