Theorem 11.40.1.133439CA2DC1 Suppose $0<\varepsilon <1$ and $(1-\varepsilon )\beta \leq \beta '\leq (1+\varepsilon )\beta $. For every bounded measurable $F$,
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11.40.2
\begin{equation} n\log (1-\varepsilon )\leq B_{\beta '}(F)-B_\beta (F)\leq n\log (1+\varepsilon ). \end{equation}
In particular, $d(B_\beta ,B_{\beta '})\leq -n\log (1-\varepsilon )$. The same conclusion holds if the two inequalities between forms are satisfied by positive representatives $\beta +\mathrm{dd}^{\mathrm{c}}h$ and $\beta '+\mathrm{dd}^{\mathrm{c}}h'$, with $h,h'$ smooth real functions.
Proof.
First let $F$ be smooth, and let $v,v'$ be smooth solutions in the two backgrounds. At a maximum point of $v'-(1+\varepsilon )v$,
\[ 0<\beta '+\mathrm{dd}^{\mathrm{c}}v'\leq (1+\varepsilon )(\beta +\mathrm{dd}^{\mathrm{c}}v). \]
Taking determinants and canceling $\mathrm{e}^ F\mathrm{d}V$ gives the upper bound. At a minimum point of $v'-(1-\varepsilon )v$, the reverse inequality between forms gives the lower bound. For bounded measurable $F$, choose smooth uniformly bounded approximations converging in $L^1$ and apply theorem 11.33.5 in each of the two backgrounds. Finally, the correspondence $v\mapsto v-h$ preserves both boundedness and the current, so $B_{\beta +\mathrm{dd}^{\mathrm{c}}h}=B_\beta $ whenever both backgrounds are positive. This proves the assertion about representatives.
$\square$
Pome
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