1 Introduction

Suppose that \(x: M^n\rightarrow N^{n+p}\) is an isometric immersion of an \(n\)-dimensional manifold \(M\) in an \((n+p)\)-dimensional Riemannian manifold \(N\). Let \(A\) denote the second fundamental form and \(H\) the mean curvature vector of the immersion \(x\). Let

$$\begin{aligned} \Phi (X,Y)=A(X,Y)-H\langle X,Y\rangle , \end{aligned}$$

for all vector fields \(X\) and \(Y\), where \(\langle ,\rangle \) is the induced metric of \(M\). We say the immersion \(x\) has finite total curvature if

$$\begin{aligned} \Vert \Phi \Vert _{L^n(M)}<+\infty . \end{aligned}$$

If \(M^n\) \((n\ge 3)\) is a complete minimal hypersurface in \(\mathbb {R}^{n+1}\) with finite index, Li and Wang [9] proved that \(M\) has finitely many ends. More generally, Zhu [12] showed that: suppose that \(N^{n+1}\) \((n\ge 3)\) is a complete simply connected manifold with non-positive sectional curvature and \(M^n\) is a complete minimal hypersurface in \(N\) with finite index. If the bi-Ricci curvature satisfies

$$\begin{aligned} b-\overline{\mathrm{Ric}}(X,Y)+\frac{1}{n}|A|^2\ge 0, \end{aligned}$$

for all orthonormal tangent vectors \(X, Y\) in \(T_pN\) for \(p\in M\), then M must has finitely many ends. Cavalcante et al. [1] considered a complete noncompact submanifold \(M^n\) \((n\ge 3)\) isometric immersed in a Hadamard manifold \(N^{n+p}\) with sectional curvature satisfying \(-k^2\le K_N\le 0\) for some constant \(k\) and obtained that if the total curvature is finite and the first eigenvalue of the Laplacian operator of \(M\) is bounded from below by a suitable constant, then the dimension of the space of the \(L^2\) harmonic \(1\)-forms on \(M\) is finite and \(M\) has finitely many non-parabolic ends. Fu and Xu [3] considered a complete submanifold \(M^n\) in a sphere \(\mathbb {S}^{n+p}\) with finite total curvature and bounded mean curvature and showed that the dimension of \(H^1(L^2(M))\) is finite and there are finitely many non-parabolic ends on \(M\) .

In this paper, we discuss a complete noncompact submanifold \(M^n\) in a sphere \(\mathbb {S}^{n+p}\) with finite total curvature and no restriction of mean curvature. We recall some relevant definitions. The Hodge operator \(*:\wedge ^k(M)\rightarrow \wedge ^{n-k}(M)\) is defined by

$$\begin{aligned} *e^{i_1}\wedge \ldots \wedge e^{i_p} =\mathrm{sgn}\sigma (i_1,i_2,\ldots ,i_n)e^{i_{p+1}}\wedge \ldots \wedge e^{i_n}, \end{aligned}$$

where \(\sigma (i_1,i_2,\ldots ,i_n)\) denotes a permutation of the set \((i_1,i_2,\ldots ,i_n)\) and \(\mathrm{sgn}\sigma \) is the sign of \(\sigma \). The operator \(d^*:\wedge ^k(M)\rightarrow \wedge ^{k-1}(M)\) is given by

$$\begin{aligned} d^*\omega =(-1)^{(nk+k+1)}*d*\omega . \end{aligned}$$

The Laplacian operator is defined by

$$\begin{aligned} \triangle \omega =-dd^*\omega -d^*d\omega . \end{aligned}$$

A \(k\)-form \(\omega \) is called \(L^2\)-harmonic if \(\triangle \omega =0\) and

$$\begin{aligned} \int \limits _M\omega \wedge *\omega <+\infty . \end{aligned}$$

We denote \(H^1(L^2(M))\) by the space of all \(L^2\) harmonic \(1\)-forms on \(M\). We obtain finiteness of non-parabolic ends for the submanifold in a sphere with finite total curvature:

Theorem 1.1

Let \(M^n\) \((n\ge 3)\) be an \(n\)-dimensional complete noncompact oriented manifold isometrically immersed in an \((n+p)\)-dimensional sphere \(\mathbb {S}^{n+p}\). If the total curvature is finite, then the dimension of \(H^1(L^2(M))\) is finite and there are finitely many non-parabolic ends on \(M\).

Remark 1.2

Theorem 1.1 generalizes Theorem 1.4 in [3] without the restriction of the mean curvature vector and is also an extension of finiteness of non-parabolic ends on submanifolds in Hadamard manifolds in [1].

2 Proof of main results

We initially introduce several results which will be used to prove Theorem 1.1.

Proposition 2.1

[8, 9] If M is a complete Riemannian manifold, then the number of non-parabolic ends of M is bounded from above by \(\dim H^1(L^2(M))+1\).

Proposition 2.2

[4, 13] Let \(M^n\) be a complete noncompact oriented manifold isometrically immersed in a sphere \(\mathbb {S}^{n+p}\). Then

$$\begin{aligned} {\left( ~\int \limits _M|f|^{\frac{2n}{n-2}}\right) }^{\frac{n-2}{n}}\le C_0\left( ~\int \limits _M|\nabla f|^2+n^2\int \limits _M(|H|^2+1)f^2\right) \end{aligned}$$

for each \(f\in C_0^1(M)\), where \(C_0\) depends only on \(n\) and \(H\) is the mean curvature vector of \(M\) in \(\mathbb {S}^{n+p}\).

Proof of Theorem 1.1

Suppose that \(\omega \in H^1(L^2(M))\). Then we have

$$\begin{aligned} \triangle |\omega |^2=2|\nabla |\omega ||^2+2|\omega |\triangle |\omega |. \end{aligned}$$
(2.1)

Note that the following Bochner’s formula holds [6]:

$$\begin{aligned} \triangle |\omega |^2&=2\langle \triangle \omega ,\omega \rangle +2|\nabla \omega |^2+2\mathrm{Ric}(\omega ^{\sharp },\omega ^\mathrm{\sharp })\nonumber \\&=2|\nabla \omega |^2+2\mathrm{Ric}(\omega ^{\sharp },\omega ^{\sharp }). \end{aligned}$$
(2.2)

Equalities (2.1) and (2.2) imply that

$$\begin{aligned} |\omega |\triangle |\omega |=|\nabla \omega |^2-|\nabla |\omega ||+\mathrm{Ric}(\omega ^{\sharp },\omega ^{\sharp }). \end{aligned}$$
(2.3)

There exists the Kato inequality [2, 11]:

$$\begin{aligned} |\nabla |\omega ||^2\le \frac{n-1}{n}|\nabla \omega |^2. \end{aligned}$$
(2.4)

Combining (2.3) and (2.4), we get that

$$\begin{aligned} |\omega |\triangle |\omega |\ge \frac{1}{n-1}|\nabla |\omega ||^2+\mathrm{Ric}(\omega ^{\sharp },\omega ^{\sharp }). \end{aligned}$$
(2.5)

Take \(h=|\omega |\). There is an estimate for the Ricci curvature of the submanifold \(M\) in [5, 10]:

$$\begin{aligned} \mathrm{Ric}(\omega ^{\sharp },\omega ^{\sharp })\ge \,&(n-1)(|H|^2+1)h^2\\&-\frac{n-1}{n}|\Phi |^2h^2-\frac{(n-2)\sqrt{n(n-1)}}{n}|H||\Phi |h^2. \end{aligned}$$

By (2.5), we obtain that

$$\begin{aligned} h\triangle h\ge \,&\frac{1}{n-1}|\nabla h|^2+(n-1)(|H|^2+1)h^2\nonumber \\&-\frac{n-1}{n}|\Phi |^2h^2-\frac{(n-2)\sqrt{n(n-1)}}{n}|H||\Phi |h^2. \end{aligned}$$
(2.6)

Suppose that \(\eta \) is a compactly supported piecewise smooth function on \(M\). Then

$$\begin{aligned} \mathrm{div}(\eta ^2h\nabla h)&=\eta ^2h\triangle h+\langle \nabla ( \eta ^2h),\nabla h\rangle \\&=\eta ^2h\triangle h+\eta ^2|\nabla h|^2+2\eta h\langle \nabla \eta ,\nabla h\rangle . \end{aligned}$$

Integrating by parts on \(M\), we obtain that

$$\begin{aligned} \int \limits _M\eta ^2h\triangle h+\int \limits _M\eta ^2|\nabla h|^2 +2\int \limits _M\eta h\langle \nabla \eta ,\nabla h\rangle =0. \end{aligned}$$

By (2.6), we get

$$\begin{aligned} -&\frac{1}{n-1}\int \limits _M\eta ^2|\nabla h|^2-(n-1)\int \limits _M\eta ^2(|H|^2 +1)h^2+\frac{n-1}{n}\int \limits _M\eta ^2|\Phi |^2h^2\nonumber \\&+\frac{(n-2)\sqrt{n(n-1)}}{n}\int \limits _M|H||\Phi |h^2\eta ^2 -\int \limits _M\eta ^2|\nabla h|^2-2\int \limits _M\eta h\langle \nabla \eta ,\nabla h\rangle \ge 0. \end{aligned}$$

That is,

$$\begin{aligned} -&2\int \limits _M\eta h\langle \nabla \eta ,\nabla h\rangle -\frac{n}{n-1}\int \limits _M\eta ^2|\nabla h|^2-(n-1) \int \limits _M\eta ^2(|H|^2+1)h^2\nonumber \\&+\frac{n-1}{n}\int \limits _M\eta ^2|\Phi |^2h^2+\frac{(n-2) \sqrt{n(n-1)}}{n}\int \limits _M|H||\Phi |h^2\eta ^2 \ge 0. \end{aligned}$$
(2.7)

Note that

$$\begin{aligned} \int \limits _M|H||\Phi |h^2\eta ^2&=\int \limits _M(|H|\eta h) \cdot (|\Phi |\eta h)\nonumber \\&\le \frac{a}{2}\int \limits _M |H|^2\eta ^2h^2+\frac{1}{2a} \int \limits _M|\Phi |^2\eta ^2h^2, \end{aligned}$$
(2.8)

for any positive real number \(a\). By (2.7) and (2.8), we obtain that

$$\begin{aligned} -&2\int \limits _M\eta h\langle \nabla \eta ,\nabla h\rangle -\frac{n}{n-1}\int \limits _M\eta ^2|\nabla h|^2\\&-(n-1)\int \limits _M\eta ^2(|H|^2+1)h^2+\frac{n-1}{n}\int \limits _M\eta ^2|\Phi |^2h^2\\&+\frac{(n-2)\sqrt{n(n-1)}}{n}\left[ \frac{a}{2}\int \limits _M |H|^2\eta ^2h^2 +\frac{1}{2a}\int \limits _M|\Phi |^2\eta ^2h^2\right] \ge 0. \end{aligned}$$

That is,

$$\begin{aligned}&-2\int \limits _M\eta h\langle \nabla \eta ,\nabla h\rangle -\frac{n}{n-1}\int \limits _M\eta ^2|\nabla h|^2+B(n,a)\int _M| \Phi |^2\eta ^2h^2\nonumber \\&\quad +\int \limits _M\left[ -(n-1)+A(n,a)|H|^2\right] \eta ^2h^2 \ge 0, \end{aligned}$$
(2.9)

where

$$\begin{aligned} A(n,a):=-(n-1)+\frac{a(n-2)\sqrt{n(n-1)}}{2n} \end{aligned}$$

and

$$\begin{aligned} B(n,a):=\frac{n-1}{n}+\frac{(n-2)\sqrt{n(n-1)}}{2an}. \end{aligned}$$

Now we estimate the term \(\int _M|\Phi |^2\eta ^2h^2\): take \(\phi (\eta )=\left( \int _{\mathrm{Supp} \eta }|\Phi |^{n}\right) ^{\frac{1}{n}}\). Then

$$\begin{aligned}&\int \limits _M|\Phi |^2\eta ^2h^2\le \left( \int \limits _{\mathrm{Supp} \eta } \left( |\Phi |^2\right) ^{\frac{n}{2}}\right) ^{\frac{2}{n}}\cdot \left( \int \limits _M\left( \eta ^2h^2\right) ^{\frac{n}{n-2}}\right) ^{ \frac{n-2}{n}}\nonumber \\&\quad =\phi (\eta )^2\cdot \left( \int \limits _M\left( \eta h\right) ^{ \frac{2n}{n-2}}\right) ^{\frac{n-2}{n}}\nonumber \\&\quad \le \phi (\eta )^2\cdot C_0\left( \int \limits _M|\nabla ( \eta h)|^2+n^2\int \limits _M(|H|^2+1){(\eta h)}^2\right) \nonumber \\&\quad \le \phi (\eta )^2\cdot C_0\left( \int \limits _M(1 +\frac{1}{b})h^2|\nabla \eta |^2 +(1+b)\eta ^2|\nabla h|^2+n^2\int _M(|H|^2+1){(\eta h)}^2\right) , \end{aligned}$$
(2.10)

for any positive real number \(b\), where the second inequality holds because of Proposition 2.2. Note that

$$\begin{aligned} -2\int \limits _M\eta h\langle \nabla \eta ,\nabla h\rangle \le c\int \limits _M\eta ^2|\nabla h|^2+\frac{1}{c}\int \limits _Mh^2|\nabla \eta |^2, \end{aligned}$$
(2.11)

for any positive real number \(c\). By (2.9)–(2.11), we have

$$\begin{aligned}&c\int \limits _M\eta ^2|\nabla h|^2+\frac{1}{c}\int \limits _Mh^2| \nabla \eta |^2-\frac{n}{n-1}\int \limits _M\eta ^2|\nabla h|^2\\&\quad +\int \limits _M\left[ -(n-1)+A(n,a)|H|^2\right] \eta ^2h^2+ C_0B(n,a)\phi (\eta )^2\\&\quad \times \left( \int \limits _M(1+\frac{1}{b})h^2| \nabla \eta |^2+(1+b)\eta ^2|\nabla h|^2 +n^2\int \limits _M(|H|^2+1){(\eta h)}^2\right) \ge 0. \end{aligned}$$

That is,

$$\begin{aligned} C\int \limits _M\eta ^2|\nabla h|^2+D\int \limits _M|H|^2\eta ^2h^2 +E\int \limits _M\eta ^2h^2\le F\int \limits _Mh^2|\nabla \eta |^2, \end{aligned}$$
(2.12)

where

$$\begin{aligned} C:&=-c+\frac{n}{n-1}-C_0B(n,a)\phi (\eta )^2(1+b),\\ D:&=-A(n,a)-n^2C_0B(n,a)\phi (\eta )^2,\\ E:&=n-1-n^2C_0B(n,a)\phi (\eta )^2 \end{aligned}$$

and

$$\begin{aligned} F:=\frac{1}{c}+\left( 1+\frac{1}{b}\right) C_0B(n,a)\phi (\eta )^2. \end{aligned}$$

Next, we prove there exists a positive constant \(\delta \) such that if \(\Vert \Phi \Vert _{L^n(M)}<\delta \) then \(C,D,E\) and \(F\) are positive. Obviously, \(\phi (\eta )\le \Vert \Phi \Vert _{L^n(M)}<\delta \). Choose \(d\in (0,\frac{1}{2})\) and let \(a=a(d)\), \(\delta =\delta (d)\) such that

$$\begin{aligned} d+\frac{(n-1)d(1+d)}{n^2}<\frac{n}{n-1},\\ \frac{a(n-2)\sqrt{n(n-1)}}{2n}<(n-1)d,\\ n^2C_0B(n,a)\delta ^2<(n-1)d. \end{aligned}$$

Choosing \(0<c<d\) and \(0<b<d\), we obtain that

$$\begin{aligned} C>\frac{n}{n-1}-d-\frac{(n-1)d(1+d)}{n^2}>0, \end{aligned}$$
$$\begin{aligned} D&=(n-1)-\frac{a(n-2)\sqrt{n(n-1)}}{2n}-n^2C_0B(n,a)\phi (\eta )^2\\&>(n-1)-2(n-1)d>0, \end{aligned}$$
$$\begin{aligned} E=n-1-n^2C_0B(n,a)\phi (\eta )^2>0 \end{aligned}$$

and

$$\begin{aligned} F>0. \end{aligned}$$

Since the total curvature \(\Vert \Phi \Vert _{L^n(M)}\) is finite, we can choose a fixed \(r_0\) such that

$$\begin{aligned} \Vert \Phi \Vert _{L^n(M-B_{r_0})}<\delta . \end{aligned}$$

Set

$$\begin{aligned} \tilde{ C}:&=-c+\frac{n}{n-1}-C_0B(n,a)\delta ^2(1+b),\\ \tilde{D}:&=-A(n,a)-n^2C_0B(n,a)\delta ^2,\\ \tilde{E}:&=n-1+\tilde{D} \end{aligned}$$

and

$$\begin{aligned} \tilde{ F}:=\frac{1}{c}+\left( 1+\frac{1}{b}\right) C_0B(n,a)\delta ^2. \end{aligned}$$

Thus,

$$\begin{aligned} \tilde{C}\int \limits _M\eta ^2|\nabla h|^2+\tilde{D}\int \limits _M|H|^2\eta ^2h^2+\tilde{E}\int \limits _M\eta ^2h^2\le \tilde{F}\int \limits _Mh^2|\nabla \eta |^2, \end{aligned}$$
(2.13)

for any \(\eta \in C^{\infty }_0(M-B_{r_0})\), where \(\tilde{C}\), \(\tilde{D}\), \(\tilde{E}\) and \(\tilde{F}\) are positive. Proposition 2.2 implies that

$$\begin{aligned} \frac{1}{C_0}&{\left( ~\int \limits _M|\eta h|^{\frac{2n}{n-2}}\right) }^{\frac{n-2}{n}} \le \int \limits _M|\nabla (\eta h)|^2+n^2\int \limits _M(|H|^2+1)(\eta h)^2\nonumber \\&\le (1+s)\int \limits _M\eta ^2|\nabla h|^2+(1+\frac{1}{s})\int \limits _Mh^2|\nabla \eta |^2+n^2\int \limits _M(|H|^2+1)\eta ^2 h^2, \end{aligned}$$
(2.14)

for any \(\eta \in C^{\infty }_0(M-B_{r_0})\) and any positive real number \(s\). Inequality (2.12) implies that

$$\begin{aligned}&(1+s)\int \limits _M\eta ^2|\nabla h|^2\le \frac{(1+s)\tilde{F}}{ \tilde{C}}\int \limits _Mh^2|\nabla \eta |^2\\&\quad -\frac{(1+s)\tilde{D}}{\tilde{C}}\int \limits _M|H|^2\eta ^2h^2 -\frac{(1+s)\tilde{E}}{\tilde{C}}\int \limits _M\eta ^2h^2. \end{aligned}$$

Combining with (2.14), we get

$$\begin{aligned} \frac{1}{C_0}&{\left( ~\int \limits _M|\eta h|^{\frac{2n}{n-2}}\right) }^{\frac{n-2}{n}} \le \left( 1+\frac{1}{s}+\frac{\tilde{F}(1+s)}{\tilde{C}}\right) \int \limits _Mh^2|\nabla \eta |^2\nonumber \\&+(n^2-\frac{(1+s)\tilde{D}}{\tilde{C}})\int \limits _M|H|^2\eta ^2 h^2 +(n^2-\frac{(1+s)\tilde{E}}{\tilde{C}})\int \limits _M\eta ^2 h^2, \end{aligned}$$
(2.15)

for any \(\eta \in C^{\infty }_0(M-B_{r_0})\). Choose a sufficiently large \(s\) such that \(n^2-\frac{(1+s)\tilde{D}}{\tilde{C}}<0\) and \(n^2-\frac{(1+s)\tilde{E}}{\tilde{C}}<0\). Then (2.15) implies that

$$\begin{aligned} {\left( ~\int \limits _M(\eta h)^{\frac{2n}{n-2}}\right) }^{\frac{n-2}{n}} \le \tilde{A}\int \limits _Mh^2|\nabla \eta |^2, \end{aligned}$$
(2.16)

for any \(\eta \in C^{\infty }_0(M-B_{r_0})\), where \(\tilde{A}\) is a positive constant depending only on \(n\). From now on, the proof follows standard techniques [for instance, see [1] after inequality (33)] and uses a Moser iteration argument and lemma 11 in [7]. We only include a concise proof here for the sake of completeness. Choose \(r>r_0+1\) and \(\eta \in C_0^{\infty }(M-B_{r_0})\) such that

$$\begin{aligned} {\left\{ \begin{array}{ll} \eta =0 \ \ on\ \ B_{r_0}\cup (M-B_{2r}),\\ \eta =1\ \ on\ \ B_r-B_{r_0+1},\\ |\nabla \eta |<c_1 \ \ on \ \ B_{r_0+1}-B_{r_0},\\ |\nabla \eta |\le c_1r^{-1} on \ \ B_{2r}-B_{r}, \end{array}\right. } \end{aligned}$$

for some positive constant \(c_1\). Then (2.16) becomes that

$$\begin{aligned} {\left( \int \limits _{B_r-B_{r_0+1}}h^{\frac{2n}{n-2}}\right) }^{\frac{n-2}{n}} \le \tilde{A}\int \limits _{B_{r_0+1}-B_{r_0}}h^2+\frac{\tilde{A}}{r^2} \int \limits _{B_{2r}-B_r}h^2. \end{aligned}$$

Letting \(r\rightarrow \infty \) and noting that \(|\omega |\in H^1(L^2(M))\), we obtain that

$$\begin{aligned} {\left( \int \limits _{M-B_{r_0+1}}h^{\frac{2n}{n-2}}\right) }^{\frac{n-2}{n}} \le \tilde{A}\int \limits _{B_{r_0+1}-B_{r_0}}h^2. \end{aligned}$$
(2.17)

Combining with Hölder inequality , we have that

$$\begin{aligned} \int \limits _{B_{r_0+2}}h^2 \le {(1+\tilde{A}Vol{(B_{r_0+2})}^{\frac{2}{n}})}\int \limits _{B_{r_0+1}}h^2. \end{aligned}$$
(2.18)

Let

$$\begin{aligned} \Psi =|(n-1)(|H|^2+1)-\frac{n-1}{n}|\Phi |^2-\frac{(n-2) \sqrt{n(n-1)}}{n}|H||\Phi ||. \end{aligned}$$

Fix \(x\in M\) and take \(\tau \in C_0^1(B_1(x))\). (2.6) implies that

$$\begin{aligned}&-2\int \limits _{B_1(x)}\tau h^{p-1}\langle \nabla \tau ,\nabla h\rangle \ge (p-1+\frac{1}{n-1})\int \limits _{B_1(x)}\tau ^2h^{p-2}|\nabla h|^2\nonumber \\&\quad -\int \limits _{B_1(x)}\tau ^2\Psi h^p. \end{aligned}$$
(2.19)

Note that

$$\begin{aligned} -2\tau h^{p-1}\langle \nabla \tau ,\nabla h\rangle&=2\tau \langle -h^{\frac{p}{2}}\nabla \tau ,h^{ \frac{p}{2}-1}\nabla h\rangle \\&\le (n-1)h^p|\nabla \tau |^2+\frac{1}{n-1}h^{p-2}\tau ^2|\nabla h|^2. \end{aligned}$$

Combining with (2.19), we obtain that

$$\begin{aligned} (p-1)\int \limits _{B_1(x)}\tau ^2h^{p-2}|\nabla h|^2\le \int \limits _{B_1(x)}\Psi \tau ^2 h^p+(n-1)\int \limits _{B_1(x)}|\nabla \tau |^2h^p. \end{aligned}$$
(2.20)

By Cauchy–Schwarz inequality and (2.20), we have

$$\begin{aligned} \int \limits _{B_1(x)}|\nabla (\tau h^{\frac{p}{2}})|^2 \le \int \limits _{B_1(x)}\mathcal {A}\Psi \tau ^2h^p +\mathcal {B}|\nabla \tau |^2h^p, \end{aligned}$$
(2.21)

where \(\mathcal {A}=\frac{p(p+1)}{4(p-1)}\le p<2n^2p\) and \(\mathcal {B}=p+1+(n-1)\mathcal {A}\le 1+np<2n^2p\). Choosing \(f=\tau h^{\frac{p}{2}}\) in Proposition 2.2 and combining with (2.21), we have

$$\begin{aligned} {\left( \int \limits _{B_1(x)}(\tau h^{\frac{p}{2}})^{\frac{2n}{n-2}}\right) }^{\frac{n-2}{2}} \le 2C_0pn^2\int \limits _{B_1(x)}(\mathcal {C}\tau ^2+|\nabla \tau |^2)h^p, \end{aligned}$$
(2.22)

where \(\mathcal {C}=|H|^2+1+\Psi \). Let \(p_k=\frac{2n^k}{(n-2)^k}\) and \(\rho _k=\frac{1}{2}+\frac{1}{2^{k+1}}\) for \(k=0,1,2,\ldots \). Take a function \(\tau _k\in C_0^\infty (B_{\rho _k(x)})\) satisfying:

$$\begin{aligned} {\left\{ \begin{array}{ll} 0\le \tau _k\le 1, \\ \tau _k=1 \ \ on\ \ B_{\rho _{k+1}}(x),\\ |\nabla \tau _k|\le 2^{k+3}. \end{array}\right. } \end{aligned}$$

Choosing \(p=p_k\) and \(\tau =\tau _k\) in (2.22), we have

$$\begin{aligned} {\left( \int \limits _{B_{\rho _{k+1}}(x)}h^{p_{k+1}}\right) }^{\frac{1}{p_{k+1}}} \le {\left( p_k4^{k+k_0}\right) }^{\frac{1}{p_k}}{\left( \int \limits _{B_{ \rho _k}(x)}h^{p_k}\right) }^{\frac{1}{p_k}}, \end{aligned}$$
(2.23)

where \(k_0\) is a positive integer such that \(2C_0n^2(4^3+\sup _{B_1(x)}\mathcal {C})\le 4^{k_0}\). By recurrence, we have

$$\begin{aligned} \Vert h\Vert _{L^{p_{k+1}}(B_{\frac{1}{2}}(x))} \le \prod _{i=0}^kp_i^{\frac{1}{p_i}}4^{ \frac{i}{p_i}}4^{\frac{k_0}{p_i}}\Vert h\Vert _{{L^2}(B_1(x))} \le \mathcal {D}\Vert h\Vert _{{L^2}(B_1(x))}, \end{aligned}$$
(2.24)

where \(\mathcal {D}\) is a positive constant depending only on \(n\) and \(\sup _{B_1(x)}{\Psi }\). Letting \(k\rightarrow \infty \), we get

$$\begin{aligned} \Vert h\Vert _{L^{\infty }(B_{\frac{1}{2}}(x))} \le \mathcal {D}\Vert h\Vert _{{L^2}(B_1(x))}. \end{aligned}$$
(2.25)

Now, choose \(y\in \overline{B}_{r_0+1}\) so that \(\sup _{B_{r_0+1}}h^2={h(y)}^2\). Note that \(B_1(y)\subset B_{r_0+2}\). (2.25) implies that

$$\begin{aligned} \sup _{B_{r_0+1}}h^2\le \mathcal {D}\Vert h\Vert _{{L^2}(B_1(y))}^2 \le \mathcal {D}\Vert h\Vert _{{L^2}(B_{r_0+2})}^2. \end{aligned}$$
(2.26)

By (2.18), we have

$$\begin{aligned} \sup _{B_{r_0+1}}h^2\le \mathcal {F}\Vert h\Vert _{{L^2}(B_{r_0+1})}^2, \end{aligned}$$
(2.27)

where \(\mathcal {F}\) depends only on \(n\), \(Vol{(B_{r_0+2})}\) and \(\sup _{B_{r_0+2}}{\Psi }\). In order to show the finiteness of the dimension of \(H^1(L^2(M))\), it suffices to prove that the dimension of any finite dimensional subspaces of \(H^1(L^2(M))\) is bounded above by a fixed constant. By (2.26) and Lemma 11 in [7], we get \(\dim H^1(L^2(M))<+\infty \). By Proposition 2.1, we obtain that the number of non-parabolic ends of \(M\) is finite.