By using Proposition 3.10, we can find a positive integer $a$ such that $L^{\otimes a}$ is very ample and

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

for all $x\in X^{\mathrm{an}}$. We set $h^{\prime}=\{|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h^{a}}(x)\}$. Then, the above inequalities means that

(9)

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

for all $x\in X^{\mathrm{an}}$. Further, by Theorem 4.1, there are a positive integer $b$ and $s_{ab}\in H^{0}(X,L^{\otimes ab})$ such that $\left.{s_{ab}}\right|_{{Y}}=l^{\otimes ab}$ and

$$\|s_{ab}\|_{{h^{\prime}}^{b}}\leq e^{ab{\epsilon^{\prime}/2}}(\|l^{\otimes a}\|_{Y,h^{\prime}})^{b}.$$

By (9),

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

Moreover, as $|\raisebox{1.72218pt}{.}|_{h^{ab}}(x)\leq|\raisebox{1.72218pt}{.}|_{{h^{\prime}}^{b}}(x)$ by (9), we have $\|s_{ab}\|_{h^{ab}}\leq\|s_{ab}\|_{{h^{\prime}}^{b}}$, so that

	$\displaystyle\|s_{ab}\|_{h^{ab}}$	$\displaystyle\leq\|s_{ab}\|_{{h^{\prime}}^{b}}\leq e^{ab{\epsilon^{\prime}/2}}(\|l^{\otimes a}\|_{Y,h^{\prime}})^{b}$
		$\displaystyle\leq e^{ab{\epsilon^{\prime}/2}}(e^{a{\epsilon^{\prime}/2}}(\|l\|_{Y,h})^{a})^{b}\leq e^{ab{\epsilon^{\prime}}}(\|l\|_{Y,h})^{ab}.$

Therefore, if we set $N=ab$, then we have the assertion of the claim. ∎