PomeLemma 11.47.2.3CD979065654 The profile $A$ is smooth and strictly convex, with $0<A'<1$, bounded $A''$ and asymptotic slopes $0,1$. Its tangents tend uniformly to $-\infty $ on each compact interval as their point of tangency tends to $-\infty $, and uniformly to the line $y\mapsto y$ as that point tends to $+\infty $. Moreover, $A(x)-x\searrow 0$ at the right end.
Lemma 11.47.23CD979065654
Proof. Put $p=\mathrm e^ x/(1+\mathrm e^ x)$, $T=1-\log p$ and $a=\alpha T^{\alpha -1}$. Differentiating (11.46.3) gives
\[ A'=p+(1-p)a,\qquad A''=p(1-p)(1-a)+\alpha (1-\alpha )(1-p)^2T^{\alpha -2}. \]
Since $T\geq 1$ and $0<a\leq \alpha <1$, these formulas give the asserted positivity and bounds. On the left, $A(x)=1-(1-x)^\alpha +o(1)$ and $A-xA'$ has leading term $-(1-\alpha )(-x)^\alpha $. On the right, $A=x+(1-\alpha )\mathrm e^{-x}+O(\mathrm e^{-2x})$ and $A'=1-(1-\alpha )\mathrm e^{-x}+O(\mathrm e^{-2x})$. The tangent at $x$ is $y\mapsto A'(x)y+A(x)-xA'(x)$, so both uniform limits follow. Finally $A'-1<0$ gives the decreasing right-end limit. $\square$
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