PomeProposition 23.1.70.A71DAFFE572E If $\sigma $ is a shrinking defect for $(\omega ,6+m)$, then so is $\mathcal A_\rho \sigma $ for every probability measure $\rho $ on $G$.
Proposition 23.1.70A71DAFFE572E
Proof. Fix a sequence $\varepsilon _ j$ as in the definition. Take a sequence $f_ k=t_ ku_ k$ realizing $\sigma $ as in (23.1.66), so that $f_ k$ has mass $6+m$, and put $H_ k=\mathrm{dd}^{\mathrm{c}}f_ k+t_ k\omega $. By lemma 23.1.68, the positive measures $\nu _ k=3\omega \wedge H_ k^2$ have masses $m+18t_ k^2$ and converge weakly to $\sigma $.
The idea is to replace $f_ k$ by a normalized sum of $N$ random translates. First suppose that $\rho $ has density at most $L$ relative to $h_ G$. Choose $N$ independent random group elements $g_1,\ldots ,g_ N$ with law $\rho $, and set
\[ W=N^{-1/2}\sum _{a=1}^ Nf_ k\circ g_ a,\qquad \tau =t_ k\sqrt N,\qquad K=\mathrm{dd}^{\mathrm{c}}W+\tau \omega =N^{-1/2}\sum _{a=1}^ Ng_ a^*H_ k\geq 0. \]
The quadratic measure of $W$ is
\[ 3\omega \wedge K^2=D+C,\qquad D=N^{-1}\sum _{a=1}^ Ng_ a^*\nu _ k,\qquad C=\frac6N\sum _{a<c}\omega \wedge g_ a^*H_ k\wedge g_ c^*H_ k\geq 0. \]
Averaging $u_ k\circ g$ against $h_ G$ gives a constant, hence $\int _ Gg^*H_ k\, \mathrm{d}h_ G=t_ k\omega $. Consequently $B=\int _ Gg^*H_ k\, \mathrm{d}\rho \leq Lt_ k\omega $, and independence gives
\[ \mathbb E[C(X)]=3(N-1)\int _ X\omega \wedge B^2 \leq 18(N-1)L^2t_ k^2. \]
For any real $\varphi \in C(X)$ with $\| \varphi \| _\infty \leq 1$, the variance of $D(\varphi )$ is at most $(m+18t_ k^2)^2/N$.
For general $\rho $, convolve it with the normalized restrictions of $h_ G$ to shrinking neighborhoods of the identity, which have positive Haar measure. This gives probability measures $\rho _ j\rightharpoonup \rho $ with bounded densities $L_ j$. Fix a countable dense set $(\varphi _ l)$ in the real unit ball of $C(X)$ and numbers $0<\delta _ j<\min (1,m/4)$ tending to zero. Choose $N_ j$ large enough that
\[ \frac{j(m+18)^2}{N_ j\delta _ j^2}<\frac14. \]
Next choose $k_ j$ large enough that $t_{k_ j}<1$, $\tau _ j=t_{k_ j}\sqrt{N_ j}<\min (\varepsilon _ j/4,1/j,1/4)$, $18N_ jt_{k_ j}^2<m/4$, and the expected cross-term mass is less than $\delta _ j/4$. Chebyshev's inequality for the first $j$ test functions and Markov's inequality for $C(X)$ show that some choice of $N_ j$ group elements satisfies
\[ |D_ j(\varphi _ l)-(\mathcal A_{\rho _ j}\nu _{k_ j})(\varphi _ l)|<\delta _ j \quad (l\leq j),\qquad C_ j(X)<\delta _ j. \]
We may also arrange $k_ j\to \infty $. For each fixed test function $\varphi $, the family $\{ \varphi \circ g^{-1}:g\in G\} $ is compact in $C(X)$, so a finite uniform net and the bounded masses of $\nu _ k$ show that $\int \varphi \circ g^{-1}\, \mathrm{d}\nu _ k\to \int \varphi \circ g^{-1}\, \mathrm{d}\sigma $ uniformly in $g$. It follows that $\mathcal A_{\rho _ j}\nu _{k_ j}\rightharpoonup \mathcal A_\rho \sigma $. The dense test functions and the bounded masses now imply $D_ j\rightharpoonup \mathcal A_\rho \sigma $, while $C_ j\to 0$ in total variation.
By the integral identity of lemma 23.1.68,
\[ d_ j=M_\omega (W_ j)-6 =m+C_ j(X)-18(N_ j-1)t_{k_ j}^2\longrightarrow m, \qquad d_ j>3m/4. \]
Put $\lambda _ j=\sqrt{m/d_ j}$, so $\lambda _ j\to 1$ and $\lambda _ j<2$. The average $a_ j=N_ j^{-1}\sum _ a u_{k_ j}\circ g_ a$ is smooth and $\omega $-plurisubharmonic. With
\[ U_ j=a_ j-\sup _ Xa_ j,\qquad r_ j=\lambda _ j\tau _ j<\varepsilon _ j, \]
the potential $r_ jU_ j$ differs from $\lambda _ jW_ j$ by a constant and has exact mass $6+m$. Its positive quadratic measure is $\lambda _ j^2(D_ j+C_ j)$. The total-variation estimate of lemma 23.1.68, with parameter $r_ j$, proves the required weak limit. $\square$
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