Admissible spectral strip profiles and the exact interpolation upper bound

Write $\mathrm{\Sigma }=X_0+X_1$ and let $V$ be the common Hausdorff topological vector space. The sum infimum is finite and nonnegative. Multiplying decompositions by a nonzero scalar and using the inverse scalar gives absolute homogeneity; adding decompositions within an arbitrary positive error of their infima gives the triangle inequality. Its value at zero is zero. If $\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\Sigma }}=0$, use AC to choose $x=a_n+b_n$ with $\mathrm{\parallel }{a}_n{\mathrm{\parallel }}_{X_0}+\mathrm{\parallel }{b}_n{\mathrm{\parallel }}_{X_1}<1\mathrm{/}n$. The endpoint embeddings imply $a_n\to 0$ and $b_n\to 0$ in $V$. Continuous addition gives $x=a_n+b_n\to 0$ in $V$. If $x\mathrm{\ne }0$, disjoint neighborhoods of $x$ and $0$ contradict this convergence of the constant sequence. Hence $x=0$, proving separation. Thus $\mathrm{\Sigma }$ is a complex normed space, and $\mathrm{\parallel }{x}_j{\mathrm{\parallel }}_{\mathrm{\Sigma }}\le \mathrm{\parallel }{x}_j{\mathrm{\parallel }}_{X_j}$ for either endpoint by its one-term decomposition.

Put $g=T^{\theta }f\in K$. Since $T\ge I$, $\mathrm{\parallel }{T}^{-z}g\mathrm{\parallel }\le \mathrm{\parallel }g\mathrm{\parallel }$ on $S$, and the sum norm of $(K,D(T))$ equals the $K$ norm: one inequality uses $\mathrm{\parallel }{x}_1{\mathrm{\parallel }}_K\le \mathrm{\parallel }T{x}_1{\mathrm{\parallel }}_K$ and the other the decomposition $x=x+0$. The complex-power formula is therefore well-defined as a bounded sum-valued function after multiplication by the Gaussian. For $z\to z_0$ in $S$, the multipliers ${\lambda }^{-z}$ converge pointwise, bounded by one, giving norm continuity by the scalar spectral norm integral. For $\mathrm{Re}{z}_0>0$, the difference quotients converge in multiplier operator norm to $-(\log T)T^{-z_0}$ locally: on $\mathrm{Re}z\ge \eta >0$ the Taylor remainder is bounded by a constant times $\mathrm{\mid }z-z_0\mathrm{\mid }(\log \lambda {)}^2{\lambda }^{-\eta }$, whose supremum for $\lambda \ge 1$ is finite. Thus the function is norm-holomorphic in the interior.

Let $F$ be admissible and put $M=\mathrm{\parallel }F{\mathrm{\parallel }}_{\mathcal{F}}$. For any $\mathrm{\ell }\in {\mathrm{\Sigma }}^{\ast }$ with $\mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }\le 1$, the scalar function $\varphi =\mathrm{\ell }\circ F$ is bounded and continuous on $S$. Indeed $\mathrm{\mid }\mathrm{\ell }(F(z))\mathrm{\mid }\le \mathrm{\parallel }F(z){\mathrm{\parallel }}_{\mathrm{\Sigma }}$. At an interior point, applying $\mathrm{\ell }$ to the norm-convergent difference quotients gives ${\varphi }^{\mathrm{\prime }}(z)=\mathrm{\ell }(F^{\mathrm{\prime }}(z))$, so $\varphi$ is scalar holomorphic. At each boundary point, step 1.1 gives $\mathrm{\mid }\varphi (j+it)\mathrm{\mid }\le \mathrm{\parallel }F(j+it){\mathrm{\parallel }}_{X_j}\le M$. Three-lines therefore yields $\mathrm{\mid }\mathrm{\ell }(F(\theta +it))\mathrm{\mid }\le M$ for every $0<\theta <1$ and $t\in \mathbb{R}$. If $F(\theta +it)\mathrm{\ne }0$, choose its norming functional by [F5]; if it is zero the bound is immediate. This proves $\mathrm{\parallel }F(\theta +it){\mathrm{\parallel }}_{\mathrm{\Sigma }}\le M$.

The admissible class is a complex vector space: sums and scalar multiples preserve boundedness, continuity, norm derivatives, endpoint continuity, and boundary decay. Its boundary quantity is homogeneous and subadditive. If that quantity vanishes, both boundary maps vanish, and step 2.1 makes every interior value zero, so it is a norm. Evaluation at $\theta$ is linear, so its range $[X_0,X_1{]}_{\theta }$ is a vector space. For nonzero scalars, multiplication is a bijection between the corresponding representing functions, giving homogeneity of the infimum. For $x,y$ in the range and $\delta >0$, choose representing functions with boundary norms less than their respective infima plus $\delta$; their sum represents $x+y$ and proves the triangle inequality after $\delta \downarrow 0$. The zero function gives norm zero at zero. Infimizing step 2.1 gives $\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\Sigma }}\le \mathrm{\parallel }x{\mathrm{\parallel }}_{[X_0,X_1{]}_{\theta }}$; separation of $\mathrm{\Sigma }$ then proves positive definiteness of the interpolation infimum.

On $z=it$, the multiplier $T^{-it}$ is unitary, by $\mathrm{\mid }{\lambda }^{-it}\mathrm{\mid }=1$. On $z=1+it$, the product/domain formula gives $T{F}_{\epsilon }(1+it)=e^{\epsilon (1+it-\theta {)}^2}{T}^{-it}g$. Both boundary maps are continuous in their respective endpoint norms by the same dominated spectral argument. Their norms equal $e^{\epsilon ((j-\theta {)}^2-t^2)}\mathrm{\parallel }g\mathrm{\parallel }$, so vanish at infinity and give the asserted upper bound. At $z=\theta$ the product identity gives $F_{\epsilon }(\theta )=f$. Taking the infimum and then $\epsilon \downarrow 0$ proves the inclusion and exact upper inequality.