Theorem 18.38.4.C55B3CFB4B5C Let $U_ m=m^{-1}\log B_{X_ n,m}$ be the partial Fubini–Study potential for the zero weight. Then
\[ U_ m\longrightarrow E_ a\circ \pi \]
in $L^1(X_ n)$, pointwise as extended-real functions, and locally uniformly off the fiber $\pi ^{-1}(b)$. The normalized kernel measures converge in total variation:
\[ \frac{q!}{m^ q}B_{X_ n,m}\, \nu _ X \longrightarrow \mathbf1_{\{ t\circ \pi >a\} }\nu _ X. \]
Proof.
Write $P_ m(t)$ for the binomial sum in proposition 18.38.2. For fixed $0<t<1$, $P_ m(t)$ is the probability that a binomial random variable with parameters $(m,t)$ is at least $r_ m$. Its variance is $mt(1-t)$, so Chebyshev's inequality and $r_ m/m\to a$ give
\[ P_ m(t)\longrightarrow \begin{cases} 0,& t<a,\\ 1,& t>a.\end{cases} \]
The pushforward of $\sigma ^ q$ under $t$ is $q t^{q-1}\, \mathrm{d}t$. To see this, note that in the affine chart the Fubini–Study measure has radial density $2q r^{2q-1}(1+r^2)^{-q-1}\, \mathrm{d}r$, and $t=r^2/(1+r^2)$. In particular, the level set $\{ t=a\} $ has measure zero. Since $0\leq P_ m\leq 1$ and $q!D_ m/m^ q\to 1$, dominated convergence gives convergence of the densities in $L^1(\sigma ^ q)$, hence in $L^1(\nu _ X)$ after lifting; this is the asserted convergence in total variation. The limit has mass $\int _ a^1 qt^{q-1}\, \mathrm{d}t=1-a^ q$.
To identify the limit of the potentials, put
\[ H(s)=-s\log s-(1-s)\log (1-s),\qquad 0\leq s\leq 1, \]
with $0\log 0=0$. Then
\[ \frac{\mathrm{e}^{mH(j/m)}}{m+1}\leq \binom mj \leq \mathrm{e}^{mH(j/m)}\quad (0\leq j\leq m). \]
Indeed, $\binom mj\mathrm{e}^{-mH(j/m)}$ is the $j$th term of the binomial distribution with parameters $(m,j/m)$. It is at most one, and it is at least $1/(m+1)$ because, as the ratio of neighboring terms shows, it is a largest of the $m+1$ terms, which sum to one. For $j=0$ and $j=m$ it equals one.
The logarithm of a sum of at most $m+1$ nonnegative terms differs from the largest of their logarithms by at most $\log (m+1)$. Hence $m^{-1}\log P_ m(t)$ differs by at most $\log (m+1)/m$ from
\[ \max _{r_ m\leq j\leq m} \left\{ H(j/m)+(j/m)\log t+(1-j/m)\log (1-t)\right\} . \]
On compact subintervals of $(0,1)$, uniform continuity and $r_ m/m\to a$ show that these maxima converge uniformly to the maximum of the same expression over $a\leq s\leq 1$. Maximizing in $s$, we find that this maximum is zero for $t\geq a$, attained at $s=t$, and equals the first expression in (18.38.1) for $t<a$, attained at $s=a$. Thus the limit is $e_ a(t)$, also at $t=a$.
The convergence is also uniform near $t=1$. Choose $c$ with $a<c<1$. The same variance estimate gives $P_ m(t)\geq 1-C/m$ uniformly for $c\leq t\leq 1$ and large $m$, so $m^{-1}\log P_ m\to 0$ uniformly there. Together with the argument on compact subintervals and $m^{-1}\log D_ m\to 0$, this proves locally uniform convergence off $\{ t=0\} $. At $t=0$ we have $r_ m>0$ for large $m$, so $P_ m(0)=0$ and the pointwise limit is $-\infty $.
Finally, $t^ m\leq P_ m(t)\leq 1$, because the term $j=m$ occurs in every binomial sum. Since $m^{-1}\log D_ m$ is nonnegative and bounded, it follows that $\log (t\circ \pi )\leq U_ m\leq C$ off the fiber over $b$. In local submersion coordinates near this fiber, $\log (t\circ \pi )=\log |z|^2+O(1)$, which is integrable against a smooth volume. The pointwise convergence and these bounds give convergence in $L^1$, by dominated convergence.
$\square$
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