Lemma 15.27.1CE5CA1071A3C

Proof. Let $v$ be a continuous solution of $\mathrm{MA}_ h(v)=c_ h(q)q\, \mathrm{d}V$, and suppose, to reach a contradiction, that $a>c_ h(q)$. Then $\tau =c_ h(q)/a<1$ and, for every real $A$,

The translate $v+A$ is bounded, so $P_{h/2}(v+A)\geq \inf _ Xv+A$ is not identically $-\infty $, and $u\preceq v+A$. Theorem 15.2.2, with $\delta =1/2$ and $c=\tau $, therefore gives $v+A\geq u$ for every real $A$. Letting $A\to -\infty $ at a point where $u$ is finite gives a contradiction. Hence $a\leq c_ h(q)$.

If $q\leq r$, a continuous solution of $\mathrm{MA}_ h(w)=c_ h(r)r\, \mathrm{d}V$ satisfies $\mathrm{MA}_ h(w)\geq c_ h(r)q\, \mathrm{d}V$, so the first assertion gives $c_ h(r)\leq c_ h(q)$. Finally, the equation $\mathrm{MA}_ h(v)=c_ h(q)q\, \mathrm{d}V$ can be written as $\mathrm{MA}_ h(v)=t^{-1}c_ h(q)\, (tq)\, \mathrm{d}V$, and uniqueness of the scalar gives $c_ h(tq)=t^{-1}c_ h(q)$. $\square$