# Unique continuation estimates on manifolds with Ricci curvature bounded
below

Christian Rose111University of Potsdam, Institute of Mathematics, Karl-
Liebknecht-Str. 24-25, D-14476 Potsdam<EMAIL_ADDRESS>and
Martin Tautenhahn222University of Leipzig, Department of Mathematics,
Augustusplatz 1, D-04103 Leipzig<EMAIL_ADDRESS>
###### Abstract

We prove quantitative unique continuation estimates for relatively dense sets
and spectral subspaces associated to small energies of Schrödinger operators
on Riemannian manifolds with Ricci curvature bounded below. The upper bound
for the energy range and the constant appearing in the estimate are given in
terms of the lower bound of the Ricci curvature and the parameters of the
relatively dense set.

##  1 Introduction

Let $M$ be a complete Riemannian manifold, $S\subset M$, and $H$ a lower semi-
bounded self-adjoint Schrödinger operator acting in $L^{2}(M)$. Quantitative
unique continuation principles on $M$ refer to inequalities of the form

$\|\varphi\|_{L^{2}(M)}\leq C\|\varphi\|_{L^{2}(S)}$ (1.1)

for functions $\varphi$ in spectral subspaces up to energy $E$. The constant
$C$ depends on the geometric properties of the manifold $M$, the set $S$, the
energy $E$, and the operator $H$, but should be independent of the function
$\varphi$. The various applications of such a unique continuation estimate
also benefit from an explicit (and optimal) dependence of $C$ on these model
parameters. Of particular interest is the study of conditions on $M$, $S$, and
$H$ such that (1.1) is true for all $I\subset(-\infty,E_{0}]$ with some
$E_{0}>\min\sigma(H)$.

Classically, the problem (1.1) is investigated in the case $M=\mathbb{R}^{n}$.
If $H=\Delta\geq 0$, and $S$ is a so-called thick set, then (1.1) follows from
the Logvinenko-Sereda theorem [LS74], see also [Pan61, Kac73, Kov00, Kov01].
Up to now, there is a huge amount of literature on quantitative unique
continuation principles on classes of subsets of $\mathbb{R}^{n}$ under
various assumptions on $H$, the geometric properties of $S$, and the length of
the energy interval $I$, see, e.g., [LR95, LZ98, JL99, RMV13, Kle13, KT16,
NTTV18, NTTV20, DRST20, Egi21, ES21] and the references therein. Recently,
(1.1) was proven for uniform elliptic operators without any regularity
conditions on the symbol in convex subsets of $\mathbb{R}^{n}$ and relatively
dense control set for subspaces related to small energies in [SS21]. The
interest in quantitative unique continuation principles stems particularly
from the enormous amount of applications in different fields. It proved to be
a powerful tool, e.g., in the control theory of the heat equation and the
theory of random Schrödinger operators, see [BK05, CHK07, GK01, GK13, BK13,
MRM22, ST20].

Our focus is on unique continuation principles on Riemannian manifolds rather
than subsets of $\mathbb{R}^{n}$, which is motivated by their strong geometric
implications. Prominent examples of utilizing unique continuation estimates in
Riemannian geometry are [DF88, DF90b, DF90a], where the authors employed
variants of (1.1) to derive vanishing order estimates for eigenfunctions of
the Laplace–Beltrami operator for a given compact manifold $M$. Since then,
unique continuation properties have been investigated for various manifolds
$M$ and $S\subset M$, see, e.g., [EV20, LM19, BM21, BM23, Mil05, DV22].
However, all known results treating (1.1) on manifolds depend explicitly on
the symbol of the Laplace–Beltrami operator. More precisely, they can only be
formulated in terms of ellipticity and Lipschitz constants of the metric
tensor $g$ or the defect with respect to a given metric. While being
quantitative, the drawback of those results is that $g$ must be known a priori
in a certain way, i.e., the full description of $M$ is needed to obtain
quantitative unique continuation estimates. It is well-known that there is no
possibility in general of recovering the metric tensor completely from local
intrinsic geometric information.

Rather than prescribing a metric tensor, it is natural to impose local
geometric restrictions to conclude global geometric and analytic properties of
manifolds such as curvature bounds. We aim at deriving (1.1) with explicit $E$
and $C$ for manifolds satisfying curvature restrictions in order to treat
_classes_ of Riemannian manifolds satisfying the same curvature conditions.
This allows in particular to consider perturbations of metrics and derive
uniform results as well apart from perturbations of Euclidean space.

Our contribution is a quantitative uncertainty principle for Schrödinger
operators on Riemannian manifolds with Ricci curvature bounded below at
energies close to the bottom of the spectrum.

Let $M=(M^{n},g)$ be a complete Riemannian manifold without boundary of
dimension $n\in\mathbb{N}$ and Ricci tensor $\operatorname{Ric}$. For a
comprehensive introduction to Riemannian geometry, see, e.g., the excellent
[GHL87]. We denote the distance function by $\mathrm{d}:M\times
M\to[0,\infty)$, and the volume element by $\operatorname{\mathrm{dvol}}$. For
$x\in M$ and $r>0$ we denote by $B_{x}(r)=\\{y\in M\colon\mathrm{d}(y,x)<r\\}$
the open ball with radius $r$ and center $x$. Given $0<\rho<R$, a set
$S\subset M$ is called _$(R,\rho)$ -relatively dense_ provided

$\forall x\in M\ \exists y\in S\colon B_{x}(R)\cap S\supset B_{y}(\rho).$

Given $x\in M$ and $R>0$, we call $R$ a _proper radius for $x$_, if
$\mathrm{diam}(M)>R$. Moreover, we call a $(R,\rho)$-relatively dense set
$S\subset M$ _proper_ , if $3R$ is a proper radius for all $x\in M$.

Let us now turn to the operator theoretic definitions. We define the quadratic
form $\mathcal{E}\colon D(\mathcal{E})\to\mathbb{R}$ by

$D(\mathcal{E})=W^{1,2}(M),\quad\mathcal{E}(f)=\int_{M}|\nabla
f|^{2}\operatorname{\mathrm{dvol}},$

and denote by $\Delta$ the Laplace-Beltrami operator of $M$, that is, the
unique self-adjoint operator in $L^{2}(M)$ associated to the form
$\mathcal{E}$. Note that $\Delta\geq 0$ by construction. Now we define
successively two form perturbations of $\mathcal{E}$. First, let
$\mathcal{V}_{+}\colon D(\mathcal{V}_{+})\to\mathbb{R}$ be a densely defined
quadratic non-negative form. We assume that the form sum
$\mathcal{E}+\mathcal{V_{+}}$ with domain $D(\mathcal{E})\cap
D(\mathcal{V}_{+})$ is densely defined and closed, and thus defines a unique
non-negative self-adjoint operator in $L^{2}(M)$, see [Kat95, Theorem VI.2.6].
This is for example the case if $\mathcal{V_{+}}$ is relatively bounded with
respect to $\mathcal{E}$ with $\mathcal{E}$-bound smaller than one, see
[Kat95, VI.1.33]. Second, let $\mathcal{V}_{-}\colon
D(\mathcal{V}_{-})\to\mathbb{R}$ be a non-negative quadratic form relatively
bounded with respect to $\mathcal{E}+\mathcal{V}_{+}$, i.e.
$D(\mathcal{V}_{-})\supset D(\mathcal{E})\cap D(\mathcal{V}_{+})$ and there
are constants $a\in(0,1)$ and $b\geq 0$ such that

$\mathcal{V}_{-}(f)\leq a(\mathcal{E}+\mathcal{V}_{+})(f)+b\lVert
f\rVert^{2},\quad f\in D(\mathcal{E})\cap D(\mathcal{V}_{+}).$ (1.2)

As a consequence, the form

$\mathcal{H}\colon D(\mathcal{E})\cap D(\mathcal{V}_{+})\to\mathbb{R},\quad
f\mapsto\mathcal{E}(f)+\mathcal{V}_{+}(f)-\mathcal{V}_{-}(f),$

is a densely defined, closed form bounded from below [Kat95, Theorem VI.1.33].
We denote the unique lower semi-bounded self-adjoint operator by $H$ and its
domain by $D(H)$, cf. [Kat95, Theorem VI.2.6]. Let us stress that our operator
theoretic setting includes nice measure perturbations, e.g. as in [SV96]), or
singular potentials.
Our main theorem is as follows.

######  Theorem 1.1.

Let $K\in\mathbb{R}$, $n\in\mathbb{N}$, $n\geq 3$, and $0<\rho<R$. There are
$\kappa=\kappa(K,R,\rho,n)>0$ and $E_{0}=E_{0}(K,R,\rho,n,a,b)\in\mathbb{R}$
such that for any complete Riemannian manifold $M$ of dimension $n$ with
$\operatorname{Ric}\geq K$, any proper $(R,\rho)$-relatively dense $S\subset
M$, and any $I\subset(-\infty,E_{0}]$, we have

$\chi_{I}(H)\mathbf{1}_{S}\chi_{I}(H)\geq\kappa\ \chi_{I}(H),$

where the inequality holds in quadratic form sense.

Note that the statement of the theorem is void if $E_{0}<\min\sigma(H)$. We
will restate Theorem 1.1 with explicit constants as Theorem 5.1 in Section 5.
In particular, we observe that $E_{0}>0$ if the negative part
$\mathcal{V}_{-}$ is small, i.e. if $a$ and $b$ are small. Moreover, the
constants naturally depend on quantities of comparison spaces
$M_{K}=M_{K}^{n}$ of constant sectional curvature $K$. More precisely,
denoting by $\mathrm{sn}_{K}$ resp. $\operatorname{Vol}_{K}$ the density and
volume of $M_{K}$, we have

$\kappa\sim\frac{\rho^{2}\mathrm{sn}_{K}(\rho)^{n-2}}{\operatorname{Vol}_{K}(R)}\left[\mathrm{e}^{\sqrt{K}R}+\rho^{2}K+\left|\ln\left(\frac{\operatorname{Vol}_{K}(R)}{\rho^{2}\mathrm{sn}_{K}(\rho)^{n-2}}\right)\right|\right]^{-2}$

and

$E_{0}\sim(1-a)\frac{\mathrm{sn}_{K}(\rho)^{n-2}}{\operatorname{Vol}_{K}(R)}-b.$

To the best of our knowledge, Theorem 1.1 is the first quantitative
uncertainty principle for manifolds where the constants $\kappa$ and $E_{0}$
only depend on Ricci curvature lower bounds instead of assumptions on a
particular given metric. In particular, we obtain uniform bounds for whole
classes of manifolds. It is also easy to see that our result yields bounds for
uniformly elliptic divergence form operators with only measurable and bounded
coefficients on manifolds without any regularity assumption on the symbol as
in [SS21].

######  Remark.

  * •

If $K=0$, we recover the results from [SS21] in the case $M=\mathbb{R}^{n}$,
that is, uncertainty principles at low energies for Schrödinger operators
defined in quadratic form sense.

  * •

In [DRST20], the authors obtained (1.1) for all energies for Schrödinger
operators where the potentials are operator perturbations. Since our operators
are defined via quadratic forms, it even holds for suitable measure
perturbations.

  * •

If a manifold $M$ satisfies $\operatorname{Ric}\geq-K$, $K\geq 0$, then

$0\leq\inf\sigma(\Delta^{M})\leq\lambda^{K,n},$

where $\lambda^{K,n}$ denotes the bottom of the spectrum of the
Laplace–Beltrami operator in the model space $M_{K}^{n}$. This implies that
the bottom of the spectrum is zero if $\operatorname{Ric}\geq 0$. Moreover, if
a non-compact manifold satisfies $\operatorname{Ric}\geq-K$, $K\geq 0$, then
there is some essential spectrum in the interval $[0,K^{2}]$, [Don81]. If
$\operatorname{Ric}\geq K>0$ the manifold is compact by the classical Bonnet-
Myers theorem such that the spectrum is purely discrete with smallest
eigenvalue zero. In this case, the smallest positive eigenvalue is at least
$\frac{\pi^{2}}{D^{2}}$ if $\mathrm{diam}(M)\leq D$.

In order to prove Theorem 5.1 we adapt the strategy developed in [LSS20, SS21]
to complete Riemannian manifolds with a lower Ricci curvature bound. While
general functional analytic principles used in these articles carry over to
our setting, our geometric assumptions reveal some new abstract insights and
technical difficulities.
The main observation used in the above articles was detected in [BL11]. More
precisely, the latter article shows that the uncertainty principle can be
drawn from the fact that the bottom of the spectrum of
$\Delta+\beta\mathbf{1}_{S}$ can be raised above a given energy interval $I$
in the large coupling limit $\beta\to\infty$.
If $B\supset S$ is slightly larger than $S$ and $\Delta^{M,S}$ denotes the
Dirichlet-Laplacian on $L^{2}(M\setminus S)$, the spectral bottoms of the
large coupling limits of $\Delta+\beta\mathbf{1}_{B}$ and
$\Delta^{M,S}+\beta\mathbf{1}_{B\setminus S}$ can be expected to be comparable
for large $\beta$ . This comparability will follow from a norm bound on the
associated heat semigroups via a generalization of the hit-and-run lemma from
[MSS95, SS21] to arbitrary Riemannian manifolds: the probability that a
particle starting in $M$ which hits $S$ does not stay too long in $B$. In
contrast to the Euclidean case, we cannot just use the reflection principle as
this is not available in general. Instead, we provide a general bound on the
semigroup differences above in terms of the first exit time of the diffusion
on $M$ and the coupling constant $\beta$.
By combining the above steps, we derive an abstract uncertainty principle
depending on the first exit time of the Brownian motion, the large coupling
constant, and the infimum of the spectrum of $\Delta^{M,S}$.
It remains to estimate the spectral bottom of $\Delta^{M,S}$ and the first
exit time depending on the geometry of $S$, $B$, and $M$. Upper and lower
bounds on the bottom of the spectrum are reduced to bound the bottom of the
Laplacian in balls inside the complement of $S$ respectively star-shaped
domains in terms of the lower Ricci curvature bound. In order to bound the
first exit time we follow the strategy of [HSC01]. Keys in these proofs are
the well-established Bishop-Gromov volume comparison estimate and appropriate
heat kernel upper bounds depending on the lower Ricci curvature bound.

The plan of the remaining article is as follows. Section 2 is devoted to the
proof of the abstract uncertainty principle in terms of the first exit time.
In Section 3 we obtain a lower bound of the Dirichlet-Laplacian of relatively
dense complements in terms of the volume density. Section 4 is reserved to
obtain all the quantitative bounds in terms of the geometric assumptions. The
main theorem is then discussed and proven in Section 5.
Acknowledgement: C.R. gratefully acknowledges support by the DFG.

##  2 Norm estimates for the semigroup at large coupling and an abstract
unique continuation principle

Let $S\subset M$ be open, $\rho\geq 0$, and $S_{\rho}=\\{y\in
M\colon\mathrm{d}(y,S)<\rho\\}$ the open tubular neighbourhood of $S$ with
radius $\rho$. Furthermore, let $(\Omega,\mathcal{A},(\mathbb{P}_{x})_{x\in
M},\allowbreak(X_{t})_{t\geq 0},(\mathcal{F}_{t})_{t\geq 0})$ be the Brownian
motion associated to $\Delta$. For $\alpha_{0}>0$ we denote the occupation
time in $S_{\rho}$ in the interval $[0,\alpha_{0}]$ by
$T^{\tau}_{S,\rho}:\Omega\to[0,\alpha_{0}]$, that is,

$T^{\alpha_{0}}_{S,\rho}(\omega)=\int_{0}^{\alpha_{0}}\mathbf{1}_{S_{\rho}}\circ
X_{s}(\omega)\mathrm{d}s.$

Moreover, we denote the first hitting time of $S_{\rho}$ by
$\sigma_{S,\rho}:\Omega\to[0,\infty)$, as well as the first exit time of the
ball with radius $r>0$ by $\tau_{r}:\Omega\to[0,\infty)$, that is,

$\sigma_{S,\rho}(\omega)=\inf\\{s\geq 0\colon X_{s}(\omega)\in
S_{\rho}\\},\quad\tau_{r}{(\omega)}=\inf\\{s\geq 0\colon
X_{s}{(\omega)}\not\in B_{r}(X_{0}(\omega))\\}.$

If $\rho=0$ we set $\sigma_{S}=\sigma_{S,0}$. The following lemma is a variant
of a result obtained for the Brownian motion in $\mathbb{R}^{n}$ in [MSS95].
It has been recaptured in [SS21]. Since we do not impose any geometric
assumption on $M$ in this section, our upper bound is given in terms of the
first exit time. If the Ricci curvature is bounded below, then we obtain an
explicit upper bound, cf. Section 4.

######  Lemma 2.1.

For all $S\subset M$ open, $\alpha_{0},\rho>0$, $\alpha\in(0,\alpha_{0})$, and
$x\in M$ we have

$\mathbb{P}_{x}(\sigma_{S}\leq\alpha_{0},T_{S,\rho}^{\alpha_{0}}\leq\alpha)\leq\sup_{z\in
M}\mathbb{P}_{z}(\tau_{\rho/2}\leq\alpha).$

###### Proof of Lemma 2.1.

Fix $\alpha_{0}$, $\rho$ and $\alpha$ as in the assumptions of the theorem.
Let $m>\alpha_{0}$ be fixed and set
$\sigma_{S,\rho/2,m}=\min\\{\sigma_{S,\rho/2},m\\}$. Note that the stopping
time $\sigma_{S,\rho/2,m}$ is $\mathbb{P}_{x}$-almost surely finite for any
$x\in M$. This will be used when applying the strong Markov property below.
Assume that $\omega\in\Omega$ is such that the corresponding sample path is
continuous and

$\forall
s\in[0,\alpha]\colon\quad\mathrm{d}(X_{\sigma_{S,\rho/2,m}(\omega)}(\omega),X_{\sigma_{S,\rho/2,m}(\omega)+s}(\omega))<\rho/2.$
(2.1)

Then we have

$\omega\not\in\\{\sigma_{S}\leq\alpha_{0},T^{\alpha_{0}}_{S,\rho}\leq\alpha\\}.$
(2.2)

Indeed, if $\sigma_{S}(\omega)>\alpha_{0}$ we conclude (2.2). If
$\sigma_{S}(\omega)\leq\alpha_{0}$ and $X_{0}(\omega)\in S_{\rho/2}$, we have
$\sigma_{S,\rho/2,m}(\omega)=\sigma_{S,\rho/2}(\omega)=0$, and by assumption
(2.1) we find $X_{s}(\omega)\in B_{\rho/2}(X_{0}(\omega))\subset S_{\rho}$ for
all $s\in[0,\alpha]$. Since $\alpha<\alpha_{0}$ we find
$T^{\alpha_{0}}_{S,\rho}(\omega)>\alpha$ which implies (2.2). Assume now
$\sigma_{S}(\omega)\leq\alpha_{0}$ and $X_{0}(\omega)\not\in S_{\rho/2}$.
Since $m>\alpha_{0}$ and $\sigma_{S,\rho/2}(\omega)<\sigma_{S}(\omega)$ we
have $\sigma_{S,\rho/2,m}(\omega)=\sigma_{S,\rho/2}(\omega)$. Thus, assumption
(2.1) implies $X_{t}(\omega)\in
B_{\rho/2}(X_{\sigma_{S,\rho/2}(\omega)}(\omega))\subset S_{\rho}$ for all
$t\in[\sigma_{S,\rho/2}(\omega),\sigma_{S,\rho/2}(\omega)+\alpha]$. Since
$B_{\rho/2}(X_{\sigma_{S,\rho/2}(\omega)}(\omega))\cap S=\emptyset$, this
implies $\sigma_{S,\rho/2}(\omega)+\alpha<\sigma_{S}(\omega)\leq\alpha_{0}$.
Thus we have $T^{\alpha_{0}}_{S,\rho}>\alpha$ which implies (2.2).

Fix now $x\in M$. Since the sample paths are almost surely continuous, the
contraposition of the implication (2.1) $\Rightarrow$ (2.2) implies

$\mathbb{P}_{x}(\\{\sigma_{S}\leq\alpha_{0},T^{\alpha_{0}}_{S,\rho}\leq\alpha\\})\leq\mathbb{P}_{x}(E),$
(2.3)

where

$E=\\{\omega\in\Omega\colon
d(X_{\sigma_{S,\rho/2,m}(\omega)}(\omega),X_{\sigma_{S,\rho/2,m}(\omega)+s}(\omega))\geq\rho/2\
\text{for some}\ s\in[0,\alpha]\\}.$

By (2.3) and the definition of the conditional expectation we have

$\mathbb{P}_{x}(\sigma_{S}\leq\alpha_{0},T^{\alpha_{0}}_{S,\rho}\leq\alpha)\leq\mathbb{E}_{x}(\mathbf{1}_{E})=\mathbb{E}_{x}(\mathbb{E}_{x}(\mathbf{1}_{E}\mid\mathcal{F}_{\sigma_{S,\rho/2,m}})),$
(2.4)

where $\mathcal{F}_{\sigma_{S,\rho/2,m}}$ denotes the $\sigma$-algebra of
$\sigma_{S,\rho/2,m}$-past, that is,

$\mathcal{F}_{\sigma_{S,\rho/2,m}}=\\{A\in\mathcal{A}\colon
A\cap\\{\sigma_{S,\rho/2,m}\leq t\\}\in\mathcal{F}_{t}\ \text{for any}\ t\geq
0\\}.$

Let $E_{0}=\\{\omega\in\Omega\colon d(X_{0}(\omega),X_{s}(\omega))\geq\rho/2\
\text{for some}\ s\in[0,\alpha]\\}$. Then, by (2.4), the fact that the
stopping time $\sigma_{S,\rho/2,m}$ is $\mathbb{P}_{x}$-almost surely finite,
and the strong Markov property, we find

$\displaystyle\mathbb{P}_{x}(\sigma_{S}\leq\alpha_{0},T^{\alpha_{0}}_{S,\rho}\leq\alpha)$
$\displaystyle\leq\mathbb{E}_{x}(\mathbb{E}_{x}(\mathbf{1}_{E}\mid\mathcal{F}_{\sigma_{S,\rho/2,m}}))$
$\displaystyle=\mathbb{E}_{x}\bigl{(}\mathbb{E}_{X_{\sigma_{S,\rho/2,m}}}(\mathbf{1}_{E_{0}})\bigr{)}\leq\sup_{z\in
M}\mathbb{P}_{z}(\tau_{\rho/2}\leq\alpha).\qed$

For a non-empty and open set $G\subset M$, and $S\subset G$ open, we define
the quadratic form $\mathcal{E}^{G,S}$ on $L^{2}(G\setminus\overline{S})$ by

$\displaystyle\mathcal{D}(\mathcal{E}^{G,S})$ $\displaystyle=\overline{\\{u\in
L^{2}(G\setminus\overline{S})\colon\hat{u}\in C^{1}(G),\ u\in
C_{\mathrm{c}}(\overline{G}\setminus\overline{S})\\}}^{W^{1,2}(G\setminus\overline{S})},$
$\displaystyle\mathcal{E}^{G,S}(f)$
$\displaystyle=\int_{G\setminus\overline{S}}|\nabla
f|^{2}\operatorname{\mathrm{dvol}}.$

Here, $\hat{u}\in L^{2}(G)$ is the extension of $u$ by zero, i.e. $\hat{u}=u$
on $G\setminus\overline{S}$ and $\hat{u}=0$ on $\overline{S}$, and $u\in
C_{\mathrm{c}}(\overline{G}\setminus\overline{S})$ means that ther is
$\overline{u}\in C_{\mathrm{c}}(\overline{G}\setminus\overline{S})$ such that
$\overline{u}=u$ on $G\setminus\overline{S}$. The form $\mathcal{E}^{G,S}$ is
densely defined, symmetric and closed, and we denote the unique self-adjoint
operator in $L^{2}(G\setminus\overline{S})$ associated with the form
$\mathcal{E}^{G,S}$ by $\Delta^{G,S}$ and its domain by
$\mathcal{D}(\Delta^{G,S})$. Note that this way we describe Dirichlet boundary
conditions on $\partial S$ and Neumann boundary conditions at the boundary of
$G$. If $G=M$, the operator $\Delta^{G,S}$ is merely the Dirichlet Laplacian
on $L^{2}(M\setminus\overline{S})$.

In order to formulate the following lemma, we denote by
$\mathbf{1}_{S_{\rho}\setminus\overline{S}}\colon
L^{2}(M\setminus\overline{S})\to L^{2}(M\setminus\overline{S})$ and
$\mathbf{1}_{S_{\rho}}\colon L^{2}(M)\to L^{2}(M)$ the multiplication operator
by the characteristic function of $S_{\rho}\setminus\overline{S}$ and
$S_{\rho}$, respectively. Moreover, if we are concerned with bounded operators
on $L^{2}(M\setminus\overline{S})$, we interpret them as operators on
$L^{2}(M)$ by setting them to zero on $L^{2}(S)$.

######  Proposition 2.2.

Let $S\subset M$ be open, $\alpha_{0},\rho>0$, $\alpha\in(0,\alpha_{0})$, and
$\beta\geq 0$. Then we have

$\displaystyle\Bigl{\|}\mathrm{e}^{-\alpha_{0}(\frac{1}{2}\Delta+\beta\mathbf{1}_{S_{\rho}})}-\mathrm{e}^{-\alpha_{0}(\frac{1}{2}\Delta^{M,S}+\beta\mathbf{1}_{S_{\rho}\setminus\overline{S}})}\Bigr{\|}^{2}\leq\mathrm{e}^{-2\beta\alpha}+\sup_{z\in
M}\mathbb{P}_{z}(\tau_{\rho/2}\leq\alpha/2).$ (2.5)

###### Proof.

We follow the proof of [SS21, Proposition 3.2]. Recall the Feynman-Kac formula
for the process $(X_{t})_{t\geq 0}$ associated to $\Delta$ and the potential
$\mathbf{1}_{S_{\rho}}$: for all $t,\beta\geq 0$, $f\in L^{2}(M)$, and all
$x\in M$ we have

$\mathrm{e}^{-t(\frac{1}{2}\Delta+\beta\mathbf{1}_{S_{\rho}})}f(x)=\mathbb{E}_{x}\left(f\circ
X_{t/2}\exp\left(-2\beta\int_{0}^{t/2}\mathbf{1}_{S_{\rho}}\circ
X_{s}\mathrm{d}s\right)\right),$

as well as

$\mathrm{e}^{-t(\frac{1}{2}\Delta^{M,S}+\beta\mathbf{1}_{S_{\rho}\setminus\overline{S}})}f(x)=\mathbb{E}_{x}\left(f\circ
X_{t/2}\exp\left(-2\beta\int_{0}^{t/2}\mathbf{1}_{S_{\rho}}\circ
X_{s}\mathrm{d}s\right)\mathbf{1}_{\\{\sigma_{S}>t/2\\}}\right).$

Specifying $t=\alpha_{0}$ implies

$\displaystyle\bigl{|}\mathrm{e}^{-\alpha_{0}(\frac{1}{2}\Delta+\beta\mathbf{1}_{S_{\rho}})}f(x)-$
$\displaystyle\mathrm{e}^{-\alpha_{0}(\frac{1}{2}\Delta^{M,S}+\beta\mathbf{1}_{S_{\rho}\setminus\overline{S}})}f(x)\bigr{|}=\bigl{|}\mathbb{E}_{x}\bigl{(}f\circ
X_{\alpha_{0}/2}\mathrm{e}^{-2\beta
T^{\alpha_{0}/2}_{S,\rho}}\mathbf{1}_{\\{\sigma_{S}\leq\alpha_{0}/2\\}}\bigr{)}\bigr{|}.$

The Cauchy-Schwarz inequality yields

$\Bigl{|}\Bigl{(}\mathrm{e}^{-\alpha_{0}(\frac{1}{2}\Delta+\beta\mathbf{1}_{S_{\rho}})}-\mathrm{e}^{-\alpha_{0}(\frac{1}{2}\Delta^{M,S}+\beta\mathbf{1}_{S_{\rho}\setminus\overline{S}})}\Bigr{)}f(x)\Bigr{|}^{2}\\\
\leq\mathbb{E}_{x}\Bigl{(}|f|^{2}\circ
X_{\alpha_{0}/2}\Bigr{)}\mathbb{E}_{x}\Bigl{(}\mathrm{e}^{-4\beta
T^{\alpha_{0}/2}_{S,\rho}}\mathbf{1}_{\\{\sigma_{S}\leq\alpha_{0}/2\\}}\Bigr{)}.$

Since $\mathbb{E}_{x}(|f|^{2}\circ
X_{\alpha_{0}/2})=\mathrm{e}^{-\alpha_{0}\Delta/4}|f|^{2}(x)$ and
$\|\mathrm{e}^{-t\Delta}\|\leq 1$ for $t\geq 0$ we conclude

$\Bigl{\|}\mathrm{e}^{-\alpha_{0}(\frac{1}{2}\Delta+\beta\mathbf{1}_{S_{\rho}})}-\mathrm{e}^{-\alpha_{0}(\frac{1}{2}\Delta^{M,S}+\beta\mathbf{1}_{S_{\rho}\setminus\overline{S}})}\Bigr{\|}\leq\Bigl{(}\sup_{x\in
M}\mathbb{E}_{x}\bigl{(}\mathrm{e}^{-4\beta
T^{\alpha_{0}/2}_{S,\rho}}\mathbf{1}_{\\{\sigma_{S}\leq\alpha_{0}/2\\}}\bigr{)}\Bigr{)}^{1/2}.$

Since

$\mathrm{e}^{-4\beta
T^{\alpha_{0}/2}_{S,\rho}}\mathbf{1}_{\\{\sigma_{S}\leq\alpha_{0}/2\\}}\leq\mathrm{e}^{-2\beta\alpha}+\mathbf{1}_{\\{\sigma_{S}\leq\alpha_{0}/2\\}\cap\\{T^{\alpha_{0}/2}_{S,\rho}\leq\alpha/2\\}},$

we conclude from Lemma 2.1

$\displaystyle\Bigl{\|}\mathrm{e}^{-\alpha_{0}(\frac{1}{2}\Delta+\beta\mathbf{1}_{S_{\rho}})}-\mathrm{e}^{-\alpha_{0}(\frac{1}{2}\Delta^{M,S}+\beta\mathbf{1}_{S_{\rho}\setminus\overline{S}})}\Bigr{\|}^{2}$
$\displaystyle\leq\mathrm{e}^{-2\beta\alpha}+\sup_{x\in
M}\mathbb{P}_{x}(\sigma_{S}\leq\alpha_{0}/2,T_{S,\rho}^{\alpha_{0}/2}\leq\alpha/2)$
$\displaystyle\leq\mathrm{e}^{-2\beta\alpha}+\sup_{z\in
M}\mathbb{P}_{z}(\tau_{\rho/2}\leq\alpha/2).\qed$

The following Lemma is a generalization of the spectral theoretic uncertainty
principle developed in [BL11] (in the case $V=0$), and refined in [Kle13,
TV20]. Let us stress that our result can be applied to operators defined via
form sums where the negative part is relatively bounded with respect to
$\Delta$.

######  Lemma 2.3.

Let $X$ be a complex Hilbert space, $h_{1}$, $h_{2}^{+}$, $h_{2}^{-}$, $h_{3}$
lower bounded quadratic forms on $X$, $h_{2}^{+},h_{2}^{-}\geq 0$,
$h_{2}:=h_{2}^{+}-h_{2}^{-}$, $a^{\prime}\in(0,1)$, $b^{\prime}\geq 0$,

$h_{2}^{-}(x)\leq a^{\prime}(h_{1}+h_{2}^{+})(x)+b^{\prime}\|x\|^{2},\quad
x\in D(h_{1})\cap D(h_{2}),$

$E_{0}\in\mathbb{R}$, $\beta>0$,

$Y:=\left\\{x\in D(h_{1})\cap D(h_{2})\cap D(h_{3})\colon
h_{1}(x)+h_{2}(x)\leq E_{0}\|x\|^{2}\right\\},$

and

$\gamma(\beta):=\inf\left\\{\frac{(h_{1}+\beta h_{3})(x)}{\|x\|^{2}}\colon
x\in D(h_{1})\cap D(h_{3})\setminus\\{0\\}\right\\}.$

Then we have

$h_{3}(x)\geq\frac{1}{\beta}\left((1-a^{\prime})\gamma\left(\frac{\beta}{1-a^{\prime}}\right)-E_{0}-b^{\prime}\right)\|x\|^{2},\quad
x\in Y.$

######  Remark.

The forms $h_{2}^{+}$ and $h_{2}^{-}$ do not necessarily have to be the
positive and negative part of $h_{2}$. It suffices that they are lower bounded
and that $h_{2}^{-}$ is relatively bounded with respect to $h_{1}+h_{2}^{+}$.

###### Proof.

We follow and modify the proof of [TV20, Lemma 3.5]. By the definition of
$\gamma(\beta)$, we have $(h_{1}+\beta h_{3})(x)\geq\gamma(\beta)\|x\|^{2}$
for all $x\in D(h_{1})\cap D(h_{3})$. Hence, for all $x\in Y$

$\displaystyle\beta h_{3}(x)$ $\displaystyle\geq\beta
h_{3}(x)-E_{0}\|x\|^{2}+h_{1}(x)+h_{2}(x)$ $\displaystyle=\beta
h_{3}(x)-E_{0}\|x\|^{2}+h_{1}(x)+h_{2}^{+}(x)-h_{2}^{-}(x)$
$\displaystyle\geq\beta
h_{3}(x)-E_{0}\|x\|^{2}+h_{1}(x)+h_{2}^{+}(x)-a^{\prime}h_{1}(x)-h_{2}^{+}(x)-b^{\prime}\|x\|^{2}$
$\displaystyle=(1-a^{\prime})\left(h_{1}(x)+\frac{\beta}{1-a^{\prime}}h_{3}(x)\right)-(E_{0}+b^{\prime})\|x\|^{2}$
$\displaystyle\geq\bigl{(}(1-a^{\prime})\gamma(\beta/(1-a^{\prime}))-E_{0}-b^{\prime}\bigr{)}\|x\|^{2}.$

Dividing by $\beta$ yields the claim. ∎

######  Theorem 2.4.

Let $\rho,\alpha_{0},\beta>0$, $\alpha\in(0,\alpha_{0})$,
$E_{0}\in\mathbb{R}$, and $I\subset(-\infty,E_{0}]$, $S\subset M$ be open,
$\mu_{0}:=\inf\sigma(\Delta^{M,S})$, and
$\lambda_{M,S_{\rho}}:=\inf\sigma(\Delta^{M,S_{\rho}})$. Then we have

$\chi_{I}(H)\mathbf{1}_{S_{\rho}}\chi_{I}(H)\geq\kappa\chi_{I}(H),$

where

$\kappa=\frac{1}{\beta}\left[(1-a)\left(\mu_{0}-\frac{2}{\alpha_{0}}\mathrm{e}^{\alpha_{0}\lambda_{M,S_{\rho}}/2}\left(\mathrm{e}^{-\frac{\beta\alpha}{2(1-a)}}+\sqrt{\sup_{z\in
M}\mathbb{P}_{z}(\tau_{\rho/2}\leq\alpha/2)}\right)\right)-E_{0}-b\right],$

and $a\in[0,1)$ and $b\geq 0$ are as in (1.2).

###### Proof.

We apply Lemma 2.3 to the quadratic forms $h_{1}=\mathcal{E}$,
$h_{2}^{+}=\mathcal{V}_{+}$, $h_{2}^{-}=\mathcal{V}_{-}$, and $h_{3}\colon
L^{2}(M)\to\mathbb{R}$, $u\mapsto\langle\mathbf{1}_{S_{\rho}}u,u\rangle$, and
obtain

$\langle\mathbf{1}_{S_{\rho}}u,u\rangle\geq\frac{1}{\beta}\left((1-a)\lambda_{\beta_{a}}-E_{0}-b\right)\langle
u,u\rangle,\quad u\in Y,$

where

$\displaystyle Y$ $\displaystyle=\left\\{u\in D(\mathcal{E})\cap
D(\mathcal{V}_{+})\cap D(\mathcal{V}_{-})\cap
D(h_{3})\colon(\mathcal{E}+\mathcal{V}_{+}-\mathcal{V}_{-})(u)\leq
E_{0}\langle u,u\rangle\right\\}$ $\displaystyle=\left\\{u\in
D(\mathcal{E})\cap D(\mathcal{V}_{+})\colon\mathcal{H}(u)\leq E_{0}\langle
u,u\rangle\right\\}.$

and

$\lambda_{t}:=\inf\sigma(\Delta+t\mathbf{1}_{S_{\rho}}),\quad t\geq
0,\quad\beta_{a}:=\frac{\beta}{1-a}.$

If $u\in\operatorname{Ran}(\chi_{I}(H))$ for some $I\subset(-\infty,E_{0}]$,
then $u\in D(H)$ and $\mathcal{H}(u)\leq E_{0}\langle u,u\rangle$. Then we
have $u\in Y$. Thus, the statement follows if

$((1-a)\lambda_{\beta_{a}}-E_{0}-b)/\beta\geq\kappa.$ (2.6)

In order to show this inequality, we set

$\displaystyle\mu_{t}:=\inf\sigma(\Delta^{M,S}+t\mathbf{1}_{S_{\rho}\setminus\overline{S}}),\quad
t\geq 0.$

By the choice of our boundary conditions we have
$\lambda_{t}\leq\mu_{t}\leq\lambda_{M,S_{\rho}}$ for all $t\geq 0$. Fix $t\geq
0$ to be chosen later. Since we have

$\left\lVert\mathrm{e}^{-\alpha_{0}(\Delta/2+t\mathbf{1}_{S_{\rho}})}\right\rVert=\mathrm{e}^{-(\alpha_{0}/2)\lambda_{2t}},\quad\text{and}\quad\left\lVert\mathrm{e}^{-\alpha_{0}(\Delta^{M,S}/2+t\mathbf{1}_{S_{\rho}\setminus\overline{S}})}\right\rVert=\mathrm{e}^{-(\alpha_{0}/2)\mu_{2t}},$

the reverse triangle inequality, $\sqrt{a+b}\leq\sqrt{a}+\sqrt{b}$ for
$a,b\geq 0$, and Proposition 2.2 imply for $\alpha\in(0,\alpha_{0})$

$\mathrm{e}^{-(\alpha_{0}/2)\lambda_{2t}}-\mathrm{e}^{-(\alpha_{0}/2)\mu_{2t}}\leq\mathrm{e}^{-t\alpha}+\sqrt{\sup_{z\in
M}\mathbb{P}_{z}(\tau_{\rho/2}\leq\alpha/2)}.$ (2.7)

By the mean value theorem there exists
$\xi\in[\alpha_{0}\lambda_{2t}/2,\alpha_{0}\mu_{2t}/2]$ such that

$\frac{\alpha_{0}}{2}\mu_{2t}-\frac{\alpha_{0}}{2}\lambda_{2t}=\mathrm{e}^{\xi}(\mathrm{e}^{-\alpha_{0}\lambda_{2t}/2}-\mathrm{e}^{-\alpha_{0}\mu_{2t}/2})\leq\mathrm{e}^{\alpha_{0}\lambda_{M,S_{\rho}}/2}(\mathrm{e}^{-\alpha_{0}\lambda_{2t}/2}-\mathrm{e}^{-\alpha_{0}\mu_{2t}/2}).$
(2.8)

From (2.7), (2.8), $\mu_{2t}\geq\mu_{0}$, and $a\in[0,1)$ we conclude for all
$\alpha\in(0,\alpha_{0})$

$(1-a)\lambda_{2t}\geq(1-a)\left(\mu_{0}-\frac{2}{\alpha_{0}}\mathrm{e}^{\alpha_{0}\lambda_{M,S_{\rho}}/2}\left(\mathrm{e}^{-t\alpha}+\sqrt{\sup_{z\in
M}\mathbb{P}_{z}(\tau_{\rho/2}\leq\alpha/2)}\right)\right).$

We choose $t=\beta_{a}/2$ and conclude (2.6). ∎

Theorem 2.4 yields a non-trivial result only if $\kappa>0$. In the following
sections we exploit certain geometric properties of $M$ and $S\subset M$
yielding the positivity of $\kappa$.

##  3 Spectral estimates for Laplacians on domains with relatively dense
complement

A set $G\subset M$ is called _star-shaped with respect to $x\in G$_ if for all
$y\in G$ distance minimizing geodesics from $x$ to $y$ are contained in $G$.
For $x\in M$ let
$\omega=\omega_{x}\colon(0,\infty)\times\mathbb{S}_{x}^{n-1}\to[0,\infty)$ be
the density of the volume form with respect to the product measure on
$(0,\infty)\times\mathbb{S}_{x}^{n-1}$, i.e.,

$\operatorname{\mathrm{dvol}}=\omega(r,\theta)\mathrm{d}r\mathrm{d}\theta=\omega_{x}(r,\theta)\mathrm{d}r\mathrm{d}\theta,$

cf. [GHL87]. Note that $\omega$ is almost everywhere positive and finite and
smooth up to the cut locus. We introduce the function

$\Lambda\colon
M\times[0,\infty)\times[0,\infty)\to[0,\infty),\quad(x,\rho,R)\mapsto\left(\operatorname*{ess\,sup}_{\theta\in\mathbb{S}_{x}^{n-1}}\int_{\rho}^{R}\int_{\rho}^{r}\frac{\omega(r,\theta)}{\omega(s,\theta)}\
\mathrm{d}s\ \mathrm{d}r\right)^{-1}.$

Then we have the following result.

######  Theorem 3.1.

Let $x\in M$, $0<\rho<R<\infty$, $G\subset M$ open and star-shaped with
respect to $x$, $B_{x}(\rho)\subset G\subset B_{x}(R)$. Then we have

$\lambda_{G,B_{x}(\rho)}:=\inf\sigma(\Delta^{G,B_{x}(\rho)})\geq\Lambda(x,\rho,R).$

###### Proof of Theorem 3.1.

Let $B=B_{x}(\rho)$. By the variational principle we have

$\lambda_{G,B}=\inf\left\\{\frac{\int_{G\setminus\overline{B}}|\nabla
f|^{2}\operatorname{\mathrm{dvol}}}{\int_{G\setminus\overline{B}}\lvert
f\rvert^{2}\operatorname{\mathrm{dvol}}}\colon f\in\mathcal{F}\right\\},$

where

$\mathcal{F}=\\{f\colon G\setminus\overline{B}\to\mathbb{C}\colon
f\in\mathcal{L}^{2}(G\setminus\overline{B}),\ \hat{f}\in C^{1}(G),\ f\in
C_{\mathrm{c}}(\overline{G}\setminus\overline{B})\\}.$

Here, $\hat{f}\in L^{2}(G)$ is the extension of $f$ by zero, and $f\in
C_{\mathrm{c}}(\overline{G}\setminus\overline{B})$ means that there is
$\overline{f}\in C_{\mathrm{c}}(\overline{G}\setminus\overline{B})$ such that
$\overline{f}=f$ on $G\setminus\overline{B}$. Thus, it suffices to prove an
($f$-independent) lower bound on $\lVert\nabla
f\rVert^{2}_{L^{2}(G\setminus\overline{B})}/\lVert
f\rVert^{2}_{L^{2}(G\setminus\overline{B})}$ for all $f\in\mathcal{F}$. We fix
$f\in\mathcal{F}$.
For $\theta\in\mathbb{S}_{x}^{n-1}$ we denote by
$\gamma_{x}^{\theta}\colon[0,\infty)\to M$ the geodesic with
$\gamma_{x}^{\theta}(0)=x$, $(\gamma_{x}^{\theta})^{\prime}(0)=\theta$, and
$L_{\theta}\geq 0$ be the distance to the cut-locus of $x$ in direction
$\theta$. Moreover, we let $R_{\theta}=\sup\\{r\in(0,R]\colon\forall
s\in[0,r]\colon\gamma_{x}^{\theta}(s)\in G\\}$. Since $G$ is star-shaped with
respect to $x$ we have $\gamma_{x}^{\theta}(r)\in G$ for all
$r\in[0,R_{\theta})$. By the Cauchy-Schwarz inequality we obtain for all
$\theta\in S_{x}^{n-1}$ and all $r\in[\rho,R_{\theta}\wedge L_{\theta})$

$\displaystyle|(f\circ\gamma_{x}^{\theta})(r)\rvert^{2}=\left\lvert\int_{\rho}^{r}\partial_{s}(f\circ\gamma_{x}^{\theta})(s)\
\mathrm{d}s\right\rvert^{2}$
$\displaystyle=\left\lvert\int_{\rho}^{r}\partial_{s}(f\circ\gamma_{x}^{\theta})(s)\omega(s,\theta)^{\frac{1}{2}}\omega(s,\theta)^{-\frac{1}{2}}\
\mathrm{d}s\right\rvert^{2}$
$\displaystyle\leq\int_{\rho}^{r}|\partial_{s}(f\circ\gamma_{x}^{\theta})|^{2}(s)\omega(s,\theta)\
\mathrm{d}s\int_{\rho}^{r}\omega(s,\theta)^{-1}\ \mathrm{d}s.$

By the chain rule, Cauchy-Schwarz, and since $\gamma_{x}^{\theta}$ is
parametrized by arc length, we have for all $s\in[0,R_{\theta})$

$|\partial_{s}(f\circ\gamma_{x}^{\theta})(s)|^{2}=|\langle\nabla
f(\gamma_{x}^{\theta}(s)),\dot{\gamma}_{x}^{\theta}(s)\rangle|^{2}\leq|\nabla
f(\gamma_{x}^{\theta}(s))|^{2}\lvert\dot{\gamma}_{x}^{\theta}(s)\rvert^{2}=|\nabla
f(\gamma_{x}^{\theta}(s))|^{2}.$

Integration along distance spheres gives

$\displaystyle\|f\|_{L^{2}(G\setminus\overline{B})}^{2}$
$\displaystyle=\int_{\mathbb{S}_{x}^{n-1}}\int_{\rho}^{L_{\theta}\wedge
R_{\theta}}|f\circ\gamma_{x}^{\theta}(r)|^{2}\omega(r,\theta)\ \mathrm{d}r\
\mathrm{d}\theta$
$\displaystyle\leq\int_{\mathbb{S}_{x}^{n-1}}\int_{\rho}^{L_{\theta}\wedge
R_{\theta}}\omega(r,\theta)\int_{\rho}^{r}|\partial_{s}(f\circ\gamma_{x}^{\theta})(s)|^{2}\omega(s,\theta)\
\mathrm{d}s\int_{\rho}^{r}\omega(s,\theta)^{-1}\ \mathrm{d}s\ \mathrm{d}r\
\mathrm{d}\theta$
$\displaystyle\leq\int_{\mathbb{S}_{x}^{n-1}}\int_{\rho}^{L_{\theta}\wedge
R_{\theta}}\omega(r,\theta)\int_{\rho}^{\rho(\theta)\wedge R_{\theta}}|\nabla
f(\gamma_{x}^{\theta}(s))|^{2}\omega(s,\theta)\
\mathrm{d}s\int_{\rho}^{r}\omega(s,\theta)^{-1}\ \mathrm{d}s\ \mathrm{d}r\
\mathrm{d}\theta$
$\displaystyle=\int_{\mathbb{S}_{x}^{n-1}}\int_{\rho}^{L_{\theta}\wedge
R_{\theta}}\int_{\rho}^{r}\frac{\omega(r,\theta)}{\omega(s,\theta)}\
\mathrm{d}s\ \mathrm{d}r\int_{\rho}^{L_{\theta}\wedge R_{\theta}}|\nabla
f(\gamma_{x}^{\theta}(s))|^{2}\omega(s,\theta)\ \mathrm{d}s\ \mathrm{d}\theta$
$\displaystyle\leq\operatorname*{ess\,sup}_{\theta\in\mathbb{S}_{x}^{n-1}}\int_{\rho}^{L_{\theta}\wedge
R_{\theta}}\int_{\rho}^{r}\frac{\omega(r,\theta)}{\omega(s,\theta)}\
\mathrm{d}s\ \mathrm{d}r\ \|\nabla f\|_{L^{2}(G\setminus\overline{B})}^{2}.$

Since $L_{\theta}\wedge R_{\theta}\leq R$ this yields the claim. ∎

######  Proposition 3.2.

Assume $0<\rho<R<\infty$ and let $S\subset M$ be $(R,\rho)$-relatively dense.
Then, there is $\Sigma\subset S$ with the following properties:

  1. (a)

$B_{\rho}(\Sigma):=\bigcup_{p\in\Sigma}B_{\rho}(p)$ is $(3R,\rho)$-relatively
dense and $B_{\rho}(\Sigma)\subset S$.

  2. (b)

$\bigcup_{p\in\Sigma}B_{3R}(p)\supset M$.

  3. (c)

If $p\in\Sigma$ and $\Sigma\setminus\\{p\\}\neq\emptyset$, then

$R\leq\mathrm{d}(p,\Sigma\setminus\\{p\\})\leq 6R.$

In particular, $\Sigma$ can be chosen uniformly discrete and
$B_{\rho}(\Sigma\setminus\\{p\\})$ is $(6R,\rho)$-relatively dense in $M$.

For a convex subset $M\subset\mathbb{R}^{d}$, Proposition 3.2 is proven in
[SS21]. The proof literally applies to our setting.
In the following we give a lower bound of the bottom of the Laplacian in the
complement of $S$ in terms of the function $\Lambda$. The idea is taken from
[SS21] with the slight difference that Voronoï cells are in general not
convex. This property however can be replaced by star-shapedness.

######  Theorem 3.3.

Let $0<\rho<R<\infty$ and $S\subset M$ be a proper $(R,\rho)$-relatively dense
subset. Then we have

$\lambda^{M,S}:=\inf\sigma(\Delta^{M,S})\geq\inf_{x\in
M}\Lambda(x,\rho/2,3R).$

###### Proof.

Let $\Sigma\subset S$ be as in Proposition 3.2. We denote the Voronoï cell
associated with $p\in\Sigma$ by

$G_{p}:=\\{x\in M\colon\forall
q\in\Sigma\setminus\\{p\\}\colon\mathrm{d}(x,p)<\mathrm{d}(x,q)\\}.$

Then, by construction

  1. (a)

$\overline{\bigcup_{p\in\Sigma}G_{p}}=M$,

  2. (b)

${G_{p}}\cap G_{q}=\emptyset$ for $p,q\in\Sigma$ with $p\neq q$,

  3. (c)

for all $p\in\Sigma$ we have $B_{\rho/2}(p)\subset G_{p}\subset B_{3R}(p)$,
and

  4. (d)

for all $p\in\Sigma$ the Voronoï cell $G_{p}$ is star-shaped with respect to
$p$.

Note that $G_{p}$ is in general not convex. In order to see the star-
shapedness, let $x\in G_{p}$ and $\gamma\colon[0,1]\to M$ be a distance
minimizing geodesic from $p$ to $x$. Then for all $t\in[0,1]$ we have

$\mathrm{d}(\gamma(t),p)=\mathrm{d}(x,p)-\mathrm{d}(\gamma(t),x)$

and consequently for all $q\in\Sigma$ with $q\not=p$

$\mathrm{d}(\gamma(t),p)=\mathrm{d}(x,p)-\mathrm{d}(\gamma(t),x)<\mathrm{d}(x,q)-\mathrm{d}(\gamma(t),x)\leq\mathrm{d}(q,\gamma(t)).$

Hence $\gamma(t)\in G_{p}$ for all $t\in[0,1]$.

Let

$\mathcal{F}=\\{f\colon
M\setminus\overline{B_{\rho/2}(\Sigma)}\to\mathbb{C}\colon
f\in\mathcal{L}^{2}(M\setminus\overline{B_{\rho/2}(\Sigma)}),\ \hat{f}\in
C^{1}(M),\ f\in C_{\mathrm{c}}(M\setminus\overline{B_{\rho/2}(\Sigma)})\\},$

where as before $\hat{f}\in L^{2}(M)$ is the extension of $f$ by zero. From
Proposition 3.2 and the above properties we conclude for all $f\in\mathcal{F}$

$\|f\|_{L^{2}(M\setminus\overline{B_{\rho/2}(\Sigma)})}^{2}=\sum_{p\in\Sigma}\|f\mathbf{1}_{G_{p}\setminus\overline{B_{\rho/2}(p)}}\|_{L^{2}(M\setminus\overline{B_{\rho/2}(\Sigma)})}^{2}=\sum_{p\in\Sigma}\|f|_{G_{p}\setminus\overline{B_{\rho/2}(p)}}\|_{L^{2}(G_{p}\setminus\overline{B_{\rho/2}(p)})}^{2}.$

Applying Proposition 3.1 to any $f|_{G_{p}\setminus\overline{B_{\rho/2}(p)}}$,
$p\in\Sigma$, and summing up yields

$\displaystyle\|f\|_{L^{2}(M\setminus\overline{B_{\rho/2}(\Sigma)})}^{2}$
$\displaystyle\leq\sup_{x\in
M}\Lambda(x,\rho/2,3R)^{-1}\sum_{p\in\Sigma}\|\nabla
f|_{G_{p}\setminus\overline{B_{\rho/2}(p)}}\|^{2}_{L^{2}(M\setminus\overline{B_{\rho/2}(p)})}$
$\displaystyle=\sup_{x\in M}\Lambda(x,\rho/2,3R)^{-1}\|\nabla
f\|^{2}_{L^{2}(M\setminus\overline{B_{\rho/2}(\Sigma)})}.\qed$

##  4 Consequences of Ricci curvature lower bounds and the main result

We assume from now on that the Ricci curvature $\operatorname{Ric}$ is bounded
below by $K\in\mathbb{R}$. Recall that $M_{K}=M_{K}^{n}$ denotes the model
space of dimension $n\in\mathbb{N}$ and constant curvature $K$, and
$\operatorname{Vol}_{K}$ its volume form. It is well-known that the volume
density of $M_{K}$ is given by $\omega_{K}=\mathrm{sn}_{K}^{n-1}$, cf. [GHL87,
p. 138], where

$\displaystyle\mathrm{sn}_{K}(r)=\begin{cases}\frac{1}{\sqrt{K}}\sin(\sqrt{K}r)&\text{if}\
K>0,\\\ \frac{1}{\sqrt{-K}}\sinh(\sqrt{-K}r)&\text{if}\ K<0,\\\ r&\text{if}\
K=0.\end{cases}$

This explicit representation yields a lower bound for the spectrum of the
Dirichlet Laplacian in the complement of the relatively dense set. In order to
obtain this, we will use the following estimate for the lifted spectrum of
star-shaped domains. Although the proof is rather elementary, we give a
complete proof for convenience of the reader.

######  Lemma 4.1.

Let $\rho$, $R$, $x$, and $G$ as in Theorem 3.1. There exists a constant
$C_{n}>0$ such that

$\Lambda(x,\rho,R)\geq
C_{n}\frac{(\mathrm{sn}_{K}(\rho)\wedge\mathrm{sn}_{K}(R))^{n-2}}{\operatorname{Vol}_{K}(R)}.$

######  Remark.

Recall that $\mathrm{sn}_{K}$ is non-decreasing for $K\leq 0$, while for $K>0$
it is non-decreasing on $[0,\pi/(2\sqrt{K})]$ and non-increasing on
$[\pi/(2\sqrt{K}),\pi/\sqrt{K}]$.

###### Proof of Lemma 4.1.

The Bishop-Gromov comparison theorem [CGT82] implies that for almost all
$\theta\in S_{x}^{n-1}$ and almost all $s,t\in(0,\infty)$ with $s\leq r$ we
have

$\frac{\omega(r,\theta)}{\omega(s,\theta)}\leq\frac{\omega_{K}(r)}{\omega_{K}(s)}.$

Hence,

$\Lambda(x,\rho,R)^{-1}=\operatorname*{ess\,sup}_{\theta\in\mathbb{S}_{x}^{n-1}}\int_{\rho}^{R}\int_{\rho}^{r}\frac{\omega(r,\theta)}{\omega(s,\theta)}\mathrm{d}s\
\mathrm{d}r\\\
\leq\int_{\rho}^{R}\omega_{K}(r)\int_{\rho}^{r}\frac{\mathrm{d}s}{\omega_{K}(s)}\
\mathrm{d}r\leq\frac{\operatorname{Vol}_{K}(R)}{\omega_{n-1}}\int_{\rho}^{R}\frac{\mathrm{d}s}{\omega_{K}(s)}.$

We are left with estimating the integral on the right-hand side. Since
$\omega_{K}(r)=\mathrm{sn}_{K}^{n-1}(r)$, we distinguish the cases of $K$
being zero, negative, or positive.

$1^{\mathrm{st}}$ case: $K=0$. In this case, we have

$\int_{\rho}^{R}\frac{\mathrm{d}s}{\omega_{K}(s)}=\int_{\rho}^{R}s^{-n+1}\mathrm{d}s=\frac{1}{n-2}\left(\frac{1}{\rho^{n-2}}-\frac{1}{R^{n-2}}\right)\leq\frac{1}{n-2}\frac{1}{\rho^{n-2}}=\frac{1}{n-2}\frac{1}{\mathrm{sn}_{0}(\rho)^{n-2}}.$

$2^{\mathrm{nd}}$ case: $K<0$. Substituting $t=\mathrm{sn}_{K}(s)$, we obtain
using the first case

$\int_{\rho}^{R}\frac{\mathrm{d}s}{\omega_{K}(s)}=\int_{\rho}^{R}\frac{\mathrm{d}s}{\mathrm{sn}_{K}(s)^{n-1}}=\int_{\mathrm{sn}_{K}(\rho)}^{\mathrm{sn}_{K}(R)}\frac{\mathrm{d}t}{\sqrt{1+t^{2}}{t^{n-1}}}\leq\int_{\mathrm{sn}_{K}(\rho)}^{\mathrm{sn}_{K}(R)}\frac{\mathrm{d}t}{{t^{n-1}}}\leq\frac{\frac{1}{n-2}}{\mathrm{sn}_{K}(\rho)^{n-2}}.$

$3^{\mathrm{rd}}$ case: $K>0$. In this case, the function
$s\mapsto\mathrm{sn}_{K}(s)$ is increasing on $[0,\pi/(2/\sqrt{K})]$, and
decreasing on $[\pi/(2/\sqrt{K}),\pi/\sqrt{K}]$. We distinguish the following
cases.

Case (i): $0<\rho<R\leq\frac{\pi}{2\sqrt{K}}$. By using the elementary
inequality $2x/\pi\leq\mathrm{sn}_{K}(x)\leq x$ and the first case, we obtain

$\int_{\rho}^{R}\frac{\mathrm{d}s}{\omega_{K}(s)}=\int_{\rho}^{R}\frac{\mathrm{d}s}{\mathrm{sn}_{K}(s)^{n-1}}\leq\left(\frac{\pi}{2}\right)^{n-1}\int_{\rho}^{R}\frac{\mathrm{d}s}{s^{n-1}}\leq\frac{\left(\frac{\pi}{2}\right)^{n-1}}{n-2}\frac{1}{\rho^{n-2}}\leq\frac{\frac{\left(\frac{\pi}{2}\right)^{n-1}}{n-2}}{\mathrm{sn}_{K}(\rho)^{n-2}}.$

Case (ii): $0<\rho\leq\frac{\pi}{2\sqrt{K}}<R$. In this case, we need to split
the domain of integration into the intervals $[\rho,\pi/(2\sqrt{K})]$ and
$[\pi/(2\sqrt{K}),R]$.
Using the symmetry around the point $\pi/(2\sqrt{K})$ of the function
$\mathrm{sn}_{K}$, and Case (i), we obtain

$\displaystyle\int_{\rho}^{R}\frac{\mathrm{d}s}{\omega_{K}(s)}$
$\displaystyle=\left(\int_{\rho}^{\frac{\pi}{2\sqrt{K}}}+\int_{\frac{\pi}{2\sqrt{K}}}^{R}\right)\frac{\mathrm{d}s}{\mathrm{sn}_{K}(s)^{n-1}}=\left(\int_{\rho}^{\frac{\pi}{2\sqrt{K}}}+\int_{\frac{\pi}{2\sqrt{K}}-R}^{\frac{\pi}{2\sqrt{K}}}\right)\frac{\mathrm{d}s}{\mathrm{sn}_{K}(s)^{n-1}}$
$\displaystyle\leq\frac{\left(\frac{\pi}{2}\right)^{n-1}}{n-2}\left(\frac{1}{\mathrm{sn}_{K}(\rho)^{n-2}}+\frac{1}{\mathrm{sn}_{K}(\pi/(2\sqrt{K})-R)^{n-2}}\right)$
$\displaystyle=\frac{\left(\frac{\pi}{2}\right)^{n-1}}{n-2}\left(\frac{1}{\mathrm{sn}_{K}(\rho)^{n-2}}+\frac{1}{\mathrm{sn}_{K}(R)^{n-2}}\right).$

Case (iii): $\frac{\pi}{2\sqrt{K}}<\rho$. Again, by the symmetry of
$\mathrm{sn}_{K}$ around $\pi/(2\sqrt{K})$ we obtain by case (i)

$\int_{\rho}^{R}\frac{\mathrm{d}s}{\omega_{K}(s)}=\int_{\pi/(2\sqrt{K})-R}^{\pi/(2\sqrt{K})-\rho}\frac{\mathrm{d}s}{\mathrm{sn}_{K}(s)^{n-1}}\leq\frac{\frac{\left(\frac{\pi}{2}\right)^{n-1}}{n-2}}{\mathrm{sn}_{K}(\pi/(2\sqrt{K})-R)^{n-2}}=\frac{\frac{\left(\frac{\pi}{2}\right)^{n-1}}{n-2}}{\mathrm{sn}_{K}(R)^{n-2}}.$

Combining the above cases yields the claim. ∎

The combination of Theorem 3.3 and Lemma 4.1 give the following theorem
describing the desired lower bound on $\lambda^{M,S}$.

######  Theorem 4.2.

Let $0<\rho<R<\infty$ and $S\subset M$ be a proper $(R,\rho)$-relatively dense
subset. There exists $C_{n}>0$ such that

$\lambda^{M,S}\geq
C_{n}\frac{(\mathrm{sn}_{K}(\rho/2)\wedge\mathrm{sn}_{K}(3R))^{n-2}}{\operatorname{Vol}_{K}(3R)}.$

The proof of the main theorem also requires an upper bound on $\lambda^{M,S}$
in terms of geometric parameters. We will obtain such a bound by using the
following variant of the Bishop-Gromov volume doubling comparison estimate,
which follows from the volume comparison estimate in [CGT82] and a Vitali
covering argument, cf. [Heb96]..

######  Proposition 4.3 ([CGT82, Heb96]).

There exists a constant $D=D(n)\geq 1$ such that for all $x\in M$ and $0<r\leq
R$ we have

$\operatorname{Vol}(B(x,2r))\leq
D\mathrm{e}^{\sqrt{K}R}\operatorname{Vol}(B(x,r)).$

This leads via a reduction argument to the following upper bound on the
spectrum of the Laplacian inside $M$ without a relatively dense subset.

######  Lemma 4.4.

Fix $0<\rho<R<\infty$ such that $\rho\leq 3R/16$ and assume that $R$ is a
proper radius for all $x\in M$. Let $\Sigma\subset M$ be uniformly discrete
such that $B_{3R}(\Sigma)\supset M$, $B_{\rho}(\Sigma)$ a
$(R,\rho)$-relatively dense subset with $M\setminus
B_{R}(\Sigma)\neq\emptyset$, and for all $p\in\Sigma$, we have
$R\leq\mathrm{d}(p,\Sigma\setminus\\{p\\})\leq 6R$. There exists a constant
$C_{D}>0$ such that

$\lambda_{M,B_{\rho}(\Sigma)}\leq\frac{C_{D}}{R^{2}}\mathrm{e}^{\sqrt{K}R}.$

In particular, we can choose $C_{D}=64D$, where $D$ is given by Proposition
4.3.

###### Proof.

Denote by $\Delta_{B}\geq 0$ the Dirichlet-Laplacian on $B=B_{R}(x)\subset M$
and set $\lambda(x,R):=\inf\sigma(\Delta_{B_{R}(x)})$. Setting
$\phi=(1-d(\cdot,x)/R)_{+}\in W_{0}^{1}(B_{R}(x))$ and noting that
$|\nabla\phi|\leq 1/R$ a.e., we obtain

$\displaystyle\lambda(x,R))\leq\frac{\int_{B_{R}(x)}|\nabla\phi|^{2}}{\int_{B_{R}(x)}\phi^{2}}\leq\frac{1}{R^{2}}\frac{\operatorname{Vol}(B_{R}(x))}{\operatorname{Vol}(B_{R/2}(x))}\frac{\operatorname{Vol}(B_{R/2}(x))}{\int_{B_{R}(x)}\phi^{2}}\leq\frac{1}{R^{2}}\frac{\operatorname{Vol}(B_{R}(x))}{\operatorname{Vol}(B_{R/2}(x))}.$

In the following, we distinguish the cases for $M$ being non-compact or
compact, i.e., $\mathrm{diam}(M)=\infty$ or $\mathrm{diam}(M)<\infty$.
First, assume that $\mathrm{diam}M=\infty$. In this case, $\Sigma$ must
contain at least two elements. Indeed, since $B_{\rho}(\Sigma)$ is
$(R,\rho)$-relatively dense, assuming that $\Sigma=\\{p\\}$ implies that for
all $x\in M$ we have $B(x,R)\cap B_{\rho}(\Sigma)=B_{\rho}(p)$. Hence, for all
$y,z\in M$, we have

$d(y,z)\leq d(y,p)+d(z,p)\leq 6R<\infty,$

a contradiction. Let $p\in\Sigma$. Since $M$ is complete,
$\overline{B_{6R}(p)}$ is compact. Since $B_{3R}(\Sigma)$ covers $M$, there
are $p_{1},\ldots,p_{N}\in\Sigma$ such that
$B_{3R}(\\{p_{1},\ldots,p_{N}\\})\supset\overline{B_{6R}(p)}$. We let $p_{1}$
the point minimizing the distance to $p$, i.e.,
$d(p,p_{1})=\min\\{d(p,p_{1})\colon p_{1},\ldots,p_{N}\\}$. Since
$p,p_{1}\in\Sigma$, we have $R\leq d(p,p_{1})$. Let $\gamma\colon[0,1]\to M$,
$\gamma(0)=p$, $\gamma(1)=p_{1}$, a distance minimizing geodesic from $p$ to
$p_{1}$. Then there exists $t_{0}\in(0,1)$ such that $q:=\gamma(t_{0})$
satisfies $d(p,q)=d(p_{1},q)=d(p,p_{1})/2$. Hence, $B_{R/2-2\rho}(q)\subset
M\setminus B_{\rho}(\Sigma)$. By assumption we have $R/2-2\rho\geq R/8$, such
that $B_{R/8}(q)\subset M\setminus B_{\rho}(\Sigma)$. By domain monotonicity
of the infimum of the spectrum of the Dirichlet Laplacian, we get

$\lambda_{M,B_{\rho}(\Sigma)}\leq\lambda(q,R/8)\leq\sup_{x\in
M}\frac{64}{R^{2}}\frac{\operatorname{Vol}(B_{R/8}(x))}{\operatorname{Vol}(B_{R/16}(x))},$

hence the claim by Proposition 4.3.

If $\mathrm{diam}(M)<\infty$ and $\Sigma$ contains at least two elements, the
argument above need not be changed. If $\mathrm{diam}M<\infty$ and
$\Sigma=\\{p\\}$, then there is $p^{\prime}\in M$ such that $M\setminus
B_{\rho}(p^{\prime})$ contains the ball
$B_{\mathrm{diam}(M)-\rho}(p^{\prime})$. Since $R$ is a proper radius for all
$x\in M$, we have that $\mathrm{diam}(M)\geq R-\delta\geq 13R/16$, such that
$B_{R/8}(p^{\prime})\subset M\setminus B_{\rho}(p)$, and the argument above
applies again. ∎

Now we turn to the problem of bounding the exit times from above. The lower
bound on the Ricci curvature particularly implies that the manifold is
stochastically complete, see [Yau78]. Recall that the first exit time for
given $x\in M$ and $\rho,\beta>0$, is given by

$\mathbb{P}_{x}(\tau_{\rho}\leq\alpha)=\int_{M\setminus
B(x,\rho)}p_{\alpha}(x,y)\operatorname{\mathrm{dvol}}(y).$

Therefore, we need quantitative estimates on the volume growth and heat kernel
behavior for large distances. To this end, we collect and reprove facts from
[HSC01] to track the constants and to obtain quantitative estimates depending
on the Ricci curvature lower bound.

First, we show that Proposition 4.3 to control the volume of balls with radius
larger than $R$. To derive such estimates, we follow [HSC01]. First of all,
Proposition 4.3 yields via a covering argument (cf. [HSC01, Eq. (2.7)])

$\operatorname{Vol}(B(x,T+R/4))\leq
D^{2}\mathrm{e}^{2\sqrt{K}R}\operatorname{Vol}(B(x,T)),\quad T>R.$

Iterating this inequality, we obtain for all $T>R$

$\displaystyle\operatorname{Vol}(B(x,T))$ $\displaystyle\leq
D_{R}\operatorname{Vol}\left(B\left(x,T-\frac{R}{4}\right)\right)\leq
D_{R}^{2}\operatorname{Vol}\left(B\left(x,T-2\frac{R}{4}\right)\right)$
$\displaystyle\leq\ldots\leq
D_{R}^{K}\operatorname{Vol}\left(B\left(x,T-K\frac{R}{4}\right)\right)\leq
D_{R}^{K}\operatorname{Vol}(B(x,R)),$

where

$K:=\min\left\\{k\in\mathbb{N}_{0}\colon T-k\frac{R}{4}\leq R\right\\}.$

Since $K\leq 4T/R$, we infer from the estimate above, the monotonicity of the
volume measure, and the definition of $D_{R}=D\mathrm{e}^{\sqrt{K}R}$ for all
$T>R$

$\operatorname{Vol}(B(x,T))\leq
D^{8\frac{T}{R}}\mathrm{e}^{8T\sqrt{K}}\operatorname{Vol}(B(x,R)).$

This yields (cf. [HSC01, Eq. (2.6)])

$\operatorname{Vol}(B(x,T))\leq D\mathrm{e}^{8(\ln
D+R\sqrt{K})\frac{T}{R}}\operatorname{Vol}(B(x,R)),\quad T>0.$

The second ingredient to bound the first exit time is an explicit form of the
heat kernel. We refrain from telling the whole story of heat kernel upper
bounds on manifolds and refer to the excellent book [Gri09] and the references
therein. Here, we will use the following estimate.

######  Proposition 4.5 ([Stu92, Corollary 2.3(a)]).

There exists $C=C(n)>0$ such that we have

$p_{t}(x,y)\leq\frac{C\mathrm{e}^{KR^{2}}}{\operatorname{Vol}(B(x,\sqrt{t}))}\exp\left(-\frac{d(x,y)^{2}}{5t}\right),\quad
x,y\in M,t\in(0,R^{2}].$

The volume comparison and heat kernel estimate deliver the following bound on
the exit time. The proof can essentially be found in [HSC01, Lemma 3.6],
however, we obtain different constants. Since we are interested in the scaling
behavior of the constants, we give the complete argument. Note that in
principle the explicit representations of volume comparison and upper heat
kernel bound suffice to get an upper bound on the exit time.

######  Lemma 4.6.

Let $K\geq 0$ and $M$ be a complete Riemannian manifold with
$\operatorname{Ric}\geq-K$. Then for all $\rho,R>0$, $x\in M$, and
$\alpha\in(0,R^{2}]$ with $\alpha\leq\frac{1}{6400}(\ln
D+R\sqrt{K})^{-2}\rho^{2}$ we have

$\mathbb{P}_{x}(\tau_{\rho}\leq\alpha)\leq
C_{0}\exp\left(C_{1}KR^{2}-\frac{\rho^{2}}{20\alpha}\right),$

where $C_{0}:=\frac{20}{\ln 2}CD\mathrm{e}^{2256(\ln D)^{2}}$, $C_{1}:=2257$.

###### Proof.

Observe that

$\displaystyle\mathbb{P}_{x}(\tau_{\rho}\leq\alpha)=\int_{M\setminus
B(x,\rho)}p_{\alpha}(x,y)\operatorname{\mathrm{dvol}}(y)=\sum_{i=1}^{\infty}\int_{\\{2^{i-1}\rho\leq
d(x,y)<2^{i}\rho\\}}p_{\alpha}(x,y)\operatorname{\mathrm{dvol}}(y).$

We estimate the summands using Proposition 4.5 and the volume comparison
estimate for large balls obtained above to obtain for all $i\in\mathbb{N}$ and
$\alpha\in(0,R^{2}]$

$\displaystyle\int\limits_{\\{2^{i-1}\rho\leq
d(x,y)<2^{i}\rho\\}}p_{\alpha}(x,y)\operatorname{\mathrm{dvol}}(y)\leq
C\mathrm{e}^{KR^{2}}\int\limits_{\\{2^{i-1}\rho\leq
d(x,y)<2^{i}\rho\\}}\frac{\mathrm{e}^{-\frac{d(x,y)^{2}}{5t}}}{\operatorname{Vol}(B(x,\sqrt{\alpha}))}\operatorname{\mathrm{dvol}}(y)$
$\displaystyle\leq
C\mathrm{e}^{KR^{2}}\frac{\operatorname{Vol}(B(x,2^{i}\rho))}{\operatorname{Vol}(B(x,\sqrt{\alpha}))}\exp\left(-\frac{4^{i}\rho^{2}}{20\alpha}\right)\leq
CD\mathrm{e}^{KR^{2}}\exp\left(8(\ln
D+R\sqrt{K})\frac{2^{i}\rho}{\sqrt{\alpha}}-\frac{4^{i}\rho^{2}}{20\alpha}\right).$

Hence, completing the square we obtain

$\mathbb{P}_{x}(\tau_{\rho}\leq\alpha)\leq
CD\mathrm{e}^{KR^{2}}\sum_{i=1}^{\infty}\exp\left(ax_{i}-bx_{i}^{2}\right)=CD\mathrm{e}^{KR^{2}+\frac{a^{2}}{4b}}\sum_{i=1}^{\infty}\exp\left(-b\left(x_{i}-\frac{a}{2b}\right)^{2}\right),$

where

$a=8(\ln D+R\sqrt{K}),\quad b=\frac{1}{20},\quad
x_{i}=\frac{2^{i}\rho}{\sqrt{\alpha}}.$

By assumption, we have $\alpha\leq
4\cdot\frac{4b^{2}}{a^{2}}\left(1-1/20b\right)^{2}\rho^{2}$, and hence

$-b\left(x_{i}-\frac{a}{2b}\right)^{2}\leq-\frac{1}{20}x_{i}^{2}.$

This yields for the exponential sum above

$\displaystyle\sum_{i=1}^{\infty}\exp\left(-b\left(x_{i}-\frac{a}{2b}\right)^{2}\right)\leq\sum_{i=1}^{\infty}\exp\left(-\frac{1}{20}x_{i}^{2}\right)=\sum_{i=1}^{\infty}\exp\left(-\left(\frac{\rho}{\sqrt{20\alpha}}2^{i}\right)^{2}\right)$
$\displaystyle\leq\int_{0}^{\infty}\exp\left(-\left(\frac{\rho}{\sqrt{20\alpha}}2^{t}\right)^{2}\right)\mathrm{d}t=\frac{1}{\ln
2}\int_{\frac{\rho}{\sqrt{20\alpha}}}^{\infty}\mathrm{e}^{-s^{2}}\frac{\mathrm{d}s}{s}\leq\frac{1}{\ln
2}\frac{\sqrt{20\alpha}}{\rho}\int_{\frac{\rho}{\sqrt{20\alpha}}}^{\infty}\mathrm{e}^{-s^{2}}\mathrm{d}s$
$\displaystyle\leq\frac{1}{\ln
2}\frac{20\alpha}{\rho^{2}}\mathrm{e}^{-\frac{\rho^{2}}{20\alpha}}\leq\frac{20}{\ln
2}\mathrm{e}^{-\frac{\rho^{2}}{20\alpha}},$

where we used in the second inequality that the summands are decreasing, in
the third the monotonicity of $s\mapsto 1/s$, in the fourth the standard tail
estimate
$\int_{u}^{\infty}\mathrm{e}^{-s^{2}}\mathrm{d}s\leq\tfrac{1}{u}\mathrm{e}^{-u^{2}}$,
$u>0$, and in the last line $\alpha\leq\rho^{2}$. ∎

##  5 Quantitative uncertainty principles and Ricci curvature

After all the preparations in the sections above, we are now able to state and
prove our main theorem.

######  Theorem 5.1.

Let $K\in\mathbb{R}$, $n\in\mathbb{N}$, and $0<\rho<R<\mathrm{diam}M$,
$\epsilon\in(0,1/2)$. There exist contants
$\kappa=\kappa(K,R,\rho,n,\epsilon)>0$ and
$E_{0}=E_{0}(K,R,\rho,n,\epsilon)>0$ such that the following holds: for any
complete Riemannian manifold $M$ of dimension $n$ with $\operatorname{Ric}\geq
K$, any proper $(R,\rho)$-relatively dense $S\subset M$, and any
$I\subset(-\infty,E_{0}]$, we have

$\chi_{I}(H)\mathbf{1}_{S}\chi_{I}(H)\geq\kappa\ \chi_{I}(H).$

The constants $E_{0}$ and $\kappa$ are given by

$E_{0}=\epsilon\
(1-a)\frac{(\mathrm{sn}_{K}(\tilde{\rho}/8)\wedge\mathrm{sn}_{K}(3R))^{n-2}}{\operatorname{Vol}_{K}(3R)}-b$

and

$\displaystyle\kappa$
$\displaystyle=C_{5}\tilde{\rho}^{2}\frac{(\mathrm{sn}_{K}(\tilde{\rho}/8)\wedge\mathrm{sn}_{K}(3R))^{n-2}}{\operatorname{Vol}_{K}(3R)}$
$\displaystyle\quad\cdot\epsilon\left[\left|\ln\left(\frac{C_{6}\operatorname{Vol}_{K}(3R)}{(1-2\epsilon)\tilde{\rho}^{2}\
(\mathrm{sn}_{K}(\tilde{\rho}/8)\wedge\mathrm{sn}_{K}(3R))^{n-2}}\right)\right|+C_{7}+C_{8}\rho^{2}K+C_{9}\mathrm{e}^{\sqrt{K}R}\right]^{-2},$

where $\tilde{\rho}=\rho\wedge 3R/16$, and $C_{5},C_{6},C_{7},C_{8},C_{9}>0$
are dimension-dependent constants.

###### Proof.

Due to domain monotonicity, we can replace $S$ by any subset. Proposition 3.2
ensures the existence of a uniformly discrete subset $\Sigma\subset S$ such
that $B_{\rho}(\Sigma)$ is $(3R,\rho)$-relatively dense. Set
$\tilde{\rho}:=\rho\wedge(3R/16)$. We choose
$S^{\prime}:=B_{\tilde{\rho}/4}(\Sigma)\subset S$, which is
$(3R,\tilde{\rho}/4)$-relatively dense. Theorem 4.6 yields for all
$R=R^{\prime}=\tilde{\rho}$ and $\alpha\leq 2\frac{1}{6400}(\ln
D+\tilde{\rho}\sqrt{K})^{-2}\tilde{\rho}^{2}$

$\mathrm{e}^{-\frac{\beta\alpha}{1-a}}+\sup_{x\in
M}\mathbb{P}_{x}(\tau_{\tilde{\rho}/8}\leq\alpha/2)\leq\mathrm{e}^{-\frac{\beta\alpha}{1-a}}+C_{0}\mathrm{e}^{C_{1}\tilde{\rho}^{2}K-\frac{\tilde{\rho}^{2}}{320\alpha}},$

where $C_{0}$ and $C_{1}$ are given in Theorem 4.6. The negative exponents on
the right-hand side are equal if we choose

$\beta=\beta_{\alpha}:=\frac{1-a}{\alpha}\left(\frac{1}{320\alpha}-C_{1}K\right)\tilde{\rho}^{2}.$

If $\alpha\leq\alpha_{1}:=1/(320C_{1}K)$ we have $\beta>0$. Further, we have

$\mathrm{e}^{-\frac{\beta\alpha}{1-a}}+\sup_{x\in
M}\mathbb{P}_{x}(\tau_{\tilde{\rho}/2}\leq\alpha/2)\leq\left(1+C_{0}\right)\mathrm{e}^{\left(-\frac{1}{320\alpha}+C_{1}K\right)\tilde{\rho}^{2}}.$

Theorem 2.4 implies

$\chi_{I}(\Delta+V)\mathbf{1}_{S^{\prime}_{\tilde{\rho}/4}}\chi_{I}(\Delta+V)\geq\tilde{\kappa}\
\chi_{I}(\Delta+V),$

where, choosing $\alpha_{0}=2\tilde{\rho}^{2}$,
$\mu_{0}=\lambda^{M,S^{\prime}}$,

$\displaystyle\tilde{\kappa}(\tilde{\rho},R,t)$
$\displaystyle:=\frac{1}{\beta_{\alpha}}\left[\left(\mu_{0}-\frac{2}{\alpha_{0}}\mathrm{e}^{\alpha_{0}\lambda_{M,S^{\prime}_{\tilde{\rho}/4}}/2}(\mathrm{e}^{-\frac{\beta}{1-a}\alpha}+\sup_{z\in
M}\mathbb{P}_{z}(\tau_{\tilde{\rho}/8}\leq\alpha/2))\right)-(E_{0}+b)\right]$
$\displaystyle\geq\frac{1}{\beta_{\alpha}}\left[(1-a)\left(\mu_{0}-p\right)-(E_{0}+b)\right],$

where we set

$p=p(\tilde{\rho},\alpha,K):=\frac{\mathrm{e}^{\tilde{\rho}^{2}\lambda_{M,S^{\prime}_{\tilde{\rho}/4}}}}{\tilde{\rho}^{2}}\left(1+C_{0}\right)\mathrm{e}^{\left(-\frac{1}{320\alpha}+C_{1}K\right)\tilde{\rho}^{2}}.$

Since $S^{\prime}$ is properly $(R,\tilde{\rho}/4)$-dense, Theorem 4.2 yields

$\mu_{0}=\lambda^{M,S^{\prime}}\geq\Lambda=\Lambda(n,\tilde{\rho},K,R):=C_{n}\frac{(\mathrm{sn}_{K}(\tilde{\rho}/8)\wedge\mathrm{sn}_{K}(3R))^{n-2}}{\operatorname{Vol}_{K}(3R)}.$

If we choose for $\tilde{\epsilon}\in(0,1)$

$\alpha\leq\alpha_{2}:=\frac{1}{320}\left[\left|\ln\left(\frac{1+C_{0}}{(1-\tilde{\epsilon})\Lambda}\frac{\mathrm{e}^{\tilde{\rho}^{2}\lambda_{M,S^{\prime}_{\tilde{\rho}/4}}}}{\tilde{\rho}^{2}}\right)\right|+C_{1}K\tilde{\rho}^{2}\right]^{-1}\tilde{\rho}^{2},$

we obtain

$\mu_{0}-p\geq\Lambda-p\geq\tilde{\epsilon}\ \Lambda.$

Hence, choosing
$\alpha=\alpha_{\ast}:=\alpha_{1}\wedge\alpha_{2}\wedge\tfrac{1}{3200}(\ln
D+\tilde{\rho}\sqrt{K})^{-2}\tilde{\rho}^{2}$ yields

$\tilde{\kappa}\geq\frac{1-a}{\beta_{\alpha_{\ast}}}\left(\mu_{0}-p(\tilde{\rho},\alpha_{\ast},K)\right)-\frac{E_{0}+b}{\beta_{\alpha_{\ast}}}\\\
\geq\frac{1}{\frac{1-a}{\alpha_{\ast}}\left(\frac{1}{320\alpha_{\ast}}-C_{1}K\right)\tilde{\rho}^{2}}\left[(1-a)\tilde{\epsilon}\
\Lambda-(E_{0}+b)\right]\geq\frac{320\alpha_{\ast}^{2}}{\tilde{\rho}^{2}}\left[\tilde{\epsilon}\
\Lambda-\frac{E_{0}+b}{1-a}\right].$

It remains to bound $\alpha_{\ast}$ from below. We have

$\displaystyle\alpha_{\ast}$
$\displaystyle=\alpha_{1}\wedge\alpha_{2}\wedge\tfrac{1}{3200}(\ln
D+\tilde{\rho}\sqrt{K})^{-2}\tilde{\rho}^{2}$
$\displaystyle=\frac{\tilde{\rho}^{2}}{20}\left[\frac{1}{16C_{1}K\tilde{\rho}^{2}}\wedge\left[\left|\ln\left(\frac{1+C_{0}}{(1-\tilde{\epsilon})\Lambda}\frac{\mathrm{e}^{\tilde{\rho}^{2}\lambda_{M,S^{\prime}_{\tilde{\rho}/4}}}}{\tilde{\rho}^{2}}\right)\right|+C_{1}K\tilde{\rho}^{2}\right]^{-1}\wedge\frac{1}{160}(\ln
D+\tilde{\rho}\sqrt{K})^{-2}\right]$
$\displaystyle\geq\tilde{\rho}^{2}\left[c_{0}\left|\ln\left(\frac{c_{1}}{(1-\tilde{\epsilon})\Lambda}\frac{\mathrm{e}^{\tilde{\rho}^{2}\lambda_{M,S^{\prime}_{\tilde{\rho}/4}}}}{\tilde{\rho}^{2}}\right)\right|+c_{1}+c_{2}\tilde{\rho}^{2}K\right]^{-1},$

where we used $(a+b)^{2}\leq 2a^{2}+b^{2}$ and set $c_{0}:=20$,
$c_{1}:=20(1+C_{0})$, $c_{1}:=12800(\ln D)^{2}$, and $c_{2}:=20(16C_{1}\vee
640)$. Lemma 4.4 applied to $\Sigma$ with the choice $\rho=\tilde{\rho}/2$
yields the estimate $\lambda_{M,S^{\prime}_{\tilde{\rho}/4}}\leq
64D\mathrm{e}^{\sqrt{K}R}/R^{2}$. Using $\tilde{\rho}\leq\rho$, we obtain

$\displaystyle\tilde{\kappa}$
$\displaystyle\geq\frac{20\alpha_{\ast}^{2}}{\tilde{\rho}^{2}}\left(\tilde{\epsilon}\
\Lambda-\frac{E_{0}+b}{1-a}\right)$
$\displaystyle\geq\tilde{\rho}^{2}\left(\tilde{\epsilon}\
\Lambda-\frac{E_{0}+b}{1-a}\right)\left[c_{0}\left|\ln\left(\frac{c_{1}}{(1-\tilde{\epsilon})\tilde{\rho}^{2}\
\Lambda}\right)\right|+\tilde{\rho}^{2}\lambda_{M,S^{\prime}_{\tilde{\rho}/4}}+c_{1}+c_{2}\tilde{\rho}^{2}K\right]^{-2}$
$\displaystyle\geq\tilde{\rho}^{2}\left(\tilde{\epsilon}\
\Lambda-\frac{E_{0}+b}{1-a}\right)\left[c_{0}\left|\ln\left(\frac{c_{1}}{(1-\tilde{\epsilon})\tilde{\rho}^{2}\
\Lambda}\right)\right|+c_{1}+c_{2}\rho^{2}K++c_{3}\mathrm{e}^{\sqrt{K}R}\right]^{-2},$

where $c_{3}:=194D/16$. The statement follows by choosing
$E_{0}=(1-a)\Lambda\tilde{\epsilon}/2-b$ and substituting
$\epsilon:=\tilde{\epsilon}/2\in(0,1/2)$. ∎

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