Let $G$ be a locally compact Hausdorff group and $H$ a closed subgroup of $G$. If $\sigma: H \to B(\mathcal{K}_\sigma)$ is a unitary representation of $G$, we can associate an "induced representation", denoted by $\operatorname{Ind}_H^G(\sigma)$.
I'll briefly recall a possible construction for this representation. All details can be found in appendix E of the book "Kazhdan's property (T)" by B. Bekka, P. de la Harpe and A. Valette or in Folland's book "A course on abstract harmonic analysis" in chapter 6. Let $p: G \to G/H$ the canonical quotient map. Start by defining the space $$\mathcal{A}:= \{f \in C(G, K_\sigma): p(\operatorname{supp}(f)) \text{ is compact}, f(gh) = \sigma(h^{-1})f(g) \text{ for } g\in G, h \in H\}.$$ Fix a strongly quasi-invariant measure $\mu$ on $G/H$ and consider the associated continuous function $$\lambda: G\times (G/H)\to (0, \infty)$$ such that $\lambda(x,yH)= \frac{d\mu_x}{d\mu}(yH).$ The space $\mathcal{A}$ becomes an inner product space for $$\langle f,g\rangle := \int_{G/H} \langle f(x), g(x)\rangle_\sigma d\mu(xH) $$ and we denote the Hilbert space completion of $\mathcal{A}$ by $\mathcal{H}_\mu$. Then the map $$\pi_\mu: \mathcal{A}\to \mathcal{A}$$ defined by $$(\pi_\mu(x)f)(y) = \sqrt{\lambda(x,yH)}f(x^{-1}y)$$ extends to a representation and this is $\operatorname{Ind}_H^G(\sigma): G \to B(\mathcal{H}_\mu)$.
Consider now the following theorem in the book "Kazhdan’s Property (T)" by B. Bekka, P. de la Harpe and A. Valette:
There are a lot of details to be filled in, but the main strategy is clear: we define $$U: \mathcal{H}_\pi\odot \mathcal{A}\to \mathcal{L}$$ (where $\mathcal{A}$ is defined as above) and we want to show that this map extends to a unitary $$U: \mathcal{H}_\pi\otimes \mathcal{H}\to \mathcal{L}$$ that implements the equivalence of the two representations. However, one thing is not clear to me in this proof: Why is $U$ surjective? I see that $U$ is isometric on the algebraic tensor product, so it extends uniquely to an isometry $$U: \mathcal{H}_\pi\otimes \mathcal{H}\to \mathcal{L}$$ and it therefore has closed range. Hence, it suffices to show that range of $U$ is dense in the codomain, but I did not manage to prove this.
Lemma E.1.3 in the aforementioned book might be relevant, but I could not make it work (I also believe that the proof of this lemma contains a mistake, but it can be fixed using a partition of unity argument).
Any help/insights is highly appreciated!
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