proposition@alttheorem lemma@alttheorem corollary@alttheorem
counterexample@alttheorem fact@alttheorem definition@alttheorem
question@alttheorem openquestion@alttheorem conjecture@alttheorem
remark@alttheorem example@alttheorem

# Weyl’s Lemma on $RCD(K,N)$ Metric Measure Spaces

Yu Peng  Department of Mathematics, Sun Yat-Sen University, Guangzhou 510275,
China<EMAIL_ADDRESS>,  Hui-Chun Zhang  Department of
Mathematics, Sun Yat-Sen University, Guangzhou 510275, China,
<EMAIL_ADDRESS>and  Xi-Ping Zhu  Department of Mathematics, Sun
Yat-Sen University, Guangzhou 510275, China<EMAIL_ADDRESS>
###### Abstract.

In this paper, we extend the classical Weyl’s lemma to $RCD(K,N)$ metric
measure spaces. As its applications, we show the local regularity of solutions
for Poisson equations and a Liouville-type result for $L^{1}$ very weak
harmonic functions on $RCD(K,N)$ spaces. Meanwhile, a byproduct is that we
obtain a gradient estimate for solutions to a class of elliptic equations with
dis-continuous coefficients.

##  1\. Introduction

The classical Weyl’s lemma states that any very weakly harmonic function in
the Euclidean spaces $\mathbb{R}^{n}$ must be smooth. Namely,

######  Theorem 1.1 (Weyl [50]).

Let $\Omega\subset\mathbb{R}^{n}$ be open. Suppose that $u\in L_{\rm
loc}^{1}(\Omega)$ and

$\int_{\Omega}u\Delta\varphi=0$

for any $\varphi\in C^{\infty}_{c}(\Omega)$. Then $u$ is smooth (I.e. there
exists a $C^{\infty}$-representative).

Extensions of Weyl’s lemma are widely studied. For examples, Burch [6]
extended it to elliptic operators with variable coefficients, Hörmander [21]
proved Weyl’s lemma for hypoelliptic operators, and Elson [16] developed a
theory of distributions on separable real Banach spaces and proved an analogy
of Weyl’s lemma. We refer to [44] for a survey. Recently, Di Fratta-Fiorenza
[14] gave a short proof for the regularity of very weak solutions of Poisson
equations via Weyl’s lemma.

In this paper, we will extend Weyl’s lemma to non-smooth settings. More
precisely, we will prove Weyl’s lemma on $RCD(K,N)$ spaces. Given a metric
measure space $(X,d,\mathfrak{m})$, the notion $RCD(K,N)$ is a synthetic
notion of lower Ricci bounds for $(X,d,\mathfrak{m})$, which has been
developed in [47, 48, 34, 2, 3, 18, 4, 17]. The parameters $K\in\mathbb{R}$
and $N\in[1,+\infty]$ play the role of “Ricci curvature $\geqslant K$ and
dimension $\leqslant N$”. We refer to the survey [1] for the improvements of
geometric analysis on $RCD(K,N)$ spaces. Important examples of $RCD(K,N)$
spaces include Ricci limit spaces [8, 9, 10] and finite-dimensional Alexandrov
spaces with curvature bounded from below with respect to their Hausdorff
measures [40, 51].

Let $(X,d,\mathfrak{m})$ be an $RCD(K,N)$ space. The Sobolev space
$W^{1,2}(X)$ is a Hilbert space, and then the inner product $\langle{\nabla
u},{\nabla v}\rangle\in L^{1}(X,\mathfrak{m})$ makes sense for any $u,v\in
W^{1,2}(X)$. Given an open subset $\Omega\subset X$, a function $u\in
W^{1,2}_{\rm loc}(\Omega)$ is called a (weakly) harmonic function on $\Omega$
if

$\int_{X}\langle{\nabla u},{\nabla\varphi}\rangle d\mathfrak{m}=0,\ \
\forall\varphi\in Lip_{c}(\Omega),$

where $Lip_{c}(\Omega)$ denotes the space of Lipschitz functions defined on
$X$ with compact supported in the interior of $\Omega$. The Lipschitz
regularity of harmonic functions on $RCD(K,N)$ spaces has been established in
[19, 24, 22].

In order to extend Weyl’s lemma to the setting of $RCD(K,N)$ spaces, let us
begin with the notion of “very weakly harmonic function”. Let
$(X,d,\mathfrak{m})$ be an $RCD(K,N)$ space. We denote by $P_{t}$ the heat
flow on $L^{2}(X,\mathfrak{m})$ and by $\Delta$ the associated infinitesimal
generator with domain $D(\Delta)$. Recall that the class of test functions on
$RCD(K,N)$ spaces is given in [20] by

$\mathrm{Test}^{\infty}(X):=\left\\{\varphi\in D(\Delta)\cap
L^{\infty}(X):\left\lvert\nabla\varphi\right\rvert\in L^{\infty}(X),\
\Delta\varphi\in L^{\infty}(X)\cap W^{1,2}(X)\right\\}.$

Given any open domain $\Omega\subset X$, denote by

$\mathrm{Test}_{c}^{\infty}(\Omega):=\left\\{\varphi\in\mathrm{Test}^{\infty}(X)\big{|}{\rm\
supp}(\varphi)\subset\Omega\ {\rm and\ supp}(\varphi)\ {\rm is\
compact}\right\\}.$ (1.1)

It is well-known [20] that the space $\mathrm{Test}^{\infty}(X)$ is dense in
$W^{1,2}(X)$ and the space $\mathrm{Test}_{c}^{\infty}(\Omega)$ is dense in
$W_{0}^{1,2}(\Omega).$

######  Definition .

Let $(X,d,\mathfrak{m})$ be an $RCD(K,N)$ space for some $1\leqslant N<\infty$
and $K\in\mathbb{R}$ and let $\Omega\subset X$ be an open domain. A function
$u\in L_{\rm loc}^{1}(\Omega)$ is called a very weakly subharmonic function on
$\Omega$ if

$\int_{\Omega}u\Delta\varphi d\mathfrak{m}\geqslant
0,\qquad\forall\varphi\in{\rm Test}^{\infty}_{c}(\Omega)\ \ {\rm and}\ \
\varphi\geqslant 0.$ (1.2)

A function $u\in L_{\rm loc}^{1}(\Omega)$ is called a very weakly
superharmonic function on $\Omega$ if $-u$ is very weakly subharmonic. A
function $u\in L_{\rm loc}^{1}(\Omega)$ is called a very weakly harmonic
function on $\Omega$ if both $u$ and $-u$ are very weakly subharmonic.

The Weyl’s lemma is concerned with the regularity of a very weak harmonic
function $u$ on $\Omega$. If $u\in W^{1,2}_{\rm loc}(\Omega)$ then it is a
(weakly) harmonic, and hence it is locally Lipschitz continuous (see [19, 24,
22]). This remains the gap between the $u\in L^{1}_{\rm loc}(\Omega)$ and the
$u\in W^{1,2}_{\rm loc}(\Omega)$.

The major obstacle of the improvement from $L^{1}_{\rm loc}(\Omega)$ to
$W^{1,2}_{\rm loc}(\Omega)$ for very weak harmonic functions on a singular
space can be understood as follows. For the convenience of the discussion, we
consider a very weak harmonic function $u:(\Omega,g)\to\mathbb{R}$ from a
domain $\Omega\\!\subset\\!\mathbb{R}^{n}$ with a _singular_ Riemannian metric
$g=(g_{ij})$. Then $u\in L^{1}_{\rm loc}(\Omega)$ is a very weak solution of
the elliptic equation of divergence form

$\sum_{i,j=1}^{n}\partial_{i}\left(a^{ij}\partial_{j}u\right)=0\quad{\rm
with}\ \ a^{ij}:=\sqrt{g}g^{ij},$ (1.3)

in the sense of distributions, where $\partial_{i}=\frac{\partial}{\partial
x_{i}}$, $g=\det(g_{ij})$ and $(g^{ij})$ is the inverse matrix of $(g_{ij})$.
It seems that Zhang-Bao’s work [53] is the only result fixing the whole gap
from $L^{1}_{\rm loc}(\Omega)$ to $W^{1,2}_{\rm loc}(\Omega)$. They proved
[53] that if the coefficients $a^{ij}\in Lip_{\rm loc}(\Omega)$, then every
$L^{1}_{\rm loc}(\Omega)$-solution of (1.3) must be in $W^{1,2}_{\rm
loc}(\Omega)$. Brezis [5] showed that if all $a^{ij}$ are Dini continuous then
every $BV_{\rm loc}(\Omega)$-solution of (1.3) must be in $W^{1,2}_{\rm
loc}(\Omega)$. Recently, Manna-Leone-Schiattarella in [35] proved that if
$a^{ij}\in C^{0}(\Omega)\cap W^{1,n}_{\rm loc}(\Omega)$ satisfying a doubling
Dini condition, then every $L^{\frac{n}{n-1}}_{\rm loc}(\Omega)$-solution of
(1.3) is in $W^{1,2}_{\rm loc}(\Omega)$. The Dini-type condition of the
coefficients $a^{ij}$ is necessary: Jin-Maz’ya-Schaftingen in [28] constructed
an equation (1.3) with $a^{ij}\in C^{0}(\Omega)$ and that it has a solution
$u\in W^{1,1}_{\rm loc}(\Omega)\setminus W^{1,2}_{\rm loc}(\Omega)$. On the
other hand, we assume even that $\Omega$ is a small neighborhood near a
regular point in a finite dimensional Alexandrov space with curvature bounded
from below. According to [38, 39], there is a $BV_{\rm loc}$-Riemannian metric
$(g_{ij})$ on $\Omega$. However, it is well-known [38] that these $g_{ij}$ may
not be continuous on a _dense_ subset of $\Omega$. Thus, It seems the
mentioned works [28, 53, 5] do not support that one can improve a $L^{1}_{\rm
loc}(\Omega)$ very weak harmonic function to be in $W^{1,2}_{\rm
loc}(\Omega)$, even on an Alexandrov space with curvature bounded from below.

The main result of this paper is the following Weyl’s lemma on RCD-setting.

######  Theorem 1.2 (Weyl’s lemma).

Let $(X,d,\mathfrak{m})$ be an $RCD(K,N)$ space for some $1\leqslant N<\infty$
and $K\in\mathbb{R}$. Let $\Omega\subset X$ be an open domain. If $u\in L_{\rm
loc}^{1}(\Omega)$ is a very weakly harmonic function on $\Omega$, then $u$ is
locally Lipschitz continuous on $\Omega$ (i.e. it has a locally Lipschitz
continuous representative). In particular, it is in $W^{1,2}_{\rm
loc}(\Omega)$.

######  Remark .

When $X=\mathbb{R}^{n}$, the regularity for very weak solutions of (1.3)
relies on the well-known Calderón-Zygmund theory which states if $\Delta u\in
L^{p}_{\rm loc}$ with $p>1$, then $u\in W^{2,p}_{\rm loc}$. In particular, if
$p>n$ then $|\nabla u|$ is in $C^{\alpha}_{\rm loc}$ for some
$\alpha\in(0,1)$. Recently, G. De Philippis and J. Núnez-Zimbrón [13] have
shown that this statement does not hold for general Alexandrov spaces with
curvature bounded from below. For example, let us consider the space $X^{2}$
constructed by Ostu-Shioya in [38], which is a two-dimensional Alexandrov
space with the property that the set of singular points is dense in $X^{2}$.
Given any non-constant harmonic function $u$ on a domain $\Omega\subset
X^{2}$, it was shown in [13, Theorem 1] that $\lim_{R\to
0}\frac{1}{\mathfrak{m}(B_{R}(x))}\int_{B_{R}(x)}|\nabla u|^{2}(x)=0$ at every
point $x\in\Omega$ with ${\rm diam}(\Sigma_{x})<\pi$, where $\Sigma_{x}$ is
the space of directions at $x$. In the two-dimensional Alexandrov space
$X^{2}$, a point $x$ satisfying ${\rm diam}(\Sigma_{x})<\pi$ is equivalent to
that $x$ is a singular point. Since the set of singular points on $X^{2}$ is
dense, if $|\nabla u|$ were assumed to be continuous, it would follow $|\nabla
u|\equiv 0$ on $\Omega$. That is a contradiction.

The first application of Theorem 1.2 is the following local regularity result
for the very weak solutions of Poisson equations on $RCD$ spaces.

######  Corollary .

Let $(X,d,\mathfrak{m})$ be an $RCD(K,N)$ space with $K\in\mathbb{R}$ and
$N\in[1,\infty)$, and let $\Omega\subset X$ be an open subset. Suppose $f\in
L^{2}_{\rm loc}(\Omega)$. If $u\in L_{\rm loc}^{1}(\Omega)$ is a very weak
solution of Poisson equation on $\Omega$ in the sense of

$\int_{\Omega}u\Delta\varphi d\mathfrak{m}=\int_{\Omega}f\varphi
d\mathfrak{m},\qquad\forall\varphi\in\mathrm{Test}_{c}^{\infty}(\Omega)\ \
{\rm and}\ \ \varphi\geqslant 0.$ (1.4)

Then $u$ is in $W^{1,2}_{\rm loc}(\Omega)$.

The second application of Theorem 1.2 is the following Liouville-type result
for $L^{1}$ very weakly subharmonic functions.

######  Corollary .

Let $(X,d,\mathfrak{m})$ be an $RCD(K,N)$ space with $K\in\mathbb{R}$ and
$1\leq N<\infty$, then any $L^{1}(X)$ very weakly subharmonic function on $X$
must be identically constant (namely, it has a constant representative).

Recall that Li-Schoen [32] and Chung [11] gave some examples that a complete
Riemannian manifold admits non-constant nonnegative $L^{1}$ harmonic
functions. Under an assumption of Ricci lower bounds, P. Li [31] proved a
Liouville-type result for $L^{1}$ subharmonic functions on complete Riemannian
manifolds. We will prove Corollary 1 along the same line in [31] and combine
Theorem 1.2.

At last, as a byproduct, we give a remark on the Lipschitz estimates for weak
solutions of a class of elliptic equations on Euclidean spaces with dis-
continuous coefficients. Let $n\geqslant 2$ and let $D\subset\mathbb{R}^{n}$
be a bounded domain. Suppose that $u\in W^{1,2}_{\rm loc}(D)$ is a weak
solution of elliptic equations of divergence form

$Lu:={\rm
div}(A(x)Du)=\sum_{i,j=1}^{n}\partial_{i}(a^{ij}\partial_{j}u)=0,\quad
1\leqslant i,j\leqslant n,$ (1.5)

where the matrix of coefficients $A(x)=(a^{ij})_{i,j=1}^{n}$ is symmetric,
bounded, measurable and uniformly elliptic, i.e.

$\|a^{ij}\|_{L^{\infty}(D)}\leqslant\Lambda,\qquad\lambda|\xi|^{2}\leqslant\sum_{i,j=1}^{n}a^{ij}(x)\xi_{i}\xi_{j},\quad{\rm
a.e.}\ x\in D,\quad\forall\xi\in\mathbb{R}^{n}$ (1.6)

for some $\lambda,\Lambda>0$. It is well known [6] that any weak solution $u$
of $Lu=0$ is in $C^{1}(D)$ provided that $A$ satisfies the Dini continuity
condition. Recently, this Dini continuity condition were weaken to others
Dini-type conditions by Y. Li [33], Dong-Kim [15] and Maz’ya-McOwen [36].
However, an example constructed by Jin-Maz’ya-Schaftingen [28] shows that,
general speaking, the continuity of the coefficient is not sufficient to
ensure $u\in Lip_{\rm loc}(D)$.

Inspired by the regularity of harmonic functions on $RCD$ spaces, we first
introduce a class of models for elliptic coefficient, which plays the role of
constant coefficient in the classical regularity theory.

######  Definition .

Let
$\overline{A}=(\overline{a}^{ij}):\mathbb{R}^{n}\to\mathbb{R}^{n}\times\mathbb{R}^{n}$
be symmetric, bounded, measurable and uniformly elliptic on the each ball
$B_{R}(0)\subset\mathbb{R}^{n}$ with $R>0$. Let $n\geqslant 3$. We first
introduce a Riemannian metric $\overline{g}_{ij}(x)$ on $\mathbb{R}^{n}$ by

$(\overline{g}_{ij})^{-1}(x)=\overline{g}^{ij}(x):=\frac{\overline{a}^{ij}(x)}{\left[{\rm
det}(\overline{a}^{ij}(x))\right]^{\frac{1}{n-2}}},\qquad 1\leqslant
i,j\leqslant n.$ (1.7)

We denote by $d_{\overline{g}}$ the distance function induced by Riemannian
metric $(\mathbb{R}^{n},\overline{g}_{ij})$.

The coefficient $\overline{A}$ is called a _conical coefficient of nonnegative
curvature_ if the metric space $(\mathbb{R}^{n},d_{\overline{g}})$ is a cone
with nonnegative curvature in the sense of Alexandrov.

We remark that if $\bar{a}^{ij}$ is constant coefficient, then the Riemannian
metric $\bar{g}_{ij}=const$, and then $(\mathbb{R}^{n},\bar{g}_{ij})$ is flat
(i.e., sectional curvature is zero).

There are also some _dis-continuous_ conical coefficient of nonnegative
curvature. For example, consider the graph $\Gamma_{f}$ of a convex function

$f(x_{1},x_{2},\cdots,x_{n})=\left(a_{1}x_{1}^{2}+a_{2}x^{2}_{2}+\cdots+a_{n}x^{2}_{n}\right)^{1/2},$

where $a_{i}\in(0,+\infty)$ for all $i=1,\cdots,n$. It is well known that
$\Gamma_{f}$ is a cone of nonnegative curvature in the sense of Alexandrov.
Under the coordinate system $(x_{1},x_{2},\cdots,x_{n})$, its Riemannian
metric is given by

$\bar{g}_{ij}=\delta_{ij}+\frac{a_{i}x_{i}}{f}\cdot\frac{a_{j}x_{j}}{f},$

for all $1\leqslant i,j\leqslant n$. Then $\bar{A}(x)=(\bar{a}^{ij}(x))$ with

$\bar{a}^{ij}:=\sqrt{{\rm
det}(\bar{g}_{ij})}\bar{g}^{ij}=\left(1+\frac{\sum_{i=1}^{n}(a_{i}x_{i})^{2}}{f^{2}}\right)^{1/2}\left(\delta_{ij}+\frac{a_{i}x_{i}}{f}\cdot\frac{a_{j}x_{j}}{f}\right)^{-1}$

is a conical coefficient of nonnegative curvature, and $\bar{a}^{ij}(x)$ is
not continuous at $0.$

######  Theorem 1.3.

Let $u\in W^{1,2}_{\rm loc}(B_{1}(0))$ be a weak solution of elliptic
equations of divergence form ${\rm div}(A(x)Du)=0,$ where the matrix of
coefficients $A(x)=(a^{ij})_{i,j=1}^{n}$ is symmetric, bounded, measurable and
uniformly elliptic. Let $n\geqslant 3.$

Suppose that $A$ is Dini asymptotic to a conical coefficient of nonnegative
curvature in the sense that there exists a conical coefficient of nonnegative
curvature $\overline{A}=(\overline{a}^{ij})$ such that

$\int_{0}^{1}\frac{\omega_{A,\overline{A}}(t)}{t}dt<\infty,\quad\ where\ \
\omega_{A,\overline{A}}(t):=\|A(x)-\overline{A}(x)\|_{L^{\infty}(B_{t}(0))}.$
(1.8)

Then

$\limsup_{r\to 0^{+}}\fint_{B_{r}(0)}|\nabla u|^{2}d\mathcal{L}^{n}\leqslant
C_{0}\fint_{B_{1}(0)}|\nabla u|^{2}d\mathcal{L}^{n},$ (1.9)

where the constant $C_{0}$ depends only on $n,\lambda,\Lambda$ and
$\omega_{A,\overline{A}}$, and $\mathcal{L}^{n}$ is the $n$-dimensional
Lebesgue measure on $\mathbb{R}^{n}$.

This extends the result in [6].

Organization of the paper. In Section 2, we will recall some ingredients on
$RCD(K,N)$ spaces. In Section 3, we will prove Weyl’s lemma (Theorem 1.2).
Corollary 1 and 1 will be proved in section 4 and 5. Finally, we discuss the
gradient estimates of solutions to elliptic equations of divergence form,
Theorem 1.3, in section 6.

Acknowledgment. The second author was partially supported by NSFC 12025109.
The third author was partially supported by NSFC 12271530.

##  2\. Preliminaries

Let $(X,d,\mathfrak{m})$ be a metric measure space, i.e. $(X,d)$ is a complete
and separable metric space endowed with a non-negative Borel metric which is
finite on bounded sets. Throughout this paper, we always assume that
$(X,d,\mathfrak{m})$ is an $RCD(K,N)$ space with $N\in[1,+\infty)$ and
$K\in\mathbb{R}$. For explicit definitions, one can refer to [18, 4, 17] and
the survey [1].

Let $(X,d,\mathfrak{m})$ be an $RCD(K,N)$ space with $K\in\mathbb{R}$ and
$N\in[1,+\infty)$. The generalized Bishop-Gromov inequality (see [34, 47])
implies a local measure doubling: for any $R>0$, there exists a constant
$C_{N,K,R}>0$ such that

$\frac{\mathfrak{m}(B_{r_{2}}(x))}{\mathfrak{m}(B_{r_{1}}(x))}\leq
C_{N,K,R}\cdot\left(\frac{r_{2}}{r_{1}}\right)^{N},\quad\forall
r_{1},r_{2}\in(0,R)\ \ {\rm with}\ \ r_{1}<r_{2}.$ (2.1)

The Sobolev spaces on $(X,d,\mathfrak{m})$ were established in [7, 43, 2, 18]
via several different approaches. We refer the readers to the textbook [20,
Chapter 2] for basic definitions and properties of
$W^{1,2}(X):=W^{1,2}(X,d,\mathfrak{m})$ and $W^{1,2}_{\rm
loc}(X):=W^{1,2}_{\rm loc}(X,d,\mathfrak{m})$. For any $f\in W^{1,2}(X),$ we
denote by $|\nabla f|\in L^{2}(X)$ the minimal weak upper gradient of $f$ (see
for example [20, Sect. 2.1]). For any $f,h\in W^{1,2}(X)$, the inner product
$\langle{\nabla f},{\nabla h}\rangle$ is given by

$\langle{\nabla f},{\nabla
h}\rangle=\frac{1}{4}\left(|\nabla(f+h)|^{2}-|\nabla(f-h)|^{2}\right)\ \in
L^{1}(X).$

The inner product $\langle{\nabla f},{\nabla h}\rangle$ is symmetric and bi-
linear, and satisfies Cauchy-Schwarz inequality, Chain rule and Leibniz rule
(see [45, Appendix] or [20]).

For any open domain $\Omega\subset X$, the space $W^{1,2}_{0}(\Omega)$ is the
closure of $Lip_{c}(\Omega)$ with support in $\Omega$ under $W^{1,2}(X)$-norm.
The space $W^{1,2}_{\rm loc}(\Omega)$ is the space of Borel functions
$f:\Omega\to\mathbb{R}$ such that $f\chi\in W^{1,2}_{\rm loc}(X)$ for any
Lipschitz function $\chi:X\to[0,1]$ such that $d\big{(}{\rm
supp}(\chi),X\setminus\Omega\big{)}>0$, where a function $f\chi$ is taken $0$
by definition on $X\setminus\Omega.$

Let $\\{P_{t}\\}_{t>0}$ be the heat flow on $X$ associated with the energy
$\int_{X}|\nabla f|^{2}d\mathfrak{m}$ for any $f\in W^{1,2}(X)$, and let
$\Delta$ be the infinitesimal generator of $P_{t}$ with domain $D(\Delta)$. It
is well know that (see for instance [20, Chapter 5])

$\int_{X}h\Delta fd\mathfrak{m}=-\int_{X}\langle{\nabla f},{\nabla h}\rangle
d\mathfrak{m},\quad\ \forall f\in D(\Delta),\ \ h\in W^{1,2}(X).$ (2.2)

For any $f\in L^{2}(X)$, it is well known that $P_{t}f\in D(\Delta)$, and that
the curve $t\mapsto P_{t}f$ is in $C^{1}\left((0,+\infty),L^{2}(X)\right)$
with $\frac{\mathrm{d}}{\mathrm{d}t}P_{t}f=\Delta P_{t}f.$ If $f\in
D(\Delta)$, then

$P_{t}(\Delta f)=\Delta(P_{t}f),\qquad\forall t>0.$

Since $P_{t}$ is a family of linear contractions on $L^{2}(X)$ for any $t>0$,
it can be uniquely extended to be a family of linear contractions on
$L^{p}(X)$ for any $p\in[1,+\infty]$.

Since $RCD(K,N)$ spaces are locally doubling and satisfy a local Poincaré
inequality (see [49, 41]), according to [45, 46, 3], there exists a heat
kernel $p_{t}(x,y)$ on $(0,+\infty)\times X\times X$ such that

$P_{t}f(x)=\int_{X}p_{t}(x,y)f(y)d\mathfrak{m}(y)\ \ \ \ \forall\ t>0$

for every $f\in L^{2}(X)$. It was proved [46, 3] that $(x,y)\mapsto
p_{t}(x,y)$ is locally Hölder continuous on $X\times X$ for each $t>0$, and
that $t\mapsto p_{t}(x,y)$ is analytic in $(0,+\infty)$ for each $x,y\in X$.
Moreover, it was showed in [26] that there exist constants $C_{1},C_{2}>0$
depending only on $N$ and $K$ such that

$\displaystyle\frac{1}{C_{1}\mathfrak{m}\left(B_{\sqrt{t}}(x)\right)}\exp$
$\displaystyle\left\\{-\frac{d^{2}(x,y)}{3t}-C_{2}t\right\\}\leq p_{t}(x,y)$
(2.3)
$\displaystyle\leq\frac{C_{1}}{\mathfrak{m}\left(B_{\sqrt{t}}(x)\right)}\exp\left\\{-\frac{d^{2}(x,y)}{5t}+C_{2}t\right\\}$

for all $t>0$ and all $x,y\in X$, and

$\left\lvert\nabla
p_{t}(\cdot,y)\right\rvert(x)\leq\frac{C_{1}}{\sqrt{t}\cdot\mathfrak{m}\left(B_{\sqrt{t}}(x)\right)}\exp\left\\{-\frac{d^{2}(x,y)}{5t}+C_{2}t\right\\}$
(2.4)

for all $t>0$ and $\mathfrak{m}$-a.e. $x,y\in X$. Combining Davies [12,
Theorem 4] and (2.3), the estimate

$\left\lvert\frac{\partial}{\partial
t}p_{t}(x,y)\right\rvert=\left\lvert\Delta
p_{t}(\cdot,y)\right\rvert(x)\leq\frac{C_{1}}{t\cdot\mathfrak{m}\left(B_{\sqrt{t}}(x)\right)}\exp\left\\{-\frac{d^{2}(x,y)}{5t}+C_{2}t\right\\}$
(2.5)

holds for all $t>0$ and $\mathfrak{m}$-a.e. $x,y\in X$. An important property
of $p_{t}(x,y)$ is the stochastic completeness (see [45]):

$\int_{X}p_{t}(x,y)d\mathfrak{m}(y)=1,\quad\forall t>0\ {\rm and}\ x\in X.$
(2.6)

The class of test functions on $RCD(K,N)$ spaces was introduced in [20]:

$\mathrm{Test}^{\infty}(X):=\left\\{f\in D(\Delta)\cap
L^{\infty}(X):\left\lvert\nabla f\right\rvert\in L^{\infty}(X),\ \Delta f\in
L^{\infty}(X)\cap W^{1,2}(X)\right\\},$

which is an algebra and is dense in $W^{1,2}(X)$ (see [20, Sect. 6.1.3]).
Given any $\varphi\in\mathrm{Test}^{\infty}(X)$, it is well-known [4] that it
has a (Lipschitz) continuous representative. Hence we always assume that every
function in $\mathrm{Test}^{\infty}(X)$ is continuous. The following lemma is
a standard fact about the heat kernel.

######  Lemma .

(1) For any $t>0$ and $x_{0}\in X$, the function $p_{t}(x_{0},\cdot)\in{\rm
Test}^{\infty}(X)$.
(2) For any $\varphi\in{\rm Test}^{\infty}(X)$ and $t>0$, it holds
$P_{t}\varphi\in{\rm Test}^{\infty}(X).$

###### Proof.

The assertion (1) follows from $p_{t/2}(x_{0},\cdot)\in L^{\infty}(X)\cap
L^{1}(X)$,

$p_{t}(x_{0},x)=\big{[}P_{t/2}\big{(}p_{t/2}(x_{0},\cdot)\big{)}\big{]}(x)$

and the estimates (2.3)-(2.5).

For (2), since $\varphi\in D(\Delta)\cap L^{\infty}(X)$ and $\Delta\varphi\in
L^{\infty}(X)$, by the contraction of $P_{t}$ on $L^{\infty}(X)$, we have
$P_{t}\varphi\in L^{\infty}(X)$ and
$\Delta(P_{t}\varphi)=P_{t}(\Delta\varphi)\in L^{\infty}(X)$. From
$\Delta\varphi\in L^{2}(X)$, we obtain
$\Delta(P_{t}\varphi)=P_{t}(\Delta\varphi)\in W^{1,2}(X).$

Finally, since $|\nabla\varphi|\in L^{\infty}(X)$, by using the Bakry-Ledoux
inequality on $RCD(K,N)$ spaces (see [17, Theorem 4.3]) and the contraction of
$P_{t}$ on $L^{\infty}(X)$ again, we conclude that

$|\nabla P_{t}\varphi|^{2}\leqslant e^{-2Kt}P_{t}(|\nabla\varphi|^{2})\in
L^{\infty}(X).$

The proof of (2) is finished. ∎

The existence of good cut-off functions is an important ingredient in
geometric analysis. For an $n$-dimensional Riemannian manifold $M^{n}$ with
Ricci curvature bounded from below, one can estimate merely the upper bounds
for the Laplacian of distance functions, by the Laplace comparison theorem. In
this aspect, Schoen-Yau in [42, Theorem 4.2 in Chapter I] constructed
distance-like functions on $M^{n}$ having $L^{\infty}$-bounds of the
Laplacian. Cheeger-Colding [8] found good cut-off functions with $L^{\infty}$
estimate on the Laplacian, which is a key technical tool in the Cheeger-
Colding theory of Ricci limit spaces (see [8, 9, 10]). The existence of a
regular cut-off function in $RCD$ setting was proved in [4, 22, 37]. We need
the following variant of [37, Lemma 3.1].

######  Lemma .

Let $(X,d,\mathfrak{m})$ be an $RCD(K,N)$ space with $K\in\mathbb{R}$ and
$1\leq N<\infty$. For each $r_{0}>0$, there exists a constant
$C_{K,N,r_{0}}>0$ such that for any ball $B_{R}(x)\subset X$ with radius
$R\geqslant r_{0}$, there exists a cut-off function
$\eta:X\to\left[0,1\right]$ such that

$\mathrm{(i)}$ $\eta\equiv 1$ on $B_{R}(x)$ and ${\rm supp}(\eta)\subset
B_{2R}(x)$,

$\mathrm{(ii)}$ $\eta\in\mathrm{Test}^{\infty}(X)$, and moreover for
$\mathfrak{m}$-a.e. $x\in X$,

$|\Delta\eta|(x)+|\nabla\eta|(x)\leqslant\frac{C_{N,K,r_{0}}}{R}.$

(Remark that the constant $C_{N,K,r_{0}}$ does not depend on $R$.)

###### Proof.

This is a slight modification of [37, Lemma 3.1]. For the convenience of
readers, we give the proof as follows. Fix a ball $B_{2R}(x)\subset X$ with
radius $R\geqslant r_{0}$ and let $\varphi$ be a Lipschitz continuous function
defined as $\varphi\equiv 1$ on $B_{R}(x)$, $\varphi\equiv 0$ on $X\setminus
B_{2R}(x)$ and $\varphi(y)=\frac{2R-d(x,y)}{R}$ on $B_{2R}(x)\setminus
B_{R}(x)$. As in [37, Lemma 3.1], we consider the heat flow regularization
$\varphi_{t}:=P_{t}\varphi$. According to [17], we can choose continuous
representatives of $\varphi_{t},|\nabla\varphi_{t}|$ and
$|\Delta\varphi_{t}|$; and moreover, by the Bakry-Ledoux’s inequality (see
[17, Theorem 4.3]), we have for everywhere

$|\nabla\varphi_{t}|^{2}+\frac{4Kt^{2}}{N(e^{2Kt}-1)}|\Delta\varphi_{t}|^{2}\leqslant
e^{-2Kt}P_{t}(|\nabla\varphi|^{2})\leqslant\frac{e^{-2Kt}}{R^{2}},$ (2.7)

where we have used $|\nabla\varphi|\leqslant 1/R$ and $L^{\infty}$-contraction
of $P_{t}$. It follows that

$|\varphi_{t}(y)-\varphi(y)|\leqslant\frac{1}{R}\int_{0}^{t}\sqrt{\frac{N(e^{2Ks}-1)e^{-2Ks}}{4Ks^{2}}}ds:=\frac{F_{K,N}(t)}{R}\leqslant\frac{F_{K,N}(t)}{r_{0}},\quad\forall
y\in X,$

by $R\geqslant r_{0}$, where $F_{K,N}(\cdot):\mathbb{R}^{+}\to\mathbb{R}^{+}$
is continuous and $\lim_{t\to 0^{+}}F_{K,N}(t)=0.$ It is clear that there
exists $t_{K,N,r_{0}}>0$ such that $F_{K,N}(t_{K,N,r_{0}})\leqslant r_{0}/4$.
Therefore we have $\varphi_{t_{K,N,r_{0}}}(y)\in[3/4,1]$ for every $y\in
B_{R}(x)$ and $\varphi_{t_{K,N,r_{0}}}(y)\in[0,1/4]$ for every $y\in
X\setminus B_{2R}(x)$. As in [37, Lemma 3.1], the desired cut-off function
$\eta$ can be given by composition with a $C^{2}$-function
$f:\mathbb{R}\to[0,1]$ such that $f\equiv 1$ on $[3/4,1]$ and $f\equiv 0$ on
$[1/4,0]$; indeed, $\eta:=f\circ\varphi_{t_{K,N,r_{0}}}$ is now identically
equal to one on $B_{R-r_{0}}(x)$, vanishes identically on $X\setminus
B_{R}(x)$ and, by (2.7) and the Chain rule (see for instance [20, Proposition
5.2.3]), it satisfies the estimate $|\nabla\eta|+|\Delta\eta|\leqslant
C(K,N,r_{0})/R$ as desired. ∎

We recall the following measure-valued Laplacian introduced in [18].

######  Definition .

Let $\Omega\subset X$ be an open subset and let $f\in W_{\rm
loc}^{1,2}(\Omega)$. It is called $f\in D(\bm{\Delta})$ on $\Omega$ if there
exists a signed Radon measure $\mu$ on $\Omega$ such that

$\int_{\Omega}\varphi d\mu=-\int_{\Omega}\langle{\nabla\varphi},{\nabla
f}\rangle d\mathfrak{m},\quad\forall\varphi\in Lip_{c}(\Omega).$

Such $\mu$ is unique and is denoted by $\bm{\Delta}f$ (which depends on
$\Omega$).

Both the Chain rule and the Leibniz rule for this ${\bf\Delta}$ were proved in
[18, Chapter 4]. When $\Omega=X$, it was showed [18, Proposition 4.24] that
for any $f\in W^{1,2}(X)$,

$f\in D(\Delta)\quad\Longleftrightarrow\quad f\in D({\bf\Delta})\ {\rm and}\
{\bf\Delta}f=g\cdot\mathfrak{m}\ {\rm for\ some}\ g\in L^{2}(X).$

In this case ${\bf\Delta}f=\Delta f\cdot\mathfrak{m}$.

The combination of [52, Theorem 1.5] and [25, Theorem 3.1] states the
following.

######  Lemma .

Let $\Omega\subset X$ be a domain and $f\in W^{1,2}_{\rm loc}(\Omega)$.
Suppose that $\mathbf{\Delta}f=g\cdot\mathfrak{m}$ on $\Omega$ for some $g\in
L^{\infty}_{\rm loc}(\Omega)$. Then $f\in Lip_{\rm loc}(\Omega)$, and moreover
for any ball $B_{R}(x)\subset\subset\Omega$, it holds

$\sup_{B_{R/2}(x)}|\nabla f|\leqslant
C_{N,K,R}\big{(}\|f\|_{L^{1}(B_{R}(x))}+\|g\|_{L^{\infty}(B_{R}(x))}\big{)}$

for some constant $C_{N,K,R}.$ In particular, any harmonic function $f$
(namely, ${\bf\Delta}f=0$ on $\Omega$) is locally Lipschitz continuous.

At last, we recall a simple fact as follows.

######  Lemma .

Let $\Omega\subset X$ be open. If $f\in D({\bf\Delta})\cap C(\Omega)$ then
${\bf\Delta}f\big{(}\Omega\setminus{\rm supp}(f)\big{)}=0$. In particular, if
$\Omega=X$ and if $f\in D(\Delta)\cap C(X)$ then $\Delta f(x)=0$ for
$\mathfrak{m}$-a.e. $x\not\in{\rm supp}(f)$.

###### Proof.

Given any $\varphi\in Lip_{c}(\Omega)$ with ${\rm
supp}(\varphi)\subset\Omega\setminus{\rm supp}(f)$, by applying [7, Corollary
2.25], we have that $|\nabla\varphi|=0$ $\mathfrak{m}$-a.e. in
$\Omega\setminus{\rm supp}(\varphi)\supset{\rm supp}(f)$ and that $|\nabla
f|=0$ $\mathfrak{m}$-a.e. in $\Omega\setminus{\rm supp}(f)$. It follows that
$\langle{\nabla f},{\nabla\varphi}\rangle=0$ $\mathfrak{m}$-a.e. in $\Omega$.
Indeed, Cauchy-Schwarz inequality in [45, Appendix] implies

$|\langle{\nabla f},{\nabla\varphi}\rangle|\leqslant|\nabla
f|\cdot|\nabla\varphi|\quad\mathfrak{m}{\rm-a.e.\ in}\ \Omega.$ (2.8)

Therefore, by the definition of ${\bf\Delta}f$, we have
${\bf\Delta}f(\varphi)=0,$ which shows the first assertion.

If $\Omega=X$ and $f\in D(\Delta)$, by [18, Proposition 4.24], we have
${\bf\Delta}f=\Delta f\cdot\mathfrak{m}.$ Applying the first assertion, we
have $\Delta f(x)=0$ for $\mathfrak{m}$-a.e. $x\in X\setminus{\rm supp}(f).$
The proof is finished. ∎

##  3\. Distributions and Weyl’s lemma

Let $(X,d,\mathfrak{m})$ be an $RCD(K,N)$ space for some $K\in\mathbb{R}$ and
$N\in[1,+\infty)$. Recall that the class of test functions

$\mathrm{Test}^{\infty}(X):=\left\\{\varphi\in D(\Delta)\cap
L^{\infty}(X):\left\lvert\nabla\varphi\right\rvert\in L^{\infty}(X),\
\Delta\varphi\in L^{\infty}(X)\cap W^{1,2}(X)\right\\},$

is dense in $W^{1,2}(X)$ and is an algebra (see [20, Sect 6.1]). We equip
$\mathrm{Test}^{\infty}(X)$ with the norm

$\|\varphi\|_{E}:=\|\varphi\|_{L^{\infty}(X)}+\||\nabla\varphi|\|_{L^{\infty}(X)}+\|\Delta\varphi\|_{L^{\infty}(X)},$

and denote by $E(X)$ the linear normed space
$\left(\mathrm{Test}^{\infty}(X),\|\cdot\|_{E}\right)$.

Let $\Omega\subset X$ be an open domain, we set

$\begin{split}\mathrm{Test}_{c}^{\infty}(\Omega)&:=\left\\{\varphi\in\mathrm{Test}^{\infty}(X)\big{|}\
{\rm supp}(\varphi)\subset\Omega\ {\rm and\ supp}(\varphi)\ {\rm is\
compact}\right\\}\\\ {\rm and}\quad\
\mathrm{Test}_{c,+}^{\infty}(\Omega)&:=\\{\varphi\in\mathrm{Test}_{c}^{\infty}(\Omega)|\
\varphi\geqslant 0\\}.\end{split}$

From Lemma 2, for each open domain $\Omega\subset X$ and every compact set
$Q\subset\Omega$, there exists $\eta\in\mathrm{Test}_{c}^{\infty}(\Omega)$
such that $\eta\equiv 1$ on $Q$ and $0\leqslant\eta\leqslant 1$.

######  Lemma .

The space $\mathrm{Test}_{c,+}^{\infty}(\Omega)$ is dense in $\\{f\in
Lip_{c}(\Omega)|f\geqslant 0\\}$. Consequently, the space
$\mathrm{Test}_{c}^{\infty}(\Omega)$ is dense in $W^{1,2}_{0}(\Omega)$.

###### Proof.

This lemma is somewhat known. For the completeness, we present a proof here.
Let $f\in Lip_{c}(\Omega)$ be nonnegative. Here $f$ is taken $0$ by the
definition on $X\setminus\Omega$. According to [20, Proposition 6.1.8]), we
take a sequence of nonnegative $f_{j}\in{\rm Test}^{\infty}(X)$ such that
$f_{j}\to f$ in $W^{1,2}(X)$, as $j\to+\infty$.

Since ${\rm supp}(f)\subset\Omega$ is compact, we have $d\big{(}{\rm
supp}(f),X\setminus\Omega\big{)}>0.$ According to Lemma 2, there exists a cut-
off function $\eta\in{\rm Test}_{c}^{\infty}(\Omega)$,
$0\leqslant\eta\leqslant 1$ and $\eta\equiv 1$, on some domain containing
${\rm supp}(f)$. Then we have $f\eta\equiv f$ on $X$ and $f_{j}\eta\in{\rm
Test}^{\infty}_{c}(\Omega)$ for all $j$. By
$|f_{j}\eta-f|=|(f_{j}-f)\eta|\leqslant|f_{j}-f|$ and

$|\nabla(f_{j}\eta-f)|^{2}\leqslant
2|\nabla(f_{j}-f)|^{2}\eta^{2}+2(f_{j}-f)^{2}|\nabla\eta|^{2},$

We conclude that $f_{j}\eta\to f$ in $W^{1,2}(X)$ as $j\to+\infty$, since
$\eta\in Lip(X)$. Notice that $f_{j}\eta\geqslant 0$. The proof of the first
assertion is finished.

The second assertion follows from the first one by noticing the facts that the
space $Lip_{c}(\Omega)$ is dense in $W^{1,2}_{0}(\Omega)$ and that for any
$f\in Lip_{c}(\Omega)$ it holds $f=f^{+}-f^{-}$ and $f^{\pm}\in
Lip_{c}(\Omega)$. ∎

######  Definition .

Let $\Omega\subset X$. A distribution  $T$ in $\Omega$ is a linear functional
on $\mathrm{Test}_{c}^{\infty}(\Omega)$ such that for every compact set
$Q\subset\Omega$ there exists a constant $C_{Q}$ such that

$|T(\varphi)|\leqslant
C_{Q}\|\varphi\|_{E},\quad\forall\varphi\in\mathrm{Test}_{c}^{\infty}(\Omega)\
{\rm with}\ {\rm supp}(\varphi)\subset Q.$

It is clear that every element $f\in E^{\prime}(X)$, the dual space of $E(X)$,
defines a distribution on $X$.

######  Definition .

Let $f\in L^{1}_{\rm loc}(\Omega)$, the _distributional Laplacian_ of $f$,
denoted by $\Delta_{\mathscr{D},\Omega}f$, is defined as a linear functional
on $\mathrm{Test}_{c}^{\infty}(\Omega)$ by

$\Delta_{\mathscr{D},\Omega}f(\varphi):=\int_{\Omega}f\Delta\varphi
d\mathfrak{m},\quad\forall\varphi\in\mathrm{Test}_{c}^{\infty}(\Omega).$

If $\Omega=X$, we denote $\Delta_{\mathscr{D}}f:=\Delta_{\mathscr{D},X}f$ for
any $f\in L^{1}_{\rm loc}(X)$.

For each $f\in L^{1}_{\rm loc}(\Omega)$, it is easy to check that
$\Delta_{\mathscr{D},\Omega}f$ is a distribution in $\Omega$. Given two
domains $\Omega\subset\Omega^{\prime}$ and a function $f\in L^{1}_{\rm
loc}(\Omega^{\prime})$, it is not hard to check that
$\Delta_{\mathscr{D},\Omega}f=\Delta_{\mathscr{D},\Omega^{\prime}}f$ in
$\Omega$. In fact, for any $\varphi\in{\rm Test}^{\infty}_{c}(\Omega)$, Lemma
2 implies that $\Delta\varphi(x)=0$ for $\mathfrak{m}$-a.e.
$x\in\Omega^{\prime}\setminus\Omega$. It follows that
$\int_{\Omega^{\prime}}f\Delta\varphi
d\mathfrak{m}=\int_{\Omega}f\Delta\varphi d\mathfrak{m}$.

First of all, we show that the distribution Laplacian is compatible with the
measure-valued Laplacian in [18], and also, when $\Omega=X$, with the
generator of $L^{p}(X)$-semigroup $P_{t}$ for each $p\in[1,\infty)$.

######  Lemma .

Let $\Omega\subset X$ be an open domain. If $f\in W^{1,2}_{\rm loc}(\Omega)$,
then $\Delta_{\mathscr{D},\Omega}f={\bf\Delta}f$ on
$\mathrm{Test}^{\infty}_{c}(\Omega).$ In particular, if $u$ is
subharmonic/superharmonic on $\Omega$ then it is very weak
subharmonic/superharmonic on $\Omega$.

###### Proof.

Take any $\varphi\in\mathrm{Test}^{\infty}_{c}(\Omega).$ Since $d\big{(}{\rm
supp}(\varphi),X\setminus\Omega\big{)}>0$, there exists a Lipschitz function
$\chi:X\to[0,1]$ such that $\chi\equiv 1$ on ${\rm supp}(\varphi)$ and ${\rm
supp}(\chi)\subset\Omega$. Noticing that ${\rm supp}(\varphi)$ is compact, we
can choose $\chi$ such that ${\rm supp}(\chi)$ is compact too. From the
definition of $W^{1,2}_{\rm loc}(\Omega)$ and $f\in W^{1,2}_{\rm
loc}(\Omega)$, we have $h:=f\cdot\chi\in W^{1,2}(X)$; moreover, since
$h(x)=f(x)$ for $\mathfrak{m}$-a.e. $x\in{\rm supp}(\varphi)$, by using [7,
Corollary 2.25], we get $|\nabla f|(x)=|\nabla h|(x)$ for $\mathfrak{m}$-a.e.
$x\in{\rm supp}(\varphi)$ and $|\nabla\varphi|(x)=0$ for $\mathfrak{m}$-a.e.
$x\not\in{\rm supp}(\varphi)$. It follows that (see (2.8))

$\langle{\nabla h},{\nabla\varphi}\rangle=\langle{\nabla
f},{\nabla\varphi}\rangle\quad\mathfrak{m}{\rm-a.e.\ in}\ \ X.$

By the definition of ${\bf\Delta}f$ (see Definition 2), we obtain

$\begin{split}{\bf\Delta}f(\varphi)&=-\int_{\Omega}\langle{\nabla
f},{\nabla\varphi}\rangle d\mathfrak{m}=-\int_{X}\langle{\nabla
f},{\nabla\varphi}\rangle d\mathfrak{m}\\\ &=-\int_{X}\langle{\nabla
h},{\nabla\varphi}\rangle d\mathfrak{m}.\end{split}$ (3.1)

Since $h\in W^{1,2}(X)$ and $\varphi\in{\rm Test}^{\infty}(X)\subset
D(\Delta)$, we have

$\int_{X}h\Delta\varphi d\mathfrak{m}=-\int_{X}\langle{\nabla
h},{\nabla\varphi}\rangle d\mathfrak{m}.$

Together this with (3.1) and the facts that $f=h$ on ${\rm supp}(\varphi)$ and
that $\Delta\varphi(x)=0$ for $\mathfrak{m}$-a.e. $x\not\in{\rm
supp}(\varphi)$ (see Lemma 2), we conclude that

$\begin{split}{\bf\Delta}f(\varphi)&=\int_{X}h\Delta\varphi
d\mathfrak{m}=\int_{{\rm supp}(\varphi)}h\Delta\varphi d\mathfrak{m}\\\
&=\int_{{\rm supp}(\varphi)}f\Delta\varphi
d\mathfrak{m}=\int_{\Omega}f\Delta\varphi
d\mathfrak{m}=\Delta_{\mathscr{D},\Omega}f(\varphi).\end{split}$

The proof is finished. ∎

######  Lemma .

For each $p\in[1,\infty)$, if $f\in D(\Delta^{(p)})$, then
$\Delta_{\mathscr{D}}f=\Delta^{(p)}f$ in the sense of distributions on $X$.
Here $\Delta^{(p)}$ is the generator of $P_{t}$ as semi-group acting on
$L^{p}(X)$. That is,

$\begin{split}D(\Delta^{(p)})&:=\left\\{f\in L^{p}(X)\Big{|}\lim_{t\to
0^{+}}\frac{P_{t}f-f}{t}\ {\rm exists\ in}\ L^{p}\right\\}\ \quad{\rm and}\\\
\Delta^{(p)}f&:=\lim_{t\to 0^{+}}\frac{P_{t}f-f}{t}.\end{split}$

###### Proof.

Take any $\varphi\in{\rm Test}_{c}^{\infty}(X)$. Since $\Delta\varphi\in
L^{2}(X)\cap L^{\infty}(X)$, we have

$\left\|\frac{P_{t}\varphi-\varphi}{t}\right\|_{L^{\infty}(X)}\leqslant\frac{1}{t}\int_{0}^{t}\|P_{s}(\Delta\varphi)\|_{L^{\infty}(X)}ds\leqslant\|\Delta\varphi\|_{L^{\infty}(X)},$

where we have used the contraction of $P_{s}$ on $L^{\infty}(X)$ for any
$s>0$. Notice that $\frac{P_{t}\varphi-\varphi}{t}\to\Delta\varphi$ in $L^{2}$
as $t\to 0^{+}$ implies that there exist a sequence $t_{j}\to 0^{+}$ such that
$\frac{P_{t_{j}}\varphi(x)-\varphi(x)}{t_{j}}\to\Delta\varphi(x)$ for
$\mathfrak{m}$-a.e. $x\in X$. Since $f\in L^{p}(X)$, the dominated converging
theorem implies that

$\int_{X}f\frac{P_{t_{j}}\varphi-\varphi}{t_{j}}d\mathfrak{m}\ \to\
\int_{X}f\Delta\varphi d\mathfrak{m},\quad{\rm as}\ \ t_{j}\to 0^{+}.$ (3.2)

On the other hand, recalling that $f\in D(\Delta^{(p)})$ states
$\frac{P_{t}f-f}{t}\to\Delta^{(p)}f$ in $L^{p}(X)$ as $t\to 0^{+}$, and by
using $p_{t}(x,y)$ is symmetric, we get

$\int_{X}f\frac{P_{t}\varphi-\varphi}{t}d\mathfrak{m}=\int_{X}\frac{P_{t}f-f}{t}\varphi
d\mathfrak{m}\ \to\ \int_{X}\Delta^{(p)}f\cdot\varphi d\mathfrak{m}\quad{\rm
as}\ \ t\to 0^{+}.$

Therefore, by combining with (3.2), we conclude that

$\int_{X}f\Delta\varphi d\mathfrak{m}=\int_{X}\Delta^{(p)}f\cdot\varphi
d\mathfrak{m},\quad\forall\varphi\in{\rm Test}^{\infty}_{c}(X).$

This is $\Delta_{\mathscr{D}}f=\Delta^{(p)}f$ in the sense of distributions. ∎

We can now show the following special case of Weyl’s lemma for $L^{1}$
functions with compact support, which is the core of this paper.

######  Lemma .

Let $v\in L^{1}(X)$ such that $v(x)=0$ a.e. $x\in X\setminus B_{R}(x_{0})$ for
some ball $B_{R}(x_{0})\subset X.$ Suppose that
$\Delta_{\mathscr{D}}v(\varphi)=0$ for any nonnegative
$\varphi\in\mathrm{Test}^{\infty}_{c}(X)$ with ${\rm supp}(\varphi)\subset
B_{R/2}(x_{0})$. Then $v\in Lip\big{(}B_{R/8}(x_{0})\big{)}$.

###### Proof.

Let $p_{t}(x,y)$ be the heat kernel on $X$. For any $t>0$, we set

$v_{t}(x):=P_{t}v(x)=\int_{X}p_{t}(x,y)v(y)d\mathfrak{m},\quad\forall t>0.$

Claim 1: For each $t\in(0,R^{2})$, we have $v_{t}\in L^{\infty}(X)\cap
W^{1,2}(X)$.

Fix any $x\in X$. Since $v(y)=0$ for $\mathfrak{m}$-a.e. $y\in X\setminus
B_{R}(x_{0})$, by the upper bound of heat kernel in (2.3), we have for any
$x\in X$ that

$|v_{t}(x)|\leqslant\int_{B_{R}(x_{0})}\frac{C_{1}|v(y)|}{\mathfrak{m}(B_{\sqrt{t}}(y))}d\mathfrak{m}(y)\leqslant\frac{C_{1}}{\inf_{y\in
B_{R}(x_{0})}\mathfrak{m}(B_{\sqrt{t}}(y))}\cdot\|v\|_{L^{1}(X)}.$ (3.3)

Here and in the following, we denote $C_{1},C_{2},\cdots$ for varying
constants depending only on $K,N$, and $R$. The local measure doubling
property (2.1) implies for $t<R^{2}$

$\inf_{y\in B_{R}(x_{0})}\mathfrak{m}(B_{\sqrt{t}}(y))\geqslant\inf_{y\in
B_{R}(x_{0})}C_{2}\mathfrak{m}(B_{2R}(y))\cdot\Big{(}\frac{\sqrt{t}}{2R}\Big{)}^{N}\geqslant
C_{3}\mathfrak{m}(B_{R}(x_{0}))\cdot t^{N/2}.$

By combining this with (3.3), we get $v_{t}\in L^{\infty}(X)$ for any
$t\in(0,R^{2}).$

Since $P_{t}$ is contraction on $L^{1}$, we have $P_{t/2}v\in L^{1}(X)$. By
combining with $P_{t/2}v\in L^{\infty}(X)$, we have $P_{t/2}v\in L^{2}(X)$.
Hence, we obtain $v_{t}=P_{t/2}(P_{t/2}v)\in W^{1,2}(X)$. Now the proof of the
Claim is completed.

Claim 2: There exists a constant $c>0$ such that for any $t\in(0,R^{2})$ it
holds

$\int_{B_{R/4}(x_{0})}\langle{\nabla v_{t}},{\nabla\varphi}\rangle
d\mathfrak{m}\leqslant
c\int_{B_{R/4}(x_{0})}|\varphi|d\mathfrak{m},\quad\forall\varphi\in
Lip_{c}\big{(}B_{R/4}(x_{0})\big{)}.$ (3.4)

Take any nonnegative
$\varphi\in\mathrm{Test}^{\infty}_{c}\big{(}B_{R/4}(x_{0})\big{)}$. Since
$\varphi\in D(\Delta)$ and $v_{t}\in W^{1,2}(X)$ (from Claim 1), by using
(2.2) and $\Delta(P_{t}\varphi)=P_{t}(\Delta\varphi)$, we have

$\begin{split}\int_{B_{R/4}(x_{0})}\langle{\nabla
v_{t}},{\nabla\varphi}\rangle d\mathfrak{m}&=\int_{X}\langle{\nabla
v_{t}},{\nabla\varphi}\rangle d\mathfrak{m}=-\int_{X}v_{t}\Delta\varphi
d\mathfrak{m}\\\
&=-\int_{X}vP_{t}(\Delta\varphi)d\mathfrak{m}=-\int_{X}v\Delta(P_{t}\varphi)d\mathfrak{m}.\end{split}$
(3.5)

According to Lemma 2, there exist two cut-off functions $\eta_{i}:X\to[0,1]$
in ${\rm Test}^{\infty}(X)$, $i=1,2$, such that:

(i) ${\rm supp}(\eta_{1})\subset B_{R/2}(x_{0})$ and $\eta_{1}\equiv 1$ on
$B_{R/3}(x_{0})$;

(ii) ${\rm supp}(\eta_{2})\subset B_{2R}(x_{0})$ and $\eta_{2}\equiv 1$ on
$B_{3R/2}(x_{0})$; and

$|\nabla\eta_{i}|+|\Delta\eta_{i}|\leqslant C_{4},\quad i=1,2.$ (3.6)

From Lemma 2, we know that $\eta_{i}\cdot P_{t}\varphi\in{\rm
Test}_{c}^{\infty}(X)$ for $i=1,2$. Since $\eta_{2}\cdot
P_{t}\varphi=P_{t}\varphi$ in $B_{3R/2}(x_{0})$, Lemma 2 implies that
$\Delta(\eta_{2}\cdot P_{t}\varphi)=\Delta(P_{t}\varphi)$ a.e. in
$B_{3R/2}(x_{0})$. Combining with the assumption that $v=0$ a.e. outside
$B_{R}(x_{0})$, we have

$\int_{X}v\Delta(P_{t}\varphi)d\mathfrak{m}=\int_{X}v\Delta(\eta_{2}\cdot
P_{t}\varphi)d\mathfrak{m}=\Delta_{\mathscr{D}}v(\eta_{2}\cdot P_{t}\varphi).$
(3.7)

Since $\eta_{1}\cdot P_{t}\varphi\in{\rm Test}_{c}^{\infty}(X)$ and ${\rm
supp}(\eta_{1}\cdot P_{t}\varphi)\subset B_{R/2}(x_{0})$, by the assumption
$\Delta_{\mathscr{D}}v=0$ on $B_{R/2}(x_{0})$, we get

$\Delta_{\mathscr{D}}v(\eta_{1}\cdot P_{t}\varphi)=0.$

Therefore, by substituting this and (3.7) into (3.5), we obtain that

$\begin{split}\int_{B_{R/4}(x_{0})}\langle{\nabla
v_{t}},{\nabla\varphi}\rangle
d\mathfrak{m}&=\Delta_{\mathscr{D}}v\big{(}(\eta_{2}-\eta_{1})\cdot
P_{t}\varphi\big{)}\\\ &=\int_{X}v\Delta\big{(}\eta\cdot
P_{t}\varphi\big{)}d\mathfrak{m},\end{split}$ (3.8)

where $\eta:=\eta_{2}-\eta_{1}\in{\rm Test}^{\infty}_{c}(X)$.

Next we want to estimate $\|\Delta(\eta\cdot P_{t}\varphi)\|_{L^{\infty}(X)}$.
Since ${\rm supp}(\eta)\subset\overline{B_{2R}(x_{0})\setminus
B_{R/3}(x_{0})}$, we know, by Lemma 2, that

$\Delta(\eta\cdot P_{t}\varphi)=0\ \ \mathfrak{m}{\rm-a.e.\ in}\
B_{R/3}(x_{0})\cup B^{c}_{2R}(x_{0}).$ (3.9)

Fix any $x\in\overline{B_{2R}(x_{0})\setminus B_{R/3}(x_{0})}$. Since
$d(x,y)\geqslant\frac{R}{12}$ for any $y\in{\rm supp}(\varphi)$, by using the
estimates (2.3) and ${\rm supp}(\varphi)\subset B_{R/4}(x_{0})$, we obtain

$\begin{split}|P_{t}\varphi(x)|&\leqslant\int_{B_{R/4}(x_{0})}\frac{C_{5}\cdot\exp\left(-\frac{R^{2}}{5t\cdot
12^{2}}\right)}{\mathfrak{m}(B_{\sqrt{t}}(y))}|\varphi(y)|d\mathfrak{m}(y)\\\
&\leqslant
C_{6}\Big{(}\frac{R}{\sqrt{t}}\Big{)}^{N}\cdot\exp\left(-\frac{R^{2}}{5t\cdot
12^{2}}\right)\int_{B_{R/4}(x_{0})}\frac{|\varphi(y)|}{\mathfrak{m}(B_{R}(y))}d\mathfrak{m}(y)\\\
&\leqslant\frac{C_{7}}{\mathfrak{m}(B_{R/2}(x_{0}))}\|\varphi\|_{L^{1}(X)},\end{split}$
(3.10)

where in the second inequality we have used (2.1), and in the last inequality
we have used $B_{R}(y)\supset B_{R/2}(x_{0})$ for any $y\in B_{R/4}(x_{0})$
and that

$\frac{1}{t^{N/2}}\cdot\exp\left(-\frac{R^{2}}{5t\cdot 12^{2}}\right)\leqslant
C_{N,R},\quad\forall t\in(0,R^{2})$

for some constant $C_{N,R}>0.$ Similarly, by (2.4) and (2.5), we get that

$\begin{split}|\nabla
P_{t}\varphi(x)|&\leqslant\int_{B_{R/4}(x_{0})}\big{|}\nabla
p_{t}(x,y)\big{|}\cdot|\varphi(y)|d\mathfrak{m}(y)\\\
&\leqslant\int_{B_{R/4}(x_{0})}\frac{C_{8}\cdot\exp\left(-\frac{R^{2}}{5t\cdot
12^{2}}\right)}{\sqrt{t}\cdot\mathfrak{m}(B_{\sqrt{t}}(y))}|\varphi(y)|d\mathfrak{m}(y)\\\
&\leqslant\frac{C_{9}}{\sqrt{t}}\Big{(}\frac{R}{\sqrt{t}}\Big{)}^{N}\cdot\exp\left(-\frac{R^{2}}{5t\cdot
12^{2}}\right)\int_{B_{R/4}(x_{0})}\frac{|\varphi(y)|}{\mathfrak{m}(B_{R}(y))}d\mathfrak{m}(y)\\\
&\leqslant\frac{C_{10}}{\mathfrak{m}(B_{R/2}(x_{0}))}\|\varphi\|_{L^{1}(X)}\end{split}$
(3.11)

and

$\begin{split}|\Delta P_{t}\varphi(x)|&=\big{|}\frac{\partial}{\partial
t}P_{t}\varphi(y)\big{|}\leqslant\int_{B_{R/4}(x_{0})}\big{|}\frac{\partial}{\partial
t}p_{t}(x,y)\big{|}\cdot|\varphi(y)|d\mathfrak{m}(y)\\\
&\leqslant\int_{B_{R/4}(x_{0})}\frac{C_{11}\cdot\exp\left(-\frac{R^{2}}{5t\cdot
12^{2}}\right)}{t\cdot\mathfrak{m}(B_{\sqrt{t}}(y))}|\varphi(y)|d\mathfrak{m}(y)\\\
&\leqslant\frac{C_{12}}{\mathfrak{m}(B_{R/2}(x_{0}))}\|\varphi\|_{L^{1}(X)},\end{split}$
(3.12)

for almost every $x\in\overline{B_{2R}(x_{0})\setminus B_{R/3}(x_{0})}$.
Noticing that $\eta,P_{t}\varphi\in{\rm Test}^{\infty}(X)$, by using Leibniz
rule (see [20, Proposition 5.2.3]) to $\Delta(\eta\cdot P_{t}\varphi)$, and
then substituting (3.6), (3.10), (3.11) and (3.12), we get that

$|\Delta(\eta\cdot
P_{t}\varphi)|(x)\leqslant\frac{C_{13}}{\mathfrak{m}(B_{R/2}(x_{0}))}\|\varphi\|_{L^{1}(X)}$

for almost all $x\in\overline{B_{2R}(x_{0})\setminus B_{R/3}(x_{0})}$. By
combining with (3.9), we conclude

$\|\Delta(\eta\cdot
P_{t}\varphi)\|_{L^{\infty}(X)}\leqslant\frac{C_{13}}{\mathfrak{m}(B_{R/2}(x_{0}))}\|\varphi\|_{L^{1}(X)},\quad\forall
t\in(0,R^{2}).$

Substituting this into (3.8), we get that for
$\varphi\in\mathrm{Test}^{\infty}_{c,+}\big{(}B_{R/4}(x_{0})\big{)}$,

$\int_{B_{R/4}(x_{0})}\langle{\nabla v_{t}},{\nabla\varphi}\rangle
d\mathfrak{m}\leqslant\frac{C_{13}\cdot\|v\|_{L^{1}(X)}}{\mathfrak{m}(B_{R/2}(x_{0}))}\|\varphi\|_{L^{1}(X)}:=c\cdot\|\varphi\|_{L^{1}(X)}.$
(3.13)

By using Lemma 3, we know that (3.13) holds too for any nonnegative
$\varphi\in Lip_{c}(\Omega)$. For any $\varphi\in Lip_{c}(\Omega)$, by
applying (3.13) to $\varphi^{\pm}$, the desired (3.4) follows. Now the proof
of Claim 2 is finished.

Now, from the Definition 2, Claim 2 states that $v_{t}\in D(\mathbf{\Delta})$
on $B_{R/4}(x_{0})$ and ${\bf\Delta}v_{t}=g_{t}\cdot\mathfrak{m}$ for some
Borel function $g_{t}$ with $\|g\|_{L^{\infty}(B_{R/4}(x_{0}))}\leqslant c$.
According to Lemma 2, we get

$\sup_{B_{R/8}(x_{0})}|\nabla v_{t}|\leqslant
C_{14}\Big{(}\|v_{t}\|_{L^{1}\big{(}B_{R/4}(x_{0})\big{)}}+c\Big{)}.$

Since $P_{t}$ is a contraction in $L^{1}(X)$, we have

$\sup_{B_{R/8}(x_{0})}|\nabla v_{t}|\leqslant
C_{14}\Big{(}\|v_{t}\|_{L^{1}(X)}+c\Big{)}\leqslant
C_{14}(\|v\|_{L^{1}(X)}+c):=c_{0}.$

Notice the constant $c_{0}$ is independent of $t$. Letting $t\to 0^{+}$, since
$v_{t}\to v$ in $L^{1}(X)$ (by $v\in L^{1}(X)$), we get that

$\sup_{B_{R/8}(x_{0})}|\nabla v|\leqslant c_{0}.$

The proof of Lemma 3 is finished. ∎

Now we are in the position to show Weyl’s lemma in $RCD$-setting.

###### Proof of Theorem 1.2.

Let $B_{R}(x_{0})\subset\subset\Omega$ be a ball. We take a Lipschitz function
$\chi:X\to[0,1]$ such that $\chi\equiv 1$ on $B_{R/2}(x_{0})$ and $\chi\equiv
0$ on $X\setminus B_{R}(x_{0})$. Let $v:=u\cdot\chi$ on $\Omega$ and $v=0$ on
$X\setminus\Omega$. Now for any nonnegative $\varphi\in{\rm
Test}^{\infty}_{c}\big{(}B_{R/2}(x_{0})\big{)}$, we have

$\Delta_{\mathscr{D}}v(\varphi)=\int_{X}v\Delta\varphi
d\mathfrak{m}=\int_{B_{R/2}(x_{0})}u\Delta\varphi d\mathfrak{m}=0,$

where we have used $\Delta\varphi=0$ a.e. in $X\setminus B_{R/2}(x_{0})$,
since ${\rm supp}(\varphi)\subset B_{R/2}(x_{0})$ and Lemma 2. By applying
Lemma 3, we conclude that $v\in Lip(B_{R/8}(x_{0})).$ It follows that $u\in
Lip(B_{R/8}(x_{0})),$ since $u=v$ on $B_{R/2}(x_{0})$. Now the proof of
Theorem 1.2 is finished. ∎

##  4\. Regularity of very weak solutions for Poisson equations

Let $(X,d,\mathfrak{m})$ be an $RCD(K,N)$ space for some $K\in\mathbb{R}$ and
$N\in[1,+\infty)$.

######  Definition .

Let $\Omega\subset X$ be an open domain. A function $u\in L_{\rm
loc}^{1}(\Omega)$ is called a very weakly subsolution (resp. supersolution) of
Poisson equation $\Delta_{\mathscr{D},\Omega}u=f$ on $\Omega$ for some $f\in
L^{1}_{\rm loc}(\Omega)$ if

$\int_{\Omega}u\Delta\varphi d\mathfrak{m}\geqslant\ ({\rm resp}.\leqslant)\
\int_{\Omega}f\varphi d\mathfrak{m},\qquad\forall\varphi\in{\rm
Test}^{\infty}_{c}(\Omega)\ {\rm and}\ \varphi\geqslant 0.$ (4.1)

A function $u\in L_{\rm loc}^{1}(\Omega)$ is called a very weakly solution of
$\Delta_{\mathscr{D},\Omega}u=f$ on $\Omega$ if it is both very weakly
subsolution and supersolution.

Next we will show Corollary 1.

###### Proof of Corollary 1.

Take any ball $B:=B_{R}(x)\subset\subset\Omega$. Since $f\in L^{2}_{\rm
loc}(\Omega)$, we have $f\in L^{2}(B)$. Let $w$ be the solution of the relaxed
Dirichlet problem (see [7, Theorem 7.12] for the solvability):

$\begin{cases}{\bf\Delta}w=f\cdot\mathfrak{m}\quad{\rm on}\ \ B,\\\ w\in
W^{1,2}_{0}(B).\end{cases}$

By Lemma 3, we get that $\Delta_{\mathscr{D},B}w={\bf\Delta}w$, i.e.,

$\int_{B}w\Delta\varphi d\mathfrak{m}={\bf\Delta}w(\varphi)=\int_{B}f\varphi
d\mathfrak{m},\quad\ \forall\varphi\in{\rm Test}^{\infty}_{c}(B)\ \ {\rm and}\
\ \varphi\geqslant 0.$

By combining with (1.4) and Lemma 2, we have

$\int_{B}(u-w)\Delta\varphi d\mathfrak{m}=0,\quad\ \forall\varphi\in{\rm
Test}^{\infty}_{c}(B)\ \ {\rm and}\ \ \varphi\geqslant 0.$

Therefore the function $u-w\in L^{1}(B)$ is a very weak harmonic function on
$B$. From Theorem 1.2, we know that $u-w\in Lip_{\rm loc}(B)$. It follows
$u=(u-w)+w\in W^{1,2}_{\rm loc}(B).$ The proof is finished. ∎

##  5\. The Liouville property for $L^{1}$ (sub)harmonic functions

In this section, we will prove the Liouville-type theorem for $L^{1}$
subharmonic/harmonic functions on $RCD(K,N)$ space, which is an extension of
Peter Li’s result in [31]. Let $(X,d,\mathfrak{m})$ be an $RCD(K,N)$ space
with $K\in\mathbb{R}$ and $N\in[1,+\infty)$.

######  Lemma .

Let $u$ and $f$ be in $L^{1}(X)$. Suppose that $u$ is a very weakly
subsolution of Poisson equation $\Delta_{\mathscr{D}}u=f$ on $X$. Then

$\int_{X}u\Delta hd\mathfrak{m}\geqslant\int_{X}fhd\mathfrak{m},\quad\ \forall
h\in\mathrm{Test}^{\infty}(X)\ {\rm and}\ h\geqslant 0.$ (5.1)

###### Proof.

If $X$ is compact, then we have
$\mathrm{Test}^{\infty}_{c}(X)=\mathrm{Test}^{\infty}(X)$. This assertion
follows from the Definition 4 with $\Omega=X$.

Suppose that $X$ is noncompact. Take any $h\in\mathrm{Test}^{\infty}(X)$ and
$h\geqslant 0$. For each $R>1$, by using Lemma 2 (with $r_{0}=1$ therein), we
have a cut-off function $\eta:X\to[0,1]$ in ${\rm Test}^{\infty}(X)$ such that
$\eta\equiv 1$ in $B_{R}(x_{0})$, $\eta\equiv 0$ outside $B_{2R}(x_{0})$ and

$|\nabla\eta|+|\Delta\eta|\leqslant\frac{C_{K,N}}{R}\leqslant
C_{K,N}\quad\mathfrak{m}{-a.e.\ in}\ X,$ (5.2)

where the constant $C_{K,N}$ is independent of $R$. We set $\varphi:=h\eta$,
and then $\varphi\in{\rm Test}^{\infty}_{c}(X)$. Since $\varphi=h$ on
$B_{R}(x_{0})$, from Lemma 2, we have

$\Delta\varphi(x)=\Delta h(x)\quad{\rm for}\ \mathfrak{m}{\rm-a.e.}\ x\in
B_{R}(x_{0}).$ (5.3)

Since $h,\eta\in{\rm Test}^{\infty}(X)$, by using Leibniz rule (see [20,
Proposition 5.2.3]) to $\varphi=h\eta$, we get that for $\mathfrak{m}$-a.e.
$x\in X\setminus B_{R}(x_{0})$,

$\begin{split}|\Delta\varphi|(x)&\leqslant
h(x)|\Delta\eta|(x)+2|\langle{\nabla h},{\nabla\eta}\rangle|(x)+\eta(x)|\Delta
h|(x)\\\ &\leqslant 2C_{K,N}\big{(}\|h\|_{L^{\infty}(X)}+\||\nabla
h|\|_{L^{\infty}(X)}\big{)}+\|\Delta
h\|_{L^{\infty}(X)}:=c_{K,N,h},\end{split}$

where we have used (5.2). By combining this with (5.3), we get

$\begin{split}\int_{X}u\Delta hd\mathfrak{m}-\int_{X}u\Delta\varphi
d\mathfrak{m}&=\int_{X\setminus B_{R}(x_{0})}u(\Delta
h-\Delta\varphi)d\mathfrak{m}\\\
&\geqslant-c^{\prime}_{K,N,h}\cdot\|u\|_{L^{1}(X\setminus
B_{R}(x_{0})},\end{split}$

where $c^{\prime}_{K,N,h}:=c_{K,N,h}+\|\Delta h\|_{L^{\infty}(X)}$. Since
$\varphi=h$ on $B_{R}(x_{0})$, we have

$\int_{X}f\varphi
d\mathfrak{m}-\int_{X}fhd\mathfrak{m}\geqslant-\|h\|_{L^{\infty}(X)}\cdot\|f\|_{L^{1}(X\setminus
B_{R}(x_{0})}.$

From the combination of these two inequalities and (4.1) with $\Omega=X$
(noticing that $\varphi\in{\rm Test}^{\infty}_{c}(X)$), we get

$\int_{X}u\Delta
hd\mathfrak{m}\geqslant\int_{X}fhd\mathfrak{m}-c^{\prime}_{K,N,h}\cdot\|u\|_{L^{1}(X\setminus
B_{R}(x_{0}))}-\|h\|_{L^{\infty}(X)}\cdot\|f\|_{L^{1}(X\setminus
B_{R}(x_{0}))}.$

Letting $R\to+\infty$, it follows that (5.1), since $u\in L^{1}(X)$ and $h\in
L^{1}(X)$. ∎

To show the $L^{1}$-Liouville property, we need the following proposition.

######  Proposition .

Let $u\in L^{1}(X)$ be a very weakly subharmonic (or superharmonic) function
on $X$. Then it is a very weak harmonic function on $X$. Furthermore, $u$ is
in $Lip_{\rm loc}(X)$ and is a harmonic function on $X$, namely, $u\in
D({\bf\Delta})$ and ${\bf\Delta}u=0$ on $X$.

###### Proof.

We first consider that $u$ is a $L^{1}$ very weakly subharmonic function on
$X$. Since by (2.5), $\frac{\partial}{\partial
t}p_{t}(x,\cdot)(y)\in{L^{\infty}(X)}$ for each $t\in(0,\infty)$, we have for
$\mathfrak{m}$-a.e. $x\in X$ that

$\displaystyle\frac{\partial}{\partial t}P_{t}u(x)$
$\displaystyle=\int_{X}\frac{\partial}{\partial
t}p_{t}(x,y)u(y)d\mathfrak{m}(y)$
$\displaystyle=\int_{X}\Delta_{y}p_{t}(x,y)u(y)d\mathfrak{m}(y).$

From Lemma 2, we have $p_{t}(x,\cdot)\in\mathrm{Test}^{\infty}(X)$. Since $u$
is a $L^{1}$ very weakly subharmonic function on $X$, by using Lemma 5, we
conclude that $P_{t}u(x)$ is a monotone nondecreasing function in $t$, for
almost all $x\in X$. On the other hand, By the stochastic completeness (2.6)
and Fubini’s theorem, we get

$\int_{X}P_{t}u(x)d\mathfrak{m}(x)=\int_{X}\int_{X}p_{t}(x,y)u(y)d\mathfrak{m}(y)d\mathfrak{m}(x)=\int_{X}u(x)\mathfrak{m}(x).$

Therefore, we have

$P_{t}u(x)=u(x)\quad\mathfrak{m}{\rm-a.e.}\ x\in X.$

This implies $u\in D(\Delta^{(1)})$ and $\Delta^{(1)}u=0$, since $u\in
L^{1}(X)$. From Lemma 3, we know that $\Delta_{\mathscr{D}}u=0$ on $X$ in the
sense of distributions. That is, $u$ is a very weak harmonic function on $X$.

On the other hand, if $u$ be a $L^{1}$ very weakly superharmonic function on
$X$, then $-u$ is a very weakly subharmonic, and hence $-u$ is a very weak
harmonic function on $X$.

Now we want to show the second statement. From the first one, $u$ is very weak
harmonic function on $X$. By applying Theorem 1.2, we get that $u\in Lip_{\rm
loc}(X)$. Hence, by Lemma 3, we get that $u\in D({\bf\Delta})$ and
${\bf\Delta}u(\varphi)=0$ for any $\varphi\in{\rm Test}^{\infty}_{c}(X)$.
Since ${\rm Test}^{\infty}_{c}(X)$ is dense in $W^{1,2}_{0}(X)$ (see Lemma 3),
we conclude that $u$ is a harmonic function on $X$. ∎

Now we are in the position to show Corollary 1

###### Proof of Corollary 1.

Let $u$ be a $L^{1}$ very weakly subharmonic function on $X$. From Proposition
5, we know in fact that $u$ is in $Lip_{\rm loc}(X)$ and is a harmonic
function on $X$.

We consider the function $u_{1}:=\min\\{u,a\\}$ for arbitrary $a>0$. Since $u$
is harmonic, according to [45, Lemma 2], we have that $u_{1}$ is
superharmonic. In particular, it is very weakly superharmonic on $X$ (by Lemma
3). Noticing that $u_{1}\in L^{1}(X)$, by using Proposition 5 again, we get
that $u_{1}$ is again harmonic on $X$. We further put
$u_{2}:=\max\\{u_{1},-a\\}$. Similar, from the harmonicity of $u_{1}$, [45,
Lemma 2] implies that $u_{2}$ is subharmonic on $X$; and then again,
Proposition 5 yields that $u_{2}$ is also harmonic on $X$. From [45, Lemma 2],
we know that $|u_{2}|$ is a nonnegative subharmonic function on $X$. Since
$-a\leqslant u_{2}\leqslant a$, we get that $u_{2}\in L^{\infty}(X)\cap
L^{1}(X)\in L^{2}(X)$. According to [45, Theorem 1], we now conclude that
$|u_{2}|$ is a constant. It follows from the arbitrariness of $a$ that $u$ is
a constant. The proof of Corollary 1 is finished. ∎

##  6\. A Remark on gradient estimates for solutions of elliptic equations

In this section, we will show Theorem 1.3.

We first deal with the harmonic functions on $RCD(0,N)$ cones. We recall that
if $N\geqslant 2$ and if $(\Sigma,d_{\Sigma},\mathfrak{m}_{\Sigma})$ is a
metric measure space, then the cone
$Cone^{N}_{0}(\Sigma):=(C,d_{c},\mathfrak{m}_{c})$ is the metric measure space
given by the warped product metric $\Sigma\times_{r^{N-1}}\mathbb{R}$ and
$d\mathfrak{m}_{c}(r,\xi):=r^{N-1}drd\mathfrak{m}_{\Sigma}(\xi)$. It was
proved [29] that $Cone^{N}_{0}(\Sigma)$ is an $RCD(0,N)$ space if and only if
$(\Sigma,d_{\Sigma},\mathfrak{m}_{\Sigma})$ is an $RCD(N-2,N-1)$ space.

######  Lemma .

Let $(C,d_{c},\mathfrak{m}_{c})$ be an $RCD(0,N)$ cone with vertex $o$, over
$(\Sigma,d_{\Sigma},\mathfrak{m}_{\Sigma})$. Suppose that $u\in
W^{1,2}(B_{R}(o))$ is a (weakly) harmonic function, then

$\fint_{B_{r_{1}}(o)}|\nabla
u|^{2}d\mathfrak{m}_{c}\leqslant\fint_{B_{r_{2}}(o)}|\nabla
u|^{2}d\mathfrak{m}_{c}$ (6.1)

for all $0<r_{1}<r_{2}<R$.

###### Proof.

According to [23, Theorem 3.1], there exist $\\{c_{i}\\}_{i}\subset\mathbb{R}$
such that

$u(r,\xi)=\sum_{i=0}^{\infty}c_{i}r^{\alpha_{i}}\varphi_{i}(\xi),$

where $\varphi_{i}(\xi)$ is an eigenfunction of $-\Delta_{\Sigma}$ with
respect to the eigenvalue $\lambda_{i}$, and $\alpha_{i}>0$ are given by
$\lambda_{i}=\alpha_{i}(N+\alpha_{i}-2)$. From [30, 27], we have the first
eigenvalue $\lambda_{1}\geqslant N-1$. It follows $\alpha_{i}\geqslant 1$ for
all $i\geqslant 1.$ A direct calculation (see [13, Eq. (2.5)]) implies that

$\frac{1}{r^{N}}\int_{B_{r}(o)}|\nabla
u|^{2}d\mathfrak{m}_{c}=\sum_{i=1}^{\infty}c^{2}_{i}\alpha_{i}\cdot
r^{2\alpha_{i}-2}.$

The desired monotonicity (6.1) follows from this and the fact
$\mathfrak{m}_{c}(B_{r}(o))=\frac{r^{N}}{N}\cdot\mathfrak{m}_{\Sigma}(\Sigma).$
∎

Let $n\geqslant 3$. Let $u\in W^{1,2}_{\rm loc}(B_{1}(0))$ be a weak solution
of elliptic equations of divergence form ${\rm div}(A(x)Du)=0,$ where the
matrix of coefficients $A(x)=(a^{ij})_{i,j=1}^{n}$ is symmetric, bounded
measurable and satisfies (1.6).

###### Proof of Theorem 1.3.

According to the assumption of $A$ in Theorem 1.3, there exists a conical
coefficient of nonnegative curvature $\overline{A}=(\overline{a}^{ij})$ such
that (1.8) holds. Since $\omega_{A,\overline{A}}(r)\to 0$ as $r\to 0$, there
is $r_{1}>0$ such that

$\|\overline{a}^{ij}\|_{L^{\infty}(B_{r_{1}}(0))}\leqslant
2\Lambda,\qquad\frac{\lambda}{2}|\xi|^{2}\leqslant\sum_{i,j=1}^{n}\overline{a}^{ij}(x)\xi_{i}\xi_{j},\quad{\rm
a.e.}\ x\in B_{r_{1}}(0),\quad\forall\xi\in\mathbb{R}^{n}.$

It is easy to check that the the distance $d_{\overline{g}}$ induced from the
Riemannian metric $\overline{g}_{ij}$ in (1.7) is bi-Lipschitz equivalent to
the Euclidean metric on $B_{r_{1}}(0)$. The bi-Lipschitz constant depends only
on $\lambda$ and $\Lambda$.

We denote by $B^{\overline{g}}_{r}(0)$ the ball with respect the metric
$d_{\overline{g}}$, for any $r>0$. Then there exists a constant
$C_{1}=C_{1}(\lambda,\Lambda)>1$ such that

$B_{r/C_{1}}(0)\subset B_{r}^{\overline{g}}(0)\subset
B_{C_{1}r}(0),\qquad\forall r>0.$ (6.2)

The bi-Lipschitz equivalence between $d_{\overline{g}}$ and the Euclidean
distance implies that

$W^{1,2}(B_{r}^{\overline{g}}(0),d_{\overline{g}},{\rm
vol}_{\overline{g}})=W^{1,2}(B_{r}(0))$

for any $r\in(0,r_{1}/C_{1})$, where the Riemannian measure

${\rm vol}_{\overline{g}}=\sqrt{G}\cdot\mathcal{L}^{n},\qquad G:={\rm
det}(\overline{g}_{ij}),$ (6.3)

and $\mathcal{L}^{n}$ is the Lebesgue measure on $\mathbb{R}^{n}$. From (1.7),
we have

$G=\left({\rm
det}{(\overline{a}^{ij})}\right)^{\frac{2}{n-2}},\qquad\overline{a}^{ij}=\overline{g}^{ij}\cdot\sqrt{G}$
(6.4)

for all $1\leqslant i,j\leqslant n.$

Denote $B^{\overline{g}}_{\ell}:=B^{\overline{g}}_{\rho^{\ell}}(0),$ where
$\rho:=1/C_{1}<1$. Let $\ell_{0}\in\mathbb{N}$, $\ell_{0}>2$, such that
$\rho^{\ell_{0}}<r_{1}/C_{1}$. For each $\ell=\ell_{0},\ell_{0}+1,\cdots,$ let
$u_{\ell}$ be the solution of Dirichlet problem

$\partial_{i}\left(\overline{a}^{ij}\partial u_{\ell}\right)=0\quad{\rm
with}\quad u-u_{\ell}\in W^{1,2}_{0}(B^{\overline{g}}_{\ell})\ \
\left(=W^{1,2}_{0}(B^{\overline{g}}_{\ell},d_{\overline{g}},{\rm
vol}_{\overline{g}})\right).$

Noticing that $\partial_{i}(a^{ij}\partial_{j}u)=0$ in the weak sense and
letting $v_{\ell}=u-u_{\ell}$, we have

$\int_{B^{\overline{g}}_{\ell}}\overline{a}^{ij}\partial_{j}v_{\ell}\partial_{i}v_{\ell}d\mathcal{L}^{n}=\int_{B^{\overline{g}}_{\ell}}\overline{a}^{ij}\partial_{j}u\partial_{i}v_{\ell}d\mathcal{L}^{n}=\int_{B^{\overline{g}}_{\ell}}(\overline{a}^{ij}-a^{ij})\partial_{j}u\partial_{i}v_{\ell}d\mathcal{L}^{n}.$

Together this, (6.2), (6.3), (6.4) and (1.8), we get

$\begin{split}\|\nabla_{\overline{g}}v_{\ell}\|^{2}_{L^{2}(B_{\ell}^{\overline{g}},{\rm
vol}_{\overline{g}})}&=\int_{B^{\overline{g}}_{\ell}}\overline{g}^{ij}\partial_{j}v_{\ell}\partial_{i}v_{\ell}d{\rm
vol}_{\overline{g}}=\int_{B^{\overline{g}}_{\ell}}\overline{a}^{ij}\partial_{j}v_{\ell}\partial_{i}v_{\ell}d\mathcal{L}^{n}\\\
&\leqslant
C_{2}\cdot\omega_{A,\overline{A}}(C_{1}\rho^{\ell})\cdot\|\nabla_{\overline{g}}u\|_{L^{2}(B_{\ell}^{\overline{g}},{\rm
vol}_{\overline{g}})}\cdot\|\nabla_{\overline{g}}v_{\ell}\|_{L^{2}(B_{\ell}^{\overline{g}},{\rm
vol}_{\overline{g}})},\end{split}$

where the constant $C_{2}$ depends only on $\lambda,\Lambda$, and we have used
that
$\|\partial_{j}u\|_{L^{2}(B_{\ell}^{\overline{g}},\mathcal{L}^{n})}\leqslant
C_{\lambda,\Lambda}\|\nabla_{\overline{g}}u\|_{L^{2}(B_{\ell}^{\overline{g}},{\rm
vol}_{\overline{g}})}$ for some constant $C_{\lambda,\Lambda}>0$. Hence, for
each $\ell=\ell_{0},\ell_{0}+1,\cdots,$ we have (noticing that
$C_{1}=\rho^{-1}$)

$\|\nabla_{\overline{g}}v_{\ell}\|_{L^{2}(B_{\ell}^{\overline{g}},{\rm
vol}_{\overline{g}})}\leqslant
C_{2}\cdot\omega_{A,\overline{A}}(\rho^{\ell-1})\cdot\|\nabla_{\overline{g}}u\|_{L^{2}(B_{\ell}^{\overline{g}},{\rm
vol}_{\overline{g}})}.$ (6.5)

On the other hand, we claim that, for each $\ell=\ell_{0},\ell_{0}+1,\cdots,$
$u_{\ell}$ is a (weakly) harmonic function on $B^{\overline{g}}_{\ell}$ in
metric measure space $(\mathbb{R}^{n},d_{\overline{g}},{\rm
vol}_{\overline{g}})$. Indeed, for any $\varphi\in
Lip_{0}(B^{\overline{g}}_{\ell})$, we have

$\int_{B^{\overline{g}}_{\ell}}\langle{\nabla_{\overline{g}}u_{\ell}},{\nabla_{\overline{g}}\varphi}\rangle
d{\rm
vol}_{\overline{g}}=\int_{B^{\overline{g}}_{\ell}}\overline{a}^{ij}\partial_{j}u_{\ell}\partial_{i}\varphi
d\mathcal{L}^{n}=0.$

Since $(\mathbb{R}^{n},d_{\overline{g}})$ has nonnegative curvature in the
sense of Alexandrov, the metric measure space
$(\mathbb{R}^{n},d_{\overline{g}},{\rm vol}_{\overline{g}})$ is an $RCD(0,n)$
cone (see [40, 51]). From Lemma 6, we have

$\fint_{B_{\ell+1}^{\overline{g}}(0)}|\nabla_{\overline{g}}u_{\ell}|^{2}d{\rm
vol}_{\overline{g}}\leqslant\fint_{B_{\ell}^{\overline{g}}(0)}|\nabla_{\overline{g}}u_{\ell}|^{2}d{\rm
vol}_{\overline{g}}$

for each $\ell=\ell_{0},\ell_{0}+1,\cdots.$ By combining with (6.5) and
$u=u_{\ell}+v_{\ell}$ on $B^{\overline{g}}_{\ell}$, we get

$\begin{split}&\left(\fint_{B_{\ell+1}^{\overline{g}}(0)}|\nabla_{\overline{g}}u|^{2}d{\rm
vol}_{\overline{g}}\right)^{1/2}\\\
\leqslant&\left(\fint_{B_{\ell+1}^{\overline{g}}(0)}|\nabla_{\overline{g}}u_{\ell}|^{2}d{\rm
vol}_{\overline{g}}\right)^{1/2}+\left(\fint_{B_{\ell+1}^{\overline{g}}(0)}|\nabla_{\overline{g}}v_{\ell}|^{2}d{\rm
vol}_{\overline{g}}\right)^{1/2}\\\
\leqslant&\left(\fint_{B_{\ell}^{\overline{g}}(0)}|\nabla_{\overline{g}}u_{\ell}|^{2}d{\rm
vol}_{\overline{g}}\right)^{1/2}+\frac{{\rm
vol}_{\overline{g}}(B^{\overline{g}}_{\ell})}{{\rm
vol}_{\overline{g}}(B^{\overline{g}}_{\ell+1})}\cdot\left(\fint_{B_{\ell}^{\overline{g}}(0)}|\nabla_{\overline{g}}v_{\ell}|^{2}d{\rm
vol}_{\overline{g}}\right)^{1/2}\\\
\leqslant&\left(\fint_{B_{\ell}^{\overline{g}}(0)}|\nabla_{\overline{g}}u|^{2}d{\rm
vol}_{\overline{g}}\right)^{1/2}+\frac{C_{2}}{\rho^{n}}\cdot\omega_{A,\overline{A}}(\rho^{\ell-1})\left(\fint_{B_{\ell}^{\overline{g}}(0)}|\nabla_{\overline{g}}u|^{2}d{\rm
vol}_{\overline{g}}\right)^{1/2}\\\
\leqslant&\left(1+C_{3}\omega_{A,\overline{A}}(\rho^{\ell-1})\right)\left(\fint_{B_{\ell}^{\overline{g}}(0)}|\nabla_{\overline{g}}u|^{2}d{\rm
vol}_{\overline{g}}\right)^{1/2},\end{split}$

for any $\ell=\ell_{0},\ell_{0}+1,\cdots,$ where we have used the
$\|\nabla_{\overline{g}}u_{\ell}\|_{L^{2}(B_{\ell}^{\overline{g}},{\rm
vol}_{\overline{g}})}\leqslant\|\nabla_{\overline{g}}u\|_{L^{2}(B_{\ell}^{\overline{g}},{\rm
vol}_{\overline{g}})}$ by the property that $u_{\ell}$ is the energy
minimizing function, and the doubling property of ${\rm vol}_{\overline{g}}$
(In fact $\frac{{\rm vol}_{\overline{g}}(B^{\overline{g}}_{\ell})}{{\rm
vol}_{\overline{g}}(B^{\overline{g}}_{\ell+1})}=\rho^{-n}$). By the elementary
inequality $\ln(1+s)\leqslant s$ for $s>0$, we have

$\ln\left(\fint_{B_{\ell+1}^{\overline{g}}(0)}|\nabla_{\overline{g}}u|^{2}d{\rm
vol}_{\overline{g}}\right)\leqslant\ln\left(\fint_{B_{\ell}^{\overline{g}}(0)}|\nabla_{\overline{g}}u|^{2}d{\rm
vol}_{\overline{g}}\right)+2C_{3}\omega_{A,\overline{A}}(\rho^{\ell-1})$

for any $\ell=\ell_{0}.\ell_{0}+1,\cdots.$ Therefore, we have

$\limsup_{\ell\to+\infty}\ln\left(\fint_{B_{\ell+1}^{\overline{g}}(0)}|\nabla_{\overline{g}}u|^{2}d{\rm
vol}_{\overline{g}}\right)\leqslant\ln\left(\fint_{B_{\ell_{0}}^{\overline{g}}(0)}|\nabla_{\overline{g}}u|^{2}d{\rm
vol}_{\overline{g}}\right)+2C_{3}\sum_{\ell=\ell_{0}}^{\infty}\omega_{A,\overline{A}}(\rho^{\ell-1}).$

Since $\omega_{A,\overline{A}}(t)$ is nondecreasing in $t>0$, we have

$\sum_{\ell=\ell_{0}-1}^{\infty}\omega_{A,\overline{A}}(\rho^{\ell})\leqslant\sum_{\ell=\ell_{0}-1}^{\infty}\int_{\rho^{\ell}}^{\rho^{\ell-1}}\frac{\omega_{A,\overline{A}}(t)}{\rho^{\ell-1}-\rho^{\ell}}\cdot\frac{\rho^{\ell-1}}{t}dt=\frac{1}{1-\rho}\int_{0}^{\rho^{\ell_{0}-2}}\frac{\omega_{A,\overline{A}}(t)}{t}dt.$

Therefore, we conclude that

$\limsup_{\ell\to+\infty}\left(\fint_{B_{\ell+1}^{\overline{g}}(0)}|\nabla_{\overline{g}}u|^{2}d{\rm
vol}_{\overline{g}}\right)\leqslant\left(\fint_{B_{\ell_{0}}^{\overline{g}}(0)}|\nabla_{\overline{g}}u|^{2}d{\rm
vol}_{\overline{g}}\right)\cdot\exp\left(\frac{2C_{3}}{1-\rho}\int_{0}^{1}\frac{\omega_{A,\overline{A}}(t)}{t}dt\right).$

It follows the desired estimate (1.9), by combining the doubling property of
${\rm vol}_{\overline{g}}$ and (6.2)-(6.4). The proof is finished. ∎

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