Monotonicity, endpoints, and asymmetry of the rate roots

Substitute $a=1+c$ in F2. This gives the stated monotonicity and ranges of $h$, with $1$ the limit at the excluded endpoint $c=-1$. Since $0<1\mathrm{/}\gamma <1$, the two ranges and strict monotonicity give exactly one root in each of $(-1,0)$ and $(0,\mathrm{\infty })$ and no others.

For $-1<x<1$ set $D(x)=h(x)-h(-x)=I(1+x)-I(1-x)$. The affine inner maps have derivatives $1$ and $-1$ by F5. Thus F2 and F5 make $D$ differentiable, hence continuous, on $(-1,1)$, with $D^{\mathrm{\prime }}(x)=\log (1+x)+\log (1-x)=\log (1-x^2)<0$ for $0<x<1$. Here both logarithm arguments are positive and $0<1-x^2<1$, so F4 applies. Also $D(0)=0$. On every segment $[0,x]$ with $0<x<1$, F4 therefore gives $D(x)<D(0)$, or $h(x)<h(-x)$.

Put $r(\gamma )=1\mathrm{/}\gamma$ on $(1,\mathrm{\infty })$. F5 gives $r^{\mathrm{\prime }}(\gamma )=-1\mathrm{/}{\gamma }^2$, and induction using its quotient and positive-integer power rules gives $r^{(k)}(\gamma )=(-1{)}^{k}k!\mathrm{/}{\gamma }^{k+1}$ for every integer $k\ge 0$. Every denominator is nonzero; each displayed derivative is itself differentiable by the same rules, hence continuous. Thus $r$ is smooth by F6, and $r((1,\mathrm{\infty }))=(0,1)$ lies in the domains of both inverse branches in F3.

Fix either smooth branch $a=a_{\pm }$. For each integer $j\ge 0$, F3 and F5 give $(a^{(j)}\circ r{)}^{\mathrm{\prime }}=(a^{(j+1)}\circ r)r^{\mathrm{\prime }}$. Every factor $a^{(j)}\circ r$ is therefore differentiable, as is every derivative $r^{(\mathrm{\ell })}$. Starting with $a\circ r$, induction using F5 expresses every iterated derivative as a finite sum of constant multiples of finite products of these factors: differentiating a factor of the first type replaces it by $(a^{(j+1)}\circ r)r^{\mathrm{\prime }}$, and differentiating a factor of the second type replaces it by $r^{(\mathrm{\ell }+1)}$. Such sums and products remain differentiable and hence continuous by F5. Consequently $a\circ r$ and $a\circ r-1$ are smooth by F6. F1 and F3 identify the latter functions with $c^{\pm }$.

Since $1\mathrm{/}\gamma$ strictly decreases, the decreasing lower inverse makes $c^{-}$ strictly increase, while the increasing upper inverse makes $c^{+}$ strictly decrease. For any $0<\delta <1$, F2 gives $I(1-\delta )>0$ and $I(1+\delta )>0$. Once $1\mathrm{/}\gamma$ is smaller than both values, branch monotonicity places $1-\delta <a_{-}(1\mathrm{/}\gamma )<1<a_{+}(1\mathrm{/}\gamma )<1+\delta$. Hence both roots tend to $0$. The assigned values at $\gamma =1$ are $-1$ and $e-1$; the definition's justification establishes their continuity as endpoint extensions.

Put $x=-c^{-}(\gamma )\in (0,1)$. Then $h(x)<h(-x)=1\mathrm{/}\gamma =h(c^{+}(\gamma ))$. Strict increase of $h$ on $[0,\mathrm{\infty })$ implies $x<c^{+}(\gamma )$, proving the asymmetry and completing the proof.