Bounded KMS strip norm and zero-edge uniqueness

The KMS convention in [F2] uses the same closed strip when $b=\beta \mathrm{\hslash }$; the following estimate only needs the stated $b>0$ strip hypotheses. Define $G(w):=F(ibw)$ on $0\le \mathrm{Re}w\le 1$. If $w=x+iy$, then $ibw=-by+ibx$, so the lines $\mathrm{Re}w=0$ and $\mathrm{Re}w=1$ map to the lower and upper edges of $S_b$. The linear map is holomorphic and maps the open strip to $S_b^{\circ }$, so $G$ is bounded and continuous on the closed unit strip and holomorphic inside.

For $0<s<b$, put $x=s\mathrm{/}b\in (0,1)$. The vertical line $\mathrm{Re}w=x$ maps onto $\{t+is:t\in \mathbb{R}\}$, while the boundary suprema of $G$ are exactly $M_0$ and $M_1$. Applying [F1] gives the asserted interpolation inequality. Both exponents $1-x$ and $x$ are positive, so the zero-supremum cases use the defined positive powers and do not invoke $0^0$.

If $M_0,M_1\le C$, step 2.1 gives $\mathrm{\mid }F(t+is)\mathrm{\mid }\le C$ for each $0<s<b$; the same bound holds on both edges by definition of $M_0,M_1$. Hence it holds on all of $S_b$.

Let $F$ and $H$ be two such functions with identical lower edges. Their difference $D=F-H$ is bounded, continuous on $S_b$, holomorphic inside, has lower supremum $0$, and has a finite upper supremum. For every $0<s<b$, step 2.1 applied to $D$ gives ${\sup }_t\mathrm{\mid }D(t+is)\mathrm{\mid }\le 0^{1-s\mathrm{/}b}{M}_1^{s\mathrm{/}b}=0$. Thus $D$ vanishes in the open strip; continuity then gives zero on the upper edge as well as the lower edge.