Theorem 5.37.22E767909834AD

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. $\square$