Meromorphic continuation and intertwining identity for A(nu)

For $r\equiv \epsilon (\mathrm{m}\mathrm{o}\mathrm{d}2)$, write the Gamma expression from [F2] as $c_r(\nu )=(-i{)}^r\sqrt{\pi }\mathrm{\Gamma }(\nu \mathrm{/}2)\mathrm{\Gamma }((\nu +1)\mathrm{/}2)/(\mathrm{\Gamma }((\nu +r+1)\mathrm{/}2)\mathrm{\Gamma }((\nu -r+1)\mathrm{/}2))$. Its numerator has no zeros and only simple poles at nonpositive integers. The denominator contributes zeros only when $\nu =-r-1-2j$ or $\nu =r-1-2j$ for some $j\ge 0$, and each such parameter belongs to ${\mathcal{W}}_{\epsilon }$; hence $c_r$ has no zeros outside that lattice. At a possible numerator pole $\nu =-m$, if $m\equiv \epsilon (\mathrm{m}\mathrm{o}\mathrm{d}2)$ the denominator arguments are half-integers and do not cancel the simple pole; if $m\equiv ̸\epsilon (\mathrm{m}\mathrm{o}\mathrm{d}2)$, the two denominator arguments are integers with at least one nonpositive, so a reciprocal-Gamma zero cancels the numerator pole. Thus all scalar poles are simple and lie in the locally finite set $P_{\epsilon }=\{-m:m\in {\mathbb{N}}_0,m\equiv \epsilon (\mathrm{m}\mathrm{o}\mathrm{d}2)\}$.

For each center ${\nu }_{\ast }\in \mathbb{C}$, choose $\rho >0$ so that the closed disc $D=\overline{D({\nu }_{\ast },\rho )}$ has boundary disjoint from $P_{\epsilon }$ and its intersection with $P_{\epsilon }$ is either empty or $\{{\nu }_{\ast }\}$; this is possible because $P_{\epsilon }$ is locally finite. Put $h(\nu )=1$ if ${\nu }_{\ast }\mathrm{\notin }{P}_{\epsilon }$ and $h(\nu )=\nu -{\nu }_{\ast }$ if ${\nu }_{\ast }\in P_{\epsilon }$. Every $h{c}_r$ is holomorphic on a neighbourhood of $D$ by step 1.1. Let $M=\mathrm{\mid }{\nu }_{\ast }\mathrm{\mid }+\rho$, so $\mathrm{\mid }\nu \mathrm{\mid }\le M$ on $D$ by the triangle inequality for the complex modulus. Choose a nonnegative $r_0$ in the parity lattice so large that $r+1\ge 2M$ for every $r\ge r_0$, and choose an integer $L\ge \max (1,4M)$. The cross-multiplied recurrence in [F2], with $h$ multiplied into both sides, gives $\mathrm{\mid }h(\nu )c_{r+2}(\nu )\mathrm{\mid }\le \frac{r+1+M}{r+1-M}\mathrm{\mid }h(\nu )c_r(\nu )\mathrm{\mid }\le (1+\frac{L}{r+1})\mathrm{\mid }h(\nu )c_r(\nu )\mathrm{\mid }$ for $\nu \in D$ and $r\ge r_0$; the denominator here is the scalar $\mathrm{\mid }r+1+\nu \mathrm{\mid }\ge r+1-M>0$, and no division by $c_r(\nu )$ is used. Since $h{c}_{r_0}$ is bounded on $D$, iteration bounds $\mathrm{\mid }h(\nu )c_r(\nu )\mathrm{\mid }$ by a constant times ${\prod }_{k=1}^N(1+L\mathrm{/}k)=\left(\genfrac{}{}{0px}{}{N+L}{L}\right)\le (N+L{)}^L$, where $N$ is the number of recurrence steps from $r_0$ to $r$; hence $\mathrm{\mid }h(\nu )c_r(\nu )\mathrm{\mid }\le C(1+\mathrm{\mid }r\mathrm{\mid }{)}^L$ uniformly. The finitely many smaller indices are bounded by holomorphy, and negative indices obey the same estimate by the symmetry in [F2].

Using the angular Haar normalization in [F3], put $\widehat{f}(r)=(2\pi {)}^{-1}{\int }_0^{2\pi }f(k_{\theta })e^{-ir\theta }d\theta$. For every positive integer $N$, periodicity removes the boundary terms in [F6], giving $\mathrm{\mid }\widehat{f}(r)\mathrm{\mid }\le p_N(f)\mathrm{\mid }r{\mathrm{\mid }}^{-N}$ for $r\mathrm{\ne }0$, where $p_N(f)={\max }_{0\le j\le N}\mathrm{\parallel }{\mathrm{\partial }}_{\theta }^{j}f{\mathrm{\parallel }}_{\mathrm{\infty }}$. For each derivative order $q$, take $N=q+2$; integration by parts gives $(ir{)}^q\widehat{f}(r)=\widehat{{\mathrm{\partial }}_{\theta }^{q}f}(r)$ and the differentiated Fourier terms are bounded by $p_N(f)\mathrm{\mid }r{\mathrm{\mid }}^{-2}$. The M-test and p-series fact [F6], applied separately to the positive and negative parity tails, give uniform convergence of the input Fourier series and every derivative series. The uniform derivative-limit theorem applied successively to real and imaginary parts shows these limits are the derivatives of the sum; their Fourier coefficients match those of $f$ and its derivatives, so completeness in [F3] and continuity identify them with those functions. Hence Fourier partial sums converge to $f$ in every $C^q$ seminorm. Next, for each angular derivative order $q$, choose $N=L+q+2$ in step 2.1; the differentiated terms of ${\sum }_{r}h(\nu )c_r(\nu )\widehat{f}(r)e^{ir\theta }$ are bounded by $C{p}_N(f)(1+\mathrm{\mid }r\mathrm{\mid }{)}^{-2}$. Applying the M-test separately to the positive and negative parity tails gives uniform convergence on the circle, locally uniformly in $\nu$, for this multiplier series and every angular derivative; applying the uniform derivative-limit theorem to real and imaginary parts shows its sum is smooth and the derivatives are the termwise derivatives. In particular each regular $A(\nu )$ defined by this series is a continuous linear map $C_{\epsilon }^{\mathrm{\infty }}(K)\to C_{\epsilon }^{\mathrm{\infty }}(K)$, with each output seminorm bounded by a constant times one input seminorm.

The same polynomial bound makes the meromorphic family holomorphic in the smooth topology after multiplication by $h$. Choose concentric parameter discs $\overline{D({\nu }_{\ast },R)}\subset D({\nu }_{\ast },S)$ with $0<R<S$ and closed radius-$S$ disc contained in the neighbourhood from step 2.1. Cauchy's coefficient estimate [F5] gives Taylor coefficients $a_{r,j}$ of $h{c}_r$ at ${\nu }_{\ast }$ with $\mathrm{\mid }{a}_{r,j}\mathrm{\mid }\le C(1+\mathrm{\mid }r\mathrm{\mid }{)}^L{S}^{-j}$. For each $j$, the Fourier multiplier $T_{j}f={\sum }_r{a}_{r,j}\widehat{f}(r)e^{ir\theta }$ is smooth by the argument of step 3.1 and satisfies $p_q(T_{j}f)\le C_q{S}^{-j}{p}_{L+q+2}(f)$. Therefore, on every smaller closed parameter disc of radius $\rho <S$, ${\sum }_{j\ge 0}{T}_{j}f(\nu -{\nu }_{\ast }{)}^j$ converges in every $C^q$ seminorm, uniformly for $f$ in bounded subsets of $C_{\epsilon }^{\mathrm{\infty }}(K)$. Its Fourier coefficients are the holomorphic values $(h{c}_r)(\nu )\widehat{f}(r)$; for regular $\nu$, completeness in [F3] makes the sum equal to $h(\nu )A(\nu )f$, and its value at ${\nu }_{\ast }$ is the removable extension when ${\nu }_{\ast }\in P_{\epsilon }$ (and the ordinary value otherwise). This is the required local power-series definition of a meromorphic family of continuous K-diagonal maps on $C_{\epsilon }^{\mathrm{\infty }}(K)$.

On $\mathrm{Re}\nu >0$, the operator from [F1] agrees with the Fourier multiplier in step 3.1: they agree on finite Fourier sums by [F2], and for a general smooth $f$ its Fourier partial sums converge uniformly by step 3.1 while the compact-picture integral satisfies $\mathrm{\parallel }A(\nu )(f-f_N){\mathrm{\parallel }}_{\mathrm{\infty }}\le C_{\nu }\mathrm{\parallel }f-f_N{\mathrm{\parallel }}_{\mathrm{\infty }}$, since for $k\in K$ the bottom-row norm in the integral is $\sqrt{1+u^2}$ and $(1+u^2{)}^{-(1+\mathrm{Re}\nu )\mathrm{/}2}$ is integrable. Thus the series is a continuation of the stated integral. Any other meromorphic K-diagonal continuation has on each K-type a scalar meromorphic coefficient agreeing with $c_r$ on this open half-plane; clearing local pole factors and applying [F9] on connected parameter discs makes the coefficients equal as meromorphic functions. The two continuous operators then agree on finite Fourier sums and, by the density from step 3.1, on all of $C_{\epsilon }^{\mathrm{\infty }}(K)$, proving uniqueness.

The recurrence in [F2], interpreted cross-multiplied at zero denominators, gives $A(\nu )L_{E_{+}}^{\nu }{f}_r=L_{E_{+}}^{-\nu }A(\nu )f_r$ on every allowed K-type; the same recurrence at index $r-2$ gives the $E_{-}$ identity, and K-diagonality gives the $W$ identity. These are meromorphic scalar equalities, so they hold at every regular parameter. By linearity they hold for the real generators $J,H,S$ on finite Fourier sums; their operators are continuous on $C^{\mathrm{\infty }}$ by [F7], and step 3.1 gives density of Fourier sums in that topology, so the derived intertwining identities hold on all smooth vectors.

Fix a real generator $X\in \{J,H,(S+J)\mathrm{/}2\}$ and a smooth $f$. The smooth orbit maps, continuity of $A(\nu )$ from step 3.1, and the real chain rule [F7] make $F(t)={\mathrm{\Pi }}_{-\nu }(\exp (-tX))A(\nu ){\mathrm{\Pi }}_{\nu }(\exp (tX))f$ continuous and differentiable in every smooth seminorm. Its derivative is ${\mathrm{\Pi }}_{-\nu }(\exp (-tX))(-L_X^{-\nu }A(\nu )+A(\nu )L_X^{\nu }){\mathrm{\Pi }}_{\nu }(\exp (tX))f=0$ by step 4.3. Evaluation at each $k_{\theta }$ gives a complex-valued function of $t$ whose real and imaginary parts satisfy [F10], so $F(t)=F(0)$ pointwise. Therefore $A(\nu ){\mathrm{\Pi }}_{\nu }(\exp (tX))={\mathrm{\Pi }}_{-\nu }(\exp (tX))A(\nu )$. Each element of $G$ is $kan$ by [F8], with $K,A,N$ generated by these one-parameter subgroups, so the asserted $G$-intertwining identity follows by multiplying the three factor identities.

Suppose $\nu \mathrm{\notin }{\mathcal{W}}_{\epsilon }$ and both $A(\nu )$ and $A(-\nu )$ are regular. Then $r+1+\nu$ and $r+1-\nu$ are nonzero for every $r\equiv \epsilon (\mathrm{m}\mathrm{o}\mathrm{d}2)$. Applying the recurrence at $\nu$ and $-\nu$ gives $c_{r+2}(\nu )c_{r+2}(-\nu )=c_r(\nu )c_r(-\nu )$; the symmetry $c_{-r}=(-1{)}^r{c}_r$ handles negative indices. Hence the product is independent of $r$ and equals $c_{n_0}(\nu )c_{n_0}(-\nu )$, with $n_0=0$ in even parity and $n_0=1$ in odd parity. Continuity and density from step 3.1 extend this K-type calculation to the composite operator on smooth vectors. The base eigenvalues are finite and nonzero at these regular parameters by step 1.1, so dividing each factor by its base eigenvalue gives $\widehat{A}(-\nu )\widehat{A}(\nu )=\mathrm{i}\mathrm{d}$ and ${\widehat{c}}_r(-\nu ){\widehat{c}}_r(\nu )=1$.