2. Some lemmas

In this section, we will recall some useful results which will be adopted in the proof of main theorems. The most basic one is the following Weitzenböck formula.

Lemma 2.1 ([Reference Li18]). Given a Riemannian manifold $M^n$ , for any 1-form $\omega$ on $M^n$ , we have

\begin{equation*}\triangle |\omega|^2=2|\nabla \omega|^2+2\langle \triangle \omega,\omega\rangle+2{\rm Ric}(\omega^{\sharp},\omega^{\sharp}),\end{equation*}

where $\omega^{\sharp}$ is the dual vector field of $\omega$ .

Besides, the Kato inequality is also a fundamental technique.

Lemma 2.2 ([Reference Calderbank, Gauduchon and Herzlich1]). Given a Riemannian manifold $M^n$ , for any closed and coclosed k-form $\omega$ on $M^n$ , we have

\begin{equation*}|\nabla \omega|^2\geq (1+C_{n,k})|\nabla |\omega||^2, \quad where \quad C_{n,k}=\begin{cases}\dfrac{1}{n-k},\quad 1\leq k\leq\dfrac{n}{2}.\\[12pt] \dfrac{1}{k},\quad \dfrac{n}{2}\leq k \leq n-1.\end{cases}\end{equation*}

What’s more, Shiohama and Xu [Reference Shiohama and Xu29] proved the following estimation on the Ricci curvature of submanifold.

Lemma 2.3 ([Reference Shiohama and Xu29]). Let M be an n-dimensional complete immersed hypersurface in a Riemannian manifold N. If all the sectional curvatures of N are bounded pointwise from below by a function k, then

(2.1) \begin{align} {\rm Ric}\geq (n-1)k-\frac{n-1}{n}|A|^2+2(n-1)H^2-\frac{(n-2)\sqrt{n(n-1)}}{n}|H|\sqrt{|A|^2-nH^2},\end{align}

where H is the mean curvature and A is the second fundamental form of M.

We should note in [Reference Shiohama and Xu29], the author assumed that all the sectional curvatures of N are bounded below by a constant k. But according to his argument, this assumption was only used in the end of the proof; hence, this method can be used to prove the above lemma without any change. Under the same assumption, the following lemma estimates the right hand side of (2.1).

Lemma 2.4 ([Reference Chao and Lv6]). Let $M^n$ be an n-dimensional orientable submanifold in Riemannian manifold N. We have

(2.2) \begin{align} 2(n-1)H^2-\frac{(n-2)\sqrt{n(n-1)}}{n}|H|\sqrt{|A|^2-nH^2}\geq \frac{2(n-1)-n\sqrt{n-1}}{2n} |A|^2. \end{align}

Definition 2.5. Let $M^n$ be an n-dimensional orientable hypersurface in a Riemannian manifold N. We say M is stable if the following inequality

(2.3) \begin{align} \int_M |\nabla \eta|^2\geq\int_M\Big(|A|^2+\overline{{\rm Ric}}(u,u)\Big)\eta^2\end{align}

holds for any $\eta\in C^{\infty}_0(M)$ , where $u$ is a unit normal vector field on M, $\overline{{\rm Ric}}$ is the Ricci curvature of N, and A is the second fundamental form of M.

Now, we will give a condition to ensure that the volume of Riemannian manifold to be infinite.

In addition, the following Hoffman-Spruck inequality generalizes the Poincaré inequality and relates it to the Sobolev inequality.

Lemma 2.7 ([Reference Hoffman and Spruck14]). Let $x\;:\;M^n\hookrightarrow N$ be an isometric immersion of a complete manifold M in a complete simply connected manifold N with nonpositive sectional curvature. Then, the following inequality holds:

\begin{equation*}\Big(\int_M h^{2n}dV\Big)^{\frac{n-2}{n}}\leq S(n)\int_M(|\nabla h|^2+(h|H|)^2)dV,\end{equation*}

for all nonnegative $C^1$ -functions $h\;:\;M^n\rightarrow \mathbb{R}$ with compact support, where S(n) is the Sobolev constant, which is positive and only depends on n.

The following Cauchy inequality gives the $L^2$ upper bound of a nonnegative sub-eigenfunction.

Lemma 2.8 ([Reference Li18]). Let M be an n-dimensional complete noncompact Riemannian manifold. For $x \in M$ and a constant $\kappa \geq 0$ , we assume that the Ricci curvature of M satisfies

\begin{equation*}\operatorname{Ric} \geq-(n-1)\kappa\end{equation*}

on the geodesic ball $B_x(4r)$ centered at p with radius 4r. Let $0 < \delta < \frac{1}{2}$ and $\lambda > 0$ be two fixed constants. Then there exists a positive constant $C = C(r, \delta, \lambda, \kappa)$ so that if any nonnegative function $\eta \in C^{\infty}(B_x(2r))$ satisfying the differential inequality $\triangle\eta\geq-\lambda\eta$ , then

\begin{equation*}\sup_{B_x((1-\delta)r)}\eta^2\leq\frac{C}{{\rm Vol}(B_x(r))}\int_{B_x(r)}\eta^2.\end{equation*}

The last lemma associates the $L^2$ and $L^{\infty}$ norms of harmonic forms with the dimension of the space of harmonic forms.

Lemma 2.9 ([Reference Li17, Reference Pigola, Rigoli and Setti24]). Let K be a finite dimensional subspace of $L^{2p}$ harmonic q-forms on an m-dimensional complete noncompact Riemannian manifold M for any $p > 0$ . Then, there exists $\eta \in K$ such that

\begin{equation*}(\dim K)^{\min\{1,p\}}\int_{B_x(r)}|\eta|^{2p}\leq \operatorname{Vol}\!(B_x(r))\cdot\min\left\{ \left(\begin{array}{c}m\\[5pt] q \end{array}\right), \dim K\right\}^{\min\{1,p\}}\cdot\sup_{B_x(r)}|\eta|^{2p},\end{equation*}

for any $x \in M$ and $r > 0$ .

3. Proofs of the theorems

Proof of Theorem 1.2. Let $\omega$ be a $L^{2p}$ harmonic 1-form. Using the Weitzenböck formula and the Kato inequality, we can get that

(3.1) \begin{align} |\omega|\triangle|\omega|\geq \frac{1}{n-1}|\nabla |\omega||^2+{\rm Ric}(\omega^{\sharp},\omega^{\sharp}).\end{align}

Under our hypothesis on the sectional curvature of N, we can estimate the Ricci curvature of M by using Lemmas 2.3 and 2.4:

\begin{align*} {\rm Ric}_M &\geq -(n-1)k^2+\frac{2(n-1)-n\sqrt{n-1}}{2n}|A|^2-\frac{n-1}{n}|A|^2 \\[5pt]&=-(n-1)k^2-\frac{\sqrt{n-1}}{2}|A|^2.\end{align*}

Thus equation (3.1) becomes

(3.2) \begin{align} |\omega|\triangle |\omega|\geq \frac{1}{n-1}|\nabla |\omega||^2-(n-1)k^2|\omega|^2-\frac{\sqrt{n-1}}{2}|A|^2|\omega|^2.\end{align}

Furthermore, using (3.2) we have that

(3.3) \begin{align} |\omega|^p\triangle |\omega|^p &=|\omega|^p\Big(p(p-1)|\omega|^{p-2}|\nabla |\omega||^2+p |\omega|^{p-1}\triangle |\omega|\Big) \\[5pt] &=\frac{p-1}{p}|\nabla |\omega|^p|^2+p|\omega|^{2p-2}|\omega|\triangle|\omega| \\[5pt] &\geq \Big(1-\frac{n-2}{(n-1)p}\Big)|\nabla |\omega|^p|^2- \frac{p\sqrt{n-1}}{2}|A|^2|\omega|^{2p}-(n-1)k^2p|\omega|^{2p}.\end{align}

Since M has finite index, there exists a compact subset $\Omega\subset M$ such that $M\backslash\Omega$ is stable ([Reference Fischer-Colbrie10, Reference Tysk31]). Without loss of generality, we assume that $\Omega = B_x(R_0)$ . Then, according to Definition 2.5, for any compactly supported Lipschitz function $\eta$ on $M\backslash\ \!B_x(R_0)$ ,

\begin{equation*}\int_{M\backslash\ \!B_x(R_0)} |\nabla \eta|^2\geq\int_{M\backslash\ \!B_x(R_0)}\Big(|A|^2+\overline{{\rm Ric}}(u,u)\Big)\eta^2.\end{equation*}

The assumption on the sectional curvature of N implies that $\overline{{\rm Ric}}(u,u)\geq -nk^2$ and

(3.4) \begin{equation} \int_{M \backslash B_x(R_0)} |\nabla \eta|^2\geq\int_{M \backslash B_x(R_0)}\Big(|A|^2-nk^2\Big)\eta^2.\end{equation}

for all compactly supported Lipschitz function $\eta$ on $M \backslash B_x(R_0)$ . Replacing $\eta$ by $\eta |\omega|^p$ in (3.4), we get

(3.5) \begin{equation}\int_{M \backslash B_x(R_0)}|A|^2\eta^2 |\omega|^{2p}-nk^2\int_{M \backslash B_x(R_0)}\eta^2|\omega|^{2p}\leq \int_{M \backslash B_x(R_0)}|\nabla(\eta |\omega|^p)|^2.\end{equation}

Moreover, the domain monotonicity of eigenvalues implies that

\begin{equation*}\lambda_1(M)\leq\lambda_1(M \backslash B_x(R_0))\leq\frac{\int_{M \backslash B_x(R_0)}|\nabla \eta|^2}{\int_{M \backslash B_x(R_0)}\eta^2}\end{equation*}

for any compactly supported Lipschitz function $\eta$ on $M \backslash B_x(R_0)$ . Replacing $\eta$ by $\eta |\omega|^p$ in this inequality and using (3.5), we have

(3.6) \begin{equation} \int_{M \backslash B_x(R_0)}\eta^2|\omega|^{2p}\leq\frac{1}{\lambda_1(M)}\int_{M \backslash B_x(R_0)}|\nabla(\eta |\omega|^p)|^2.\end{equation}

and

(3.7) \begin{align}\int_{M \backslash B_x(R_0)}&|A|^2\eta^2 |\omega|^{2p}\leq \Big(1+\frac{nk^2}{\lambda_1(M)}\Big)\int_{M \backslash B_x(R_0)}|\nabla(\eta |\omega|^p)|^2 \\[5pt]&=\Big(1+\frac{nk^2}{\lambda_1(M)}\Big)\int_{M \backslash B_x(R_0)}\!\Big(\eta^2|\nabla|\omega|^p|^2+|\nabla\eta|^2|\omega|^{2p}+2\eta|\omega|^p\langle\nabla\eta,\nabla|\omega|^p\rangle\Big).\end{align}

Applying the divergence theorem, we get

(3.8) \begin{equation} 2\int_{M \backslash B_x(R_0)}\eta|\omega|^p\langle\nabla\eta,\nabla|\omega|^p\rangle =-\int_{M \backslash B_x(R_0)}\!\Big(\eta^2|\nabla|\omega|^p|^2+\eta^2|\omega|^p\triangle |\omega|^p\Big).\end{equation}

Therefore,

(3.9) \begin{equation}\int_{M \backslash B_x(R_0)}|A|^2\eta^2 |\omega|^{2p}\leq\Big(1+\frac{nk^2}{\lambda_1(M)}\Big)\int_{M \backslash B_x(R_0)}\Big(|\nabla\eta|^2|\omega|^{2p}-\eta^2|\omega|^p\triangle |\omega|^p\Big).\end{equation}

From (3.3) and (3.9), we have

\begin{align*}\int_{M \backslash B_x(R_0)}|A|^2\eta^2 |\omega|^{2p}&\leq \Big(1+\frac{nk^2}{\lambda_1(M)}\Big)\int_{M \backslash B_x(R_0)}|\nabla\eta|^2|\omega|^{2p} \\[5pt]&\;\;\;-\Big(1+\frac{nk^2}{\lambda_1(M)}\Big)\Big(1-\frac{n-2}{(n-1)p}\Big)\int_{M \backslash B_x(R_0)}\eta^2|\nabla |\omega|^p|^2 \\[5pt]&\;\;\;+\Big(1+\frac{nk^2}{\lambda_1(M)}\Big)\int_{M \backslash B_x(R_0)}\Big(\frac{p\sqrt{n-1}}{2}|A|^2+(n-1)k^2p\Big)\eta^2|\omega|^{2p}.\end{align*}

Therefore, the assumption on $\lambda_1(M)$ implies

(3.10) \begin{align}\Big(1-\frac{n-2}{(n-1)p}\Big)&\int_{M \backslash B_x(R_0)}\eta^2|\nabla |\omega|^p|^2 \\[5pt]&\leq \int_{M \backslash B_x(R_0)}|\nabla\eta|^2|\omega|^{2p}+(n-1)k^2p\int_{M \backslash B_x(R_0)}\eta^2|\omega|^{2p} \\[5pt]&\quad +\left(\frac{p\sqrt{n-1}}{2}-\frac{1}{1+\frac{nk^2}{\lambda_1(M)}}\right)\int_{M \backslash B_x(R_0)}|A|^2\eta^2|\omega|^{2p} \\[5pt]&\leq \int_{M \backslash B_x(R_0)}|\nabla\eta|^2|\omega|^{2p}+(n-1)k^2p\int_{M \backslash B_x(R_0)}\eta^2|\omega|^{2p}.\end{align}

On the other hand, applying Young’s inequality in (3.6), we obtain

\begin{equation*}\int_{M \backslash B_x(R_0)}\eta^2|\omega|^{2p}\leq\frac{1}{\lambda_1(M)}\cdot\frac{1+\varepsilon}{\varepsilon}\int_{M \backslash B_x(R_0)}|\nabla\eta|^2|\omega|^{2p}+\frac{1+\varepsilon}{\lambda_1(M)}\int_{M \backslash B_x(R_0)}\eta^2|\nabla |\omega|^p|^2\end{equation*}

for any $\varepsilon>0$ . Combining this with (3.10), we get

\begin{align*}\Big(1-\frac{n-2}{(n-1)p}&-\frac{(n-1)k^2p(1+\varepsilon)}{\lambda_1(M)}\Big)\int_{M \backslash B_x(R_0)}\eta^2|\nabla |\omega|^p|^2\\[5pt]&\leq\Big(1+\frac{(n-1)k^2p}{\lambda_1(M)}\cdot\frac{1+\varepsilon}{\varepsilon}\Big)\int_{M \backslash B_x(R_0)}|\nabla\eta|^2|\omega|^{2p}.\end{align*}

Using the assumption on $\lambda_1(M)$ , we choose a sufficiently small $\varepsilon > 0$ such that

\begin{equation*}1-\frac{n-2}{(n-1)p}-\frac{(n-1)k^2p(1+\varepsilon)}{\lambda_1(M)}>0.\end{equation*}

Then we have

(3.11) \begin{equation}\int_{M \backslash B_x(R_0)}\eta^2|\nabla |\omega|^p|^2\leq C_1\int_{M \backslash B_x(R_0)}|\nabla\eta|^2|\omega|^{2p},\end{equation}

for some positive constant $C_1$ which depends only on p, n, k and $\lambda_1(M)$ . Moreover, from Lemma 2.7 and Hölder inequality, we obtain

(3.12) \begin{align}\Big(\int_{M \backslash B_x(R_0)}(\eta|\omega|^p)^{\frac{2n}{n-2}}\Big)^{\frac{n-2}{n}}&\leq S(n)\int_{M \backslash B_x(R_0)}\big(|\nabla (\eta|\omega|^p)|^2+\eta^2|\omega|^{2p}|H|^2\big) \\[5pt]&\leq 2S(n)\int_{M \backslash B_x(R_0)}\eta^2|\nabla |\omega|^p|^2 \\[5pt]&\quad +2S(n)\int_{M \backslash B_x(R_0)}|\nabla\eta|^2|\omega|^{2p} \\[5pt]&\quad +S(n)\Big(\int_{M \backslash B_x(R_0)} |H|^n\Big)^{\frac{2}{n}}\Big(\int_{M \backslash B_x(R_0)} (\eta|\omega|^p)^{\frac{2n}{n-2}}\Big)^{\frac{n-2}{n}},\end{align}

i.e.

(3.13) \begin{align}\Big(1-S(n)&\|H\|^2_{L^n(M)}\Big)\Big(\int_{M \backslash B_x(R_0)} (\eta|\omega|^p)^{\frac{2n}{n-2}}\Big)^{\frac{n-2}{n}} \\[5pt]&\leq 2S(n)\int_{M \backslash B_x(R_0)}\eta^2|\nabla |\omega|^p|^2+2S(n)\int_{M \backslash B_x(R_0)}|\nabla\eta|^2|\omega|^{2p}.\end{align}

Combining (3.11) and (3.13) and the assumption $1-S(n)\|H\|^2_{L^n(M)}>0$ , we get

(3.14) \begin{equation}\Big(\int_{M \backslash B_x(R_0)} (\eta|\omega|^p)^{\frac{2n}{n-2}}\Big)^{\frac{n-2}{n}}\leq C_2 \int_{M \backslash B_x(R_0)}|\nabla\eta|^2|\omega|^{2p},\end{equation}

for some positive constant $C_2$ . Now we choose our test function $0 \leq \eta \leq 1$ as in [Reference Choi and Seo7]: given $R > R_0 + 1$ ,

\begin{equation*}\eta=\left\{ \begin{array}{ll} 1, & on\;B_x(R)\backslash B_x(R_0+1) \\[5pt] 0, & on\;B_x(R_0)\cup(M\backslash B_x(2R)), \end{array} \right.\end{equation*}

$|\nabla\eta|\leq C_3$ on $B_x(R_0+1)\backslash B_x(R_0)$ and $|\nabla\eta|\leq \frac{C_3}{R}$ on $B_x(2R)\backslash B_x(R)$ . Applying this test function $\eta$ to (3.14), we get

\begin{equation*}\Big(\int_{B_x(R) \backslash B_x(R_0+1)} |\omega|^{\frac{2np}{n-2}}\Big)^{\frac{n-2}{n}}\leq C_4 \int_{B_x(R_0+1) \backslash B_x(R_0)}|\omega|^{2p}+\frac{C_4}{R^2}\int_{B_x(2R) \backslash B_x(R)}|\omega|^{2p}.\end{equation*}

Letting $R\rightarrow\infty$ and using the assumption that $\displaystyle \int_M |\omega|^{2p}<\infty$ , we obtain

(3.15) \begin{equation} \Big(\int_{M \backslash B_x(R_0+1)} |\omega|^{\frac{2np}{n-2}}\Big)^{\frac{n-2}{n}}\leq C_4 \int_{B_x(R_0+1) \backslash B_x(R_0)}|\omega|^{2p}.\end{equation}

Then, using Hölder inequality and (3.15), we conclude that

(3.16) \begin{align}\int_{B_x(R_0+2) \backslash B_x(R_0+1)}|\omega|^{2p} &\leq \big({\rm Vol}(B_x(R_0+2))\big)^{\frac{2}{n}}\Big(\int_{B_x(R_0+2) \backslash B_x(R_0+1)} |\omega|^{\frac{2np}{n-2}}\Big)^{\frac{n-2}{n}} \\[5pt]&\leq C_4\cdot \big({\rm Vol}(B_x(R_0+2))\big)^{\frac{2}{n}} \int_{B_x(R_0+1) \backslash B_x(R_0)}|\omega|^{2p}.\end{align}

Adding $\displaystyle \int_{B_x(R_0+1) \backslash B_x(R_0)}|\omega|^{2p}$ to both sides of (3.16), we get

\begin{equation*} \int_{B_x(R_0+2) \backslash B_x(R_0)}|\omega|^{2p}\leq \Big(1+C_4\cdot \big({\rm Vol}(B_x(R_0+2))\big)^{\frac{2}{n}}\Big)\int_{B_x(R_0+1) \backslash B_x(R_0)}|\omega|^{2p}.\end{equation*}

Again adding $\displaystyle \int_{B_x(R_0)}|\omega|^{2p}$ to both sides infers

(3.17) \begin{equation}\int_{B_x(R_0+2)}|\omega|^{2p}\leq C_5 \int_{B_x(R_0+1)}|\omega|^{2p}.\end{equation}

On the other hand, since $|\omega|$ satisfies the differential inequality (3.3), Lemma 2.8 implies that

\begin{equation*}\sup_{B_x((1-\delta)(R_0+2))}|\omega|^{2p}\leq C_6\int_{B_x(R_0+2)}|\omega|^{2p}\end{equation*}

for some positive constant $C_6=C_6(\delta, n, k, {\rm Vol}(B_x(R_0+2)), \sup_{B_x(R_0+2)}|A|^2)$ . For a sufficiently small $\delta > 0$ such that $(1-\delta)(R_0+2) > R_0 + 1$ ,

\begin{equation*}\sup_{B_x(R_0+1)}|\omega|^{2p}\leq C_6\int_{B_x(R_0+2)}|\omega|^{2p}.\end{equation*}

Together with (3.17), we have

(3.18) \begin{equation} \sup_{B_x(R_0+1)}|\omega|^{2p}\leq C_7\int_{B_x(R_0+1)}|\omega|^{2p}.\end{equation}

for some positive constant $C_7=C_7(n,p,k,R_0,\lambda_1(M),S(n),{\rm Vol}(B_x(R_0+2)))$ . In what follows, as in [Reference Choi and Seo7], we consider any finite dimensional subspace $K\subset H^1(L^{2p}(M))$ . According to Lemma 2.9, we see that there exists an $L^{2p}$ harmonic 1-form $\omega\subset K$ such that

\begin{equation*}(\!\dim K)^{\min\{1,p\}}\int_{B_x(R_0+1)}|\omega|^{2p}\leq {\rm Vol}(B_x(R_0+1))\cdot\min\!\left\{n, \dim K\right\}^{\min\{1,p\}}\cdot\sup_{B_x(R_0+1)}|\omega|^{2p}.\end{equation*}

From (3.18), we have

\begin{equation*}\dim K\leq \Big(C_7\cdot {\rm Vol}(B_x(R_0+1))\Big)^{\frac{1}{\min\{1,p\}}}\cdot\min\!\left\{n, \dim K\right\},\end{equation*}

which implies that $\dim K$ is bounded by a fixed constant. Since K is an arbitrary subspace of finite dimension, we get that $\dim H^1(L^{2p}(M))<\infty$ .

Proof of Theorem 1.4. Let $\omega$ be a $L^{2p}$ harmonic 1-form. Using Weitzenböck formula and Kato inequality, we can get that

(3.19) \begin{align} |\omega|\triangle|\omega|\geq \frac{1}{n-1}|\nabla |\omega||^2+{\rm Ric}(\omega^{\sharp},\omega^{\sharp}).\end{align}

Under our hypothesis on the sectional curvature of N, we can estimate the Ricci curvature of M by using Lemmas 2.3 and 2.4:

\begin{align*} {\rm Ric}_M &\geq -(n-1)\frac{(1-\tau)\rho}{(2n-1)(n-1)}+(n-1)H^2-\frac{n-1}{n}|\Phi|^2-\frac{(n-2)\sqrt{n(n-1)}}{n}|H||\Phi| \\[5pt] &=-\frac{(1-\tau)\rho}{2n-1}+2(n-1)H^2-\frac{(n-2)\sqrt{n(n-1)}}{n}|H|\sqrt{|A|^2-nH^2}-\frac{n-1}{n}|A|^2 \\[5pt] &\geq -\frac{(1-\tau)\rho}{2n-1}+\frac{2(n-1)-n\sqrt{n-1}}{2n}|A|^2-\frac{n-1}{n}|A|^2 \\[5pt]&=-\frac{(1-\tau)\rho}{2n-1}-\frac{\sqrt{n-1}}{2}|A|^2.\end{align*}

Thus, equation (3.19) becomes

(3.20) \begin{align} |\omega|\triangle |\omega|\geq \frac{1}{n-1}|\nabla |\omega||^2-\frac{(1-\tau)\rho}{2n-1}|\omega|^2-\frac{\sqrt{n-1}}{2}|A|^2|\omega|^2.\end{align}

Given any $\alpha>0$ , using (3.20) we have that

(3.21) \begin{align} |\omega|^\alpha\triangle |\omega|^\alpha &=|\omega|^\alpha\Big(\alpha(\alpha-1)|\omega|^{\alpha-2}|\nabla |\omega||^2+\alpha |\omega|^{\alpha-1}\triangle |\omega|\Big) \\[5pt] &=\frac{\alpha-1}{\alpha}|\nabla |\omega|^\alpha|^2+\alpha|\omega|^{2\alpha-2}|\omega|\triangle|\omega| \\[5pt] &\geq \Big(1-\frac{n-2}{(n-1)\alpha}\Big)|\nabla |\omega|^\alpha|^2- \frac{\alpha\sqrt{n-1}}{2}|A|^2|\omega|^{2\alpha}-\frac{(1-\tau)\rho\alpha}{2n-1}|\omega|^{2\alpha}.\end{align}

Since M has finite index, as in the proof of Theorem 1.2, we assume that $M\backslash B_x(R_0)$ is stable. In other words,

(3.22) \begin{equation} \int_{M \backslash B_x(R_0)} |\nabla \eta|^2\geq\int_{M \backslash B_x(R_0)}\Big(|A|^2+\overline{{\rm Ric}}(u,u)\Big)\eta^2.\end{equation}

for all compactly supported Lipschitz function $\eta$ on $M \backslash B_x(R_0)$ . Replacing $\eta$ by $|\omega|^{(s+1)\alpha}\eta$ in (3.22) and applying the lower bound of sectional curvature of N allow us to conclude that

(3.23) \begin{align} \int_{M \backslash B_x(R_0)}& |A|^2|\omega|^{2(s+1)\alpha}\eta^2 \\[5pt]&\leq \int_{M \backslash B_x(R_0)} |\nabla (|\omega|^{(s+1)\alpha}\eta)|^2+\frac{n(1-\tau)}{(2n-1)(n-1)}\int_{M \backslash B_x(R_0)}\rho |\omega|^{2(s+1)\alpha}\eta^2 \\[5pt] &=(s+1)^2\int_{M \backslash B_x(R_0)} |\omega|^{2s\alpha}|\nabla |\omega|^\alpha|^2\eta^2+\int_{M \backslash B_x(R_0)} |\omega|^{2(s+1)\alpha}|\nabla \eta|^2\\[5pt] &\quad+2(s+1)\int_{M \backslash B_x(R_0)} |\omega|^{(2s+1)\alpha}\eta\langle \nabla \eta,\nabla |\omega|^\alpha\rangle \\[5pt] &\quad+\frac{n(1-\tau)}{(2n-1)(n-1)}\int_{M \backslash B_x(R_0)}\rho |\omega|^{2(s+1)\alpha}\eta^2.\end{align}

On the other hand, for $s>0$ and a smooth function $\eta$ with compactly support in M, multiplying both sides of the inequality (3.21) by $|\omega|^{2s\alpha}\eta^2$ and integrating over M, we obtain that

\begin{align*} &\quad\Big(1-\frac{n-2}{(n-1)\alpha}\Big)\int_{M \backslash B_x(R_0)}|\omega|^{2s\alpha}|\nabla |\omega|^\alpha|^2\eta^2 \\[5pt] &\leq\int_{M \backslash B_x(R_0)}|\omega|^{(2s+1)\alpha}\eta^2\triangle|\omega|^\alpha+ \frac{\alpha \sqrt{n-1}}{2}\int_{M \backslash B_x(R_0)}|A|^2|\omega|^{2(s+1)\alpha}\eta^2\\[5pt]&\quad+\frac{\alpha (1-\tau)}{2n-1}\int_{M \backslash B_x(R_0)}\rho|\omega|^{2(s+1)\alpha}\eta^2 \\[5pt] &=-(2s+1)\int_{M \backslash B_x(R_0)} |\omega|^{2s\alpha}|\nabla |\omega|^\alpha|^2\eta^2-2\int_{M \backslash B_x(R_0)} \eta|\omega|^{(2s+1)\alpha}\langle \nabla \eta,\nabla |\omega|^\alpha\rangle \\[5pt] &\quad+\frac{\alpha\sqrt{n-1}}{2}\int_{M \backslash B_x(R_0)}|A|^2|\omega|^{2(s+1)\alpha}\eta^2+\frac{\alpha (1-\tau)}{2n-1}\int_{M \backslash B_x(R_0)}\rho|\omega|^{2(s+1)\alpha}\eta^2,\end{align*}

i.e.

(3.24) \begin{align} &\quad\Big(2(s+1)-\frac{n-2}{(n-1)\alpha}\Big)\int_{M \backslash B_x(R_0)}|\omega|^{2s\alpha}|\nabla |\omega|^\alpha|^2\eta^2 \\[5pt] &\leq-2\int_{M \backslash B_x(R_0)} \eta|\omega|^{(2s+1)\alpha}\langle \nabla \eta,\nabla |\omega|^\alpha\rangle+\frac{\alpha\sqrt{n-1}}{2}\int_{M \backslash B_x(R_0)}|A|^2|\omega|^{2(s+1)\alpha}\eta^2 \\[5pt] &\quad+\frac{\alpha (1-\tau)}{2n-1}\int_{M \backslash B_x(R_0)}\rho|\omega|^{2(s+1)\alpha}\eta^2.\end{align}

Combining (3.24) with (3.23), we obtain that

(3.25) \begin{align} &\quad \Big(2(s+1)-\frac{n-2}{(n-1)\alpha}-\frac{\alpha\sqrt{n-1}}{2}\cdot (s+1)^2\Big)\int_{M \backslash B_x(R_0)} |\omega|^{2s\alpha}|\nabla |\omega|^\alpha|^2\eta^2 \\[5pt] &\leq \frac{\alpha \sqrt{n-1}}{2}\int_{M \backslash B_x(R_0)} |\omega|^{2(s+1)\alpha}|\nabla \eta|^2+E\int_{M \backslash B_x(R_0)} \rho |\omega|^{2(s+1)\alpha}\eta^2 \\[5pt] &\quad +\big(\alpha\sqrt{n-1}\cdot (s+1)-2\big)\int_{M \backslash B_x(R_0)}|\omega|^{(2s+1)\alpha}\eta\langle \nabla \eta,\nabla |\omega|^\alpha\rangle,\end{align}

where

\begin{equation*}E=\Big(\frac{ n\sqrt{n-1}}{2}+n-1\Big)\cdot\frac{\alpha(1-\tau)}{(2n-1)(n-1)}.\end{equation*}

From the assumption of weighted Poincaré inequality (1.1), we obtain that

(3.26) \begin{align} \int_{M \backslash B_x(R_0)} \rho (|\omega|^{2(s+1)\alpha}\eta^2)&\leq \int_{M \backslash B_x(R_0)} |\nabla (|\omega|^{(s+1)\alpha}\eta)|^2 \\[5pt] &=(s+1)^2\int_{M \backslash B_x(R_0)} |\omega|^{2s\alpha}|\nabla |\omega|^\alpha|^2\eta^2+\int_{M \backslash B_x(R_0)} |\omega|^{2(s+1)\alpha}|\nabla \eta|^2 \\[5pt] &\quad+2(s+1)\int_{M \backslash B_x(R_0)}|\omega|^{(2s+1)\alpha}\eta\langle \nabla \eta,\nabla |\omega|^\alpha\rangle.\end{align}

Plugging (3.26) into (3.25) implies that

(3.27) \begin{align} B\int_{M \backslash B_x(R_0)} |\omega|^{2s\alpha}|\nabla |\omega|^\alpha|^2\eta^2\leq C \int_{M \backslash B_x(R_0)} |\omega|^{2(s+1)\alpha}|\nabla \eta|^2+2D\int_{M \backslash B_x(R_0)} |\omega|^{(2s+1)\alpha}\eta\langle \nabla \eta,\nabla |\omega|^\alpha\rangle,\end{align}

where

\begin{align*} &B=2(s+1)-\frac{n-2}{(n-1)\alpha}-\frac{\alpha\sqrt{n-1}}{2}\cdot(s+1)^2-E(s+1)^2, \\[5pt] &C=\frac{\alpha \sqrt{n-1}}{2}+E, \\[5pt] &D=\frac{\alpha\sqrt{n-1}}{2}\cdot(1+s)-1+E(s+1). \end{align*}

For any $\varepsilon>0$ , using Cauchy-Schwarz inequality, we can rewrite equation (3.27) as

(3.28) \begin{align} \Big(B -|D|\varepsilon\Big)\int_{M \backslash B_x(R_0)} |\omega|^{2s\alpha}|\nabla |\omega|^\alpha|^2\eta^2\leq \Big(C+|D|\frac{1}{\varepsilon}\Big)\int_{M \backslash B_x(R_0)} |\omega|^{2(s+1)\alpha}|\nabla \eta|^2.\end{align}

Now let $p=(s+1)\alpha$ , we see that

(3.29) \begin{align}&\quad B=2(s+1)-\frac{n-2}{(n-1)\alpha}-\frac{\alpha\sqrt{n-1}}{2}\cdot(s+1)^2-E(s+1)^2 \\[5pt]&=\frac{1}{\alpha}\Big\{2p-\frac{n-2}{n-1}-\frac{p^2\sqrt{n-1}}{2}-\left(\frac{n\sqrt{n-1}}{2}+n-1\right)\frac{(1-\tau)p^2}{(2n-1)(n-1)}\Big\} \\[5pt]&=\frac{1}{\alpha}\Big\{2p-\frac{n-2}{n-1}-\frac{\sqrt{n-1}}{2}\Big[1+\Big(n+2\sqrt{n-1}\Big)\frac{(1-\tau)}{(2n-1)(n-1)}\Big]p^2\Big\}.\end{align}

Let

\begin{equation*} f(p)=-\frac{\sqrt{n-1}}{2}\Big[1+\Big(n+2\sqrt{n-1}\Big)\frac{(1-\tau)}{(2n-1)(n-1)}\Big]p^2+2p-\frac{n-2}{n-1},\end{equation*}

then the discriminant of f(p) is

(3.30) \begin{align}\Delta=4\Big(1-\frac{n-2}{2\sqrt{n-1}}\Big[1+\frac{(n+2\sqrt{n-1})(1-\tau)}{(2n-1)(n-1)}\Big]\Big)>0\end{align}

when $2\leq n\leq 6$ and $\frac{122-51\sqrt{5}}{12+4\sqrt5}< \tau\leq 1$ . Consequently, the condition $C_1<p<C_2$ allows us to conclude that $f(p)>0$ , or equivalently $B>0$ . Therefore, for a sufficiently small $\varepsilon>0$ , we have $B -|D|\varepsilon>0.$ Then, the inequality (3.28) becomes

(3.31) \begin{equation} \int_{M \backslash B_x(R_0)} |\nabla |\omega|^p|^2\leq C_0 \int_{M \backslash B_x(R_0)} |\omega|^{2p}\end{equation}

for some positive constant $C_0=C_0(n,p,\tau)$ . Then following the same method as in Theorem 1.2, we can obtain $\dim H^1(L^{2p}(M))<\infty$ .

Proof of Theorem 1.6. Let $K_N\geq -k\rho$ , where $k<\frac{4p(n-1)-2(n-2)-(n-1)\sqrt{n-1}p^2}{p^2(n-1)(2n-2+n\sqrt{n-1})}$ . Similarly as in the proof of Theorem 1.4, we can obtain that

\begin{align*} \widetilde{B}\int_{M \backslash B_x(R_0)} |\omega|^{2s\alpha}|\nabla |\omega|^\alpha|^2\eta^2&\leq \widetilde{C} \int_{M \backslash B_x(R_0)} |\omega|^{2(s+1)\alpha}|\nabla \eta|^2\\[5pt]&\quad+2\widetilde{D}\int_{M \backslash B_x(R_0)} |\omega|^{(2s+1)\alpha}\eta\langle \nabla \eta,\nabla |\omega|^\alpha\rangle\end{align*}

for any compactly supported nonconstant Lipschitz function $\eta$ on $M \backslash B_x(R_0)$ , where

\begin{align*} &\widetilde{B}=2(s+1)-\frac{n-2}{(n-1)\alpha}-\frac{\alpha\sqrt{n-1}}{2}\cdot(s+1)^2-\widetilde{E}(s+1)^2, \\[5pt] &\widetilde{C}=\frac{\alpha \sqrt{n-1}}{2}+\widetilde{E}, \\[5pt] &\widetilde{D}=\frac{\alpha\sqrt{n-1}}{2}\cdot(s+1)-1+\widetilde{E}(s+1),\\[5pt] &\widetilde{E}=\Big(\frac{n\sqrt{n-1}}{2}+n-1\Big)k\alpha.\end{align*}

For any $\varepsilon>0$ , applying Cauchy-Schwarz inequality, we have that

\begin{equation*}\Big(\widetilde{B}-|\widetilde{D}|\varepsilon\Big)\int_{M \backslash B_x(R_0)} |\omega|^{2s\alpha}|\nabla |\omega|^\alpha|^2\eta^2\leq \Big(\widetilde{C}+|\widetilde{D}|\frac{1}{\varepsilon}\Big) \int_{M \backslash B_x(R_0)} |\omega|^{2(s+1)\alpha}|\nabla \eta|^2.\end{equation*}

Let $p=(s+1)\alpha,$ then we have

\begin{align*} \widetilde{B}=\frac{1}{\alpha}\Big\{2p-\frac{n-2}{n-1}-\frac{\sqrt{n-1}}{2}p^2-\Big(\frac{n\sqrt{n-1}}{2}+n-1\Big)kp^2\Big\}.\end{align*}

Let $\widetilde{f}(p)=-(n-1)\sqrt{n-1}p^2+4(n-1)p-2(n-2)$ , then the discriminant of $\widetilde{f}(p)$ is

\begin{align*} \Delta=16(n-1)^2\Big(1-\frac{n-2}{2\sqrt{n-1}}\Big)>0,\end{align*}

which is satisfied when $3\leq n\leq 6$ . Thus from the assumption on p, we see that $\widetilde{f}(p)>0$ . Moreover, the condition $k<\frac{4p(n-1)-2(n-2)-(n-1)\sqrt{n-1}p^2}{p^2(n-1)(2n-2+n\sqrt{n-1})}$ allow us to conclude that

\begin{align*} \widetilde{B}&=\frac{1}{\alpha}\Big\{2p-\frac{n-2}{n-1}-\frac{p^2\sqrt{n-1}}{2}-\Big(\frac{n\sqrt{n-1}}{2}+n-1\Big)kp^2\Big\} \\[5pt] &=\frac{1}{\alpha}\Big\{\frac{4(n-1)p-2(n-2)-(n-1)\sqrt{n-1}p^2}{2(n-1)}-\Big(\frac{n\sqrt{n-1}}{2}+n-1\Big)kp^2\Big\} \\[5pt] &=\frac{1}{\alpha}\Big\{\frac{\widetilde{f}(p)}{2(n-1)}-\Big(\frac{n\sqrt{n-1}}{2}+n-1\Big)kp^2\Big\}>0.\end{align*}

Therefore, for a sufficiently small $\varepsilon>0$ , we have $\widetilde{B}-|\widetilde{D}|\varepsilon>0$ . Using same argument as before, we can complete the proof of Theorem 1.6.