The standard intertwining operator A(nu)

Let $C$ be a compact coordinate box in $G$, write $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, and set $B={\max }_{g\in C}\mathrm{\parallel }(a,b)\mathrm{\parallel }$ and $B_1={\max }_{g\in C}\mathrm{\parallel }(c,d)\mathrm{\parallel }$; these are finite and positive by [F8] and $\det g=1$. For any two real rows $x,y$, direct expansion gives $\mathrm{\parallel }x{\mathrm{\parallel }}^2\mathrm{\parallel }y{\mathrm{\parallel }}^2-\det (x,y{)}^2=(x\cdot y{)}^2\ge 0$. Thus the rows of $g$ give $1\le \mathrm{\parallel }(a,b)\mathrm{\parallel }\mathrm{\parallel }(c,d)\mathrm{\parallel }$ and hence $\mathrm{\parallel }(c,d)\mathrm{\parallel }\ge B^{-1}$. The rows of $w{n}_{u}g$ are $(-c,-d)$ and $(a+uc,b+ud)$, so its determinant also gives $\mathrm{\parallel }(a+uc,b+ud)\mathrm{\parallel }\ge B_1^{-1}$. Set $R:=\max (1,2B^2)$. If $\mathrm{\mid }u\mathrm{\mid }\ge R$, the triangle inequality gives $\mathrm{\parallel }(a+uc,b+ud)\mathrm{\parallel }\ge \mathrm{\mid }u\mathrm{\mid }\mathrm{/}(2B)\ge (1+\mathrm{\mid }u\mathrm{\mid })\mathrm{/}(4B)$; for $\mathrm{\mid }u\mathrm{\mid }\le R$ the determinant bound gives $\mathrm{\parallel }(a+uc,b+ud)\mathrm{\parallel }\ge (1+\mathrm{\mid }u\mathrm{\mid })\mathrm{/}(B_1(1+R))$. Thus for $c_C:=\min ((4B{)}^{-1},(B_1(1+R){)}^{-1})>0$, the bottom-row norm $r(u,g)$ is at least $c_C(1+\mathrm{\mid }u\mathrm{\mid })$ throughout $C$. The compact-picture formula [F2] therefore gives $\mathrm{\mid }\phi (w{n}_{u}g)\mathrm{\mid }\le \mathrm{\parallel }\phi {\mathrm{\mid }}_K{\mathrm{\parallel }}_{\mathrm{\infty }}{c}_C^{-1-\sigma }(1+\mathrm{\mid }u\mathrm{\mid }{)}^{-1-\sigma },\sigma =\mathrm{Re}\nu >0.$ The majorant is integrable: for $T>0$, the antiderivative supplied by [F4] gives ${\int }_0^T(1+u{)}^{-1-\sigma }du=(1-(1+T{)}^{-\sigma })\mathrm{/}\sigma \to 1\mathrm{/}\sigma$; reflection $u\mapsto -u$ in [F5] gives the same finite integral on $(-\mathrm{\infty },0)$, and [F5] combines the two pieces. Taking $C$ to contain any fixed $g$ proves absolute convergence there.

For each fixed $g$, step 1.1 makes $u\mapsto \phi (w{n}_{u}g)$ integrable; the integrand is continuous, hence measurable by [F8]. Thus the displayed formula defines a value for every $g$. Applying [F6] to the integrands for ${\phi }_1,{\phi }_2$ and their linear combination, which are all integrable by step 1.1, shows $A(\nu )$ is complex-linear.

In a smooth coordinate chart $g=\gamma (z)$ and on any compact sub-box, the map $(u,z)\mapsto \phi (w{n}_u\gamma (z))$ is smooth by [F7]. Every coordinate derivative is a finite sum of right derivatives of $\phi$ at $w{n}_u\gamma (z)$ with smooth coefficients bounded on that sub-box. Each such right derivative remains left-$P$-covariant with character ${\chi }_{\epsilon ,\nu }$, and its restriction to compact $K$ is bounded; therefore the estimate of step 1.1 gives one integrable majorant $C_{\alpha }(1+\mathrm{\mid }u\mathrm{\mid }{)}^{-1-\sigma }$ for each coordinate derivative ${\mathrm{\partial }}_z^{\alpha }$, uniformly on the sub-box. These derivatives are continuous in $u$ and hence measurable by [F8]. Applying the differentiation-under-the-integral theorem [F6] successively to the coordinates gives ${\mathrm{\partial }}_z^{\alpha }(A(\nu )\phi )(\gamma (z))={\int }_{\mathbb{R}}{\mathrm{\partial }}_z^{\alpha }[\phi (w{n}_u\gamma (z))]du$; dominated convergence [F6] makes each such derivative continuous in $z$. All coordinate derivatives therefore exist and are continuous, so $A(\nu )\phi$ is smooth.

For $n_x\in N$, $n_u{n}_x=n_{u+x}$, so translation change of variables [F5] gives $A(\nu )\phi (n_{x}g)=A(\nu )\phi (g)$. For $a_s\in A$, use $n_u{a}_s=a_s{n}_{e^{-s}u}$ and $w{a}_s=a_{-s}w$; covariance contributes $e^{-(1+\nu )s\mathrm{/}2}$, and the change of variables for integrable complex functions $v=e^{-s}u$ contributes $e^s$. Hence $A(\nu )\phi (a_{s}g)=e^{(1-\nu )s\mathrm{/}2}A(\nu )\phi (g)\mathrm{.}$ For $m\in M=\{\pm I\}$, centrality gives $w{n}_{u}mg=mw{n}_{u}g$, so $A(\nu )\phi (mg)={\sigma }_{\epsilon }(m)A(\nu )\phi (g)$. These are exactly the $P=MAN$ covariance rules for $I_{\epsilon ,-\nu }$. For every $g_0\in G$, $A(\nu ){\mathrm{\Pi }}_{\nu }(g_0)\phi (g)={\int }_{\mathbb{R}}\phi (w{n}_{u}g{g}_0)du=({\mathrm{\Pi }}_{-\nu }(g_0)A(\nu )\phi )(g),$ so the operator intertwines right translations. Together with step 2.2 this proves that its target is the smooth model $I_{\epsilon ,-\nu }$.

The compact picture identifies source and target with the same parity space and its K-types with the one-dimensional lines $\mathbb{C}{f}_n$. Since step 3.1 intertwines every right translation, $A(\nu )$ maps each finite-dimensional right-$K$ orbit span into a finite-dimensional right-$K$ orbit span. If $f_n$ is a K-type vector, its image has the same right-$K$ character; the corresponding target character space is exactly $\mathbb{C}{f}_n$ by [F3]. Hence $A(\nu )f_n=c_n(\nu )f_n$ for a scalar $c_n(\nu )$ and every allowed $n$, proving the K-finite-target and diagonalization assertions.

For $\epsilon =0$, the compact vector $f_0=1$ extends by [F2]. At $g=I$, the bottom row of $w{n}_u$ is $(1,u)$, so its norm is $r=\sqrt{1+u^2}$. The positive-real-log convention for complex powers gives $\phi (w{n}_u)=r^{-1-\nu }=(1+u^2{)}^{-(1+\nu )\mathrm{/}2}$ and $c_0(\nu )=(A(\nu )f_0)(I)={\int }_{\mathbb{R}}(1+u^2{)}^{-(1+\nu )\mathrm{/}2}du\mathrm{.}$ For real $\nu >0$ the integrand is positive and on $[0,1]$ is at least $2^{-(1+\nu )\mathrm{/}2}$; that interval has measure $1$ by [F5], so $c_0(\nu )>0$ by monotonicity [F4]. For complex $\mathrm{Re}\nu >0$, the integrand is even and absolutely integrable by step 1.1. Splitting off the null endpoint and reflecting the negative half-line by [F5] gives twice its integral on $(0,\mathrm{\infty })$. Under the C1 diffeomorphism $t=u^2\mathrm{/}(1+u^2):(0,\mathrm{\infty })\to (0,1)$, one has $dt\mathrm{/}du=2u\mathrm{/}(1+u^2{)}^2$ and $t^{-1\mathrm{/}2}(1-t{)}^{\nu \mathrm{/}2-1}\frac{dt}{du}=2(1+u^2{)}^{-(1+\nu )\mathrm{/}2}\mathrm{.}$ The change-of-variables formula for integrable complex functions [F5] therefore yields $c_0(\nu )={\int }_0^1{t}^{-1\mathrm{/}2}(1-t{)}^{\nu \mathrm{/}2-1}dt=B(1\mathrm{/}2,\nu \mathrm{/}2),$ as claimed by [F10]. For $\epsilon =1$, the same integral defines the formal scalar $c_0(\nu )$ but is not an eigenvalue because $f_0$ is not an allowed K-type; its scalar continuation is proved by K-type eigenvalues of A(nu): recurrence, closed form and nonvanishing ↗, while the full smooth-operator continuation is supplied by Meromorphic continuation and intertwining identity for A(nu) ↗.