Lemma 10.6.294EC30DAF1A3

Proof. (1) Let $M$ be the supremum in the statement. If $\overline{B(z,\delta )}\subset U$, then $|f_ j(z)|^2\leq |B(z,\delta )|^{-1}M$ by (6.8.7). Hence the $f_ j$ are uniformly bounded on every compact subset of $U$: if $K\subset U$ is compact and $\delta >0$ is smaller than the distance from $K$ to $\mathbb C^ n\setminus U$ (any $\delta >0$ if $U=\mathbb C^ n$), then $|f_ j|^2\leq |B(0,\delta )|^{-1}M$ on $K$.

We next show that the sequence is equicontinuous on compact sets. Let $K\subset U$ be compact, and choose $r>0$ such that the compact set $K'$ of points at distance at most $2\sqrt n\, r$ from $K$ lies in $U$; let $M'$ be a bound for all $|f_ j|$ on $K'$. For $z$ at distance at most $\sqrt n\, r$ from $K$, the closed polydisc of radius $r$ centered at $z$ lies in $K'$, and the coefficient bound in (5.2.1) gives $|\partial _{z_ l}f_ j(z)|\leq M'/r$ for $1\leq l\leq n$. Since $\mathrm df_ j=\sum _ l\partial _{z_ l}f_ j\, \mathrm dz_ l$, the $f_ j$ are Lipschitz with a common constant on every ball of radius $\sqrt n\, r$ centered at a point of $K$, so the sequence is equicontinuous on $K$. By proposition 2.7.4, applied on the compact sets $K_ m=\{ z\in U:|z|\leq m,\ \operatorname {dist}(z,\mathbb C^ n\setminus U)\geq 1/m\} $, which exhaust $U$, and a diagonal argument, a subsequence converges locally uniformly on $U$ to a continuous function $f$. Then $\bar\partial f=0$ in the sense of distributions, as a limit of $\bar\partial f_ j=0$, so $f$ agrees almost everywhere with a holomorphic function by lemma 6.4.10(1); both being continuous, $f$ is holomorphic.

(2) Choose finitely many coordinate balls $B_ k$, with closures in the domains of their charts, such that the concentric balls $B'_ k$ of half the radius cover $X$, and let $B''_ k$ be the concentric balls of three quarters of the radius. By lemma 10.6.1, $E$ has a holomorphic frame $e_ k$ over $B_ k$; write $s_ j=f_{j,k}e_ k$ on $B_ k$. On the compact set $\overline{B''_ k}$ the function $|e_ k|_{h_ E}^2$ is bounded below by a positive constant and $\mathrm dV$ is bounded below by a positive multiple of Lebesgue measure in the chart, so the integrals of $|f_{j,k}|^2$ over $B''_ k$ with respect to Lebesgue measure in the chart are bounded independently of $j$. Applying (1) successively for the finitely many $k$, we obtain a subsequence along which $f_{j,k}$ converges locally uniformly on $B''_ k$, hence uniformly on $\overline{B'_ k}$, to a holomorphic function $f_ k$, for every $k$. On $B''_ k\cap B''_ l$ the sections $f_ ke_ k$ and $f_ le_ l$ are both the pointwise limit of $s_ j$, so they define a holomorphic section $s$ of $E$ over $X$, and $\sup _{B'_ k}|s_ j-s|_{h_ E}\leq \sup _{\overline{B'_ k}}|e_ k|_{h_ E}\cdot \sup _{\overline{B'_ k}}|f_{j,k}-f_ k|\to 0$ for every $k$.

(3) Fix a smooth quadratic metric $h_ E$ on $E$ (lemma 6.14.10(1)). The integral $\| s\| ^2=\int _ X|s|_{h_ E}^2\, \mathrm dV$ defines an inner product on $H^0(X,E)$: if $\| s\| =0$, then the continuous function $|s|_{h_ E}$ vanishes almost everywhere, hence everywhere. If $H^0(X,E)$ were infinite dimensional, the Gram–Schmidt process would give an orthonormal sequence $(s_ j)$. By (2), a subsequence converges uniformly on $X$, hence in $L^2$, since $X$ has finite volume; but $\| s_ i-s_ j\| ^2=2$ for $i\neq j$. This contradiction proves (3).

The proofs of (2) and (3) use only that $X$ is a compact complex manifold of positive dimension and that $\mathrm dV$ is a smooth positive volume form; hence they also prove the last assertion of the lemma. $\square$