Upper-semicircle imaginary-dilation identity on lowest-weight vectors

The data are: Exactly the unitary lowest-weight representation, self-adjoint generator, Fourier convention and compact upper-semicircle support just specified. The quasi-primary supplier [F1], proof steps 1.1, 2.1, and 3.1, proves that every finite-energy vector of this genuine positive-energy Möbius representation is Möbius smooth, without any finite-multiplicity hypothesis. In particular $a$ and all its $L_{-1}$ descendants are in the common smooth domain, so the ladder and adjoint calculations beginning in 2.1 have their required domains.

The zero-vector case is immediate. If $d=0$, then $\mathrm{\parallel }{L}_{-1}a{\mathrm{\parallel }}^2=⟨a,L_1{L}_{-1}a⟩=2d\mathrm{\parallel }a{\mathrm{\parallel }}^2=0$. Hence $L_{-1}a=L_{1}a=0$, $Ka=0$ and $a(f)={\widehat{f}}_{0}a$. Since ${\widehat{f\circ j}}_0={\widehat{f}}_0$, the asserted identity follows. Suppose henceforth $d>0$ and put $m=2d$. Repeated use of the Lie relations gives the mutually orthogonal vectors $v_n=L_{-1}^{n}a\mathrm{/}n!$ with $\mathrm{\parallel }{v}_n{\mathrm{\parallel }}^2=\mathrm{\parallel }a{\mathrm{\parallel }}^2\frac{(m{)}_n}{n!},(m{)}_n=m(m+1)\cdots (m+n-1)\mathrm{.}$ Indeed $L_1{L}_{-1}^{n}a=n(2d+n-1)L_{-1}^{n-1}a$, and induction gives this norm formula. Fourier coefficients of a smooth circle function decay faster than any power, whereas $(m{)}_n\mathrm{/}n!$ grows polynomially of degree $m-1$, so both $a(f)$ and $a(f\circ j)$ converge in Hilbert norm.

For $\mathrm{\mid }z\mathrm{\mid }<1$ define the coherent vector $v(z)={\sum }_{n\ge 0}{z}^n{v}_n$. Its series converges normally in Hilbert norm on compact subdisks and the binomial series gives $⟨v(z),v(w)⟩=\mathrm{\parallel }a{\mathrm{\parallel }}^2(1-z\bar{w}{)}^{-m}\mathrm{.}$ The ladder relations also give, as Hilbert-vector identities, $L_{-1}v(z)={\mathrm{\partial }}_{z}v(z),L_{1}v(z)=(z^2{\mathrm{\partial }}_z+2dz)v(z)\mathrm{.}$ These are genuine domain identities: truncating the power series, the images under each ladder operator converge in Hilbert norm, since the additional ladder coefficients are polynomial in $n$ and $\mathrm{\mid }z{\mathrm{\mid }}^n$ decays exponentially. Closedness of the representation generators therefore puts $v(z)$ in their domains; the same estimates apply uniformly on every compact subdisk. In particular $v(z)\in \mathrm{Dom}K$ with the displayed generator action.

For $0<s<1\mathrm{/}2$, set $c_s=\cos (\pi s),h_s=\sin (\pi s),A_s(z)=c_s-i{h}_{s}z,N_s(z)=c_{s}z-i{h}_s,g_s(z)=\frac{N_s(z)}{A_s(z)}\mathrm{.}$ For $z=e^{it}$ in the upper semicircle, $\mathrm{\mid }{A}_s(z){\mathrm{\mid }}^2=1+\sin (2\pi s)\sin t,\mathrm{\mid }{N}_s(z){\mathrm{\mid }}^2=1-\sin (2\pi s)\sin t\mathrm{.}$ Hence $\mathrm{\mid }{g}_s(z)\mathrm{\mid }<1$ and $A_s(z)\mathrm{\ne }0$. Define a Hilbert-valued Bochner integral $v_s=\frac{1}{2\pi }{\int }_0^{\pi }f(e^{it})e^{idt}{A}_s(e^{it}{)}^{-m}v(g_s(e^{it}))dt\mathrm{.}$ For $s$ in a compact subinterval of $(0,1\mathrm{/}2)$ and $t$ in the compact support of $f$, the transformed points belong to a fixed compact subdisk. Step 3.1 therefore justifies the integral, its norm derivatives in $s$, and integration in the graph norm of $K$.

Differentiate the explicit functions: ${\mathrm{\partial }}_s{g}_s=i\pi (g_s^2-1),{\mathrm{\partial }}_s(A_s^{-m})=2di\pi g_s{A}_s^{-m}\mathrm{.}$ Using step 3.1 in the integrand yields the exact local differential equation $\frac{d}{ds}{v}_s=K{v}_s(0<s<1\mathrm{/}2)\mathrm{.}$ This equation has not yet been identified with an unbounded exponential at the endpoints; that identification will follow from a uniform norm bound and spectral cutoffs below.

We now prove the required uniform Hardy bound directly. Put $q=\sin (2\pi s)>0$ and $r=\cos (2\pi s)$, so $q^2+r^2=1$. The coherent kernel in step 3.1 gives $\mathrm{\parallel }{v}_s{\mathrm{\parallel }}^2=\frac{\mathrm{\parallel }a{\mathrm{\parallel }}^2}{(2\pi {)}^2}\iint f(e^{it})\overline{f(e^{iu})}\frac{e^{id(t-u)}}{(A_s(e^{it})\overline{A_s(e^{iu})}-N_s(e^{it})\overline{N_s(e^{iu})}{)}^m}dtdu\mathrm{.}$ Direct multiplication, with $z=e^{it},w=e^{iu}$, gives $A_s(z)\overline{A_s(w)}-N_s(z)\overline{N_s(w)}=2e^{i(t-u)\mathrm{/}2}(q\sin \frac{t+u}{2}-ir\sin \frac{t-u}{2})\mathrm{.}$ Since $m=2d$, its phase cancels the numerator phase exactly. In the real coordinates $x=\tan (t\mathrm{/}2)>0$, $y=\tan (u\mathrm{/}2)>0$, put $F(x)=f(e^{2i\arctan x})(1+x^2{)}^{d-1}\mathrm{.}$ Compact upper-semicircle support means $F\in C_c^{\mathrm{\infty }}(0,\mathrm{\infty })$. The identities for the sine of the sum/difference of arctangents and $dt=2dx\mathrm{/}(1+x^2)$ then give $\mathrm{\parallel }{v}_s{\mathrm{\parallel }}^2=\frac{\mathrm{\parallel }a{\mathrm{\parallel }}^2}{2^m{\pi }^2}\iint \frac{F(x)\overline{F(y)}}{(q(x+y)-ir(x-y){)}^m}dxdy\mathrm{.}$ All integrals here are ordinary convergent integrals for fixed $0<s<1\mathrm{/}2$.

The elementary Laplace identity $z^{-m}=\mathrm{\Gamma }(m{)}^{-1}{\int }_0^{\mathrm{\infty }}{u}^{m-1}{e}^{-uz}du$, valid for $\mathrm{Re}z>0$, gives $\mathrm{\parallel }{v}_s{\mathrm{\parallel }}^2=\frac{\mathrm{\parallel }a{\mathrm{\parallel }}^2}{2^m{\pi }^2\mathrm{\Gamma }(m)}{\int }_0^{\mathrm{\infty }}{u}^{m-1}{\mid {\int }_0^{\mathrm{\infty }}F(x)e^{-u(q-ir)x}dx\mid }^{2}du\mathrm{.}$ For fixed $q>0$, Fubini is justified absolutely by the positive lower bound on $x+y$ on the compact support; the Laplace identity itself follows by $m-1$ integrations by parts from the exponential integral and holomorphy in the right half-plane. Crucially $\mathrm{\mid }q-ir\mathrm{\mid }=1$. Integrating by parts $N$ times in the inner integral, with no boundary terms because $F$ is compactly supported inside $(0,\mathrm{\infty })$, proves $\mid \int F(x)e^{-u(q-ir)x}dx\mid \le \mathrm{\parallel }F{\mathrm{\parallel }}_1(0<u\le 1),\mid \int F(x)e^{-u(q-ir)x}dx\mid \le u^{-N}\mathrm{\parallel }{F}^{(N)}{\mathrm{\parallel }}_1(u\ge 1)\mathrm{.}$ The estimate $\mathrm{\mid }{e}^{-uqx}\mathrm{\mid }\le 1$ uses $x>0$. Choosing $N>d$ proves the explicit bound ${\sup }_{0<s<1\mathrm{/}2}\mathrm{\parallel }{v}_s{\mathrm{\parallel }}^2\le \frac{\mathrm{\parallel }a{\mathrm{\parallel }}^2}{2^m{\pi }^2\mathrm{\Gamma }(m)}(\frac{\mathrm{\parallel }F{\mathrm{\parallel }}_1^2}{m}+\frac{\mathrm{\parallel }{F}^{(N)}{\mathrm{\parallel }}_1^2}{2N-m})<\mathrm{\infty }\mathrm{.}$ This is uniform at BOTH endpoints and does not appeal to a boundary distribution kernel theorem.

For each fixed $n$, the coefficient of $v_n$ in $v_s$ is $b_n(s)=\frac{1}{2\pi }\int f(e^{it})e^{idt}{A}_s(e^{it}{)}^{-m}{g}_s(e^{it}{)}^{n}dt\mathrm{.}$ Since $\mathrm{\mid }{A}_s(e^{it})\mathrm{\mid }\ge 1$ on the support, these finite coefficients have the endpoint limits obtained by direct bounded convergence. At $s=0$, $A_0=1$ and $g_0(z)=z$, so $b_n(0)={\widehat{f}}_{-n-d}$. At $s=1\mathrm{/}2$, $A_{1\mathrm{/}2}(z)=-iz$ and $g_{1\mathrm{/}2}(z)=z^{-1}$, so $b_n(1\mathrm{/}2)=(-i{)}^{-2d}\frac{1}{2\pi }\int f(e^{it})e^{-i(n+d)t}dt=(-1{)}^d{\widehat{f\circ j}}_{-n-d}\mathrm{.}$ The spans of the $v_n$ are dense in the cyclic subrepresentation. Together with the uniform norm bound and the Hilbert convergence established in step 2.1, these coefficient limits imply $v_s\rightharpoonup a(f)(s\downarrow 0),v_s\rightharpoonup (-1{)}^{d}a(f\circ j)(s\uparrow 1\mathrm{/}2)\mathrm{.}$ This also fixes the phase without an independent choice of square-root or a suppressed antilinear reflection: the multiplier is precisely $(-iz{)}^{-2d}$ and the external density factor is $z^d$.

Let $E_R=1_{[-R,R]}(K)$. By step 5.1, $E_R{v}_s$ solves the bounded-generator equation $(E_R{v}_s{)}^{\mathrm{\prime }}=K{E}_R{v}_s$. Therefore $e^{-sK}{E}_R{v}_s$ is constant on $(0,1\mathrm{/}2)$, where the exponent is restricted to $E_{R}H$ and is bounded. Taking its weak initial limit using step 8.1 gives $E_R{v}_s=e^{sK}{E}_{R}a(f)\mathrm{.}$ Letting $R\to \mathrm{\infty }$, the uniform finite bound on $\mathrm{\parallel }{v}_s\mathrm{\parallel }$ proves $a(f)\in \mathrm{Dom}(e^{sK})$ and $v_s=e^{sK}a(f)$ for every $0<s<1\mathrm{/}2$, by the spectral theorem. Finally, if $\mu$ is the $K$-spectral measure of $a(f)$, Fatou's lemma gives ${\int }_{\mathbb{R}}{e}^{\lambda }d\mu (\lambda )\le {\mathrm{lim\hspace{0.17em}inf}}_{s\uparrow 1\mathrm{/}2}{\int }_{\mathbb{R}}{e}^{2s\lambda }d\mu (\lambda )\le {\sup }_{0<s<1\mathrm{/}2}\mathrm{\parallel }{v}_s{\mathrm{\parallel }}^2<\mathrm{\infty }\mathrm{.}$ Hence $a(f)\in \mathrm{Dom}(e^{K\mathrm{/}2})$. Spectral dominated convergence also gives $e^{sK}a(f)\to e^{K\mathrm{/}2}a(f)$ strongly: on positive spectrum use the integrable majorant $e^{\lambda }$, and on negative spectrum use one. Comparing with the weak final limit in step 8.1 proves $e^{K\mathrm{/}2}a(f)=(-1{)}^{d}a(f\circ j)\mathrm{.}$ This is the full asserted domain and vector identity for every allowed complex test function.