Lemma 10.24.9.0A057E71255D There is $F\in C^\infty (\mathbb R)$ with $0<p:=F'<1$ and $p'>0$ such that, for $c=1/100$, some $B\in \mathbb R$ and $K=(9c^2/8)\mathrm{e}^{-B}>0$,
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10.24.10
\begin{align} F(t)& =B-3\log (1-c\mathrm{e}^{t/2}),& pp'& =K\mathrm{e}^{F+t} & & (t\leq -2),\\ F(t)& =t+\log H(p(t)),& pp'& =\mathrm{e}^{F-t} & & (t\geq 2),\tag{10.24.11} \end{align}
where $H(p)=(1-p)^2(1+2p)/6$. At the two ends,
\begin{align*} F(t)& =B+3c\mathrm{e}^{t/2}+O(\mathrm{e}^ t),& p'(t)& \sim (3c/4)\mathrm{e}^{t/2} & & (t\to -\infty ),\\ 1-p(t)& \sim 2/t,& p'(t)& \sim 2/t^2 & & (t\to +\infty ), \end{align*}
and $F(t)-t=-2\log t+\log 2+o(1)$ as $t\to +\infty $.
Proof.
For $t\leq -2$ put $L(t)=-3\log (1-c\mathrm{e}^{t/2})$. With $q=c\mathrm{e}^{t/2}$,
\[ p_ L=L'=\frac{3q}{2(1-q)},\qquad p_ L'=\frac{3q}{4(1-q)^2},\qquad p_ Lp_ L'=\frac{9c^2}{8}\mathrm{e}^{t}(1-c\mathrm{e}^{t/2})^{-3}. \]
This proves the left identity for every additive constant $B$.
Since $H'(p)=(p-1)p$, the increasing function
\[ \mathcal T(p)=2+\int _{3/4}^{p}\frac{q}{H(q)}\, \mathrm{d}q \]
tends to $+\infty $ as $p\to 1$, because its integrand is asymptotic to $2/(1-p)^2$ there. Hence it has an inverse $p_ R(t)$, defined for $t\geq 2$ and on a neighborhood of $2$, with $p_ R(2)=3/4$ and $p_ R'=H(p_ R)/p_ R$. The function $F_ R=t+\log H(p_ R)$ has derivative $p_ R$ and satisfies the right identity. For $q=1-p_ R$,
\[ (1/q)'=\frac{3-2q}{6(1-q)}\longrightarrow \frac12. \]
Hence $q\sim 2/t$ and $H(p_ R)\sim 2/t^2$, and the asymptotics at the right end follow.
It remains to interpolate the derivatives. Set $d=p_ R(2)-p_ L(-2)>0$. Choose disjoint short collars in $[-2,2]$ at the two endpoints, and cutoffs $\alpha _ L,\alpha _ R$ equal to one on smaller collars. For small $\varepsilon >0$, the function
\[ Q_0=\alpha _ Lp_ L'+\alpha _ Rp_ R' +\varepsilon (1-\alpha _ L-\alpha _ R) \]
is positive and agrees with the derivative germs at the endpoints. The collars and $\varepsilon $ can be chosen so that $\int _{-2}^2Q_0<d$. Adding a nonnegative smooth bump in the interior, with integral $d-\int _{-2}^2Q_0$, gives a positive function $Q$, and we put
\[ p(t)=p_ L(-2)+\int _{-2}^ tQ(x)\, \mathrm{d}x. \]
This function matches both endpoint germs, is increasing and takes values in $(0,1)$. Integrating $F'=p$ with $F(2)=F_ R(2)$ gives the required $F$ and fixes the constant $B$ at the left end. The left expansion follows directly.
$\square$
Pome
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