# Type problem, the first eigenvalue and Hardy inequalities

Gilles Carron, Bo-Yong Chen, Yuanpu Xiong

► To cite this version:

Gilles Carron, Bo-Yong Chen, Yuanpu Xiong. Type problem, the first eigenvalue and Hardy inequalities. Journal of Spectral Theory, 2025, 15 (1), pp.279 - 303. 10.4171/jst/542. hal-05224447

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Submitted on 26 Aug 2025

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# Type problem, the first eigenvalue and Hardy inequalities

Gilles Carron, Bo-Yong Chen, and Yuanpu Xiong

**Abstract.** In this paper, we study the relationship between the type problem and the asymptotic behaviour of the first (Dirichlet) eigenvalues $\lambda_1(B_r)$ of “balls” $B_r := \{\rho < r\}$ on a complete Riemannian manifold $M$ as $r \to +\infty$, where $\rho$ is a Lipschitz continuous exhaustion function with $|\nabla\rho| \le 1$ a.e. on $M$. We obtain several sharp results. First, if $r^2\lambda_1(B_r) \ge \gamma > 0$ for all $r > r_0$, we obtain a sharp estimate of the volume growth: $|B_r| \ge cr^{\mu(\gamma)}$. Moreover, when $\gamma > j_0^2 \approx 5.784$, where $j_0$ denotes the first positive zero of the Bessel function $J_0$, then $M$ is non-parabolic and we have a Hardy-type inequality. In the case where $r_0 = 0$, a sharp Hardy-type inequality holds. These spectral conditions are satisfied if one assumes that $\Delta\rho^2 \ge 2\mu(\gamma) > 0$. In particular, when $\inf_M \Delta\rho^2 > 4$, $M$ is non-parabolic and we get a sharp Hardy-type inequality. Related results for finite volume case are also studied.

## 1. Introduction

Let $(M, g)$ be a complete, non-compact Riemannian manifold with $\dim M \ge 2$, and denote by $\Delta$ the Laplace operator associated to $g$. An upper semicontinuous function $u$ on $M$ is called subharmonic if $\Delta u \ge 0$ holds in the sense of distributions. If every negative subharmonic function on $M$ has to be a constant, then $M$ is said to be *parabolic*; otherwise $M$ is called *non-parabolic*. It is well known that $M$ is parabolic (resp. non-parabolic) if and only if the Green function $G_M(x, y)$ is infinite (resp. finite) for all $x \ne y$; or the Brownian motion on $M$ is recurrent (resp. transient).

The type problem is how to decide the parabolicity and non-parabolicity through intrinsic geometric conditions. The case of surfaces is classical, for the type of $M$ depends only on the conformal class of $g$, i.e., the complex structure determined by $g$. Ahlfors [1] and Nevanlinna [23] first showed that $M$ is parabolic whenever

$$ \int_{1}^{+\infty} \frac{dr}{|\partial B(x_0, r)|} = +\infty, \qquad (1.1) $$

<footnote>*Mathematics Subject Classification 2020:* 31C12 (primary); 58C40 (secondary).</footnote>

<footnote>*Keywords:* non-parabolic Riemannian manifold, first eigenvalue, Hardy inequality.</footnote>
where $B(x_0, r)$ is the geodesic ball with center $x_0 \in M$ and radius $r$. The same conclusion was extended to high dimensional cases by Lyons and Sullivan [21] and Grigor'yan [12, 13]. Moreover, (1.1) can be relaxed to

$$ \int_{1}^{+\infty} \frac{rdr}{|B(x_0, r)|} = +\infty $$

(cf. Karp [17], Varopolous [26], and Grigor'yan [12, 13]; see also Cheng and Yau [9]). We refer to the excellent survey [14] of Grigor'yan for other sufficient conditions of parabolicity.

On the other side, it seems more difficult to find sufficient conditions for non-parabolicity. Yet, there is a classical result stating that $M$ is non-parabolic whenever the first (Dirichlet) eigenvalue $\lambda_1(M)$ of $M$ is positive. Recall that

$$ \lambda_1(M) := \lim_{j \to +\infty} \lambda_1(\Omega_j) $$

for some/any increasing sequence of precompact open sets $\{\Omega_j\}$ in $M$, such that $M = \bigcup \Omega_j$. Here, given an open set $\Omega \subset \subset M$, define

$$ \lambda_1(\Omega) := \inf \left\{ \frac{\int_{\Omega} |\nabla \phi|^2 dV}{\int_{\Omega} \phi^2 dV} : \phi \in \text{Lip}_0(M), \text{supp} \phi \subset \bar{\Omega}, \phi \neq 0 \right\}, $$

where $\text{Lip}_0(M)$ denotes the set of Lipschitz continuous functions on $M$ with compact supports.

**Remark.** We use the convention that $\lambda_1(\emptyset) = +\infty$.

The main focus of this paper is to determine the non-parabolicity in the case $\lambda_1(M) = 0$. Grigor'yan showed that $M$ is non-parabolic if the following Faber-Krahn-type inequality holds:

$$ \lambda_1(\Omega) \geq f(|\Omega|) \quad \text{for all } \Omega \subset \subset M \text{ such that } |\Omega| \geq v_0 > 0, $$

where $f$ is a positive decreasing function on $(0, +\infty)$ such that

$$ \int_{v_0}^{+\infty} \frac{dv}{v^2 f(v)} < +\infty $$

(see, e.g., [14, Theorem 10.3]). We shall use certain quantity measuring the asymptotic behaviour of $\lambda_1(B_r)$ for certain “balls” $B_r$ as $r \to +\infty$, which seems to be easier to analyse. More precisely, let us first fix a nonnegative locally Lipschitz continuous function $\rho$ on $M$, which is an exhaustion function (i.e., $B_r := \{\rho < r\} \subset \subset M$ for any
$r > 0$), such that $|\nabla \rho| \le 1$ holds a.e. on $M$. Note that if $\rho$ is the distance $\mathrm{dist}_M(x_0, \cdot)$ from some $x_0 \in M$, then $B_r$ is precisely the geodesic ball $B(x_0, r)$.

To state the main result, we denote by $\lambda_\mu$ the first eigenvalue of the Laplace operator on $[0, 1)$ for the Dirichlet condition at $s = 1$ and with respect to the measure $s^{\mu-1}ds$, i.e.,

$$ \lambda_{\mu} = \inf \left\{ \frac{\int_{0}^{1} |\psi'(s)|^2 s^{\mu-1} ds}{\int_{0}^{1} |\psi(s)|^2 s^{\mu-1} ds} : \psi \in \mathrm{Lip}([0, 1]), \psi(1) = 0, \psi \neq 0 \right\}. $$

It is known that $\lambda_\mu = j_{\mu/2-1}^2$, where $j_\nu$ is the first positive zero of the Bessel function $J_\nu$, with the infimum $\lambda_\mu$ realised by

$$ \psi_{\mu}(s) := s^{1-\mu/2} J_{\mu/2-1}(\sqrt{\lambda_{\mu}s}). \quad (1.2) $$

We have the following result.

**Theorem 1.1.** Suppose that

$$ \lambda_1(B_r) \ge \frac{\lambda_\mu}{r^2}, \quad (1.3) $$

holds for all $r \ge r_0$. Then the following properties hold.

(1) There is a constant $c > 0$ such that

$$ |B_r| \ge c r^\mu \quad \text{for all } r \ge r_0. \quad (1.4) $$

Here, the constant $c$ might depend on the geometry of $B_{r_0}$.

(2) If $\mu > 2$, then $M$ is non-parabolic.

(3) If $\mu > 2$, then the Hardy-type inequality

$$ C \int_M \frac{u^2}{1 + \rho^2} dV \le \int_M |\nabla u|^2 dV \quad (1.5) $$

holds for some $C > 0$ and for any $u \in \mathrm{Lip}_0(M)$. If (1.3) holds with $\mu > 2$ for all $r > 0$, then we have the sharp Hardy-type inequality

$$ \left(\frac{\mu-2}{2}\right)^2 \int_M \frac{u^2}{\rho^2} dV \le \int_M |\nabla u|^2 dV \quad \text{for all } u \in \mathrm{Lip}_0(M). \quad (1.6) $$

**Remark.** (a) It is well known that if $M = \mathbb{R}^n$ and $\rho(x) = |x|$ then

$$ \lambda_1(B(0, r)) = \frac{\lambda_n}{r^2} = \frac{j_{n/2-1}^2}{r^2} \quad \text{for all } r > 0. $$
Thus, the volume growth in (1.4) is sharp for $\mu = n$. Moreover, (1.6) is also sharp, since on the Euclidean space $\mathbb{R}^n$, with $n \ge 3$, we have the classical Hardy inequality

$$ \left(\frac{n-2}{2}\right)^2 \int_{\mathbb{R}^n} \frac{u^2}{|x|^2} dV \le \int_{\mathbb{R}^n} |\nabla u|^2 dV, \quad \text{for all } u \in \text{Lip}_0(\mathbb{R}^n). $$

and our hypothesis holds with $\mu = n$.

(b) (1.4) also follows from the sharp Hardy-type inequality (cf. Carron [7, Proposition 2.26]). Indeed, (1.6) implies the following reverse doubling property of order $\mu$ (cf. Lansade [18, Proposition 5.2]):

$$ \frac{|B_R|}{|B_r|} \ge c_\mu \frac{R^\mu}{r^\mu} \quad \text{for all } R > r > 0. $$

Define

$$ \Lambda_* := \liminf_{r \to +\infty} \{r^2 \lambda_1(B_r)\}. $$

Note that $\mu \mapsto \lambda_\mu$ is a strict increasing continuous function when $\mu \ge 2$ (cf. [11]). Thus, a direct consequence of Theorem 1.1 is the following.

**Corollary 1.2.** If $\Lambda_* > \lambda_2 = j_0^2 \approx 5.784$, then $M$ is non-parabolic. In other words, if $M$ is parabolic, then $\Lambda_* \le j_0^2$.

**Remark.** We have already said that if $M = \mathbb{R}^2$ and $\rho(x) = |x|$, then

$$ \lambda_1(B(0, r)) = \frac{\lambda_2}{r^2} = \frac{j_0^2}{r^2}. $$

so that $\Lambda_* = j_0^2$. Since $\mathbb{R}^2$ is parabolic, so the above result turns out to be the best possible.

Conversely, it is natural to ask whether there exists a universal constant $c_0$ such that $M$ is parabolic whenever $\Lambda_* \le c_0$. The answer is, however, negative (see Example 5.5 in Section 5).

Theorem 1.1 (1) allows us to estimate $\Lambda_*$ through volume growth conditions. Cheng and Yau [9] showed that $\lambda_1(M) = 0$ if $M$ has polynomial volume growth. This was extended by Brooks [3], who showed that if the volume $|M|$ of $M$ is infinite, then

$$ \lambda_1(M) \le \frac{\mu^{*2}}{4}, \quad \mu^* := \limsup_{r \to +\infty} \frac{\log |B(x_0, r)|}{r}. $$

Refined results for ends of complete Riemannian manifolds are obtained by Li and Wang [20] (see also Carron [7, Section 2.4]). The following consequence of Theorem 1.1 may be viewed as a quantitative version of the theorem of Cheng and Yau.
**Corollary 1.3.** If $v_* := \lim \inf_{r \to +\infty} \log |B_r| / \log r$, then $\Lambda_* \le \lambda_{v_*} = j_{v_* / 2 - 1}^2$. In particular, we have

$$ (1) \quad \Lambda_* = 0 \text{ if } v_* = 0; $$

$$ (2) \quad \Lambda_* \lesssim v_* \text{ if } 0 < v_* \ll 1; $$

$$ (3) \quad \Lambda_* \lesssim v_*^2 \text{ if } v_* \gg 1. $$

When $|M| < +\infty$, we have $v_* = 0$, so that $\lambda_1(B_r)$ decays faster than quadratically as $r \to +\infty$. More precisely, we have the following.

**Proposition 1.4.** If $|M| < \infty$, then

$$ \tilde{\Lambda}_* := \liminf_{r \to +\infty} \frac{-\log \lambda_1(B_r)}{r} \ge \alpha_* := \liminf_{r \to +\infty} \frac{-\log |M \setminus B_r|}{r}. \quad (1.7) $$

It follows from Proposition 1.4 that $\lambda_1(B_r)$ decays exponentially if $|M \setminus B_r|$ decays exponentially as $r \to +\infty$. On the other hand, the relationship of $\lambda_1(M \setminus B_r)$ and $|M \setminus B_r|$ is studied by Brooks [4], who proved that if $|M| < \infty$, then

$$ \lambda_1(M \setminus B_r) \le \frac{\alpha^{*2}}{4} \quad \text{for all } r > 0, $$

where

$$ \alpha^* := \limsup_{r \to +\infty} \frac{-\log |M \setminus B_r|}{r}. $$

A more precise version of Brooks' result is also proved by Li and Wang [20].

Motivated by a result of Dodziuk, Pignataro, Randol, and Sullivan [10], we present an example in Section 5 showing that the estimate in Proposition 1.4 is sharp. Some other examples such that $\lambda_1(B_r)$ have various decaying behaviours are also given in Section 5. In particular, Example 5.5 shows that $\tilde{\Lambda}_* > 0$ does not necessarily imply $|M| < \infty$, i.e., the assumption that $M$ has finite volume in Proposition 1.4 cannot be removed.

We also show that (1.3) holds under suitable condition on $\rho$.

**Proposition 1.5.** Suppose that $\rho$ is a nonnegative locally Lipschitz continuous exhaustion function on $M$ such that $|\nabla \rho| \le 1$ a.e. and $\Delta \rho^2 \ge 2\mu$ in the sense of distributions. Then

$$ \lambda_1(B_r) \ge \frac{\lambda_\mu}{r^2} \quad \text{for all } r > 0. $$

Proposition 1.5 and Theorem 1.1, immediately yield the following.

**Corollary 1.6 (cf. [5]).** If $\Delta \rho^2 \ge 2\mu > 4$, then $M$ is non-parabolic and the sharp Hardy-type inequality (1.6) holds.
Proposition 1.5 implies several well-known results. First, if $M$ is an $n$-dimensional Cartan–Hadamard manifold and $\rho$ is the geodesic distance function, then Proposition 1.5, together with the Hessian comparison theorem, yields $\Lambda_* \ge \lambda_n$. In particular, it follows from Theorem 1.1 that $M$ is non-parabolic when $n \ge 3$ (cf. Ichihara [15, 16], see also Grigor'yan [14, Theorem 15.3]). Next, if $M$ is a complete $n$-dimensional minimal submanifold in $\mathbb{R}^N$ and $\rho$ is the restriction of Euclidean distance to a given point $x_0 \in \mathbb{R}^N$, then $\Delta\rho^2 \ge 2n$, so that

$$ \lambda_1(B^N(x_0, r) \cap M) \ge \frac{\lambda_n}{r^2}, \quad (1.8) $$

in view of Proposition 1.5. Here $B^N(x_0, r)$ is a Euclidean ball in $\mathbb{R}^N$. (1.8) was first proved by Cheng, Li, and Yau [8] by heat kernel method. In particular, $M$ is non-parabolic for $n \ge 3$, which can also be proved by the isoperimetric inequality of Michael and Simon [22] and [14, Theorem 8.2].

## Comments

In an early version of this paper, the last two authors obtained the conclusion of Corollary 1.2 under a worse condition $\Lambda_* > 18.624...$. Shortly afterwards, the first author suggested some ideas for improving certain results. We then decided to write a joint paper on the subject, and the improvements found in this paper are the result of this collaboration.

# 2. Proof of Theorem 1.1

## 2.1. Bessel's functions

Assume that $\nu > -1$ and $\mu > 0$. The Bessel function $J_\nu$ is given by

$$ J_\nu(t) = \sum_{m=0}^{\infty} \frac{(-1)^m}{m!\Gamma(m+\nu+1)} \left(\frac{t}{2}\right)^{2m+\nu}, $$

which is a solution of the ODE

$$ t^2 J_{\nu}''(t) + t J_{\nu}'(t) + (t^2 - \nu^2) J_{\nu}(t) = 0. $$

Thus, $\psi_\mu(s) = s^{1-\mu/2} J_{\mu/2-1}(\sqrt{\lambda_\mu s})$ satisfies

$$ \psi_{\mu}''(s) + \frac{\mu - 1}{s} \psi_{\mu}'(s) + \lambda_{\mu} \psi_{\mu}(s) = 0, \quad (2.1) $$
where $\sqrt{\lambda_{\mu}} = j_{\mu/2-1}$ is the first positive zero of $J_{\mu/2-1}$. In particular, $\psi_{\mu}(1) = 0$.
Moreover, we have

$$ \psi_{\mu}(s) = \sum_{m=0}^{\infty} \frac{(-1)^m}{m!\Gamma(m + \frac{\mu}{2})} \left(\frac{\sqrt{\lambda_{\mu}}s}{2}\right)^{2m}, $$

so that $\psi'_{\mu}(0) = 0$.

Let us first verify the following.

**Lemma 2.1.** *The following properties hold.*

(1) For any $[a, b] \subset [0, 1]$, we have

$$ \int_a^b (\lambda_\mu \psi_\mu(s)^2 - \psi'_\mu(s)^2) s^{\mu-1} ds = -\psi'_\mu(b) \psi_\mu(b) b^{\mu-1} + \psi'_\mu(a) \psi_\mu(a) a^{\mu-1}. \quad (2.2) $$

(2) $\psi'_{\mu}(s) < 0$ for all $s \in (0, 1]$. More precisely,

$$ \psi'_{\mu}(s)s^{\mu-1} = -\lambda_{\mu} \int_{0}^{s} \psi_{\mu}(t)t^{\mu-1}dt. \quad (2.3) $$

*Proof.* By (2.1), we have

$$ \begin{align*} (\lambda_\mu \psi_\mu(s)^2 - \psi'_\mu(s)^2) s^{\mu-1} &= \left(-\psi''_\mu(s) \psi_\mu(s) - \frac{\mu-1}{s} \psi'_\mu(s) \psi_\mu(s) - \psi'_\mu(s)^2\right) s^{\mu-1} \\ &= -(\psi'_\mu(s) \psi_\mu(s))' s^{\mu-1} - (\mu-1) \psi'_\mu(s) \psi_\mu(s) s^{\mu-2} \\ &= -(\psi'_\mu(s) \psi_\mu(s) s^{\mu-1})' \end{align*} $$

and

$$ -\lambda_{\mu} \psi_{\mu}(s) s^{\mu-1} = \psi''_{\mu}(s) s^{\mu-1} + (\mu-1) \psi'_{\mu}(s) s^{\mu-2} = (\psi'_{\mu}(s) s^{\mu-1})', $$

from which (2.2) and (2.3) follow immediately. ■

## 2.2. The volume growth

*Proof of Theorem 1.1* (1). We first assume that $\rho > 0$. Set $\phi = \psi_{\mu}(\rho/r)$, where $\psi_{\mu}$ is given as (1.2). Then the variational characterization of eigenvalue gives

$$ \lambda_{\mu} \int_{B_r} \psi_{\mu} \left(\frac{\rho}{r}\right)^2 dV \le r^2 \lambda_1(B_r) \int_{B_r} \phi^2 dV \le r^2 \int_{B_r} |\nabla \phi|^2 dV \le \int_{B_r} \psi'_{\mu} \left(\frac{\rho}{r}\right)^2 dV. $$
By using the co-area formula, this can be rewritten as

$$ \lambda_{\mu} \int_{0}^{r} \psi_{\mu} \left(\frac{t}{r}\right)^{2} d\sigma(t) \leq \int_{0}^{r} \psi'_{\mu} \left(\frac{t}{r}\right)^{2} d\sigma(t), $$

where $d\sigma(t) := (\rho)_{\#}(dV)$ is a Lebesgue-Stieltjes measure on $(0, +\infty)$. Divide this inequality by $r^{\mu+1}$ and integrate on $r \in [r_0, \bar{r}]$, we obtain

$$ \lambda_{\mu} \int_{0}^{\bar{r}} \int_{\max\{r_0, t\}}^{\bar{r}} \psi_{\mu} \left(\frac{t}{r}\right)^2 \frac{dr}{r^{\mu+1}} d\sigma(t) \leq \int_{0}^{\bar{r}} \int_{\max\{r_0, t\}}^{\bar{r}} \psi'_{\mu} \left(\frac{t}{r}\right)^2 \frac{dr}{r^{\mu+1}} d\sigma(t), $$

in view of Fubini's theorem. By using the new variable $s = t/r$, we get

$$ \lambda_{\mu} \int_{0}^{\bar{r}} \left( \int_{t/\bar{r}}^{\min\{1, t/r_0\}} \psi_{\mu}(s)^2 s^{\mu-1} ds \right) \frac{d\sigma(t)}{t^{\mu}} \leq \int_{0}^{\bar{r}} \left( \int_{t/\bar{r}}^{\min\{1, t/r_0\}} \psi'_{\mu}(s)^2 s^{\mu-1} ds \right) \frac{d\sigma(t)}{t^{\mu}}. \quad (2.4) $$

Take $a = t/\bar{r}$ and $b = \min\{1, t/r_0\}$ when $t \le r_0$ in (2.2); we infer from (2.4) and the facts $\psi_\mu(1) = 0$ and $\psi'_\mu \le 0$ that

$$ \begin{aligned} & \int_0^{\bar{r}} -\psi'_{\mu}\left(\frac{t}{\bar{r}}\right) \psi_{\mu}\left(\frac{t}{\bar{r}}\right) \left(\frac{t}{\bar{r}}\right)^{\mu-1} \frac{d\sigma(t)}{t^{\mu}} \\ & \geq \int_0^{\bar{r}} -\psi'_{\mu}\left(\min\{1, \frac{t}{r_0}\}\right) \psi_{\mu}\left(\min\{1, \frac{t}{r_0}\}\right) \left(\min\{1, \frac{t}{r_0}\}\right)^{\mu-1} \frac{d\sigma(t)}{t^{\mu}} \\ & \geq \int_0^{r_0} -\psi'_{\mu}\left(\frac{t}{r_0}\right) \psi_{\mu}\left(\frac{t}{r_0}\right) \left(\frac{t}{r_0}\right)^{\mu-1} \frac{d\sigma(t)}{t^{\mu}}, \end{aligned} $$

i.e.,

$$ \frac{1}{\bar{r}^{\mu}} \int_{B_{\bar{r}}} -\frac{\psi'_{\mu}(\frac{\rho}{\bar{r}}) \psi_{\mu}(\frac{\rho}{\bar{r}})}{(\frac{\rho}{\bar{r}})} dV \geq \frac{1}{r_0^{\mu}} \int_{0}^{r_0} -\frac{\psi'_{\mu}(\frac{\rho}{r_0}) \psi_{\mu}(\frac{\rho}{r_0})}{(\frac{\rho}{r_0})} dV. \quad (2.5) $$

If we merely have $\rho \ge 0$, then we may also apply the above argument to

$$ \rho_{\varepsilon} := \sqrt{\rho^2 + \varepsilon^2}. $$

Note that we still have

$$ |\nabla \rho_{\varepsilon}| = \frac{\rho |\nabla \rho|}{(\rho^2 + \varepsilon^2)^{1/2}} \le 1. $$
Since

$$B_r^\epsilon := \{\rho_\epsilon < r\} = B_{\sqrt{r^2 - \epsilon^2}},$$

it follows that

$$\lambda_1(B_r^\epsilon) = \lambda_1(B_{\sqrt{r^2-\epsilon^2}}) \ge \frac{\lambda_\mu}{r^2-\epsilon^2} \ge \frac{\lambda_\mu}{r^2} \quad \text{for all } r \ge r_{0,\epsilon} := \sqrt{r_0^2 + \epsilon^2}, \quad (2.6)$$

so that (2.5) still holds with $\rho$ and $r_0$ replaced by $\rho_\epsilon$ and $r_{0,\epsilon}$, respectively. Moreover, since $\psi_\mu$ is a $C^2$ function on $[0, 1]$ with $\psi'_\mu(0) = 0$, there exists a constant $A > 0$ such that

$$-\psi'_{\mu}(s)\psi_{\mu}(s) \le As \quad \text{for all } s \in [0, 1].$$

Letting $\epsilon \to 0+$, we see that (2.5) remains valid for $\rho$ and $r_0$ in view of the dominated convergence theorem, which yields

$$\frac{|B_{\bar{r}}|}{\bar{r}^{\mu}} \ge \frac{1}{A r_0^{\mu}} \int_{0}^{r_0} - \frac{\psi'_{\mu}(\frac{t}{r_0}) \psi_{\mu}(\frac{t}{r_0})}{(\frac{t}{r_0})} d\sigma(t) =: c \quad \text{for all } \bar{r} \ge r_0. \quad \blacksquare$$

*Proof of Corollary 1.3.* Suppose on the contrary that $\Lambda_* > \lambda_{v_*}$. Since the function $\mu \mapsto \lambda_\mu$ is continuous, we have $\Lambda_* > \lambda_{v_*+\epsilon}$ for $\epsilon \ll 1$, so that $\lambda_1(B_r) \ge \lambda_{v_*+\epsilon}/r^2$ for $r \gg 1$. By Theorem 1.1 (1), we conclude that $|B_r| \gtrsim r^{v_*+\epsilon}$. But this implies $v_* \ge v_* + \epsilon$, which is impossible.

Since $j_\nu \sim 2\sqrt{\nu} + 1$ as $\nu \to -1+$ (cf. Piessens [25]), we have

$$\lambda_{v_*} = j_{v_* / 2 - 1}^2 \sim 2v_*, \quad v_* \to 0+,$$

from which assertions (1) and (2) immediately follow. On the other hand, since $j_\nu \sim \nu$ as $\nu \to +\infty$ (cf. Watson [27, pp. 521], see also Elbert [11, Section 1.4]), we have

$$\lambda_{v_*} = j_{v_* / 2 - 1}^2 \sim v_*^2 / 4, \quad v_* \to +\infty,$$

which implies (3).

## 2.3. The Hardy-type inequalities

Recall that the capacity cap(K) of a compact set $K \subset M$ is given by

$$\text{cap}(K) := \inf_M |\nabla \psi|^2 dV,$$

where the infimum is taken over all $\psi \in \text{Lip}_0(M)$ with $0 \le \psi \le 1$ and $\psi|_K = 1$. The following criterion is of fundamental importance (cf. Grigor'yan [14, Theorem 5.1], and Ancona [2, pp. 46–47], see also Carron [6, Definition 2.13]).
**Theorem 2.2.** Let $(M, g)$ be a complete Riemannian manifold. Then the following properties are equivalent:

(1) $M$ is non-parabolic;

(2) $\mathrm{cap}(K) > 0$ for some/any compact set $K \subset M$ with non-empty interior;

(3) given some/any pen subset $U \subset M$, there exists a constant $C = C(U)$ such that

$$ \int_U u^2 dV \le C(U) \int_M |\nabla u|^2 dV $$

for any $u \in \mathrm{Lip}(M)$ with a compact support.

By Theorem 2.2, Theorem 1.1 (2) is a direct consequence of the Hardy-type inequality (1.5). We shall take a unified approach to proving the Hardy inequalities (1.5) and (1.6). To begin with, set

$$ \Phi(s) = J_{\mu/2-1}(\sqrt{\lambda_{\mu}}s). $$

The function $\Phi$ satisfies $\Phi(0) = \Phi(1) = 0$ and

$$ \Phi''(s) + \frac{1}{s}\Phi'(s) + \left(\lambda_{\mu} - \frac{\left(\frac{\mu}{2} - 1\right)^2}{s^2}\right)\Phi(s) = 0. \quad (2.7) $$

We shall make use the following property of $\Phi$.

**Lemma 2.3.** For any $s \in [0, 1]$, we have

$$ \int_0^x \Phi'(s)^2 s ds = \lambda_\mu \int_0^x \Phi(s)^2 s ds - \left(\frac{\mu-2}{2}\right)^2 \int_0^x \Phi(s)^2 \frac{ds}{s} + \Phi'(x)\Phi(x)x. \quad (2.8) $$

In particular,

$$ \int_0^1 \Phi'(s)^2 s ds = \lambda_\mu \int_0^1 \Phi(s)^2 s ds - \left(\frac{\mu-2}{2}\right)^2 \int_0^1 \Phi(s)^2 \frac{ds}{s}. \quad (2.9) $$

*Proof.* By (2.7), we have

$$ \begin{aligned} \lambda_{\mu} \Phi(s)^2 s - \nu^2 \Phi(s)^2 \frac{1}{s} &= \Phi''(s) \Phi(s)s + \Phi'(s) \Phi(s) \\ &= (\Phi'(s) \Phi(s))' - \Phi'(s)^2 s, \end{aligned} $$

from which (2.8) and (2.9) follow immediately. ■
*Proof of Theorem 1.1 (3).* As in the proof of (1), we first consider the case $\rho > 0$. Given $u \in \text{Lip}_0(M)$, define

$$ \phi(x) := u(x)\Phi\left(\frac{\rho(x)}{r}\right), \quad x \in M. $$

Since $|\nabla \rho| \le 1$, we have

$$ |\nabla \phi|^2 \le \Phi\left(\frac{\rho}{r}\right)^2 |\nabla u|^2 + \frac{1}{r^2} u^2 \Phi'\left(\frac{\rho}{r}\right)^2 + 2 \frac{u}{r} \langle \nabla u, \nabla \rho \rangle \Phi'\left(\frac{\rho}{r}\right) \Phi\left(\frac{\rho}{r}\right). $$

The variational characterization of eigenvalue gives

$$ \begin{aligned} \frac{\lambda_\mu}{r^2} \int_{B_r} \phi^2 dV &\le \lambda_1(B_r) \int_{B_r} \phi^2 dV \le \int_{B_r} |\nabla \phi|^2 dV \\ &\le \int_{B_r} |\nabla u|^2 \Phi\left(\frac{\rho}{r}\right)^2 dV + \frac{1}{r^2} \int_{B_r} u^2 \Phi'\left(\frac{\rho}{r}\right)^2 dV \\ &\quad + 2 \int_{B_r} \frac{u}{r} \langle \nabla u, \nabla \rho \rangle \Phi'\left(\frac{\rho}{r}\right) \Phi\left(\frac{\rho}{r}\right) dV \quad \text{for all } r > r_0. \end{aligned} \quad (2.10) $$

We then divide (2.10) by $r$ and integrate for $r \in (r_0, +\infty)$. (Under the condition of (3) we set $r_0 = \inf_M \rho > 0$.) First,

$$ \begin{aligned} & \int_{r_0}^{+\infty} \int_{B_r} u(x)^2 \Phi\left(\frac{\rho(x)}{r}\right)^2 dV(x) \frac{dr}{r^3} \\ &= \int_M \left( \int_{\max\{r_0, \rho(x)\}}^{+\infty} \Phi\left(\frac{\rho(x)}{r}\right)^2 \frac{dr}{r^3} \right) u(x)^2 dV(x) \\ &= \int_M \left( \int_{\min\{1, \rho(x)/r_0\}}^{+\infty} \Phi(s)^2 s ds \right) \frac{u(x)^2}{\rho(x)^2} dV(x) \ge \left( \int_0^1 \Phi(s)^2 s ds \right) \int_{M \setminus B_{r_0}} \frac{u^2}{\rho^2} dV, \end{aligned} $$

where we used the new variable $s = \rho(x)/r$ in the second step. Analogously, we have

$$ \begin{aligned} & \int_{r_0}^{+\infty} \int_{B_r} |\nabla u(x)|^2 \Phi\left(\frac{\rho(x)}{r}\right)^2 dV(x) \frac{dr}{r} \\ &= \int_M \left( \int_{\max\{r_0, \rho(x)\}}^{+\infty} \Phi^2\left(\frac{\rho(x)}{r}\right) \frac{dr}{r} \right) |\nabla u(x)|^2 dV(x) \\ &= \int_M \left( \int_{\min\{1, \rho(x)/r_0\}}^{+\infty} \Phi^2(s) \frac{ds}{s} \right) |\nabla u(x)|^2 dV(x) \le \left( \int_0^1 \Phi^2(s) \frac{ds}{s} \right) \int_M |\nabla u|^2 dV \end{aligned} $$
and

$$
\begin{align*}
& \int_{r_0}^{+\infty} \int_{B_r} u(x)^2 \Phi'\left(\frac{\rho(x)}{r}\right)^2 dV(x) \frac{dr}{r^3} \\
&= \int_M \left( \int_{\max\{r_0, \rho(x)\}}^{+\infty} \Phi'\left(\frac{\rho(x)}{r}\right)^2 \frac{dr}{r^3} \right) u(x)^2 dV(x) \\
&= \int_M \left( \int_{0}^{\min\{1, \rho(x)/r_0\}} \Phi'(s)^2 s ds \right) \frac{u^2(x)}{\rho(x)^2} dV(x) \\
&\leq \left( \int_{0}^{1} \Phi'(s)^2 s ds \right) \int_{M \setminus B_{r_0}} \frac{u^2}{\rho^2} dV + \int_{B_{r_0}} \left( \int_{0}^{\rho(x)/r_0} \Phi'(s)^2 s ds \right) \frac{u^2}{\rho^2} dV \\
&=: \left( \int_{0}^{1} \Phi'(s)^2 s ds \right) \int_{M \setminus B_{r_0}} \frac{u^2}{\rho^2} dV + I_1(r_0),
\end{align*}
$$

where

$$
I_1(r_0) \le \frac{\|\Phi'\|_{L^\infty([0,1])}^2}{2r_0^2} \int_{B_{r_0}} u^2 dV \quad \text{for all } r_0 \ge \inf_M \rho, \quad (2.11)
$$

so that $I_1(r_0) = 0$ if $r_0 = \inf_M \rho$. Eventually,

$$
\begin{align*}
& \int_{r_0}^{+\infty} \int_{B_r} \frac{u}{r} \langle \nabla u, \nabla \rho \rangle \Phi'\left(\frac{\rho}{r}\right) \Phi\left(\frac{\rho}{r}\right) dV \frac{dr}{r} \\
&= \int_M \left( \int_{\max\{r_0, \rho(x)\}}^{+\infty} \Phi'\left(\frac{\rho(x)}{r}\right) \Phi\left(\frac{\rho(x)}{r}\right) \frac{dr}{r^2} \right) u(x) \langle \nabla u(x), \nabla \rho(x) \rangle dV(x) \\
&= \int_M \left( \int_{0}^{\min\{1, \rho(x)/r_0\}} \Phi'(s) \Phi(s) ds \right) \frac{u(x)}{\rho(x)} \langle \nabla u(x), \nabla \rho(x) \rangle dV(x) \\
&= \int_{B_{r_0}} \Phi\left(\frac{\rho}{r_0}\right)^2 \frac{u}{\rho} \langle \nabla u, \nabla \rho \rangle dV =: I_2(r_0),
\end{align*}
$$

where we have used the fact

$$
2 \int_{0}^{1} \Phi'(s) \Phi(s) ds = \Phi^{2}(1) - \Phi^{2}(0) = 0.
$$
We also have $I_2(r_0) = 0$ if $r_0 = \inf_M \rho$. Moreover, since $\Phi(0) = 0$, we get

$$ \Phi^2(s) \leq \|\Phi'\|_{L^\infty([0,1])}^2 s^2 $$

for all $s \geq 0$, so that,

$$ I_2(r_0) \leq \frac{\|\Phi'\|_{L^\infty([0,1])}^2}{r_0} \int_{B_{r_0}} |\nabla u| |u| dV \quad \text{for all } r_0 \geq \inf_M \rho \quad (2.12) $$

These together with (2.10) yield

$$ \begin{aligned} & \lambda_\mu \left( \int_0^1 \Phi(s)^2 s ds \right) \int_{M \setminus B_{r_0}} \frac{u^2}{\rho^2} dV \\ & \leq \left( \int_0^1 \Phi^2(s) \frac{ds}{s} \right) \int_M |\nabla u|^2 dV + \left( \int_0^1 \Phi'(s)^2 s ds \right) \int_{M \setminus B_{r_0}} \frac{u^2}{\rho^2} dV \\ & \qquad + I_1(r_0) + I_2(r_0). \end{aligned} $$

Thus,

$$ \left(\frac{\mu - 2}{2}\right)^2 \int_{M \setminus B_{r_0}} \frac{u^2}{\rho^2} dV \leq \int_M |\nabla u|^2 dV + \frac{I_1(r_0) + I_2(r_0)}{A} \quad (2.13) $$

in view of (2.9), where $A = \int_0^1 \Phi^2(s) ds/s$. If (1.3) holds for all $r > 0$, we may take $r_0 = \inf_M \rho$, so that (1.6) follows immediately from (2.13).

Under the condition of (2), we infer from (2.11)–(2.13) that

$$ \left(\frac{\mu - 2}{2}\right)^2 \int_M \frac{u^2}{1 + \rho^2} dV \leq (1 + \delta) \int_M |\nabla u|^2 dV + C_\delta \int_{B_{r_0}} u^2 dV \quad (2.14) $$

holds for any $u \in \text{Lip}_{\text{loc}}(M)$ with a compact support and $\delta > 0$, where

$$ C_{\delta} = \left(\frac{\mu - 2}{2}\right)^2 + \frac{\|\Phi'\|_{L^{\infty}([0,1])}^2}{A r_0^2} + \frac{\|\Phi'\|_{L^{\infty}([0,1])}^4}{A^2 r_0^2 \delta}. \quad (2.15) $$

Note that $C_\delta$ only depends on $\mu$ and $r_0$ if $\delta$ is fixed.

The above proofs of inequalities (1.6) and (2.14) require an additional condition $\rho > 0$. In general, if $\inf_M \rho = 0$, we consider $\rho_\varepsilon := \sqrt{\rho^2 + \varepsilon^2}$. Analogously to the proof of Theorem 1.1 (1), we have (2.6), so that (2.14) becomes

$$ \left(\frac{\mu - 2}{2}\right)^2 \int_M \frac{u^2}{1 + \varepsilon^2 + \rho^2} dV \leq (1 + \delta) \int_M |\nabla u|^2 dV + C_{\delta, \varepsilon} \int_{B_{r_0^\varepsilon}} u^2 dV, $$
where $C_{\delta,\varepsilon}$ is given by (2.15) with $r_0^2$ replaced by $r_0^2 + \varepsilon^2$. Thus, (2.14) follows by letting $\varepsilon \to 0$. Moreover, if (1.3) holds for all $r > 0$, then $\lambda_1(B_r^\varepsilon) \ge \lambda_\mu/r^2$ when $r > \varepsilon$. Take $r_0 = \varepsilon = \inf_M \rho_\varepsilon$ in (2.13) with $\rho$ replaced by $\rho_\varepsilon$; we have

$$ \left(\frac{\mu-2}{2}\right)^2 \int_M \frac{u^2}{\rho_\varepsilon^2} dV \le \int_M |\nabla u|^2 dV. $$

We obtain (1.6) immediately by letting $\varepsilon \to 0$, which completes the proof of (3).

To prove (2), we shall first derive the non-parabolicity of $M$ from (2.14) when $\mu > 2$. By Theorem 1.1 (1), there exist some constants $c > 0$ and $\mu > 2$ such that $|B_r| \ge cr^\mu$ for $r \gg 1$. Thus,

$$ \int_M \frac{dV}{1+\rho^2} \ge \limsup_{r \to +\infty} \frac{|B_r|}{1+r^2} = +\infty. $$

Choose $\bar{r} \gg r_0$ so that

$$ \left(\frac{\mu-2}{2}\right)^2 \int_{B_{\bar{r}}} \frac{dV}{1+\rho^2} \ge C_{\delta}|B_{r_0}| + 1 + \delta. $$

Then

$$ \int_M |\nabla u|^2 dV \ge 1 $$

whenever $u \in \text{Lip}_0(M)$ and $u = 1$ on $\bar{B}_{\bar{r}}$. Thus, $\text{cap}(\bar{B}_{\bar{r}}) \ge 1$ and $M$ is non-parabolic in view of Theorem 2.2 (2).

Finally, it follows from Theorem 2.2 (3) that

$$ \int_{B_{r_0}} u^2 dV \le C(B_{r_0}) \int_M |\nabla u|^2 dV $$

for any $u \in \text{Lip}_0(M)$. This, together with (2.14), gives (1.5) with

$$ C = \left(\frac{\mu-2}{2}\right)^2 \frac{1}{1 + \delta + C_{\delta}C(B_{r_0})}. $$

If we fix some $\delta > 0$, then $C$ only depends on $\mu$, $r_0$ and the geometry of $M$. $\blacksquare$

### 3. Proof of Proposition 1.4

By definition, there exists a sequence $\{r_k\}$ with $\lim_{k \to +\infty} r_k = +\infty$, such that

$$ \lambda_1(B_{r_k}) > e^{-(\tilde{\Lambda}_*+\varepsilon)r_k} $$
for some $0 < \varepsilon \ll 1$. Again, for $k \ge 1$ and $0 < \delta < 1$, we take a cut-off function $\phi_k: M \to [0, 1]$ such that $\phi_k|_{B_{\delta r_k}} = 1$, $\phi_k|_{M \setminus B_{r_k}} = 0$ and $|\nabla \phi_k| \le (1-\delta)^{-1}r_k^{-1}$. Then

$$ 
\begin{aligned}
e^{-(\tilde{\Lambda}_*+\varepsilon)r_k} |B_{\delta r_k}| &\le \lambda_1(B_{r_k}) \int_M \phi_k^2 dV \le \int_M |\nabla \phi_k|^2 dV \\
&\le \frac{1}{(1-\delta)^2 r_k^2} |B_{r_k} \setminus B_{\delta r_k}| \\
&\le \frac{1}{(1-\delta)^2 r_k^2} |M \setminus B_{\delta r_k}|.
\end{aligned}
$$

That is,

$$ |M| \ge (1 + e^{-(\tilde{\Lambda}_*+\varepsilon)r_k} (1 - \delta)^2 r_k^2) |B_{\delta r_k}|, $$

which is equivalent to

$$ |M \setminus B_{\delta r_k}| \ge \frac{e^{-(\tilde{\Lambda}_*+\varepsilon)r_k} (1-\delta)^2 r_k^2}{1 + e^{-(\tilde{\Lambda}_*+\varepsilon)r_k} (1-\delta)^2 r_k^2} |M|. $$

Thus,

$$ \alpha_* \le \lim_{k \to \infty} \frac{-\log |M \setminus B_{\delta r_k}|}{\delta r_k} \le \frac{\tilde{\Lambda}_* + \varepsilon}{\delta}. $$

Letting $\delta \to 1-$ and $\varepsilon \to 0+$, we conclude that $\tilde{\Lambda}_* \ge \alpha_*.$

## 4. Proof of Proposition 1.5

Let $w \in \text{Lip}_{\text{loc}}(M)$ and $v \in L_0^1(M)$; by $\Delta w \ge v$ in the sense of distributions (or weakly) for some locally integrable function $v$, we mean

$$ \int_M \langle \nabla w, \nabla \varphi \rangle dV \le - \int_M v \varphi dV \quad (4.1) $$

for any nonnegative function $\varphi \in C_0^\infty(M)$. Since $C_0^\infty(M)$ is dense in $\text{Lip}_0(M)$ with respect to the Sobolev norm $\| \cdot \|_{W^{1,2}}$, we see that $\Delta w \ge v$ weakly if and only if (4.1) holds for all nonnegative $\varphi \in \text{Lip}_0(M)$. One can define $\Delta w \le v$ in a similar way.

We first prove the following technical lemma.

**Lemma 4.1.** Suppose that $\Delta w \ge v$ weakly and $w > 0$. Let $f: (0, +\infty) \to (0, +\infty)$ be a smooth function. Then the following properties hold:

(1) if $f$ is increasing, then $\Delta f(w) \ge f'(w)v + f''(w)|\nabla w|^2$ weakly;

(2) if $f$ is decreasing, then $\Delta f(w) \le f'(w)v + f''(w)|\nabla w|^2$ weakly.
*Proof.* (1) Let $\varphi \in \text{Lip}_0(M)$ and $\varphi \ge 0$. If $f$ is increasing, then we also have $f'(w)\varphi \in \text{Lip}_0(M)$ and $f'(w)\varphi \ge 0$. Thus,

$$ 
\begin{align*}
\int_M \langle \nabla f(w), \nabla \varphi \rangle dV &= \int_M \langle f'(w) \nabla w, \nabla \varphi \rangle dV \\
&= \int_M \langle \nabla w, \nabla (f'(w)\varphi) \rangle dV - \int_M f''(w) |\nabla w|^2 \varphi dV \\
&\le - \int_M (f'(w)v + f''(w) |\nabla w|^2) \varphi dV,
\end{align*}
$$

which proves the first assertion.

(2) Analogously, we have

$$ 
\begin{align*}
\int_M \langle \nabla f(w), \nabla \varphi \rangle dV &= \int_M \langle f'(w) \nabla w, \nabla \varphi \rangle dV \\
&= - \int_M \langle \nabla w, \nabla (-f'(w)\varphi) \rangle dV - \int_M f''(w) |\nabla w|^2 \varphi dV \\
&\ge - \int_M (f'(w)v + f''(w) |\nabla w|^2) \varphi dV. \quad \blacksquare
\end{align*}
$$

*Proof of Proposition 1.5.* We follow the ideas in [5]. Assume for a moment that $\rho > 0$. Apply Lemma 4.1 (1) with $w = \rho^2$ and $f(t) = t^{1/2}$; we get

$$ \Delta \rho \ge \frac{\mu - |\nabla \rho|^2}{\rho} \qquad (4.2) $$

weakly.

Let $\phi$ be a Lipschitz continuous function which is positive on $B_r$ and fix $u \in \text{Lip}_0(B_r)$. With $v := u/\phi$, we have

$$ 
\begin{align}
\int_M |\nabla u|^2 dV &= \int_M \phi^2 |\nabla v|^2 dV + \int_M v^2 |\nabla \phi|^2 dV + 2 \int_M \phi v \langle \nabla \phi, \nabla v \rangle dV \nonumber \\
&= \int_M \phi^2 |\nabla v|^2 dV + \int_M \langle \nabla (\phi v^2), \nabla \phi \rangle dV. \tag{4.3}
\end{align}
$$

Let us choose $\phi = \psi_\mu(\rho/r)$, where

$$ \psi_\mu(s) = s^{1-\mu/2} J_{\mu/2-1}(\sqrt{\lambda_\mu} s). $$
Recall that $\psi_\mu$ is a solution of the ODE (2.1) and $\psi'_\mu \le 0$. Thus, (4.2) and (2.1) together with Lemma 4.1 (2) yield

$$
\begin{align*}
\Delta\phi &\le \psi'_{\mu}\left(\frac{\rho}{r}\right) \frac{\mu - |\nabla\rho|^2}{\rho r} + \psi''_{\mu}\left(\frac{\rho}{r}\right) \frac{|\nabla\rho|^2}{r^2} \\
&= \psi'_{\mu}\left(\frac{\rho}{r}\right) \frac{\mu - 1}{\rho r} + \psi''_{\mu}\left(\frac{\rho}{r}\right) \frac{1}{r^2} + \psi'_{\mu}\left(\frac{\rho}{r}\right) \frac{1 - |\nabla\rho|^2}{\rho r} - \psi''_{\mu}\left(\frac{\rho}{r}\right) \frac{1 - |\nabla\rho|^2}{r^2} \\
&= -\frac{\lambda_{\mu}}{r^2}\phi + \frac{1 - |\nabla\rho|^2}{r^2}\left(\frac{\psi'_{\mu}\left(\frac{\rho}{r}\right)}{\rho/r} - \psi''_{\mu}\left(\frac{\rho}{r}\right)\right). \tag{4.4}
\end{align*}
$$

We have (cf. Lebedev [19, formula (5.3.5)])

$$
\psi'_{\mu}(s) = -s\psi_{\mu+2}(s),
$$

from which it follows that

$$
\frac{\psi'_{\mu}(s)}{s} - \psi''_{\mu}(s) = -\psi_{\mu+2}(s) - (-s\psi_{\mu+2}(s))' = -s^2\psi_{\mu+4}(s) \le 0.
$$

Then (4.4) implies

$$
\Delta\phi \leq -\frac{\lambda_{\mu}}{r^2}\phi,
$$

which gives

$$
\int_M \langle \nabla(\phi v^2), \nabla\phi \rangle dV \geq - \int_M \phi v^2 \left(-\frac{\lambda_\mu}{r^2} \phi\right) dV = \frac{\lambda_\mu}{r^2} \int_M u^2 dV.
$$

This, together with (4.3), yields

$$
\frac{\lambda_{\mu}}{r^2} \int_{B_r} u^2 dV \leq \int_{B_r} |\nabla u|^2 dV \quad \text{for all } u \in \text{Lip}_{\text{loc}}(B_r) \quad (4.5)
$$

under the additional condition $\rho > 0$.

In general, we use $\rho_\varepsilon := (\rho^2 + \varepsilon^2)^{1/2}$ instead of $\rho$. Note that $|\nabla\rho_\varepsilon| \le 1$ and

$$
\Delta \rho_{\varepsilon}^{2} = \Delta \rho^{2} \geq 2\mu.
$$

Thus, (4.5) holds if $B_r$ is replaced by $B_r^\varepsilon$ in view of (2.6). The proposition follows by letting $\varepsilon \to 0$. ■

**5. Examples**

Let $M = \mathbb{R} \times \mathbb{S}^1$ be equipped with the following Riemannian metric:

$$
g = dt^2 + \eta'(t)^2 d\theta^2, \quad t \in \mathbb{R}, \quad e^{i\theta} \in \mathbb{S}^1,
$$
where $\eta: \mathbb{R} \to \mathbb{R}$ is a smooth function such that $\eta'(t) > 0$ and $\lim_{t \to -\infty} \eta(t) = 0$. Dodziuk, Pigmataro, Randol, and Sullivan [10, Proposition 3.1] showed that if $\eta(t) = e^t$, then $\lambda_1(M) \ge 1/4$, so that $(M, g)$ is non-parabolic.

In general, we consider

$$h(t) := \int_{0}^{t} \frac{ds}{\eta'(s)}, \quad t \in \mathbb{R},$$

which gives a function on $M$. A straightforward calculation shows

$$\Delta h = \frac{\partial^2 h}{\partial t^2} + \frac{\eta''(t)}{\eta'(t)} \frac{\partial h}{\partial t} + \frac{1}{\eta'(t)^2} \frac{\partial^2 h}{\partial \theta^2} = 0,$$

i.e., $h$ is a harmonic function on $M$. Moreover, in the new coordinate system $(\tilde{t}, \theta) := (h(t), \theta)$, we may write

$$g = \eta'(t)^2(h'(t)^2 dt^2 + d\theta^2) = \eta'(t)^2(d\tilde{t}^2 + d\theta^2),$$

which indicates that $(M, g)$ is conformally equivalent to the cylinder $(\inf h, \sup h) \times \mathbb{S}^1$ equipped with a flat metric. In conclusion, we have the following result.

**Proposition 5.1.** $(M, g)$ is non-parabolic if and only if $\inf h > -\infty$ or $\sup h < +\infty$.

Let $\rho(t, \theta) = |t|$. Clearly, $\rho$ is an exhaustion function which satisfies $|\nabla \rho|_g \le 1$. Indeed, $\rho$ is the geodesic distance to the circle $\{0\} \times \mathbb{S}^1$. The goal of this section is to investigate the asymptotic behaviour of $\lambda_1(B_r)$ as $r \to +\infty$ for different choices of $\eta$. We start with the following elementary lower estimate.

**Proposition 5.2.** We have

$$\lambda_1(B_r) \ge \frac{1}{4} \inf_{|t| \le r} \frac{\eta'(t)^2}{\eta(t)^2}.$$

*Proof.* The idea is essentially implicit in [10]. Since $dV = \eta'(t)dt d\theta$, we have

$$\int_{-r}^{r} \phi^2 \eta'(t) dt = -2 \int_{-r}^{r} \phi \frac{\partial \phi}{\partial t} \eta(t) dt, \quad \text{for all } \phi \in C_0^\infty(B_r),$$

so that

$$\begin{align*}
\int_{-r}^{r} \phi^2 \eta'(t) dt &\le 4 \int_{-r}^{r} \left| \frac{\partial \phi}{\partial t} \right|^2 \frac{\eta(t)^2}{\eta'(t)} dt \\
&\le 4 \int_{-r}^{r} |\nabla \phi|^2 \frac{\eta(t)^2}{\eta'(t)} dt \\
&\le 4 \sup_{|t| \le r} \frac{\eta(t)^2}{\eta'(t)^2} \int_{-r}^{r} |\nabla \phi|^2 \eta'(t) dt
\end{align*}$$
in view of the Cauchy–Schwarz inequality. Thus,

$$ \int_{B_r} \phi^2 dV = \int_0^{2\pi} \int_{-r}^r \phi^2 \eta'(t) dt d\theta \le 4 \sup_{|t| \le r} \frac{\eta(t)^2}{\eta'(t)^2} \int_{B_r} |\nabla \phi|^2 dV, $$

from which the assertion immediately follows.

We give the following test example for Proposition 5.1 and 5.2.

**Example 5.3.** Given $\alpha > 0$, take $\eta$ such that

$$ \eta(t) = \begin{cases} (-t)^{-\alpha}, & t < -1, \\ 2t^{\alpha}, & t > 1. \end{cases} $$

We claim that $\Lambda_* \sim \alpha^2/4$ as $\alpha \to +\infty$. To see this, first note that

$$ \Lambda_* = \liminf_{r \to +\infty} \{r^2 \lambda_1(B_r)\} \ge \frac{\alpha^2}{4}. $$

in view of Proposition 5.2. Then, by using

$$ |B_r| = 2\pi \int_{-r}^{r} \eta'(t) dt = 2\pi(\eta(r) - \eta(-r)) = 4\pi r^{\alpha} - 2\pi r^{-\alpha} \quad \text{for all } r \gg 1, $$

we see that

$$ \nu_* = \liminf_{r \to +\infty} \frac{\log |B_r|}{\log r} = \alpha. $$

This, together with Corollary 1.3, gives

$$ \frac{\alpha^2}{4} \le \Lambda_* \le \lambda_\alpha = j_{\alpha/2-1}^2 \sim \frac{\alpha^2}{4}, $$

which verifies the claim.

In particular, $M$ is non-parabolic provided $\alpha \gg 1$, in view of Corollary 1.2. More precisely, Proposition 5.1 implies that $M$ is non-parabolic if and only if $\alpha > 2$.

The next example shows that estimate (1.7) in Proposition 1.4 is sharp.

**Example 5.4.** Given $\alpha > 0$, take $\eta$ such that

$$ \eta'(t) = e^{-\alpha|t|} \quad \text{for all } |t| > 1. \tag{5.1} $$

Then, we have

$$ \liminf_{r \to +\infty} \frac{-\log \lambda_1(B_r)}{r} = \alpha = \liminf_{r \to +\infty} \frac{-\log |M \setminus B_r|}{r}. \tag{5.2} $$

Namely, (1.7) is sharp.
Indeed, since

$$|M \setminus B_r| = 4\pi \int_r^\infty e^{-\alpha t} dt \asymp e^{-\alpha r}, \quad r \gg 1,$$

we have

$$\liminf_{r \to +\infty} \frac{-\log |M \setminus B_r|}{r} = \alpha, \qquad (5.3)$$

which implies

$$\liminf_{r \to +\infty} \frac{-\log \lambda_1(B_r)}{r} \ge \alpha, \qquad (5.4)$$

in view of Proposition 1.4. On the other hand, we have the following Hardy-type inequality (cf. Opic and Kufner [24, pp. 100–103]):

$$\int_{-r}^{r} \phi(t)^2 \eta'(t) dt \lesssim e^{\alpha r} \int_{-r}^{r} \phi'(t)^2 \eta'(t) dt \quad \text{for all } \phi \in C_0^\infty((-r, r)), \qquad (5.5)$$

where the implicit constant is independent of $r$. By (5.5), we immediately see that

$$\lambda_1(B_r) \gtrsim e^{-\alpha r},$$

which combined with (5.3) and (5.4) gives (5.2).

For reader's convenience, we include here a rather short proof. Since $\int_{-\infty}^{+\infty} \eta'(t) dt$ is finite in view of (5.1), it follows that

$$\int_{-r}^{r} \phi(t)^2 \eta'(t) dt \le \sup_{-r<t<r} \phi(t)^2 \int_{-r}^{r} \eta'(t) dt \lesssim \sup_{-r<t<r} \phi(t)^2.$$

On the other hand, by setting $|\phi(t_0)| = \sup_{-r<t<r} |\phi(t)|$, we have

$$\int_{-r}^{r} |\phi'(t)| dt \ge \int_{-r}^{t_0} |\phi'(t)| dt \ge \left| \int_{-r}^{t_0} \phi'(t) dt \right| = |\phi(t_0)| = \sup_{-r<t<r} |\phi(t)|.$$

This, together with the Cauchy–Schwarz inequality, yields

$$\sup_{-r<t<r} \phi(t)^2 \le \left( \int_{-r}^{r} \phi'(t)^2 \eta'(t) dt \right) \left( \int_{-r}^{r} \frac{1}{\eta'(t)} dt \right) \lesssim e^{\alpha r} \int_{-r}^{r} \phi'(t)^2 \eta'(t) dt.$$

**Remark.** By Proposition 5.2, we only obtain a weaker conclusion

$$\lambda_1(B_r) \ge \frac{1}{4} \frac{\eta'(r)^2}{\eta(r)^2} \ge e^{-2\alpha r}.$$
The following example shows that the converse of Theorem 1.1 (2) (or Corollary 1.2) does not hold, i.e., M can be non-parabolic with $\lambda_1(B_r)$ decaying so quickly that $\Lambda_* = 0$.

**Example 5.5.** Take $\eta$ such that

$$ \eta'(t) = \begin{cases} n^2, & 2^n < t < 2^n + 1, \\ e^t, & 2^n + 2 < t < 2^{n+1} - 1, \end{cases} $$

and

$$ \eta'(t) \ge n^2, \quad 2^n - 1 \le t \le 2^n + 2 $$

for $n = 2, 3, \dots$. Then M is non-parabolic but $\lambda_1(B_r)$ shall decay exponentially, so that $\Lambda_* = 0$.

Actually, the non-parabolicity of M is a direct consequence of Proposition 5.1, for $\sup h = \int_0^{+\infty} \frac{dt}{\eta'(t)} < +\infty$. To study the behaviour of $\lambda_1(B_r)$, consider the test function,

$$ \phi(t) := \begin{cases} t, & 0 \le t \le 1, \\ 1, & 1 \le t \le r-1, \\ r-t, & r-1 \le t \le r, \\ 0, & \text{otherwise.} \end{cases} $$

We obtain

$$ \lambda_1(B_r) \le \frac{\int_{-r}^{r} \phi'(t)^2 \eta'(t) dt}{\int_{-r}^{r} \phi(t)^2 \eta'(t) dt} \le \frac{\eta(r) - \eta(r-1) + \eta(1) - \eta(0)}{\eta(r-1) - \eta(1)}. \quad (5.6) $$

For $r = 2^n + 1$, where $n \in \mathbb{Z}^+$ and $n \ge 2$, we have $\eta(r) - \eta(r-1) = n^2$ and

$$ \eta(r-1) = \eta(2^n) \ge \int_{2^{n-1}+2}^{2^n-1} e^t dt \asymp e^{2^n}, $$

so that (5.6) gives

$$ \lambda_1(B_{2^n+1}) \lesssim e^{-2^n} n^2. $$

In general, for $r \gg 1$, take $n \in \mathbb{Z}^+$ such that $2^n + 1 \le r < 2^{n+1} + 1$. Then

$$ \lambda_1(B_r) \le \lambda_1(B_{2^n+1}) \lesssim e^{-2^n} n^2 \lesssim e^{-r/2} (\log r)^2 $$

and hence $\Lambda_* = 0$.

**Remark.** We have

$$ \Lambda^* := \limsup_{r \to +\infty} r^2 \lambda_1(B_r) = 0 $$
and

$$ \tilde{\Lambda}_* = \liminf_{r \to +\infty} \frac{-\log \lambda_1(B_r)}{r} \ge \frac{1}{2} > 0. $$

On the other hand, $M$ is non-parabolic with an infinite volume.

Our last example shows that $\lambda_1(B_r)$ can also decay to 0 very slowly.

**Example 5.6.** Let $\mu$ be a positive, smooth and decreasing function on $[1, +\infty)$ satisfying

(1) $\lim_{t \to +\infty} \mu(t) = 0,$

(2) $\int_{1}^{+\infty} \mu(s)ds = +\infty,$

(3) $t\mu(t)$ is increasing on $[c, +\infty)$ for some $c \gg 1$.

Take $\eta$ such that

$$ \eta(t) = \begin{cases} e^{-\int_{1}^{-t} \mu(s)ds}, & t < -1, \\ 2e^{\int_{1}^{t} \mu(s)ds}, & t > 1. \end{cases} $$

We claim that

$$ \lambda_1(B_r) \asymp \mu(r)^2. \quad (5.7) $$

To see this, first note that $\eta'(t)/\eta(t) = \mu(-t)$ for $t < -1$ and $\eta'(t)/\eta(t) = \mu(t)$ for $t > 1$, which implies

$$ \lambda_1(B_r) \ge \frac{1}{4} \inf_{|t| \le r} \frac{\eta'(t)^2}{\eta(t)^2} = \frac{\mu(r)^2}{4}. \quad (5.8) $$

in view of Proposition 5.2. On the other hand, we have $r\mu(r) \ge c\mu(c) > 0$ for $r \ge c \gg 1$ in view of the condition (3). Thus, we may take $0 < \varepsilon \le c\mu(c)/2$ so that

$$ r_{\varepsilon} := r - \varepsilon\mu(r)^{-1} = r(1 - \varepsilon r^{-1}\mu(r)^{-1}) \ge \frac{r}{2} \quad \text{for all } r \ge c. \quad (5.9) $$

Set $I_r := (-r, -r_\varepsilon)$. Since $\eta''(t) = -\mu'(-t)\eta(t) + \mu(-t)\eta'(t) \ge 0$, i.e., $\eta'(t)$ is increasing on $(-\infty, -1]$, it follows that

$$ \begin{align*} \lambda_1(B_r) &\le \lambda_1(\{(t, \theta) \in M : -r \le t \le -r_\varepsilon\}) \\ &\le \inf_{\phi \in C_0^\infty(I_r)} \left\{ \frac{\int_{I_r} \phi'(t)^2 \eta'(t) dt}{\int_{I_r} \phi(t)^2 \eta'(t) dt} \right\} \\ &\le \inf_{\phi \in C_0^\infty(I_r)} \left\{ \frac{\int_{I_r} \phi'(t)^2 dt}{\int_{I_r} \phi(t)^2 dt} \right\} \cdot \frac{\eta'(-r_\varepsilon)}{\eta'(-r)} \\ &= \lambda_1(I_r) \cdot \frac{\eta'(-r_\varepsilon)}{\eta'(-r)}. \end{align*} $$
Since $\lambda_1(I_r) \lesssim |I_r|^{-2} \asymp \mu(r)^2$, we obtain

$$ \lambda_1(B_r) \lesssim \mu(r)^2 \cdot \frac{\eta'(-r_\varepsilon)}{\eta'(-r)}. \qquad (5.10) $$

We have

$$ \frac{\eta'(-r_\varepsilon)}{\eta'(-r)} = \frac{\mu(r_\varepsilon)}{\mu(r)} \exp\left(\int_{r_\varepsilon}^{r} \mu(s) ds\right) \le \frac{\mu(r_\varepsilon)}{\mu(r)} \exp\left(\varepsilon \frac{\mu(r_\varepsilon)}{\mu(r)}\right), $$

for $\mu$ is decreasing and $r - r_\varepsilon = \varepsilon\mu(r)^{-1}$. By condition (3) and (5.9), we have

$$ \frac{\mu(r_\varepsilon)}{\mu(r)} \le \frac{r}{r_\varepsilon} \le 2. $$

Thus,

$$ \frac{\eta'(-r_\varepsilon)}{\eta'(-r)} = O(1) \quad \text{as } r \to +\infty. $$

This, together with (5.8) and (5.10), gives (5.7).

Particular choices of $\mu$ give the following:

(1) for $\mu(t) = t^{-1}(\log t)^\beta$ with $\beta > 0$, $\lambda_1(B_r) \asymp r^{-2}(\log r)^{2\beta}$;

(2) for $\mu(t) = t^{-\alpha}$ with $0 < \alpha < 1$, $\lambda_1(B_r) \asymp r^{-2\alpha}$;

(3) for $\mu(t) = (\log(t+1))^{-\gamma}$ with $\gamma > 0$, $\lambda_1(B_r) \asymp (\log r)^{-2\gamma}$.

In all three cases, we have

$$ \Lambda_* = \liminf_{r \to +\infty} \{r^2 \lambda_1(B_r)\} = +\infty. $$

Thus, these Riemannian manifolds $(M, g)$ are non-parabolic in view of Corollary 1.2.

**Acknowledgements.** G. Carron thanks the Centre Henri Lebesgue ANR-11-LABX-0020-01 for creating an attractive mathematical environment.

**Funding.** B.-Y. Chen and Y. Xiong are supported by National Natural Science Foundation of China, No. 12271101.

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Received 11 July 2024; revised 5 December 2024.

**Gilles Carron**

CNRS, Laboratoire de Mathématiques Jean Leray, LMJL, Nantes Université, UMR 6629, 44000 Nantes, France; gilles.carron@univ-nantes.fr

**Bo-Yong Chen**

School of Mathematical Sciences, Fudan University, Handan Road 220, 200433 Shanghai, P. R. China; boychen@fudan.edu.cn

**Yuanpu Xiong**

School of Mathematical Sciences, Fudan University, Handan Road 220, 200433 Shanghai, P. R. China; ypxiong@fudan.edu.cn