Pome

Corollary 11.40.39B8101BC6CE4

Corollary 11.40.3.9B8101BC6CE4 Let $\beta _ j$ be Hermitian forms with $(1-\varepsilon _ j)\beta \leq \beta _ j\leq (1+\varepsilon _ j)\beta $, where $0<\varepsilon _ j<1$ and $\varepsilon _ j\to 0$. Suppose that $F_ j\to F$ in $L^1(\mathrm{d}V)$ with a common $L^\infty $ bound. Then $B_{\beta _ j}(F_ j)\to B_\beta (F)$. For any $\nu _ j\in \mathcal A_{\beta _ j}(F_ j)$, the densities are uniformly integrable, and every subsequence has a further subsequence converging weakly in $L^1$ to an element of $\mathcal A_\beta (F)$. If this set is a singleton, which is the case in particular when $F$ is smooth, the whole sequence converges to it weakly in $L^1$ and as probability measures.

Proof. Put $\delta _ j=-n\log (1-\varepsilon _ j)$. The scalar assertion follows from (11.40.2) and theorem 11.33.5 in the fixed background $\beta $. Taking suprema over tests gives the inequalities, valid also for infinite values,

\[ H_{\beta _ j}\leq H_\beta +\delta _ j,\qquad H_\beta \leq H_{\beta _ j}+\delta _ j. \]

In particular, the two duals are finite on the same measures, and the identity $H_{\beta _ j}(\nu _ j)=B_{\beta _ j}(F_ j)+\int _ XF_ j\, \mathrm{d}\nu _ j$ gives

\[ 0\leq H_\beta (\nu _ j)-B_\beta (F_ j)-\int _ XF_ j\, \mathrm{d}\nu _ j\leq 2\delta _ j. \]

If $\| F_ j\| _\infty \leq M$, this puts the measures in the fixed sublevel set $\{ H_\beta \leq B_\beta (0)+2M+2\sup _ j\delta _ j\} $, to which theorem 11.35.4 applies. For a weak $L^1$ cluster point, uniform integrability gives $\int _ X|F_ j-F|\, \mathrm{d}\nu _ j\to 0$, and lower semicontinuity in the last inequality, together with the minimum formula of theorem 11.36.1, identifies the limit as a supporting probability at $F$. Compactness gives convergence of the whole sequence when that supporting probability is unique, and theorem 11.34.9 gives uniqueness for smooth $F$. $\square$

Cite this page

Comments

Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.