General dual-Banach-space and weak-star isometry-group hypotheses are retained; no algebra, positivity or faithful-weight premise is added.

Bounded strip comparison from scalar proofs

CC0-1.0 to the extent of rights held.

This proof supplies only the analytic comparison needed for operator-valued weight transport. All dual spaces below have a specified predual. A group of isometries means a group of complex-linear surjective isometries whose individual maps are weak-star continuous and whose real orbits are weak-star continuous. No algebra or positivity property of that group is assumed.

The exact earlier written proofs are SF4 scalar Cauchy, maximum, Morera, identity and zero-edge proofs; FF-1 Gaussian normalization and transform; CP1, CP2, CP3, CP4, CP5, CP6 for the concrete predual and its dual; scalar interchange, SC4 monotone convergence and SC5 dominated convergence for the scalar integrals. The proofs below do not import three-lines, Carlson, Paley–Wiener, Stone, or a theorem about closed analytic generators.

The comparison theorem is discussed in F. Hiai's free author manuscript, printed pp.75–77. Here a fourth-power sine kernel is normalized by direct Gaussian integration, and comparison is confined to the upper half-plane, where an elementary zero theorem applies.

AS1. Two elementary analytic bounds

Let $h>0$, and let $f$ be scalar, continuous on $-h\leq\operatorname{Im}z\leq0$, holomorphic inside, with a finite bound $M$ on the entire strip. If both boundary lines have modulus at most $C$, then

$|f(z)|\leq C\qquad(-h\leq\operatorname{Im}z\leq0). \tag{AS1}$ Indeed, multiply by $e^{-\varepsilon z^2}$, not by a function growing in the horizontal direction. On the horizontal sides of the rectangle $[-R,R]\times[-h,0]$ the modulus is at most $Ce^{\varepsilon h^2}$; on the vertical sides it is at most $Me^{-\varepsilon R^2+\varepsilon h^2}$. SF4 maximum modulus, followed by $R\to\infty$, gives $|f(z)e^{-\varepsilon z^2}|\leq Ce^{\varepsilon h^2}.$ Now let $\varepsilon\downarrow0$. Translation gives the same statement on every horizontal strip. This argument needs the finite bound on the whole strip; a bound on the two edges alone is not the premise.

We also need the following zero statement:

$\begin{gathered} g\text{ holomorphic and bounded on }\operatorname{Im}z>0,\qquad g(in)=0\quad(n=1,2,\ldots)\\ \Longrightarrow\quad g=0. \tag{AS2} \end{gathered}$ If $g\neq0$, the identity theorem gives a noninteger $y>0$ with $g(iy)\neq0$. Set $H(w)=g\!\left(iy\,\frac{1+w}{1-w}\right),\qquad |w|<1.$ The displayed map has positive imaginary part $y(1-|w|^2)/|1-w|^2$, so $H$ is bounded by some $M$. It vanishes at $a_n=\frac{n-y}{n+y}\in(-1,1),\qquad n\geq1.$ For any finite selection of these zeros, divide $H$ by the product of the factors $B_{a_n}(w)=\frac{w-a_n}{1-a_nw}.$ The quotient has removable singularities at the selected zeros. The maximum principle on $|w|\leq r$, followed by $r\uparrow1$, bounds its modulus by $M$, because each finite product has boundary modulus tending uniformly to $1$. Consequently $|H(0)|\leq M\prod_{n=1}^{N}|a_n|.$ The finite factors are nonzero. For $n>y$, $\log a_n=\log\!\left(1-\frac{2y}{n+y}\right) \leq-\frac{2y}{n+y},$ so the product tends to zero. This contradicts $H(0)\neq0$, proving (AS2). The disk maximum principle here has the same elementary proof as SF4's rectangle version: the local power series makes $|H|^2$ twice differentiable with nonnegative Laplacian; adding $\varepsilon|w|^2$ excludes an interior maximum on a closed disk, and then $\varepsilon\downarrow0$ gives the boundary maximum.

It follows in particular that an entire $f$ satisfying

$|f(z)|\leq K e^{c|\operatorname{Im}z|},\qquad f(in)=0\quad(n\geq1), \tag{AS3}$ vanishes: $e^{icz}f(z)$ is bounded in the upper half-plane, so (AS2) applies there and the identity theorem finishes. The horizontal-band exponent, rather than a radial exponential bound, is essential to this proof.

AS2. Weak-star analytic maps and integration

For a dual Banach space $X=E^*$, a bounded weak-star continuous function of a real variable can be integrated against any $L^1$ scalar kernel: its value is the functional $e\longmapsto \int k(s)\langle F(s),e\rangle\,ds,\qquad e\in E.$ This defines an element of $X$, of norm at most $\int|k(s)|\,\|F(s)\|\,ds$. The same construction defines circle integrals for weak-star continuous, norm-bounded curves.

A locally norm-bounded weak-star holomorphic map is norm holomorphic. To see this without an additional vector-valued theorem, take a closed circle of radius $r$ in its domain, use the preceding integrals to form its Cauchy coefficients $A_n$, and apply the scalar circle formula to every $e\in E$. If the bound on the circle is $M$, then $\|A_n\|\leq Mr^{-n}$. The norm-convergent power series on every smaller disk equals the original map after pairing with all $e$, hence equals it in $X$. The same argument proves the norm Cauchy formula. Scalar continuity on a closed strip, uniform norm bounds, and scalar holomorphy inside are sufficient for every use of this observation below.

For bounded scalar functions, two adjacent closed-strip holomorphic functions that agree on the shared line glue holomorphically across it. Subdivide a small rectangle meeting the line into upper and lower rectangles with a gap of width $2\delta$. Their boundary integrals vanish; continuity makes the gap contributions cancel as $\delta\downarrow0$. SF4 Morera applies. Testing by every $e$, and then using the preceding local norm bounds, proves the corresponding assertion for dual-valued maps.

AS3. Contractive approximation by directly integrated entire kernels

Let $(\alpha_t)_{t\in\mathbb R}$ be a group of isometries of $X=E^*$ in the opening sense. We construct a bounded approximate identity whose entire translates have an exponential bound in the imaginary direction. An element $a$ is called exponential if its real orbit has a norm-entire extension $A$ with

$A(t)=\alpha_t(a),\qquad \|A(z)\|\leq K e^{c|\operatorname{Im}z|} \quad(z\in\mathbb C) \tag{AS4}$ for finite $K,c\geq0$. Scalar identity gives uniqueness and $A(z+t)=\alpha_t(A(z))$ for real $t$. Consequently $A(z_0)$ is again exponential, with extension $z\mapsto A(z+z_0)$.

A positive kernel with an elementary normalization. Put $k(t)=\frac3{2\pi}\left(\frac{\sin t}{t}\right)^4,\qquad k_R(z)=Rk(Rz)\quad(R>0),$ using the removable value $k(0)=3/(2\pi)$. We first prove $\int_{\mathbb R}k(t)\,dt=1$.

For real $a$, set $J(a)=\int_{\mathbb R}(1-\cos(at))/t^2\,dt$. The integral is finite by the quadratic bound near zero and the $t^{-2}$ bound at infinity. Since $t^{-2}=\int_0^\infty e^{-ut^2}\,du$ for $t\ne0$, nonnegative scalar interchange and the Gaussian transform from FF-1 give $J(a)=\sqrt\pi\int_0^\infty u^{-1/2} (1-e^{-a^2/(4u)})\,du.$ For $a\ne0$, substitute $v=|a|/(2\sqrt u)$. Integration by parts, with both boundary terms zero, gives $J(a)=\sqrt\pi\,|a|\int_0^\infty\frac{1-e^{-v^2}}{v^2}\,dv =2\sqrt\pi\,|a|\int_0^\infty e^{-v^2}\,dv =\pi|a|.$ For $a=0$ the equality is immediate. Now let $f(t)=\sin^4t$. Since $f''(t)=2\cos(2t)-2\cos(4t)$, two integrations by parts on intervals avoiding zero yield

$\int_{\mathbb R}\frac{\sin^4t}{t^4}\,dt =\frac16\int_{\mathbb R}\frac{f''(t)}{t^2}\,dt =\frac{J(4)-J(2)}3=\frac{2\pi}3. \tag{AS5}$ The boundary terms vanish at infinity by boundedness of $f,f'$, and at zero by $f(t)=O(t^4)$, $f'(t)=O(t^3)$. Thus $k$ is a nonnegative probability density. Also $k(t)\leq3/(2\pi t^4)$ for $t\ne0$. Scaling gives

$k_R(s)\geq0,\qquad \int_{\mathbb R}k_R(s)\,ds=1,\qquad \int_{|s|>\delta}k_R(s)\,ds\leq\frac1{\pi R^3\delta^3}. \tag{AS6}$

Entire translates and their integrable bounds. The exponential formula for sine gives $|\sin(t+iv)|\leq e^{|v|}$. The identity $\sin z=z\int_0^1\cos(uz)\,du$ also gives $|\sin z/z|\leq e^{|\operatorname{Im}z|}$. Combining these estimates and using $|t+iv|\geq|t|$ yields $|k(t+iv)|\leq\frac3{2\pi}e^{4|v|} \min(1,|t|^{-4}).$ The right side is integrable in $t$. Its integral, followed by scaling, proves

$\int_{\mathbb R}|k_R(s+iv)|\,ds \leq\frac4\pi e^{4R|v|}. \tag{AS7}$ On every compact set of complex shifts, the kernels $k_R(s-z)$ have a common integrable bound of the form $C/(1+|s|^4)$. Cauchy estimates on a slightly larger compact set give such bounds for their derivatives and Taylor remainders. The map $z\mapsto k_R(\,\cdot-z)$ is therefore holomorphic in $L^1(\mathbb R)$.

Approximation of the orbit. Define the weak-star integral from AS2 by

$A_R(z)=\int_{\mathbb R}k_R(s-z)\alpha_s(a)\,ds,\qquad a_R=A_R(0). \tag{AS8}$ The integral map from $L^1(\mathbb R)$ into $X$ has norm at most $\|a\|$, so $A_R$ is norm entire. Real substitution and weak-star continuity of each $\alpha_t$ give

$A_R(t)=\alpha_t(a_R),\qquad \|A_R(z)\|\leq\frac4\pi\|a\|e^{4R|\operatorname{Im}z|},\qquad \|a_R\|\leq\|a\|. \tag{AS9}$ For $e\in E$, split $\langle a_R-a,e\rangle=\int k_R(s)\langle\alpha_s(a)-a,e\rangle\,ds$ at $|s|=\delta$. Continuity of the scalar orbit makes the first part small when $\delta$ is small; the explicit tail in AS6 makes the second part tend to zero as $R\to\infty$. Thus $a_R\to a$ weak-star. The exponential elements are weak-star dense, with contractive approximation and no separability hypothesis.

AS4. Comparison from one-strip continuations

Let $Y\subset X$ be a weak-star closed subspace with its inherited predual topology. Let $\alpha_t$ on $X$ and $\gamma_t$ on $Y$ be groups as above. Suppose that for every norm-entire $\gamma$-orbit $A(z)=\gamma_z(a)$ there is an $X$-valued function $B_a$, weak-star continuous on the closed lower unit strip, holomorphic inside, bounded on the whole strip, with

$B_a(t)=\alpha_t(a),\qquad B_a(t-i)=\alpha_t(\gamma_{-i}(a)) \quad(t\in\mathbb R). \tag{AS10}$

Then

$\alpha_t(y)=\gamma_t(y)\qquad(y\in Y,\ t\in\mathbb R). \tag{AS11}$

It suffices, by AS3 and weak-star continuity, to prove this for an exponential $y\in Y$. Write $A$ for its entire $\gamma$-orbit, with $\|A(z)\|\leq K e^{c|\operatorname{Im}z|}$, and put $a_k=A(ik)$ for $k\geq0$. These are exponential elements and $\gamma_{-i}(a_k)=a_{k-1}$ for $k\geq1$.

We construct only an upper-half-plane comparison. On $k-1\leq\operatorname{Im}z\leq k$, for $k=1,2,\ldots$, put

$F(z)=B_{a_k}(z-ik). \tag{AS12}$ At each seam $z=t+ik$, the two definitions are both $\alpha_t(a_k)$, by AS10. AS2's local rectangle argument glues their scalar pairings holomorphically. Consequently $F$ is weak-star continuous on the closed upper half-plane, holomorphic in its interior, and $F(t)=\alpha_t(y)$.

Apply AS1 to each scalar pairing on the $k$-th strip. Its two boundary norms are at most $\|a_k\|$ and $\|a_{k-1}\|$, because the real action is isometric. Duality gives $\|F(t+iv)\|\leq K e^{ck}\leq K e^{c(v+1)} \qquad(k-1\leq v\leq k).$ Also $F(in)=a_n=A(in)$ for every positive integer $n$.

Fix $e\in E$. The scalar function $g_e(z)=e^{icz}\langle F(z)-A(z),e\rangle \qquad(\operatorname{Im}z>0)$ is holomorphic and bounded by $K(e^c+1)\|e\|$. It vanishes at every $in$, $n\geq1$. AS1's bounded-half-plane zero theorem therefore makes $g_e=0$. Continuity to the real boundary gives $\langle\alpha_t(y)-\gamma_t(y),e\rangle=0$ for every $e$, proving AS11 on exponential elements. The contractive weak-star approximation in AS3 and continuity of the two real maps extend it to all $Y$.

Only the continuations of the positive imaginary translates $A(ik)$ were used. No negative-half-plane construction, global entire extension of the competing orbit, or closed analytic-generator calculus is required.