Hadamard three-lines theorem

Fix $\delta >0$ and define $G_{\delta }(z):=f(z)\exp ((z-1)\log (M(0)+\delta ))\exp (-z\log (M(1)+\delta ))\mathrm{.}$ This is the positive-base principal-power normalization of [L2] and [L3]; it is bounded and continuous on $S$ and holomorphic inside.

On $z=iy$, the two exponential factors have moduli $(M(0)+\delta {)}^{-1}$ and $1$, so $\mathrm{\mid }{G}_{\delta }(iy)\mathrm{\mid }\le M(0)\mathrm{/}(M(0)+\delta )\le 1$. On $z=1+iy$, their moduli are $1$ and $(M(1)+\delta {)}^{-1}$, so the same bound holds. By [L1], $\mathrm{\mid }{G}_{\delta }(z)\mathrm{\mid }\le 1$ throughout $S$.

At $z=\theta +iy$ with $0<\theta <1$, step 2.1 rearranges to $\mathrm{\mid }f(z)\mathrm{\mid }\le (M(0)+\delta {)}^{1-\theta }(M(1)+\delta {)}^{\theta }$. Taking the supremum over $y$ and letting $\delta$ decrease to $0$ gives $M(\theta )\le M(0{)}^{1-\theta }M(1{)}^{\theta }$. Both exponents $1-\theta$ and $\theta$ are strictly positive, so the limit is correct including either zero boundary supremum, where the convention of [L2] reads the vanishing factor as $0$.

For $0\le x_0<x_1\le 1$, apply step 3.1 to the rescaled strip function $w\mapsto f(x_0+(x_1-x_0)w)$. Its boundary suprema are $M(x_0)$ and $M(x_1)$, so the resulting inequality is the asserted log-convexity at $(1-t)x_0+t{x}_1$.