Pome76376AD88152Theorem 8.8.3.76376AD88152 Let $A<B$ and let $f$ be smooth on a neighborhood of $X\times [A,B]$. Suppose that, for every $x\in X$,
There is a unique continuous admissible solution with values in $[A,B]$ of
This solution is smooth and strictly admissible.
Proof. Choose $a\geq L$ and let $\mathcal C=\{ v\in C(X):A\leq v\leq B\} $, a complete subset of $C(X)$ in the uniform norm. For $v\in \mathcal C$ put $H_ v(x)=f(x,v(x))-av(x)$ and let $T(v)$ be the continuous proper solution with exponent $aT(v)+H_ v$, furnished by 8.8.1. The function $f(x,t)-at$ is nonincreasing in $t$, so
The zero-order bounds in that proposition give $A\leq T(v)\leq B$. The mean-value theorem also gives
Since $0\leq 1-\kappa /a<1$, the contraction theorem gives a unique fixed point $w\in \mathcal C$. It is admissible because it equals $T(w)$, and its equation is exactly the stated equation. Every continuous admissible solution in this interval is a fixed point, by uniqueness of the continuous proper equation with its prescribed $H_ w$. This proves uniqueness in the stated class. 8.8.2 makes $w$ smooth. Its form lies in the closed $m$-cone and has positive $m$th symmetric polynomial; Maclaurin's inequalities then put it in the open cone. Thus it is strictly admissible. $\square$
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