The proof of the claim follows from basically from Weyl’s law. For the partial sum of $v$,

(4.162)

$$V_{N}\equiv\sum\limits_{j=1}^{N}v_{j}\varphi_{j}=\sum\limits_{j=1}^{N}\Big(\int_{Y^{3}}v\cdot\varphi_{j}\dvol_{h_{0}}\Big)\varphi_{j}=\sum\limits_{j=1}^{N}\Big(\int_{Y^{3}}v\cdot\frac{(-\Delta_{h_{0}})^{K_{0}}\varphi_{j}}{(\Lambda_{j})^{K_{0}}}\dvol_{h_{0}}\Big)\varphi_{j}.$$

Applying integration by parts,

	$\displaystyle\|V_{N}\|_{L^{\infty}(B_{1}(p_{0}))}$	$\displaystyle\leq\sum\limits_{j=1}^{N}\Big(\frac{1}{(\Lambda_{j})^{K_{0}}}\int_{Y^{3}}|\Delta_{h_{0}}^{K_{0}}v|\cdot|\varphi_{j}|\dvol_{h_{0}}\Big)\|\varphi_{j}\|_{L^{\infty}(B_{1}(p_{0}))}$
(4.163)			$\displaystyle\leq V_{0}\cdot\|v\|_{C^{2K_{0}}(Y^{3}\times\{z_{0}\})}\cdot\sum\limits_{j=1}^{N}\frac{1}{(\Lambda_{j})^{K_{0}-1}},$

where $V_{0}=\Vol_{h_{0}}(Y^{3})$. Notice that, the spectrum $\{\Lambda_{j}\}_{j=1}^{\infty}$ satisfies the Weyl’s law on $(Y^{3},h_{0})$, so in particular for sufficiently large $j$,

(4.164)

$$C_{0}^{-1}j^{\frac{2}{3}}\leq|\Lambda_{j}|\leq C_{0}j^{\frac{2}{3}}.$$

Since $K_{0}\geq 3$,

(4.165)

$$\|V_{N}\|_{L^{\infty}(B_{1}(p_{0}))}\leq C\|v\|_{C^{2K_{0}}(Y^{3}\times\{z_{0}\})}\cdot\sum\limits_{j=1}^{N}\frac{1}{j^{\frac{4}{3}}}\leq C.$$

The proof of the claim is done. ∎