5.37.17 The optimal constant in dimension oneA5A50BF6024B

Proof. With the normalization of $\mathrm{dd}^{\mathrm{c}}$ fixed in the conventions, $\mu =\mathrm{dd}^{\mathrm{c}}u$ is a positive Radon measure and $a=\mu (\{ x\} )=\nu (u,x)$. After removal of this atom, the distribution $u-a\log |z-x|^2$ has positive complex Hessian, so it has a subharmonic representative $v$, and $\mathrm{dd}^{\mathrm{c}}v$ has no atom at $x$. Since subharmonic functions that agree almost everywhere agree everywhere,

We claim that, for each $t>0$, the function $\mathrm e^{-tv}$ is integrable on some neighborhood of $x$. Indeed, since $\mathrm{dd}^{\mathrm{c}}v$ has no atom at $x$, there is a disk $D=D(x,R)\Subset \Omega $ with $M=(\mathrm{dd}^{\mathrm{c}}v)(D)<1/t$. On $D$ write

where $h$ is harmonic: the integral is locally integrable by Fubini's theorem and has complex Hessian $\mathrm{dd}^{\mathrm{c}}v$ on $D$, so it differs from $v$ by a harmonic distribution. If $M=0$, the claim follows from harmonicity. For $M>0$, Jensen's inequality for the probability measure $(\mathrm{dd}^{\mathrm{c}}v)/M$ gives

Since $tM<1$, the integral of $|z-y|^{-2tM}$ over $z\in D(x,R/2)$ is bounded uniformly in $y\in D$. The harmonic function $h$ is bounded below on this smaller disk, so Tonelli's theorem proves the claim. (Truncating the logarithmic kernel justifies Jensen's inequality at its poles.)

A nonzero holomorphic germ of order $k$ at $x$ is a unit times $(z-x)^ k$. Since $v$ is locally bounded above, weighted integrability requires $k+1>a$; equality is excluded by the logarithmic divergence of the radial integral. Conversely, if $k+1>a$, choose $b>1$ such that $b(a-k)<1$. Then $|z-x|^{2k-2a}$ belongs locally to $L^ b$, and $\mathrm e^{-v}$ belongs locally to $L^{b/(b-1)}$ by the claim, so Hölder's inequality proves integrability. Thus the germ belongs to $\mathcal I(u)_ x$ exactly when $k\geq \lfloor a\rfloor $.

The points $y$ with $\nu (u,y)\geq 1$ form a locally finite set, since each of them carries an atom of mass at least one of the locally finite measure $\mathrm{dd}^{\mathrm{c}}u$. Near any point, therefore, the ideal sheaf is generated by a single power of a coordinate, which proves the last assertion. $\square$

Proof. For $r=1$ there is nothing to prove, so let $r\geq 2$. Put $J_ i=\mathcal I(u_ i)$ and $J=\sum _ iJ_ i$. By lemma 5.37.18, these sheaves are locally free of rank one. At each point, the order of $J$ is the smallest of the orders of the $J_ i$, and a local generator of an ideal $J_ i$ attaining this order also generates $J$. Consequently the sum map $\bigoplus _ iJ_ i\to J$ is locally split, and its kernel $E$ is a holomorphic vector bundle of rank $r-1$.

Fix $f\in \Gamma (\Omega ,J)$. Gluing local holomorphic lifts of $f$ with a smooth partition of unity gives a smooth section $s$ of $\bigoplus _ iJ_ i$ whose image is $f$. Since $f$ is holomorphic, $\beta =\bar\partial s$ is an $E$-valued $(0,1)$-form. It suffices to solve $\bar\partial v=\beta $ for a section $v$ of $E$, for then $s-v$ is a holomorphic lift of $f$.

We solve this equation using a bundle metric of positive curvature. Choose a smooth strictly subharmonic exhaustion function $\rho \geq 1$ of $\Omega $. The Kähler metric

is complete: it dominates $2\mathrm i\partial \rho \wedge \bar\partial \rho $, so $\rho $ has bounded gradient with respect to it, and therefore every curve leaving all compact subsets has infinite length.

Equip $E$ with a smooth Hermitian metric $h$. Choose a smooth increasing convex function $\chi $ on $[1,+\infty )$ so that the metric $h_\chi =h\mathrm e^{-\chi (\rho )}$ has

These two requirements can be imposed successively on the compact sublevel sets of $\rho $: increase $\chi '$ so that $\chi '(\rho )$ times the positive Hessian of $\rho $ dominates the negative part of the curvature, and increase $\chi $ so that the integral over each shell $\{ m\leq \rho <m+1\} $ is at most $2^{-m}$. Increasing the slopes preserves the inequalities already achieved.

Use the global frame $\mathrm{d}z$ of the canonical bundle $K_\Omega $ to identify $E$-valued $(0,q)$-forms with $(E\otimes K_\Omega ^{-1})$-valued $(1,q)$-forms, and give $K_\Omega ^{-1}$ its flat coordinate metric. Its curvature is zero, and $\omega _*$ dominates the Euclidean metric, so the identified form $\beta $ remains $L^2$. The Bochner–Kodaira inequality for compactly supported forms $w$ of type $(1,1)$ gives

Here we used that there are no $(1,2)$-forms. By completeness, cutoffs and smoothing extend this inequality to the domain of the adjoint. Since $\beta $ is $L^2$, the Cauchy–Schwarz inequality bounds the functional $w\mapsto \langle w,\beta \rangle $ by $\| \beta \| \| \bar\partial ^*w\| $, and the Riesz representation theorem gives a distributional solution $v$ of $\bar\partial v=\beta $. Thus $s-v$ is distributionally holomorphic, hence holomorphic in every bundle chart, and it is the required global lift. No global weighted norm condition is imposed on its components. $\square$

Proof. Put $J_ i=\mathcal I(p_ i\phi _ i+\psi )$ and $J=\sum _ iJ_ i$. At a point $x$, write $A=\nu (\psi ,x)$ and $B_ i=\nu (\phi _ i,x)$. By lemma 5.37.18, the order of $J_ x$ is $\min _ i\lfloor A+p_ iB_ i\rfloor $.

Suppose first that $s<1$, and put $P=1/s>1$. The numbers $P/p_ i$ sum to one, so

Taking integer parts and using lemma 5.37.18 gives $\mathcal I(P\sum _ i\phi _ i+\psi )\subseteq J$. By lemma 5.37.20, every global section of this smaller ideal sheaf lies in $\sum _ i\Gamma (\Omega ,J_ i)$, so every remainder admissible for the one-weight problem is admissible for $c$. Proposition 5.37.14 with $\phi =\sum _ i\phi _ i$ and $p=P$ therefore gives $c\leq (1-1/P)I=(1-s)I$.

Now let $s\geq 1$. Since $I<+\infty $, we have $f\in \Gamma (\Omega ,\mathcal I(\psi +\sum _ i\phi _ i))$. If $f\equiv 0$ there is nothing to prove; otherwise let $k$ be its order at $x$. Then $k+1>A+\sum _ iB_ i$. If its germ did not belong to $J_ x$, we would have $k+1\leq A+p_ iB_ i$ for every $i$. With $L=k+1-A>0$, this would give $\sum _ iB_ i\geq sL\geq L$, a contradiction. Hence $f\in \Gamma (\Omega ,J)$, and lemma 5.37.20 shows that $g=0$ is admissible.

For sharpness when $s<1$, take the unit disk, $\psi =0$, $f=1$, and $\phi _ i=p_ i^{-1}\log |z|^2$. Then $J_ i=(z)$ for all $i$, so an admissible $g$ has $g(0)=1$. Orthogonality of the monomials on the disk gives $c=\pi $, attained by $g=1$, whereas $I=\pi /(1-s)$. If $s\geq 1$, keep these weights and take $f=z^ q$ with an integer $q\geq 0$ satisfying $q+1>s$. Then $I=\pi /(q+1-s)>0$ and $c=0$, so the constant $0$ cannot be replaced by a negative number. $\square$