Induction in stages for closed subgroup chains

Write $\tau ={\mathrm{Ind}}_L^H\sigma$ and, for $F\in C_c(G,L;V)$, define $(UF)(x)(h)=r_x(hL{)}^{1\mathrm{/}2}F(xh),$ using the continuous covariant representative on $H$ for the inner section. Since $r_x$ is right-$L$ invariant and $F(xhl)=\sigma (l{)}^{-1}F(xh)$, this is an $L$-covariant inner section. For $k\in H$, the identity $r_{xk}(hL)=\frac{{\rho }_{HL}(kh)}{{\rho }_{HL}(h)}{r}_x(khL)$ and the induced action formula show $UF(xk)=\tau (k{)}^{-1}UF(x)$. Its outer support is contained in the image in $G\mathrm{/}H$ of the compact support of $F$ in $G\mathrm{/}L$. To check continuity in the inner norm near $x_0$, choose a compact neighborhood $C$ of $x_0$ and a compact lift $K_0\subset G$ of the quotient support of $F$. If $x\in C$ and $xhL\in {\mathrm{supp}}_{G\mathrm{/}L}F$, then $xh=zl$ for some $z\in K_0,l\in L$; hence $x^{-1}z=h{l}^{-1}\in H$ and $hL=x^{-1}zL$. The image in $H\mathrm{/}L$ of the closed subset $\{(x,z)\in C\times K_0:x^{-1}z\in H\}$ is a fixed compact set containing all these inner supports. Lift that compact set to a compact subset of $H$. On this lift, continuity of the rho ratios and of $F$ gives uniform convergence as $x\to x_0$; its quotient measure is finite, so the inner $L^2$ norm also converges. Thus $UF$ belongs to the continuous outer model.

Apply [F1] to the continuous compactly supported scalar function $q\mapsto \mathrm{\parallel }F(q){\mathrm{\parallel }}^2$. It gives $\mathrm{\parallel }UF{\mathrm{\parallel }}^2={\int }_{G\mathrm{/}H}{\int }_{H\mathrm{/}L}{r}_x(hL)\mathrm{\parallel }F(xh){\mathrm{\parallel }}^{2}d{\mu }_{HL}d{\mu }_{GH}={\int }_{G\mathrm{/}L}\mathrm{\parallel }F(q){\mathrm{\parallel }}^{2}d{\mu }_{GL}=\mathrm{\parallel }F{\mathrm{\parallel }}^2\mathrm{.}$ Hence $U$ extends to an isometry. The rho ratios also give $U{\mathrm{\Pi }}_{GL}(g)={\mathrm{\Pi }}_{GH}(g)U$: after expanding both sides, the only required cancellation is ${\rho }_{GH}(g^{-1}xh)\mathrm{/}{\rho }_{GH}(g^{-1}x)={\rho }_{GH}(xh)\mathrm{/}{\rho }_{GH}(x)$, which is exactly the rho covariance under $h\in H$.

By [F2], outer generators ${\mathrm{\Xi }}_f(v)(x)={\int }_{H}f(xk)\tau (k)vdk$ with $f\in C_c(G)$ and $v\in W={\mathrm{Ind}}_L^H\sigma$ have dense span. For fixed $f$, this generator depends continuously on $v$: choose a compact lift $C$ of $p_{GH}(\mathrm{supp}f)$; then $\mathrm{\parallel }{\mathrm{\Xi }}_f(v)(x)\mathrm{\parallel }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}dh(C^{-1}\mathrm{supp}f\cap H)\mathrm{\parallel }v\mathrm{\parallel }$, and its quotient support lies in the compact set $p_{GH}(\mathrm{supp}f)$. Since that set has finite measure, replacing $v$ by a dense inner compactly supported covariant section approximates ${\mathrm{\Xi }}_f(v)$ in the outer norm. It therefore suffices to treat such $v$. The resulting $\mathrm{\Xi }(x)$ has a continuous covariant representative $H\to V$, so evaluation at each $h\in H$ is defined. Outer covariance gives $\mathrm{\Xi }(xh)(e)={\left(\frac{{\rho }_{HL}(h)}{{\rho }_{HL}(e)}\right)}^{1\mathrm{/}2}\mathrm{\Xi }(x)(h)\mathrm{.}$ Set $F(y)=r_y(eL{)}^{-1\mathrm{/}2}\mathrm{\Xi }(y)(e)$. From the definition of $r$, $r_{xh}(eL)=r_x(hL)\frac{{\rho }_{HL}(h)}{{\rho }_{HL}(e)},$ so the displayed covariance identity gives $r_x(hL{)}^{1\mathrm{/}2}F(xh)=\mathrm{\Xi }(x)(h)$ for all $h\in H$. For $l\in L$, inner covariance gives $\mathrm{\Xi }(y)(l)=\sigma (l{)}^{-1}\mathrm{\Xi }(y)(e)$, and the same identities imply $F(yl)=\sigma (l{)}^{-1}F(y)$. The function $F$ is continuous: on a compact neighborhood of $y_0$, the $k$-integral defining $\mathrm{\Xi }(y)(e)$ is supported in a fixed compact subset of $H$, and its integrand is jointly continuous, so [F4] applies. Its support modulo $L$ is compact: nonzero values require $yk\in \mathrm{supp}f$ and $k^{-1}L\in \mathrm{supp}v$, hence lie in the image of a product of compact lifts of these supports. Thus $F\in C_c(G,L;V)$ and $UF=\mathrm{\Xi }$. The dense outer generators lie in the range, so the closed isometric range is the whole target. ∎