Proof. [026N]

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Proof.

Clearly, we may assume that $l\not=0$ . Let $\mathscr{Y}$ be the Zariski closure of $Y$ in $\mathscr{X}$ (cf. §1.1.7).

Claim 3.2.1.

There are a positive integer $a$ and $\alpha\in k^{\times}$ such that

e^{-a\epsilon/2}\leq\|\alpha l^{\otimes a}\|_{Y,h^{a}}\leq 1.

Proof.

First we assume that $|\raisebox{1.72218pt}{.}|$ is discrete. We take a positive integer $a$ such that $e^{-\epsilon a/2}\leq|\varpi|$ . We also choose $\alpha\in k^{\times}$ such that

|\alpha^{-1}|=\min\{|\gamma|\mid\text{$\gamma\in k^{\times}$ and $\|l^{\otimes a}\|_{Y,h^{a}}\leq|\gamma|$}\}.

Then, as $\|l^{\otimes a}\|_{Y,h^{a}}\leq|\alpha^{-1}|\leq|\varpi|^{-1}\|l^{\otimes a}\|_{Y,h^{a}}$ , we have

e^{-a\epsilon/2}\leq|\varpi|\leq\|\alpha l^{\otimes a}\|_{Y,h^{a}}\leq 1.

Next we assume that $|\raisebox{1.72218pt}{.}|$ is not discrete. In this case, $|k^{\times}|$ is dense in $\mathbb{R}_{>0}$ by Lemma 1.15, so that we can choose $\beta\in k^{\times}$ such that

e^{-\epsilon/2}\leq\|l\|_{Y,h}/|\beta|\leq 1.

Thus if we set $\alpha=\beta^{-1}$ and $a=1$ , we have the assertion. ∎

By Corollary 2.2, there is $\beta\in\mathfrak{o}_{K}\setminus\{0\}$ such that

\beta(\alpha l^{\otimes a})^{\otimes m}\in H^{0}(\mathscr{Y},\left.{\mathscr{L}^{\otimes am}}\right|_{{\mathscr{Y}}})

for all $m\geq 0$ . We choose a positive integer $m$ such that $|\beta|^{-1}\leq e^{am\epsilon/2}$ and

H^{0}(\mathscr{X},\mathscr{L}^{\otimes am})\to H^{0}(\mathscr{Y},\left.{\mathscr{L}^{\otimes am}}\right|_{{\mathscr{Y}}})

is surjective, so that we can find $l_{m}\in H^{0}(\mathscr{X},\mathscr{L}^{\otimes am})$ such that $\left.{l_{m}}\right|_{{\mathscr{Y}}}=\beta(\alpha l^{\otimes a})^{\otimes m}$ . Note that $\|l_{m}\|_{h^{am}}\leq 1$ . Thus, if we set $s=\beta^{-1}\alpha^{-m}l_{m}$ , then $\left.{s}\right|_{{\mathscr{Y}}}=l^{\otimes am}$ and

	$\displaystyle\\|s\\|_{h^{am}}$	$\displaystyle=\|\beta\|^{-1}\|\alpha\|^{-m}\\|l_{m}\\|_{h^{am}}\leq e^{am\epsilon/2}\|\alpha\|^{-m}$
		$\displaystyle\leq e^{am\epsilon/2}\|\alpha\|^{-m}\left(e^{a\epsilon/2}\\|\alpha l^{\otimes a}\\|_{Y,h^{a}}\right)^{m}=e^{am\epsilon}\left(\\|l\\|_{Y,h}\right)^{am},$

as required. ∎