Parameter bounds giving the signs of shifted entropy roots

The parameter bounds give $0<t<1$ and $t>(1+\epsilon ){\epsilon }^2\mathrm{/}(1+{\epsilon }^2)>{\epsilon }^2$, the last inequality using $\epsilon >{\epsilon }^2$. F1 gives $I(1-\epsilon )<{\epsilon }^2$. In particular $(1+\epsilon )\mathrm{/}I(1-\epsilon )>(1+\epsilon )\mathrm{/}{\epsilon }^2>1+{\epsilon }^{-2}$.

For $-1<u<1$, set $B(u)=I(1-u)-I(1+u)$; both arguments of $I$ are positive throughout this open interval, and $B(0)=0$. The affine functions $1-u$ and $1+u$ have difference quotients $-1$ and $1$, respectively. By F2 and F6, $B$ is differentiable, hence continuous, on $(-1,1)$ and $B\mathrm{\prime }(u)=-\log (1-u)-\log (1+u)$. For $0<u<1$, F4 gives $B\mathrm{\prime }(u)=-\log (1-u^2)>0$, since $0<1-u^2<1$ and $\log 1=0$.

The restriction of $B$ to $[0,\epsilon ]$ is continuous, and its derivative is positive at every point of $(0,\epsilon )$ by step 1.2. F3 therefore gives $B(\epsilon )>B(0)=0$, so $I(1+\epsilon )<I(1-\epsilon )<t$ by step 1.1. F5 gives $I(a_{-}(t))=I(a_{+}(t))=t$, with $a_{-}(t)\in (0,1)$ and $a_{+}(t)>1$. Since $I$ strictly decreases on $(0,1)$ and strictly increases on $(1,\mathrm{\infty })$ by F2, these inequalities imply $a_{-}(t)<1-\epsilon$ and $a_{+}(t)>1+\epsilon$. Together with $a_{-}(t)>0$, these prove both shifted signs and the asserted lower bound; step 1.1 gives the remaining parameter comparison.