10.31.46 Stability under changes of reference4D69742DF215

Proof. Write $M_0=\overline{\operatorname {vol}}_\theta (V_\theta )$. Every solution above has mass between $\delta $ and $M_0$. In the first case let $s_ j=\sup _ Xu_ j$. Then

which is a uniform lower bound since $\mu _ j(X)\to \mu (X)>0$. To bound $s_ j$ from above, suppose that $s_ j\to +\infty $ along a subsequence. By normalized compactness, after a further extraction, $v_ j=u_ j-s_ j\to v$ in $L^1$ with $\sup _ Xv=0$. The fixed-measure lemma lemma 5.44.2 and $f_ j\to f$ in $L^1(\nu )$ imply

The limit is positive because $f\nu $ is nonzero and does not charge the poles of $v$. But the left-hand side equals $\mathrm{e}^{-\lambda s_ j}\mu _{u_ j}(X)\leq M_0\mathrm{e}^{-\lambda s_ j}$, which tends to zero, a contradiction. Thus $s_ j$ is bounded, and the $u_ j$ form a relatively compact family in $L^1$ whose limits are not identically $-\infty $.

Start with any subsequence and extract $u_ j\to u$ in $L^1$. Extract again so that $\sum _ j\| f_ j-f\| _{L^1(\nu )}<\infty $. The function

dominates all retained $f_ j$. If $C$ is a common upper bound for the $u_ j$, then $\mu _{u_ j}\leq \mathrm{e}^{\lambda C}g\nu $. By theorem 10.31.30, $u\in \mathcal E_\varphi $, the potentials converge in capacity, and the measures converge weakly to $\mu _ u$. On the other hand,

Indeed, the error coming from $f_ j-f$ is at most $\mathrm{e}^{\lambda C}\| f_ j-f\| _{L^1(\nu )}$, and lemma 5.44.2 for the fixed measure $f\nu $ controls the remaining term. Hence $\mu _ u=\mathrm{e}^{\lambda u}\mu $, with convergence in total variation. By the uniqueness in theorem 10.31.44, every cluster limit is this solution, and the subsequence criterion gives all three convergences for the original sequence.

In the second case, integrating the equation gives

Since $\mu _ j(X)\to \mu (X)>0$, these scalars stay in a compact subinterval of $(0,+\infty )$. After extraction, normalized compactness gives $u_ j\to u$ in $L^1$ with $\sup _ Xu=0$, and $c_ j\to c'>0$. Extract a summable density error as above. Then $\mu _{u_ j}\leq Cg\nu $ for a fixed $C$, so theorem 10.31.30 gives $u\in \mathcal E_\varphi $, convergence in capacity and weak convergence to $\mu _ u$. Since $c_ jf_ j\to c'f$ in $L^1(\nu )$, the limiting equation is $\mu _ u=c'\mu $, and the uniqueness of the scalar in theorem 10.31.45 gives $c'=c$. Hence the whole sequence of scalars converges to $c$, and $c_ j\mu _ j\to c\mu $ in total variation. Applying the same argument within any $L^1$-convergent subsequence proves its convergence in capacity and the equation for its limit. $\square$