PomeProposition 15.35.12.B2DB6BD3AC4D If $\lambda _0\in \Lambda _ h(f)$, then $[\lambda _0,+\infty )\subset \Lambda _ h(f)$. Consequently, $\Lambda _ h(f)$ is empty, or equals $(a,+\infty )$ for some $a\geq 0$, or equals $[a,+\infty )$ for some $a>0$. If $b_ h(f)>0$, then $\Lambda _ h(f)=(0,+\infty )$.
Proposition 15.35.12B2DB6BD3AC4D
Proof. Let $u$ be a solution at $\lambda _0$, and put $s=u-\max (0,\sup _ Xu)$. Then $s\leq \min (u,0)$, so for $\lambda \geq \lambda _0$,
\begin{align*} \mathrm{MA}^{\mathrm{ext}}_ h(s) & =\mathrm e^{\lambda _0u}f\, \mathrm dV\\ & \geq \mathrm e^{\lambda _0s}f\, \mathrm dV \geq \mathrm e^{\lambda s}f\, \mathrm dV. \end{align*}
Thus $s$ is an exponential subsolution at every $\lambda \geq \lambda _0$, and theorem 15.35.2 gives a solution. Hence $\Lambda _ h(f)$ contains $[\lambda _0,+\infty )$ together with each of its points $\lambda _0$. A nonempty subset of $(0,+\infty )$ with this property is determined by its infimum $a$ and by whether $a$ belongs to it, which gives the stated forms. If $b_ h(f)>0$, corollary 15.35.3 gives $\Lambda _ h(f)=(0,+\infty )$. $\square$
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