First we assume that $|\raisebox{1.72218pt}{.}|$ is non-trivial. Let us begin with the following:

Let $\mathscr{X}$ be the Zariski closure of $X$ in $\mathbb{P}(\mathscr{H})$ (cf. §1.1.7) and $\mathscr{L}=\left.{\mathscr{O}_{\mathbb{P}(\mathscr{H})}(1)}\right|_{{\mathscr{X}}}$. Moreover, let $h^{\prime}$ be a continuous metric of $(L^{\otimes a})^{\mathrm{an}}$ given by

$$\big\{|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{(H,\|\raisebox{1.20552pt}{.}\|_{\mathscr{H}})}(x)\big\}_{x\in X^{\mathrm{an}}}.$$

Then, by Proposition 3.8 and Remark 3.9, $|\raisebox{1.72218pt}{.}|_{h^{\prime}}=|\raisebox{1.72218pt}{.}|_{\mathscr{L}}$. Therefore, by virtue of Theorem 3.2, there are a positive integer $m$ and $s\in H^{0}(X,L^{\otimes am})$ such that $\left.{s}\right|_{{Y}}=l^{\otimes am}$ and

(5)

$$\|s\|_{{h^{\prime}}^{m}}\leq e^{am\epsilon/2}(\|l^{\otimes a}\|_{Y,h^{\prime}})^{m}.$$

As $\|\raisebox{1.72218pt}{.}\|_{h^{a}}\leq\|\raisebox{1.72218pt}{.}\|_{\mathscr{H}}\leq e^{a\epsilon/2}\|\raisebox{1.72218pt}{.}\|_{h^{a}}$, we have

$$|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h^{a}}(x)\leq|\raisebox{1.72218pt}{.}|_{h^{\prime}}(x)\leq e^{a\epsilon/2}|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h^{a}}(x)$$

for all $x\in X^{\mathrm{an}}$. Therefore, by Proposition 3.6,

(6)

$$|\raisebox{1.72218pt}{.}|_{h^{a}}(x)\leq|\raisebox{1.72218pt}{.}|_{h^{\prime}}(x)\leq e^{a\epsilon/2}|\raisebox{1.72218pt}{.}|_{h^{a}}(x)$$

for all $x\in X^{\mathrm{an}}$. In particular, $|\raisebox{1.72218pt}{.}|_{h^{am}}(x)\leq|\raisebox{1.72218pt}{.}|_{{h^{\prime}}^{m}}(x)$. Therefore,

(7)

$$\|s\|_{h^{am}}\leq\|s\|_{{h^{\prime}}^{m}}.$$

On the other hand, by using (6),

(8)

$$\|l^{\otimes a}\|_{Y,h^{\prime}}\leq e^{a\epsilon/2}\sup\{|l^{\otimes a}|_{h^{a}}(y)\mid y\in Y^{\mathrm{an}}\}\leq e^{a\epsilon/2}(\|l\|_{Y,h})^{a}.$$

Thus the assertion follows from (5), (7) and (8).

Next we assume that $|\raisebox{1.72218pt}{.}|$ is trivial. Clearly we may assume that $l\not=0$. Let $k^{\prime}$ be the field $k(\!(T)\!)$ of formal Laurent power series over $k$, that is, the quotient field of the ring $k[\![T]\!]$ of formal power series over $k$. We set

$$\Sigma:=\bigcup_{i=0}^{\infty}\left(\bigcup_{s,s^{\prime}\in H^{0}(X,L^{\otimes i})\setminus\{0\}}\mathbb{Q}\left(\log\|s\|_{h^{i}}-\log\|s^{\prime}\|_{h^{i}}\right)\right).$$

As $\left\{\|s\|_{h^{i}}\mid s\in H^{0}(X,L^{\otimes i})\setminus\{0\}\right\}$ is a finite set by (1) in Lemma 1.12, we have $\#(\Sigma)\leq\aleph_{0}$. Therefore, we can find $\alpha\in\mathbb{R}_{>0}\setminus\Sigma$. Here we consider an absolute value $|\raisebox{1.72218pt}{.}|^{\prime}$ of $k^{\prime}$ given by

$$|\phi(T)|^{\prime}:=\exp(-\alpha\operatorname{ord}(\phi(T)))\quad(\phi(T)\in k^{\prime}).$$

We set

$$X^{\prime}:=X\times_{\operatorname{Spec}(k)}\operatorname{Spec}(k^{\prime}),\quad Y^{\prime}:=Y\times_{\operatorname{Spec}(k)}\operatorname{Spec}(k^{\prime})\quad\text{and}\quad L^{\prime}=L\otimes_{k}k^{\prime}.$$

Note that $H^{0}(X^{\prime},L^{\prime})=H^{0}(X,L)\otimes_{k}k^{\prime}$. Let $h^{\prime}$ be a continuous metric of ${L^{\prime}}^{\mathrm{an}}$ given by the scalar extension of $h$. Then, by Lemma 3.7, $h^{\prime}$ is given by

$$\big\{|\raisebox{1.72218pt}{.}|_{(H^{0}(X^{\prime},L^{\prime}),\|\raisebox{1.20552pt}{.}\|_{k^{\prime}})}^{\mathrm{quot}}(x^{\prime})\big\}_{x^{\prime}\in{X^{\prime}}^{\mathrm{an}}},$$

where $\|\raisebox{1.72218pt}{.}\|_{k^{\prime}}$ is the scalar extension of $\|\raisebox{1.72218pt}{.}\|$. Moreover, for $s\in H^{0}(X,L)$, $|s|_{h^{\prime}}(x^{\prime})=|s|_{h}(p^{\mathrm{an}}(x^{\prime}))$ for $x^{\prime}\in{X^{\prime}}^{\mathrm{an}}$, where $p:X^{\prime}\to X$ is the projection. Note that $p^{\mathrm{an}}:{X^{\prime}}^{\mathrm{an}}\to X^{\mathrm{an}}$ is surjective. Therefore, $\|s\|_{h^{\prime}}=\|s\|_{h}$ for all $s\in H^{0}(X,L)$.

By the previous observation, there are a positive integer $n$ and $s^{\prime}\in H^{0}(X^{\prime},{L^{\prime}}^{\otimes n})$ such that

$$\left.{s^{\prime}}\right|_{{Y^{\prime}}}={l}^{\otimes n}\quad\text{and}\quad\|s^{\prime}\|_{{h^{\prime}}^{n}}\leq e^{n\epsilon}(\|l\|_{Y^{\prime},h^{\prime}})^{n}=e^{n\epsilon}(\|l\|_{Y,h})^{n}.$$

Note that, for a positive integer $d$,

$${s^{\prime}}^{\otimes d}\in H^{0}(X^{\prime},{L^{\prime}}^{\otimes dn}),\quad\left.{{s^{\prime}}^{\otimes d}}\right|_{{Y^{\prime}}}={l}^{\otimes dn}\quad\text{and}\quad\|{s^{\prime}}^{\otimes d}\|_{{h^{\prime}}^{dn}}\leq e^{dn\epsilon}(\|l\|_{Y,h})^{dn}.$$

Thus we may assume that $H^{0}(X,L^{\otimes n})\to H^{0}(Y,\left.{L}\right|_{{Y}}^{\otimes n})$ is surjective. Let $(e_{1},\ldots,e_{r})$ be an orthogonal basis of $H^{0}(X,L^{\otimes n})$ with respect to $\|\raisebox{1.72218pt}{.}\|_{h^{n}}$ such that $(e_{t+1},\ldots,e_{r})$ forms a basis of $\operatorname{Ker}(H^{0}(X,L^{\otimes n})\to H^{0}(Y,\left.{L}\right|_{{Y}}^{\otimes n}))$ (cf. Proposition 1.3). We set

$$s^{\prime}=a_{1}(T)e_{1}+\cdots+a_{t}(T)e_{t}+a_{t+1}(T)e_{t+1}+\cdots+a_{r}(T)e_{r}$$

for some $a_{1}(T),\ldots,a_{r}(T)\in k^{\prime}=k(\!(T)\!)$. As $\left.{s^{\prime}}\right|_{{Y^{\prime}}}=l^{\otimes n}\in H^{0}(Y,\left.{L}\right|_{{Y}}^{\otimes n})$ and $(\left.{e_{1}}\right|_{{Y}},\ldots,\left.{e_{t}}\right|_{{Y}})$ forms a basis of $H^{0}(Y,\left.{L}\right|_{{Y}}^{\otimes n})$, we have $a_{1}(T),\ldots,a_{t}(T)\in k$. Note that

$$\alpha\not\in\bigcup_{s,s^{\prime}\in H^{0}(X,L^{\otimes n})\setminus\{0\}}\mathbb{Q}\left(\log\|s\|_{h^{n}}-\log\|s^{\prime}\|_{h^{n}}\right),$$

so that, by (2) in Lemma 1.12 and Remark 1.13, $(e_{1},\ldots,e_{r})$ forms an orthogonal basis of $H^{0}(X^{\prime},{L^{\prime}}^{\otimes n})$ with respect to $\|\raisebox{1.72218pt}{.}\|_{{h^{\prime}}^{n}}$. Therefore, if we set $s=a_{1}e_{1}+\cdots+a_{t}e_{t}$, then $s\in H^{0}(X,L^{\otimes n})$, $\left.{s}\right|_{{Y}}=l^{\otimes n}$ and

	$\displaystyle\|s\|_{h^{n}}$	$\displaystyle=\max\{|a_{1}|\|e_{1}\|_{h^{n}},\ldots,|a_{t}|\|e_{t}\|_{h^{n}}\}$
		$\displaystyle\leq\max\left\{|a_{1}|\|e_{1}\|_{h^{n}},\ldots,|a_{t}|\|e_{t}\|_{h^{n}},|a_{t+1}(T)|^{\prime}\|e_{t+1}\|_{h^{n}},\ldots,|a_{r}(T)|^{\prime}\|e_{r}\|_{h^{n}}\right\}$
		$\displaystyle=\|s^{\prime}\|_{{h^{\prime}}^{n}}\leq e^{n\epsilon}(\|l\|_{Y,h})^{n},$

as required. ∎

Clearly we may assume that $l\not=0$. Let us begin with the following claim:

Since $L$ is ample, by Corollary 1.2, the above claim is actually equivalent to the assertion of the theorem. Thus the theorem is proved. ∎