2. Preliminaries

Throughout the paper, we write $\operatorname {AMV}$ as a shortform for “Asymptotic Mean Value’ and $\operatorname {SAMV}$ for ‘Symmetrized Asymptotic Mean Value’. We keep a positive integer $n$ fixed and we denote by $0_n$ the origin of the vector space $\mathbb {R}^n$, by $\mathsf {d}_{e}$ the Euclidean distance on $\mathbb {R}^n$, by $\omega _n$ the Lebesgue measure of the unit Euclidean ball and by $\sigma _{n-1}$ the total surface measure of the unit Euclidean sphere.

We say that a triple $(X,\,\mathsf {d},\,\mu )$ is a metric measure space if $(X,\,\mathsf {d})$ is a metric space and $\mu$ is a fully supported Borel measure on $(X,\,\mathsf {d})$ such that $0<\mu (B_r(x))<+\infty$ for any $x \in X$ and $r>0$. In case $X$ is an open subset of a space equipped with a norm $\|\cdot \|$ and $\mathsf {d}$ is the canonical distance associated with $\|\cdot \|$, we write $(X,\,\|\cdot \|,\,\mu )$.

Let $(X,\,\mathsf {d},\,\mu )$ be a metric measure space. We use classical notation to denote function spaces on $X$, like $C(X)$ for the set of continuous functions, $\operatorname {Lip}(X)$ for the set of Lipschitz functions, and so on. We set $\overline {\mathbb {R}}:= \mathbb {R}\cup \{\pm \infty \}$. We shall often identify a $\mu$-measurable function $f : X \to \overline {\mathbb {R}}$ with its equivalence class under $\mu$-a.e. equality. For any $p>0$, we let $L^p(X,\,\mu )$ be the set of $p$-integrable functions, that is, $\mu$-measurable functions $f : X \to \overline {\mathbb {R}}$ such that $\int _X |f|^p \mathop {}\!\mathrm {d} \mu < +\infty$. We also let $L^p_{\text {loc}}(X,\,\mu )$ be the set of locally $p$-integrable functions, that is, $\mu$-measurable functions $f : X \to \overline {\mathbb {R}}$ such that any $x \in X$ admits a neighbourhood $V$ such that $\int _V |f|^p \mathop {}\!\mathrm {d} \mu < +\infty$. We let $L^\infty (X,\,\mu )$ be the set of essentially bounded $\mu$-measurable functions on $X$, and $L_\text {loc}^\infty (X,\,\mu )$ be the set of locally essentially bounded $\mu$-measurable functions on $X$, that is, the $\mu$-measurable functions $f : X \to \overline {\mathbb {R}}$ such that $\mu$-a.e. $x \in X$ admits a neighbourhood $V$ such that the essential supremum of $f$ on $V$ is bounded from above. In case $X$ is an open subset of $\mathbb {R}^n$ and $\mu$ is the Lebesgue measure, we simply write $L^p(X)$, $L^p_{ {{\textrm {loc}}}}(X)$, $L^\infty (X)$, and $L^\infty _{{{\textrm {loc}}}}(X)$ respectively. If $A$ is a bounded Borel subset of $X$, we write $\mathbf {1}_A$ for the characteristic function of $A$, and for any locally integrable function $u:X \to {\mathbb {R}}$, we set

\[ {\unicode{x2A0F}}_A u \mathop{}\!\mathrm{d} \mu := \frac{1}{\mu(A)} \int_A u \mathop{}\!\mathrm{d} \mu. \]

Assume that $(X,\,\mathsf {d})$ is locally compact. Under this assumption, we shall often make use of compact sets. We use the subscript $c$ to denote the restriction of a function space to compactly supported functions: for instance, by $C_c(X)$ we mean the set of compactly supported continuous functions. We recall that a Radon measure $\nu$ on $(X,\,\mathsf {d})$ is a Borel measure which is locally finite and inner regular, meaning that any $x \in X$ admits a neighbourhood $V$ such that $\nu (V)<+\infty$, and any measurable subset $A \subset X$ satisfies $\mu (A)=\sup \{\mu (K) \, : \, K \subset A \, \, \text {compact}\}$, respectively. The Riesz–Markov–Kakutani theorem states that the space of Radon measures $\mathrm {Rad}(X)$ is the topological dual of $C_c(X)$. Therefore, we equip $\mathrm {Rad}(X)$ with the associated weak topology, and we denote by $\rightharpoonup$ the convergence in this topology. Lastly, for any $f \in L^1_{\text {loc}}(X,\,\mu )$, we denote by $f \mu$ the Radon measure defined by

\[ (f \mu)(A) := \int_A f \mathop{}\!\mathrm{d} \mu \]

for any measurable $A \subset X$.

Lebesgue points. Let $(X,\,\mathsf {d},\,\mu )$ be a metric measure space. For any $u \in L^1_{\text {loc}}(X,\,\mu )$, we recall that $x \in X$ is a Lebesgue point of $u$ if there exists some real number $u^{*}(x)$ such that

(2.1)\begin{equation} \lim\limits_{r\downarrow 0}{\unicode{x2A0F}}_{B_r(x)} |u(y)-u^*(x)|\mathop{}\!\mathrm{d} \mu(y) = 0. \end{equation}

We write $\mathrm {Leb}(u)$ for the set of Lebesgue points of $u$. It is easily checked that neither the set $\mathrm {Leb}(u)$ nor the real number $u^*(x)$ for any $x \in \mathrm {Leb}(u)$ depend on the choice of a representative in the equivalence class of $u$. Moreover, the following hold.

1. (1) If $(X,\,\mathsf {d})$ is locally compact, then for any $u \in C(X)$ a simple argument based on the Heine–Cantor theorem shows that $\mathrm {Leb}(u)=X$ and $u^*(x)=u(x)$ for any $x \in X$.

2. (2) If $(X,\,\mathsf {d},\,\mu )$ is infinitesimally doubling, meaning that

   (2.2)\begin{equation} \limsup_{r \downarrow 0} \frac{\mu(B_{2r}(x))}{\mu(B_r(x))}<{+}\infty \end{equation}
   for $\mu$-a.e. $x \in X$, then $\mu (X\backslash \mathrm {Leb}(u))=0$ for any $u \in L^1_{\text {loc}}(X,\,\mu )$ and $u^*(x)=u(x)$ for $\mu$-a.e. $x \in X$. This follows from the Lebesgue Differentiation Theorem for infinitesimally doubling metric measure spaces: see [Reference Heinonen, Koskela, Shanmugalingam and Tyson15, Theorem 3.4.3 and page 77].

Comparability conditions. We now introduce a couple of definitions which originate at least from [Reference Naor and Tao23]. They have also appeared in a different way in [Reference Aldaz4,   Reference Aldaz5].

Observe that: $(3) \Rightarrow (2) \Rightarrow (1)$.

Remark 2.2 If $(X,\,\mathsf {d})$ is locally compact, then the neighbourhoods in the previous definition can be chosen compact. In particular, if $\mu$ satisfies a locally uniform comparability condition, it is not difficult to show that for any compact set $K \subset X$ there exists $r_K,\,C_K>0$ such that

\[ \mu(B_r(x)) \le C_K \mu(B_r(y)) \]

for $\mu$-a.e. $x \in K$, any $r \in (0,\,r_K)$ and $\mu$-a.e. $y \in B_r(x)$.

Example 2.3 A metric measure space is doubling at scale $r_0>0$ if there exists a constant $C\ge 1$ such that

\[ \mu(B_{2r}(x)) \le C \mu(B_r(x)) \]

for any $x \in X$ and $r \in (0,\,r_0/2)$. Any such space trivially satisfies a uniform comparability condition (3), since for any $y \in B_r(x)$ where $x,\,r$ are as above,

\[ \mu(B_r(x)) \le \mu(B_{2r}(y)) \le C \mu(B_r(y)). \]

Example 2.4 For any $Q>0$, we say that a metric measure space $(X,\,\mathsf {d},\,\mu )$ is Ahlfors $Q$-regular if there exist $C\ge 1$ and $r_0\in (0,\,+\infty ]$ such that

(2.4)\begin{equation} C^{{-}1} r^Q \le \mu(B_r(x)) \le C r^Q \end{equation}

for any $x \in X$ and $r \in (0,\,r_0)$. Non-trivial examples including fractal-type ones may be found in [Reference Bourdon and Pajot9, Reference Kleiner and Schioppa18, Reference Laakso20], see also [Reference Barlow8] where they are called fractional metric measure spaces. Ahlfors regular spaces are obviously doubling, hence they satisfy a uniform comparability condition (3). Moreover, we say that $(X,\,\mathsf {d},\,\mu )$ is locally Ahlfors $Q$-regular if any $x_0 \in X$ admits a neighbourhood $V$ and two constants $r_0=r_0(x_0),\,C=C(x_0)>0$ such that

\[ C^{{-}1} r^Q \le \mu(B_r(x)) \le C r^Q \]

for any $x \in V$ and $r \in (0,\,r_0)$. In this case $\mu$ trivially satisfies a locally uniform comparability condition (2).

Averaging operator. We recall now the definition and some properties of the averaging operator.

Definition 2.6 Averaging operator

Let $(X,\,\mathsf {d},\,\mu )$ be a metric measure space. For any $r>0$, the averaging operator $A_r$ is defined by setting, for all $f \in L^1_{\text {loc}}(X,\,\mu )$ and $\mu$-a.e. $x \in X$,

\[ A_rf(x):={\unicode{x2A0F}}_{B_r(x)} f \mathop{}\!\mathrm{d} \mu. \]

Remark 2.7 Obviously $\sup _{r>0}\|A_r\|_{L^\infty (X,\mu )\to L^\infty (X,\mu )} \le 1$. In addition, if $f \in L^\infty _{\text {loc}}(X,\,\mu )$, then any $x \in X$ admits a neighbourhood $V$ and a radius $r_0=r_0(x)$ such that $\|A_r f\|_{L^\infty (V,\mu )}<+\infty$ for any $r \in (0,\,r_0)$. Indeed, if $W$ is a neighbourhood of $x$ such that $\|f\|_{L^\infty (W,\mu )}<+\infty$, then there exists $r_0>$ such that $B_{2r_0}(x) \subset W$, and for any $r \in (0,\,r_0)$ and $\mu$-a.e. $y \in B_{r_0}(x)$,

\[ |A_rf(y)| \le {\unicode{x2A0F}}_{B_r(y)} |f(z)| \mathop{}\!\mathrm{d} \mu(z) \le \|f\|_{L^\infty(B_r(y),\mu)} \le \|f\|_{L^\infty(W,\mu)}<{+}\infty, \]

hence we can take $V=B_{r_0}(x)$.

Proof. $\textit {(1)}$ If $\mu$ satisfies a uniform comparability condition with parameters $r_0,\,C$, then for any $x \in X$ and $r \in (0,\,r_0)$,

\[ a_r(x) := \int_{B_r(x)} \frac{\mathop{}\!\mathrm{d} \mu(y)}{\mu(B_r(y))} \le C. \]

This implies boundedness of $A_r$ from $L^1(X,\,\mu )$ to itself with norm at most $C$, see [Reference Aldaz4, Theorem 3.3] for instance. Then, the Jensen inequality implies the case $p >1$, see e.g. [Reference Aldaz5, Theorem 2.9].

$\textit {(2)}$ Consider $x \in X$ and $f \in L^p_{\text {loc}}(X,\,\mu )$. By (4.2), there exist $r_0=r_0(x,\,f), C=C(x,\,f)>0$ and a neighbourhood $V$ of $x$ also depending of $f$ such that $\int _V |f|^p \mathop {}\!\mathrm {d} \mu < +\infty$, $B_{2r_0}(x) \subset V$ and $\mu (B_r(y))\le C \mu (B_r(z))$ for $\mu$-a.e. $y \in V$, any $r \in (0,\,r_0)$ and $\mu$-a.e. $z \in B_{r}(y)$. The Hölder inequality yields

\[ \left( {\unicode{x2A0F}}_{B_r(z)} |f(y)| \mathop{}\!\mathrm{d} \mu(y) \right)^{p} \le {\unicode{x2A0F}}_{B_r(z)} |f(y)|^p \mathop{}\!\mathrm{d} \mu(y). \]

Then,

\begin{align*} \int_V |A_r f|^p \mathop{}\!\mathrm{d} \mu \le\! \int_V {\unicode{x2A0F}}_{B_r(z)} |f(y)|^p \mathop{}\!\mathrm{d} \mu(y) \mathop{}\!\mathrm{d} \mu(z) & =\! \int_V \int_V\! 1_{B_r(z)}(y) \frac{|f(y)|^p}{\mu(B_r(z))} \mathop{}\!\mathrm{d} \mu(y) \mathop{}\!\mathrm{d} \mu(z)\\ & \le \int_V |f(y)|^p \int_{B_r(y)} \frac{\mathop{}\!\mathrm{d} \mu(z)}{\mu(B_r(z))} \mathop{}\!\mathrm{d} \mu(y)\\ & \le C \int_V |f(y)|^p \mathop{}\!\mathrm{d} \mu(y). \end{align*}

Remark 2.9 Adjoint of $A_r$

Assume that a uniform comparability condition holds with parameters $r_0,\,C$. Then, for any $r \in (0,\,r_0)$, the operator $A_r^{*}$ defined by

\[ A_r^{*} f(x):=\int_{B_r(x)} \frac{f(y)\mathop{}\!\mathrm{d} \mu(y)}{\mu(B_r(y))} \]

for $\mu$-a.e. $x \in X$, is bounded from $L^p(X,\,\mu )$ into itself for any $p \ge 1$. Indeed,

\[ A_r^{*} f(x) = {\unicode{x2A0F}}_{B_r(x)} \frac{f(y)\mu(B_r(x))\mathop{}\!\mathrm{d} \mu(y)}{\mu(B_r(y))} \le C {\unicode{x2A0F}}_{B_r(x)} f(y)\mathop{}\!\mathrm{d} \mu(y) = CA_r f(x). \]

Moreover, the Fubini theorem and the symmetry of the distance easily imply that

\[ A_r^{*} f(x):=\int_{B_r(x)} \frac{f(y)\mathop{}\!\mathrm{d} \mu(y)}{\mu(B_r(y))} \]

for $\mu$-a.e. $x \in X$, for all $f \in L^2(X,\,\mu )$. This means that $A_r^*$ is the adjoint of $A_r$ with respect to the $L^2$ scalar product. More generally, we have that for any $p,\,q \ge 1$ such that $1/p+1/q=1$, one has

\[ \int_X (A_rg)h \mathop{}\!\mathrm{d} \mu = \int_X g (A_r^*h) \mathop{}\!\mathrm{d} \mu \]

for all $g \in L^p(X,\,\mu )$ and $h \in L^q(X,\,\mu )$.

Distortion. We finally introduce a couple of notions which helps us formalizing the infinitesimal relation between the metric and the measure of a metric measure space. Compare with [Reference Kapovitch, Lytchak and Petrunin16].

Definition 2.10 Let $(X,\,\mathsf {d},\,\mu )$ be a metric measure space. For any $r>0$, we define:

1. (1) for any $x,\,y \in X$,

   \[ \delta_r (x,y) := 1-\frac{\mu(B_r(x))}{\mu(B_r(y))} , \]

2. (2) for any $x \in X$,

   \[ z_r(x) := {\unicode{x2A0F}}_{B_r(x)} |\delta_r(x,y)| \mathop{}\!\mathrm{d} \mu(y) . \]

We say that $\delta _r$ is the $r$-distortion function of $(X,\,\mathsf {d},\,\mu )$.

Remark 2.11 In [Reference Kapovitch, Lytchak and Petrunin16], the authors considered the $r$-deviation functions

\[ v_r(x) := 1 - \frac{\mu(B_r(x))}{\omega_n r^n} \]

which measure in a very rough way how much a metric measure space differs from the Euclidean $\mathbb {R}^n$. These deviation functions may behave quite differently from the distortion ones. For instance, consider a smooth Riemannian manifold equipped with its canonical Riemannian distance and volume measure. At an interior point $x$ of such a space,

\[ v_r(x) = c S(x) r^2 + O(r^4) \quad \text{as $r \downarrow 0$,} \]

where $S(x)$ is the scalar curvature at $x$ and $c$ is a dimensional constant (see (4.8)). However, we show in the proof of proposition 4.3 that if $y=y(r) \in B_r(x)$ for any $r>0$, then

\[ \delta_r(x,y) = c(S(y)-S(x))r^2 + O(r^4) \quad \text{as $r \downarrow 0$,} \]

where $O(r^4)$ does not depend on $y$. In particular, if the scalar curvature is constant but non-zero, then $v_r(x) = c r^2 + O(r^4)$ but $\delta _r(x,\,y) = O(r^4)$.

3. Definitions

In this section, we refine the definitions of AMV Laplacian proposed in [Reference Adamowicz, Kijowski and Soultanis1, Reference Minne and Tewodrose21] and we put forward similar definitions for the SAMV Laplacian.

Pointwise AMV and SAMV Laplacians. In order to define the pointwise $\operatorname {AMV}$ Laplacian of a locally integrable function $u$ over a metric measure space $(X,\,\mathsf {d},\,\mu )$, in [Reference Minne and Tewodrose21, Definition 1.1] we made the convention of choosing the representative $\tilde {u}$ of $u$ defined by

\[ \tilde{u}(x) := \lim\limits_{r \downarrow 0} {\unicode{x2A0F}}_{B_r(x)} u(y) \mathop{}\!\mathrm{d} \mu(y) \]

for any $x \in X$ where this limit exist and then to set

(3.1)\begin{equation} \Delta^{\mathsf{d}}_{\mu}u(x) := \frac{1}{r^2}{\unicode{x2A0F}}_{B_r(x)} u(y) -\tilde{u}(x)\mathop{}\!\mathrm{d} \mu(y). \end{equation}

However, this definition has a couple of disadvantages which we point out in remark 3.2. Therefore, we propose a new definition, formalized with the aid of Lebesgue points, along with a similar definition for the $\operatorname {SAMV}$ Laplacian.

Strong AMV/SAMV Laplacian. In [Reference Adamowicz, Kijowski and Soultanis1] the authors proposed a definition of strongly $\operatorname {AMV}$-harmonic functions by appealing on compact sets. In order to work in a more general setting, we define here a strong $\operatorname {AMV}$ Laplacian and a strong $\operatorname {SAMV}$ Laplacian without involving compact sets. Let $(X,\,\mathsf {d},\,\mu )$ be a metric measure space. Consider $u \in L^\infty _{\text {loc}}(X,\,\mu )$. According to remark 2.7, any $x \in X$ admits a neighbourhood $V$ and a radius $r_0>0$ such that

\[ \|\Delta^{\mathsf{d}}_{\mu,r}u\|_{L^\infty(B_{r_0}(x),\mu)} \le \frac{2\|u\|_{L^\infty(V,\mu)}}{r^2} \]

for any $r \in (0,\,r_0)$. This observation guarantees the well-posedness of the following definition.

Definition 3.3 Strong $\operatorname {AMV}$ Laplacian

Let $(X,\,\mathsf {d},\,\mu )$ be a metric measure space. We say that $u \in L^\infty _{\text {loc}}(X,\,\mu )$ admits a strong $\operatorname {AMV}$-Laplacian if there exists a function $v \in L^\infty _{\text {loc}}(X,\,\mu )$ such that any $x \in X$ admits a neighbourhood $V$ such that

\[ \lim\limits_{r \downarrow 0} \|\Delta^{\mathsf{d}}_{\mu,r} u -v\|_{L^\infty(V,\mu)} = 0, \]

in which case we say that $\Delta ^{\mathsf {d}}_{\mu } u := v$ is the strong $\operatorname {AMV}$-Laplacian of $u$.

Let us provide a analogous definition in the symmetrized case. To this aim, we assume that $\mu$ satisfies a local comparability condition. Under this condition, for any $u \in L^\infty _{\text {loc}}(X,\,\mu )$ and $x \in X$, there exist $C,\,r_0>0$ and a neighbourhood $V$ of $x$ such that

\[ \|\tilde{\Delta}^{\mathsf{d}}_{\mu,r} u\|_{L^\infty(B_{r_0}(x))} \le \frac{2C\|u\|_{L^\infty(V,\mu)}}{r^2} \]

for any $r\in (0,\,r_0)$.

Definition 3.4 Strong $\operatorname {SAMV}$ Laplacian

Let $(X,\,\mathsf {d},\,\mu )$ be a metric measure space. Assume that $\mu$ satisfies a local comparability condition. We say that $u \in L^\infty _{\text {loc}}(X,\,\mu )$ admits a strong $\operatorname {SAMV}$-Laplacian if there exists a function $v \in L^\infty _{\text {loc}}(X,\,\mu )$ such that any $x \in X$ admits a neighbourhood $V$ such that

\[ \lim\limits_{r \downarrow 0} \|\tilde{\Delta}^{\mathsf{d}}_{\mu,r} u - v \|_{L^\infty(V,\mu)} = 0, \]

in which case we say that $\tilde {\Delta }^{\mathsf {d}}_{\mu } u:= v$ is the strong $\operatorname {SAMV}$-Laplacian of $u$.

$\boldsymbol{L}^p/\boldsymbol{L}^p_{\boldsymbol{loc}}$-AMV/SAMV Laplacian. Let us consider $p\in [1,\,+\infty )$. Recall that lemma 2.8 ensures that under a uniform comparability condition, the operator $A_r$ is bounded from $L^p$ to itself for any sufficiently small $r$. As a consequence, also $\Delta _{\mu,r}^\mathsf {d}$ is bounded from $L^p$ to itself. Then, the following definition is well-posed.

Remark 3.6 Let $u,\,v \in L^2(X,\,\mu )$ be both admitting an $L^2$-$\operatorname {SAMV}$ Laplacian. Since

\[ \tilde{\Delta}^{\mathsf{d}}_{\mu,r} u(x) = \int_X k_r(x,y) (u(y)-u(x))\mathop{}\!\mathrm{d} \mu(y) \]

for any $r>0$ and $\mu$-a.e. $x \in X$, where

\[ k_r(x,y) := \frac{1}{2r^2} \mathbf{1}_{[0,r)}(\mathsf{d}(x,y)) \left( \frac{1}{\mu(B_r(x))} + \frac{1}{\mu(B_r(y))} \right), \]

the Fubini theorem implies

\[ \int_X (\tilde{\Delta}^{\mathsf{d}}_{\mu,r}u) v \mathop{}\!\mathrm{d} \mu = \int_X u(\tilde{\Delta}^{\mathsf{d}}_{\mu,r}v) \mathop{}\!\mathrm{d} \mu, \]

hence from letting $r \downarrow 0$ we obtain

(3.4)\begin{equation} \int_X (\tilde{\Delta}^{\mathsf{d}}_{\mu}u) v \mathop{}\!\mathrm{d} \mu = \int_X u(\tilde{\Delta}^{\mathsf{d}}_{\mu}v) \mathop{}\!\mathrm{d} \mu. \end{equation}

Let us now provide a definition of $L^p_{\text {loc}}$-AMV Laplacian and $L^p_{\text {loc}}$-SAMV Laplacian. In this case we assume a locally uniform comparability condition.

Weak AMV/SAMV Laplacian. Let us now define a notion of weak $\operatorname {AMV}$ Laplacian and weak $\operatorname {SAMV}$ Laplacian. We focus on locally compact metric measure spaces $(X,\,\mathsf {d},\,\mu )$ such that $\mu$ satisfies a locally uniform comparability condition. Under these assumptions, remark 2.2 implies that if $K$ is the support of a function $\varphi \in C_c(X)$, then there exists $r_K>$ such that for any $r \in (0,\,r_K)$,

\[ \left| \int_K \Delta^\mathsf{d}_{\mu,r}\varphi \mathop{}\!\mathrm{d} \mu \right| + \left| \int_K \tilde{\Delta}^\mathsf{d}_{\mu,r}\varphi \mathop{}\!\mathrm{d} \mu \right| <{+}\infty. \]

Then, the following definition is well-posed.

We point out that if $u$ admits a weak $\operatorname {SAMV}$ Laplacian and $\varphi \in C_c(X)$ admits an $L^2$-$\operatorname {SAMV}$ Laplacian, it follows from (3.4) that

\[ \int_X \varphi \mathop{}\!\mathrm{d} \bf{\tilde{\Delta}}^\mathsf{d}_{\mu}u = \int_X (\tilde{\Delta}^\mathsf{d}_{\mu,r} \varphi) u \mathop{}\!\mathrm{d} \mu. \]

The connection between the weak and $L_{\text {loc}}^{p}$ Laplacians is the following.

Proof. Consider $\varphi \in C_{c}(X)$ with support $K$. Let $q$ be the conjugate exponent of $p$, i.e. $1/p+1/q=1$. Then, for any sufficiently small $r>0$,

\begin{align*} \left|\int_{X}\varphi(\Delta_{\mu,r}^{\mathsf{d}}u-\Delta_{\mu}^{\mathsf{d}}u)\mathop{}\!\mathrm{d} \mu\right| & \le\int_{K}\left|\varphi\right|\left|\Delta_{\mu,r}^{\mathsf{d}}u-\Delta_{\mu}^{\mathsf{d}}u\right|\mathop{}\!\mathrm{d} \mu\\ & \le\left\Vert \varphi\right\Vert _{L^{q}(K,\mu)}\left\Vert \Delta_{\mu,r}^{\mathsf{d}}u-\Delta_{\mu}^{\mathsf{d}}u\right\Vert _{L^{p}(K,\mu)}. \end{align*}

The last term tends to zero as $r\downarrow 0$, hence $\lim _{r\downarrow 0}\int _{X}\varphi \Delta _{\mu,r}^{\mathsf {d}}u\,\mathop {}\!\mathrm {d}\mu =\int _{X}\varphi \Delta _{\mu }^{\mathsf {d}}u\,\mathop {}\!\mathrm {d}\mu$. The $\operatorname {SAMV}$ case is proved the same way.

Some functions admit a distributional or weak $\operatorname {AMV}$ Laplacian but no $L_{\text {loc}}^{p}$-$\operatorname {AMV}$ Laplacian. Here is an example.

Example 3.13 Consider $(\mathbb {R},\,\mathsf {d}_{\text {e}},\,\mathscr {L}^1)$ and $u(x)=x/|x|$, defined to be zero for ${x=0}$. Then, an easy computation shows that

\[ \Delta_{\mathscr{L}^1,r}^{\mathsf{d}_{\text{e}}}u(x)=\begin{cases} \dfrac{x-r\textrm{sgn} x}{r^{3}}, & r>|x|,\\ 0, & r\le|x|. \end{cases} \]

Let $V$ be any open neighbourhood around the origin on which $v$ is $p$-integrable. Then, there exists an $\epsilon >0$ such that $(-\epsilon,\,\epsilon )\subseteq V$, and we can assume without loss of generality that $3r<\epsilon$. Then, for $p\in (1,\,+\infty )$,

\begin{align*} \lVert\Delta_{\mathscr{L}^1,r}^{\mathsf{d}_{\text{e}}} u-v\rVert_{L^{p}(V)} & \ge\lVert\Delta_{\mathscr{L}^1,r}^{\mathsf{d}_{\text{e}}} u-v\rVert_{L^{p}([-\epsilon,\epsilon])}\\ & \ge\lVert\Delta_{\mathscr{L}^1,r}^{\mathsf{d}_{\text{e}}} u\rVert_{L^{p}([-\epsilon,\epsilon])}-\lVert v\rVert_{L^{p}([-\epsilon,\epsilon])}\\ & \ge\frac{1}{r^{3}}\left(\int_{2r}^{\epsilon}(x-r)^{p}dx\right)^{1/p}-\lVert v\rVert_{L^{p}([-\epsilon,\epsilon])}\\ & \ge\frac{1}{r^{3}}\left((\epsilon-2r)(2r-r)^{p}\right)^{1/p}-\lVert v\rVert_{L^{p}([-\epsilon,\epsilon])}\\ & \ge\frac{r^{1/p}\cdot r}{r^{3}}-\lVert v\rVert_{L^{p}([-\epsilon,\epsilon])}\to+\infty,\qquad r\downarrow 0. \end{align*}

This shows that $u$ does not admit an $L_{\text {loc}}^{p}$-AMV Laplacian for any $p\in (1,\,+\infty )$. This result trivially extends to $p=\infty$. Instead, $u$ has a distributional $\operatorname {AMV}$ Laplacian which can be shown to be the distributional derivative of the Dirac delta at the origin divided by three [Reference Minne and Tewodrose21, p. 21].

4. Equality between the $\operatorname {AMV}$ and $\operatorname {SAMV}$ Laplacians

In this section, we study some contexts where the equality $\Delta ^{\mathsf {d}}_{\mu }=\tilde {\Delta }^{\mathsf {d}}_{\mu }$ holds in a suitable sense.

Topological groups. We recall that a topological group is a set $G$ equipped with a group law $\cdot$ and a topology $\mathcal {T}$ with respect to which the maps $(g,\,h) \mapsto g \cdot h$ and $g \mapsto g^{-1}$ are continuous. A left-invariant distance $\mathsf {d}$ on a topological group $(G,\,\cdot,\,\mathcal {T})$ is a distance on $G$ which induces the same topology as $\mathcal {T}$ and such that

\[ \mathsf{d}(g \cdot x, g \cdot y) = \mathsf{d}(x,y) \]

for any $g,\,x,\,y \in G$. When a topological group $(G,\,\cdot,\,\mathcal {T})$ is locally compact, it admits a unique—up to a positive multiplicative constant—non-zero Borel measure $\mu$, called Haar measure, which is finite on compact sets, quasi-regular, and left-invariant in the sense that

\[ \mu(g\cdot A)=\mu(A) \]

for any $g \in G$ and $A \in \mathcal {T}$.

Proof. For any $x \in G$, $r>0$ and $y \in B_r(x)$, since $\mathsf {d}$ is left-invariant,

\[ y\cdot x^{{-}1} \cdot B_r(x) = B_r(y), \]

and since $\mu$ is left-invariant, this implies that

(4.3)\begin{equation} \mu(B_r(x)) = \mu(B_r(y)). \end{equation}

Then, for any $u \in L^1(G,\,\mu )$ and any $x \in \mathrm {Leb}(u)$, one has $\Delta ^{\mathsf {d}}_{\mu,r}u(x)=\tilde {\Delta }^{\mathsf {d}}_{\mu,r}u(x)$ for any $r>0$, from which (4.1) follows as $r \downarrow 0$. Moreover, (4.3) ensures that $\mu$ trivially satisfies a uniform comparability condition.

Riemannian and sub-Riemannian manifolds. We provide in the next proposition a general criterion to get that $\Delta ^\mathsf {d}_{\mu }$ and $\tilde {\Delta }^\mathsf {d}_{\mu }$ coincide. We phrase this result in terms of the $r$-distortion function $\delta _r (x,\,y)$ introduced in definition 2.10.

Proposition 4.2 Let $(X,\,\mathsf {d},\,\mu )$ be a metric measure space. For any $x \in X$ and $r>0$, set

(4.4)\begin{equation} \varepsilon_x(r):= \sup_{y \in B_r(x)} |\delta_r(x,y)|. \end{equation}

Let $u \in L^1_{\text {loc}}(X,\,\mu )$ be given. Then, for any $x\in \mathrm {Leb}(u)$ such that

(4.5)\begin{equation} \varepsilon_x(r)=O(r^2),\quad r \downarrow 0, \end{equation}

it holds that

(4.6)\begin{equation} \lim\limits_{r \downarrow 0} |\Delta^\mathsf{d}_{\mu,r}u(x) - \tilde{\Delta}^\mathsf{d}_{\mu,r} u(x)| = 0. \end{equation}

In particular, if $x \in \mathrm {Leb}(u)$ satisfies (4.5), then $\Delta ^\mathsf {d}_{\mu,r}u(x)$ converges if and only if $\tilde {\Delta }^\mathsf {d}_{\mu,r} u(x)$ does, in which case $\Delta ^\mathsf {d}_{\mu }u(x)= \tilde {\Delta }^\mathsf {d}_{\mu } u(x)$.

Proof. For any $x \in \mathrm {Leb}(u)$ and $r>0$,

(4.7)\begin{equation} \Delta^\mathsf{d}_{\mu,r}u(x) - \tilde{\Delta}^\mathsf{d}_{\mu,r} u(x) = \frac{1}{2r^2} {\unicode{x2A0F}}_{B_r(x)} \delta_r(x,y) (u(y) - u^*(x)) \mathop{}\!\mathrm{d} \mu(y). \end{equation}

Then,

\[ |\Delta^\mathsf{d}_{\mu,r}u(x) - \tilde{\Delta}^\mathsf{d}_{\mu,r} u(x)| \le \frac{\varepsilon_x(r)}{2r^2} {\unicode{x2A0F}}_{B_r(x)} |u(y)-u^*(x)| \mathop{}\!\mathrm{d} \mu(y). \]

The conclusion follows since ${\unicode{x2A0F}} _{B_r(x)} |u(y)-u^*(x)| \mathop {}\!\mathrm {d} \mu (y) \to 0$ as $r \downarrow 0$.

Thanks to proposition 4.2 we can immediately show that $\Delta ^\mathsf {d}_{\mu }$ and $\tilde {\Delta }^\mathsf {d}_{\mu }$ coincide on regular enough Riemannian manifolds. Indeed, let $M$ be a smooth $n$-dimensional manifold equipped with a $C^2$ Riemannian metric. Let $\mu$ be the associated Riemannian volume measure. As is well-known (see e.g. [Reference Gallot, Hulin and Lafontaine13]), the scalar curvature of $(M,\,g)$ is a continuous function $S_g : M \to \mathbb {R}$ such that

(4.8)\begin{equation} \frac{\mu(B_r(x))}{\omega_n r^n} = 1 - c_n S_g(x) r^2 + O(r^4) \quad \text{as $r \downarrow 0$} \end{equation}

for any $x \in M$, where $c_n:= (n+2)^{-1}6^{-1}$ and $O(r^4)$ depends only on curvature terms of $g$ at $x$ — in particular, if $K$ is a compact subset of $M$, then $O(r^4)$ can be made independent of $x \in K$.

Proposition 4.3 Let $M$ be a smooth manifold equipped with a $C^2$ Riemannian metric $g$. Let $\mu$ be the associated Riemannian volume measure. Let $u \in L^1_{\text {loc}}(M,\,\mu )$ be given. Then, for $\mu$-a.e. $x \in M$,

\[ \lim\limits_{r \downarrow 0} |\Delta_{\mu,r}^{\mathsf{d}}u(x) - \tilde{\Delta}_{\mu,r}^{\mathsf{d}} u(x)| = 0. \]

In particular, for $\mu$-a.e. $x \in M$, one has that $\Delta ^\mathsf {d}_{\mu,r}u(x)$ converges if and only if $\tilde {\Delta }^\mathsf {d}_{\mu,r} u(x)$ does, in which case $\Delta ^\mathsf {d}_{\mu }u(x)= \tilde {\Delta }^\mathsf {d}_{\mu } u(x)$.

Proof. Let $n$ be the dimension of $M$. For any $x \in M$ and $y \in B_r(x)$, thanks to (4.8) we find that

\begin{align*} \frac{\mu(B_r(y))}{\mu(B_r(x))} & = \frac{1-c_nS_g(y)r^2 + O(r^4)}{1-c_nS_g(x)r^2 + O(r^4)}\\ & = (1-c_nS_g(y)r^2 + O(r^4))(1+c_nS_g(x)r^2 + O(r^4))\\ & = 1 - c_n(S_g(y)-S_g(x))r^2 + O(r^4) \end{align*}

as $r \downarrow 0$, where we choose $O(r^4)$ uniform over points in the closed ball $\overline {B}_1(x)$. Since $S_g$ is continuous it is uniformly continuous on $\overline {B}_1(x)$ hence it admits a non-decreasing modulus of continuity $\omega$. Therefore,

\[ \frac{\varepsilon_x(r)}{r^2} \le c_n \left( \sup_{y \in B_r(x)} |S_g(y)-S_g(x)| \right) + O(r^2) \le c_n \omega(r) + O(r^2) \]

as $r \downarrow 0$. This shows that (4.5) holds at any $x \in M$.

Let us show now that $(M,\,g)$ satisfies the infinitesimally doubling condition (2.2). Thanks to (4.8), for any $x \in M$,

\begin{align*} \frac{\mu(B_{2r}(x))}{\mu(B_r(x))} & = 2^n \frac{\mu(B_{2r}(x))}{\omega_n (2r)^n}\frac{\omega_n r^n}{\mu(B_r(x))} = 2^n \,\frac{1 - c_n S_g(z) (2r)^2 + O(r^4)}{1 - c_n S_g(z) r^2 + O(r^4)}\,, \end{align*}

as $r \downarrow 0$, where we choose $O(r^4)$ uniform over points in the closed ball $\overline {B}_1(x)$, hence

\[ \lim_{r \downarrow 0} \frac{\mu(B_{2r}(x))}{\mu(B_r(x))} = 2^n. \]

From (2.2) we get that $\mu$-a.e. point $x \in M$ is a Lebesgue point of $u$, hence the result follows from proposition 4.2.

Let us pass now to the context of sub-Riemannian manifolds. We recall that a sub-Riemannian structure on a manifold $M$ is a pair $(D,\, g)$ where $D$ is a bracket generating subbundle of the tangent bundle $TM$, i.e. Lie brackets of vector fields tangent to $D$ span the full tangent bundle, and $g$ is a smooth metric defined on $D$. Such a structure yields a well-defined distance $\mathsf {d}$ called the sub-Riemannian (or Carnot–Carathéodory) distance. More precisely, $\mathsf {d}(p,\, q)$ is the infimum of the length of Lipschitz curves tangent to $D$ (also called horizontal curves) joining two points $p$ and $q$. Here the length of the curve is computed with respect to the metric $g$. We refer to [Reference Agrachev, Barilari and Boscain3] for a complete introduction to sub-Riemannian geometry.

In this context, the analysis made in [Reference Barilari, Beschastnyi and Lerario7] yields the following.

Proposition 4.4 Let $(M,\,D,\,g)$ be a $3$-dimensional, contact sub-Riemannian manifold equipped with the Popp volume $\mu$. Let $u \in L^1_{\text {loc}}(M,\,\mu )$ be given. Then, for $\mu$-a.e. $x \in M$, it holds that

\[ \lim\limits_{r \downarrow 0} |\Delta_{\mu,r}^{\mathsf{d}}u(x) - \tilde{\Delta}_{\mu,r}^{\mathsf{d}} u(x)| = 0. \]

In particular, for $\mu$-a.e. $x \in M$, one has that $\Delta ^\mathsf {d}_{\mu,r}u(x)$ converges if and only if $\tilde {\Delta }^\mathsf {d}_{\mu,r} u(x)$ does, in which case $\Delta ^\mathsf {d}_{\mu }u(x)= \tilde {\Delta }^\mathsf {d}_{\mu } u(x)$.

Proof. From [Reference Barilari, Beschastnyi and Lerario7, Theorem 1] we know the following for any $x \in M$,

(4.9)\begin{equation} \frac{\mu(B_r(x))}{c_o r^4} = 1 - c_1 \kappa(x) r^2 + O(r^3) \quad \text{as $r \downarrow 0$}, \end{equation}

where $c_o$ and $c_1$ are positive constants and $\kappa$ depends smoothly of $x$. Then, the proof of proposition 4.3 carries over and yields the result.

Spaces with suitably vanishing distortion. We investigate now the case of spaces where the distortion functions $\delta _r$ satisfy more subtle assumptions. We begin with the following elementary lemma. Recall that $z_r(x)$ is defined in definition 2.10 as the average of $|\delta _r(x,\,y)|$.

Lemma 4.5 Let $(X,\,\mathsf {d},\,\mu )$ be a locally compact metric measure space. Then, for any $u \in \operatorname {Lip}(X)$, $\varphi \in C_c(X)$, and $r>0$,

\[ \left| \int_X \varphi (\Delta^{\mathsf{d}}_{\mu,r}u-\tilde{\Delta}^{\mathsf{d}}_{\mu,r} u) \mathop{}\!\mathrm{d} \mu\right| \le \frac{\textrm{Lip}(u)}{2}\int_X |\varphi| \frac{z_r}{r} \mathop{}\!\mathrm{d} \mu. \]

Proof. The result follows from a direct computation:

\begin{align*} \left| \int_X \varphi (\Delta^{\mathsf{d}}_{\mu,r}u-\tilde{\Delta}^{\mathsf{d}}_{\mu,r} u) \mathop{}\!\mathrm{d} \mu\right| & \le \frac{1}{2}\int_X |\varphi(x)| {\unicode{x2A0F}}_{B_r(x)} |\delta_r(x,y)| \frac{|u(y)-u(x)|}{r^2}\mathop{}\!\mathrm{d} \mu(y) \mathop{}\!\mathrm{d} \mu(x)\\ & \le \frac{\textrm{Lip}(u)}{2}\int_X |\varphi(x)| {\unicode{x2A0F}}_{B_r(x)} |\delta_r(x,y)| \frac{\mathsf{d}(x,y)}{r^2}\mathop{}\!\mathrm{d} \mu(y) \mathop{}\!\mathrm{d} \mu(x)\\ & \le \frac{\textrm{Lip}(u)}{2}\int_X |\varphi(x)| {\unicode{x2A0F}}_{B_r(x)} \frac{ |\delta_r(x,y)| }{r}\mathop{}\!\mathrm{d} \mu(y) \mathop{}\!\mathrm{d} \mu(x)\\ & = \frac{\textrm{Lip}(u)}{2}\int_X |\varphi(x)| \frac{z_r(x)}{r} \mathop{}\!\mathrm{d} \mu(x). \end{align*}

As an immediate corollary, we obtain the following result by letting $r \downarrow 0$.

Corollary 4.6 Let $(X,\,\mathsf {d},\,\mu )$ be a locally compact metric measure space such that $z_r \in L^1_\text {loc}(X,\,\mu )$ for all $r>0$ small enough. Assume that

(4.10)\begin{equation} \frac{z_r}{r} \mu \rightharpoonup 0 \quad \text{as }r\downarrow 0.\end{equation}

Then, $u \in \operatorname {Lip}(X)$ admits a weak $\operatorname {AMV}$ Laplacian if and only if $u$ admits a weak $\operatorname {SAMV}$ Laplacian, in which case $\bf {\Delta }^{\mathsf {d}}_\mu u=\bf {\tilde {\Delta }}^{\mathsf {d}}_\mu u$.

Remark 4.7 If $\mu$ satisfies a locally uniform comparability condition, then $z_r \in L^1_\text {loc}(X,\,\mu )$. Indeed, this assumption and the local compactness of $(X,\,\mathsf {d})$ imply that there exist $r_0,\,C>0$ such that for any $x \in X$, $r \in (0,\,r_0)$ and any compact neighbourhood $K$ of $x$,

\[ \int_K z_r \mathop{}\!\mathrm{d} \mu \le (1+C) \mu\left(\cup_{\bar{x} \in K}B_r(\bar{x})\right)<{+}\infty. \]

We are going to apply corollary 4.6 in the context of locally Ahlfors regular spaces (see example 2.4 for the definition). Let $Q$ be a fixed positive number. We set

\[ \omega_Q := \frac{\pi^{Q/2}}{\Gamma(Q/2+1)} \]

where $\Gamma$ is the classical Gamma function; if $Q$ is an integer $n$, then $\omega _Q$ coincides with the Lebesgue measure of the unit Euclidean ball in $\mathbb {R}^n$. For a locally Ahlfors $Q$-regular metric measure space $(X,\,\mathsf {d},\,\mu )$, we define

\[ \theta_r (x) := \frac{\mu(B_r(x))}{\omega_Q r^Q} \]

for any $x \in X$ and $r>0$, and

\[ \mu_r = \frac{1-\theta_r}{r} \, \mu . \]

Each $\mu _r$ is a signed Radon measure. We let

\[ |\mu_r| = \frac{|1-\theta_r|}{r} \, \mu \]

be the associated total variation measure. Here is the main result of this paragraph.

Proposition 4.8 Let $(X,\,\mathsf {d},\,\mu )$ be a locally compact, locally Ahlfors $Q$-regular metric measure space satisfying

(4.11)\begin{equation} |\mu_r| \rightharpoonup 0 \qquad \text{as $r \downarrow 0$.} \end{equation}

Then, $u \in \operatorname {Lip}(X)$ admits a weak $\operatorname {AMV}$ Laplacian if and only if $u$ admits a weak $\operatorname {SAMV}$ Laplacian, in which case $\bf {\Delta }^{\mathsf {d}}_\mu u=\bf {\tilde {\Delta }}^{\mathsf {d}}_\mu u$.

To prove proposition 4.8, we need the following lemma.

Lemma 4.9 Let $(X,\,\mathsf {d},\,\mu )$ be a locally compact metric measure space satisfying a locally uniform comparability condition. For any compact set $K \subset X,$ there exist $r_K,\, C_K>0$ depending only on $K,$ and another compact set $K' \subset X$ containing $K$ such that for any $f \in L^\infty _{\text {loc}}(X,\,\mu ),$

(4.12)\begin{equation} \sup_{r \in (0,r_K)} \|A_r^*f - f \|_{L^\infty(K,\mu)} \le C_K \|f\|_{L^\infty(K',\mu)}. \end{equation}

Proof. Let $K \subset X$ be a compact set. The local compactness of $(X,\,\mathsf {d})$ ensures that there exist $r_K>0$ and another compact set $K' \subset X$ containing $K$ such that

\[ \bigcup_{x \in K} B_{r_K}(x) \subset K'. \]

Consider $f \in L^\infty _{\text {loc}}(X,\,\mu )$. For $\mu$-a.e. $x\in K$ and any $r \in (0,\,r_K)$,

\begin{align*} |A_r^* f(x) - f(x)| & = \left| \int_{B_r(x)} \left(\frac{f(y)}{\mu(B_r(y))}-\frac{f(x)}{\mu(B_r(x))}\right) \mathop{}\!\mathrm{d} \mu(y) \right|\\ & \le \left| \int_{B_r(x)} \frac{f(y)-f(x)}{\mu(B_r(y))} \mathop{}\!\mathrm{d} \mu(y) \right|\\ & \quad + \left| \int_{B_r(x)} \left(\frac{f(x)}{\mu(B_r(y))}-\frac{f(x)}{\mu(B_r(x))}\right) \mathop{}\!\mathrm{d} \mu(y) \right|\\ & \le \int_{B_r(x)} \frac{|f(y)-f(x)|}{\mu(B_r(y))} \mathop{}\!\mathrm{d} \mu(y) + |f(x)| \left| \int_{B_r(x)}\frac{\mathop{}\!\mathrm{d} \mu(y)}{\mu(B_r(y))}-1\right|\\ & \le 2 \|f\|_{L^\infty(K',\mu)} \int_{B_r(x)} \frac{\mathop{}\!\mathrm{d} \mu(y)}{\mu(B_r(y))}\\ & \quad + \|f\|_{L^\infty(K',\mu)} \left( \int_{B_r(x)} \frac{\mathop{}\!\mathrm{d} \mu(y)}{\mu(B_r(y))}+1\right)\\ & = \|f\|_{L^\infty(K',\mu)} \left( 3 \int_{B_r(x)} \frac{\mathop{}\!\mathrm{d} \mu(y)}{\mu(B_r(y))} + 1\right). \end{align*}

By remark 2.2, the locally uniform comparability condition implies that

\[ \int_{B_r(x)} \frac{\mathop{}\!\mathrm{d} \mu(y)}{\mu(B_r(y))} \le C'_K \]

for some $C'_K>0$ depending only on $K$. We obtain the desired result by setting $C_K:= 3 C'_K +1$.

We are now in a position to prove proposition 4.8.

Proof of proposition 4.8 Proof of proposition 4.8

By example 2.4 our space fulfils a locally uniform comparability condition, and by remark 4.7 we infer that $z_r\in L^1_{\text {loc}}(X,\,\mu )$. Hence corollary 4.6 shows that we only need to prove that

\[ \frac{z_r}{r} \mu \rightharpoonup 0 \]

as $r \downarrow 0$. Take $\varphi \in C_c(X)$. Up to decomposing $\varphi$ as $\varphi ^+ - \varphi ^-$, we do not lose any generality in assuming that $\varphi$ is non-negative, what we do from now on. Let $K$ be the support of $\varphi$. Let $r_K,\, C_K$ and $K'$ be given by lemma 4.9. Consider $r \in (0,\,r_K)$. Observe that if $x \in K$ and $y \in B_r(x)$, then

\begin{align*} |\delta_r(x,y)| & = \frac{|\mu(B_r(y))-\mu(B_r(x))|}{\mu(B_r(y))} = \frac{|\theta_r(y)-\theta_r(x)|}{\theta_r(y)} \\ & \le \frac{1}{\theta_r(y)} (|\theta_r(y)-1| + |1-\theta_r(x))|)\\ & \le \omega_Q^{{-}1} C_{K'} \bigg( |\theta_r(y) -1| + |1-\theta_r(x)| \bigg) \end{align*}

where we have multiplied and divided by $\omega _Q r^Q$ to get the second equality and we have used the local Ahlfors regularity property on $K'$ to get the last one. Then,

\begin{align*} \int_X \varphi \, \frac{z_r}{r} \mathop{}\!\mathrm{d} \mu & = \int_X \varphi(x) {\unicode{x2A0F}}_{B_r(x)} \frac{|\delta_r(x,y)|}{r} \mathop{}\!\mathrm{d} \mu(y) \mathop{}\!\mathrm{d} \mu(x)\\ & \le \omega_Q^{{-}1} C_{K'} \left( \int_X \varphi(x) {\unicode{x2A0F}}_{B_r(x)} \frac{|\theta_r(y)-1|}{r} \mathop{}\!\mathrm{d} \mu(y) \mathop{}\!\mathrm{d} \mu(x)\right.\\ & \quad + \left. \int_X \varphi(x) \frac{|\theta_r(x)-1|}{r} \mathop{}\!\mathrm{d} \mu(x) \right) \end{align*}

By (4.11), the second term on the right-hand side converges to $0$ when $r \downarrow 0$. Let us show that also the first one converges to 0. We have

\begin{align*} \int_X \varphi(x) {\unicode{x2A0F}}_{B_r(x)} \frac{|\theta_r(y)-1|}{r} \mathop{}\!\mathrm{d} \mu(y) \mathop{}\!\mathrm{d} \mu(x) & = \int_X \varphi(x) A_r \left[ \frac{|\theta_r({\cdot})-1|}{r} \right](x) \mathop{}\!\mathrm{d} \mu(x)\\ & = \int_X A_r^*\varphi(x) \frac{|\theta_r(x)-1|}{r} \mathop{}\!\mathrm{d} \mu(x). \end{align*}

Since $x \notin K'$ yields $B_r(x) \cap \operatorname {supp} \varphi = \emptyset$ which in turn implies that $A_r^*\varphi (x)=0$, we may replace the previous integral over $X$ with an integral over $K'$. Then,

\begin{align*} & \int_X \varphi(x) {\unicode{x2A0F}}_{B_r(x)} \frac{|\theta_r(y)-1|}{r} \mathop{}\!\mathrm{d} \mu(y) \mathop{}\!\mathrm{d} \mu(x)\\ & \quad\le \int_{K'} |A_r^*\varphi(x)- \varphi(x)| \frac{|\theta_r(x)-1|}{r} \mathop{}\!\mathrm{d} \mu(x) + \int_{K'} \varphi \frac{|\theta_r-1|}{r} \mathop{}\!\mathrm{d} \mu \\ & \quad \le C \|\varphi\|_{L^\infty(X,\mu)}\int_{K'} \frac{|\theta_r-1|}{r} \mathop{}\!\mathrm{d} \mu\\ & \quad \le C \|\varphi\|_{L^\infty(X,\mu)}\int_{X} \psi \frac{|\theta_r-1|}{r} \mathop{}\!\mathrm{d} \mu \end{align*}

where we have used (4.12) to get the second inequality and we have set $\psi (\cdot ):= (1-\mathsf {d}(\cdot,\,K')/\rho )^+$ for $\rho >0$ small enough to ensure that $\psi \in C_c(X)$. By (4.11), we get

\[ \lim\limits_{ r\downarrow 0} \int_{X} \psi \frac{|\theta_r-1|}{r} \mathop{}\!\mathrm{d} \mu = 0. \]

□

5. Weighted Lebesgue measures

In this section, we study the context of weighted Lebesgue measures where the two notions of pointwise Laplacian do not coincide. We first focus on points where the weight is positive, before tackling points where the weight vanishes.

5.1. Positive weights

We begin this subsection with a straightforward result.

Proposition 5.1 Let $\Omega \subset \mathbb {R}^n$ be an open set equipped with the Euclidean distance $\mathsf {d}_e$, and $\mu := w \mathscr {L}^n {\unicode{x02EA}} \Omega$ where $w \in C^1(\Omega )$. Consider $(\Omega,\, \mathsf {d}_e,\,\mu )$. Then, for any $x \in \Omega$ such that $w(x)>0$, any $u \in C^2(\Omega )$ admits a pointwise $\operatorname {AMV}$ Laplacian and a pointwise $\operatorname {SAMV}$ Laplacian at $x$, and

(5.1)\begin{align} \Delta_{\mu}^{\mathsf{d}_{e}}u(x) & = \frac{1}{2(n+2)}(\Delta u(x) + 2\langle\nabla \ln w,\nabla u\rangle(x)), \end{align}

(5.2)\begin{align} \tilde{\Delta}_{\mu}^{\mathsf{d}_{e}}u(x) & = \frac{1}{2(n+2)}(\Delta u(x) + \langle\nabla \ln w,\nabla u\rangle(x)) \, \cdot \end{align}

Proof. The first equality was proved in [Reference Minne and Tewodrose21, Proposition 2.3]. Let us prove the second equality following the same lines. Let $r_0>0$ be such that $B_{r_0}(x) \subset \Omega$ and $w(y) > w(x)/2$ for any $y \in B_{r_0}(x)$. Then, for any $r \in (0,\,r_0)$,

(5.3)\begin{equation} \tilde{\Delta}_{\mu,r}^{\mathsf{d}_{e}}u(x) = \frac{1}{2}\left(\Delta_{\mu,r}^{\mathsf{d}_{e}}u(x)+I(r)\right). \end{equation}

where

\[ I(r):= \frac{1}{r^{2}}\int_{B_{r}(x)}(u(y)-u(x))\frac{w(y)}{\mu(B_{r}(y))}\mathop{}\!\mathrm{d} y. \]

We claim that

(5.4)\begin{equation} I(r) \to \frac{\Delta u(x)}{2(n+2)}\qquad \text{as }r\downarrow 0. \end{equation}

Once this claim is proved, (5.2) follows from letting $r \downarrow 0$ in (5.3) and using (5.1). We now prove (5.4). Since $w$ is $C^1$, a first-order Taylor expansion shows that for any $0< r< r_0/2$, $y \in B_r(x)$, and $z \in B_r(y)$,

\[ w(z) = w(y) + \langle R_y(z),y-z \rangle \]

for some $R_y(z) \in \mathbb {R}^n$ such that

\[ |R_y(z)| \le C := \sup_{\xi \in \overline{B}_{r_0}(x)} |\nabla w (\xi)|. \]

Then,

\[ \mu(B_{r}(y))=\int_{B_{r}(y)}w(z)\mathop{}\!\mathrm{d} z = \mathscr{L}^{n}(B_{r}(y))\left( w(y)+ {\unicode{x2A0F}}_{B_r(y)}\langle R_y(z),y-z \rangle \mathop{}\!\mathrm{d} z \right) \]

so that

(5.5)\begin{equation} \frac{w(y)}{\mu(B_{r}(y))} = \frac{1}{\mathscr{L}^{n}(B_{r}(y))\left( 1+ \frac{1}{w(y)}{\unicode{x2A0F}}_{B_r(y)}\langle R_y(z),y-z \rangle \mathop{}\!\mathrm{d} z \right)}\, \cdot \end{equation}

Now

(5.6)\begin{equation} \left| \frac{1}{w(y)}{\unicode{x2A0F}}_{B_r(y)}\langle R_y(z),y-z \rangle \mathop{}\!\mathrm{d} z \right| \le \frac{2C }{w(x)} \, r = O(r) \end{equation}

and $\mathscr {L}^{n}(B_{r}(y))=\mathscr {L}^{n}(B_{r}(x))$, hence we get

\[ I(r) = \frac{1}{r^{2}(1+O(r))}{\unicode{x2A0F}}_{B_{r}(x)}u(y)-u(x)\,\mathop{}\!\mathrm{d} y\to\frac{\Delta u(x)}{2(n+2)}\qquad \text{as $r\downarrow 0$}. \]

Remark 5.2 Proposition 5.1 shows that if $w>0$ everywhere in $\Omega$, then $\tilde {\Delta }_{\mu }^{\mathsf {d}_{e}}$ coincides with a dimensional constant times the classical weighted Laplacian of $(\mathbb {R}^n,\,\mathsf {d}_e,\,w\mathscr {L}^n)$ (also known as drifted Laplacian, or $f$-Laplacian, or Witten Laplacian, see e.g. [Reference Colbois, El Soufi and Savo11, Reference Munteanu and Wang22] and the references therein), namely

\[ \Delta u + \langle\nabla \ln w,\nabla u\rangle. \]

Moreover, working in exponential coordinates, it is not difficult to show that (5.1) and (5.2) extend to the setting of weighted Riemannian manifolds $(M,\,g,\,w\mu )$ where $\mu$ is the canonical Riemannian measure and $w : M \to (0,\,+\infty )$ is a smooth map.

We now study the case where the Euclidean distance on $\Omega$ is replaced with a more general distance $\mathsf {d}$. To this aim, for any $x \in \Omega$ and $r>0$, we set

\[ M^{r}(x) := \left( M_{ij}^{r}(x) := \frac{1}{r^2} {\unicode{x2A0F}}_{B_r(x)} (y-x)_i (y-x)_j \mathop{}\!\mathrm{d} y \right)_{1 \le i< j\le n} . \]

In case $\mathsf {d}$ is associated with a norm, the change of variable $\xi =(y-x)/r$ shows that the matrices $M^{r}(x)$ are all equal to the second-moment matrix

\[ M(0) := \left( M_{ij}(0) := {\unicode{x2A0F}}_{B_1(0)} \xi_i \xi_j \mathop{}\!\mathrm{d} \xi \right)_{1 \le i< j\le n}. \]

We will also work under the assumption that balls for $\mathsf {d}$ are symmetric with respect to the vector space structure of $\mathbb {R}^n$, in the sense that for any $x \in \Omega$ and $r>0$ such that $B_r(x) \subset \Omega$, for any $v \in \mathbb {R}^n$,

(5.7)\begin{equation} x+v \in B_r(x) \iff x-v \in B_r(x). \end{equation}

This is trivially satisfied when $\mathsf {d}$ is associated with a norm. But it may fail in general. For instance, set

\[ \mathsf{d}(x,y):=\begin{cases} \lvert x-y\rvert, & x,y\le0,\\ \dfrac{1}{2}|x-y|, & x,y>0,\\ \dfrac{1}{2}y-x, & x\le0,\,y>0,\\ \dfrac{1}{2}x-y, & y\le0,\,x>0. \end{cases} \]

Then, $\mathsf {d}$ is a metric on $\mathbb {R}$ such that $\frac {1}{2}\mathsf {d}_{\text {e}}\le \mathsf {d}\le 2\mathsf {d}_{\text {e}}$ and $B_{r}(0)=(-r,\,2r)$ for any $r>0$.

A large class of distances $\mathsf {d}$ which satisfy (5.7) without being associated with a norm is given by the following: For $\alpha =(\alpha _1,\,\ldots,\,\alpha _n) \in \{1,\,2\}^n$, set

\[ \Phi_\alpha (x) := (x_1^{\alpha_1},\ldots,x_n^{\alpha_n}) \]

for any $x=(x_1,\,\ldots,\,x_n)\in (0,\,+\infty )^n$. Set

\[ \mathsf{d}_\alpha (x,y) := \|\Phi_\alpha(x)-\Phi_\alpha(y)\| \]

for any $x,\,y \in (0,\,+\infty )^n$, where $\|\cdot \|$ is any $l^p$-norm on $\mathbb {R}^n$. Then, $\mathsf {d}_\alpha$ is a distance on $(0,\,+\infty )^n$ which always satisfies (5.7); it is associated with a norm if and only if $\alpha _1 = \ldots = \alpha _n=1$, because otherwise homogeneity fails.

Our next result makes use of $z^{\mathscr {L}^{n}}_r$, i.e. the average of the absolute value of the distortion function, defined for any $r>0$ by

(5.8)\begin{equation} z^{\mathscr{L}^n}_r(x) := {\unicode{x2A0F}}_{B_r(x)}\left|1-\frac{\mathscr{L}^n(B_{r}(x))}{\mathscr{L}^n(B_{r}(y))}\right| \mathop{}\!\mathrm{d} y. \end{equation}

for any $x \in \mathbb {R}^n$.

Proposition 5.3 Let $\Omega \subset \mathbb {R}^n$ be open with respect to a distance $\mathsf {d}$ satisfying (5.7). For $w \in C^1(\Omega )$ set $\mu := w \mathscr {L}^n {\unicode{x02EA}} \Omega$. Consider $(\Omega,\, \mathsf {d},\,\mu )$. Let $x \in \Omega$ be such that $w(x)>0$.

1. (1) Assume that the limit $M(x):= \lim _{r \downarrow 0^+} M^r(x)$ exists. Then, any $u \in C^2(\Omega )$ admits a pointwise $\operatorname {AMV}$ Laplacian at $x$, and

   (5.9)\begin{equation} \Delta_{\mu}^{\mathsf{d}}u(x) =\frac{1}{2}\textrm{Tr}(M(x)\nabla^2 u(x))+\frac{1}{w(x)}\langle\nabla w(x),M(x)\nabla u(x)\rangle. \end{equation}

2. (2) If in addition

   (5.10)\begin{equation} z^{\mathscr{L}^{n}}_r(x) =o(r) \end{equation}
   as $r \downarrow 0$, then $u$ admits a pointwise $\operatorname {SAMV}$ Laplacian at $x$, and
   (5.11)\begin{equation} \tilde{\Delta}_{\mu}^{\mathsf{d}}u(x) =\frac{1}{2w(x)}\textrm{Tr}(M(x)\nabla(w\nabla u)(x)). \end{equation}

Proof. Step 1. We first prove (5.9). To this aim, we claim that for any $r >0$,

(5.12)\begin{align} \Delta_{\mu,r}^{\mathsf{d}}u(x)& = \frac{\mathscr{L}^{n}(B_{r}(x))w(x)}{\mu(B_{r}(x))} \left( \frac{1}{2}\textrm{Tr}\left(\nabla^{2}u(x)M^{r}(x)\right)\right.\nonumber\\ & \quad +\left. \frac{1}{w(x)} \langle M^{r}(x)\nabla u(x),\nabla w(x)\rangle \right) + O(r). \end{align}

Once this is proved, (5.9) follows immediately since $\mu (B_{r}(x))/\mathscr {L}^{n}(B_{r}(x)) \to w(x)$ as $r \downarrow 0$. Let us prove (5.12). By a second-order Taylor expansion of $u$ and a first-order Taylor expansion of $w$, for any $y \in B_r(x)$ we infer that

\begin{align*} (u(y)-u(x))w(y) & = \sum_{i=1}^n [wu_i](x) (y-x)_i\\ & \quad + \sum_{i,j=1}^{n}\left[\frac{wu_{ij}}{2} + w_i u_j\right](x) \, (y-x)_i (y-x)_j + O(r^3) \end{align*}

where $O(r^3)$ is independent of $y$ and where we have denoted by $u_i$ the partial derivatives of $u$, by $u_{ij}$ the second-order partial derivatives of $u$, and similarly for $w$. Integrate with respect to $y \in B_{r}(x)$ and then divide by $r^2 \mu (B_r(x))$. Then, the left-hand side becomes $\Delta _{\mu,r}^{\mathsf {d}}u(x)$. Due to antisymmetry, assumption (5.7) implies that the first term on the right-hand side vanishes when integrating w.r.t. $y$ over $B_{r}(x)$. Then, we get

\begin{align*} \Delta_{\mu,r}^{\mathsf{d}}u(x) & = \sum_{i,j=1}^{n} \left[\frac{wu_{ij}}{2} + w_i u_j\right](x) \frac{1}{r^{2}\mu(B_{r}(x))}\\ & \quad \times \int_{B_r(x)}\ \, (y-x)_i (y-x)_j \mathop{}\!\mathrm{d} y + \frac{\mathscr{L}^n(B_r(x))}{\mu(B_{r}(x))} O(r)\\ & = \frac{\mathscr{L}^n(B_r(x))}{\mu(B_{r}(x))}\sum_{i,j=1}^{n} \left[\frac{wu_{ij}}{2} + w_i u_j\right](x) M^r_{ij}(x) + O(r)\\ & =\frac{\mathscr{L}^{n}(B_{r}(x))}{\mu(B_{r}(x))} \left( \frac{w(x)}{2}\textrm{Tr}\left(\nabla^{2}u(x)M^{r}(x)\right)\right.\\ & \quad +\left. \langle M^{r}(x)\nabla u(x),\nabla w(x)\rangle \right) + O(r). \end{align*}

Step 2. Now we prove (5.11). Observe that

(5.13)\begin{equation} \tilde{\Delta}^\mathsf{d}_{\mu,r}u(x)= \frac{1}{2}\left( \Delta^\mathsf{d}_{\mu,r}u(x) +I(r) \right) \end{equation}

where

\[ I(r):= \frac{1}{r^{2}}\int_{B_{r}(x)}(u(y)-u(x))\frac{w(y)}{\mu(B_{r}(y))} \mathop{}\!\mathrm{d} y. \]

We claim that

(5.14)\begin{equation} I(r) \to \Delta_{\mathscr{L}^n}^\mathsf{d} u(x) = \frac{1}{2}\textrm{Tr} (M(x) \nabla^2u(x)). \end{equation}

Once this claim is proved, letting $r \downarrow 0$ in (5.13) yields (5.11), since a direct computation gives

(5.15)\begin{equation} \frac{1}{2w(x)}\textrm{Tr}(M(x)\nabla(w\nabla u)(x))= \frac{1}{2}\textrm{Tr}(M(x)\nabla^{2}u(x))+\frac{1}{2w(x)}\langle M(x)\nabla u(x),\nabla w(x)\rangle.\end{equation}

Thus, we prove (5.14). Thanks to (5.5) and (5.6), we get

(5.16)\begin{align} I(r) & =\frac{1}{r^{2}(1+O(r))}\int_{B_{r}(x)}(u(y)-u(x))\frac{1}{\mathscr{L}^{n}(B_{r}(y))} \mathop{}\!\mathrm{d} y \nonumber\\ & = \frac{1}{r^{2}(1+O(r))}\Bigg[{\unicode{x2A0F}}_{B_{r}(x)}(u(y)-u(x)) \mathop{}\!\mathrm{d} y \nonumber\\ & \qquad-{\unicode{x2A0F}}_{B_{r}(x)}(u(y)-u(x))\left(1-\frac{\mathscr{L}^{n}(B_{r}(x))}{\mathscr{L}^{n}(B_{r}(y))}\right) \mathop{}\!\mathrm{d} y\Bigg] \nonumber\\ & = \frac{1}{1+O(r)}\Bigg[\Delta_{\mathscr{L}^n,r}^\mathsf{d} u(x)-\frac{1}{r^2}{\unicode{x2A0F}}_{B_{r}(x)}(u(y)-u(x))\left(1-\frac{\mathscr{L}^{n}(B_{r}(x))}{\mathscr{L}^{n}(B_{r}(y))}\right) \mathop{}\!\mathrm{d} y\Bigg]. \end{align}

Now using that $u$ is Lipschitz, we get

\begin{align*} & \frac{1}{r^2} \left|{\unicode{x2A0F}}_{B_{r}(x)}(u(y)-u(x))\left(1-\frac{\mathscr{L}^{n}(B_{r}(x))}{\mathscr{L}^{n}(B_{r}(y))}\right) \mathop{}\!\mathrm{d} y \right| \\ & \quad \le \frac{\textrm{Lip}(u)}{r^2} {\unicode{x2A0F}}_{B_r(x)} \mathsf{d}(x,y)\left|1-\frac{\mathscr{L}^{n}(B_{r}(x))}{\mathscr{L}^{n}(B_{r}(y))}\right| \mathop{}\!\mathrm{d} y \\ & \quad \le\frac{\textrm{Lip}(u)}{r} {\unicode{x2A0F}}_{B_r(x)} \left|1-\frac{\mathscr{L}^{n}(B_{r}(x))}{\mathscr{L}^{n}(B_{r}(y))} \right| \mathop{}\!\mathrm{d} y\\ & \quad = \textrm{Lip}(u) \frac{z^{\mathscr{L}^{n}}_r(x)}{r} \to 0 \qquad \text{as $r \downarrow 0$} \end{align*}

by assumption (5.10). Then, (5.14) follows from letting $r\downarrow 0$ in (5.16).

Remark 5.4 It follows from the previous proof that

(5.17)\begin{equation} \tilde{\Delta}_{\mu}^{\mathsf{d}}u(x) = \frac{1}{2} \left(\Delta_{\mu}^{\mathsf{d}}u(x) + \Delta_{\mathscr{L}^n}^{\mathsf{d}}u(x) \right). \end{equation}

Remark 5.5 Let $\Vert \cdot \Vert$ denote the infinity matrix norm. By the Jensen inequality,

\[ \left\| M^{r}(x) \right\| \le {\unicode{x2A0F}}_{B_r(x)} \left\| \left( \frac{(y-x)_i(y-x)_j}{r^2}\right)_{1 \le i \le j \le n} \right\| \mathop{}\!\mathrm{d} y \le 1 \]

for any $r>0$, so the sequence $\{M^{r}(x)\}_{r>0}$ is bounded and therefore it admits a set of accumulation points $\{M_\alpha \}$ as $r \downarrow 0$. This set is a singleton if and only if the limit $M(x):= \lim _{r \downarrow 0} M^r(x)$ exists. If this is not the case, then $\Delta _{\mu }^{\mathsf {d}}u(x)$ and $\tilde {\Delta }_{\mu }^{\mathsf {d}}u(x)$ may be understood as multivalued:

\begin{align*} \Delta_{\mu}^{\mathsf{d}}u(x) & =\left\{\frac{1}{2}\textrm{Tr}(M_\alpha(x)\nabla^2 u(x))+\frac{1}{w(x)}\langle\nabla w(x),M_\alpha(x)\nabla u(x)\rangle\right\}, \\ \tilde{\Delta}_{\mu}^{\mathsf{d}}u(x) & =\left\{\frac{1}{2w(x)}\textrm{Tr}(M_\alpha(x)\nabla(w\nabla u)(x))\right\}. \end{align*}

5.2. Vanishing weights

We are now interested in the pointwise $\operatorname {AMV}$ and $\operatorname {SAMV}$ Laplacians at a point where the weight vanishes. With no loss of generality, we assume that this point is the origin $0_n$. We let $\|\cdot \|$ be a norm on $\mathbb {R}^n$.

Definition 5.6 Let $w$ be a Lebesgue integrable function defined on a neighbourhood $\Omega$ of $0_n$. We say that $w$ is a vanishing weight at $0_n$ if $w(x) > 0$ for $\mathscr {L}^n$-a.e. $x \in \Omega$ and $0_n \in \mathrm {Leb}(w)$ with $w^*(0_n)=0$.

For any Radon measure $\nu$ on $\mathbb {R}^n$, we define the second-moment matrix of $\nu$ as

(5.18)\begin{equation} M_\nu := \left( \int_{B_1(0_n)} y_i y_j \mathop{}\!\mathrm{d} \nu(y) \right)_{1 \le i,j \le n}. \end{equation}

5.2.1. Infinitesimally even weights

Definition 5.7 We say that a weight $w$ is infinitesimally even at $0_n$ if it is $\mathscr {L}^n$-essentially bounded in a neighbourhood of $0_n$ and such that

\[ \mathscr{L}^n-\mathop{{\rm ess\,sup}}_{x \in B_r(0_n)} |w(x)-w({-}x)|=o\left( r {\unicode{x2A0F}}_{B_r(0_n)} w \mathop{}\!\mathrm{d} \mathscr{L}^n \right) \qquad \text{as $r \downarrow 0$.} \]

Our first result is the following.

Proposition 5.8 Consider $(\Omega,\,\|\cdot \|,\,w \mathscr {L}^n)$ where $w$ is a vanishing weight at $0_n$ of domain $\Omega$ which is infinitesimally even at $0_n$. Set $\mu := w \mathscr {L}^n$. Assume that the blow-up probability measures

\[ \nu_r := \left( \frac{w(r\, \cdot)r^n}{\mu(B_r(0_n))} \right) \mathscr{L}^n {\unicode{x02EA}} B_1(0_n) \]

weakly converge to some Radon measure $\nu$ supported in $B_1(0_n)$.Footnote ¹ Then, the pointwise $\operatorname {AMV}$ Laplacian at $0_n$ of any function $u : \Omega \to \mathbb {R}$ two times differentiable at $0_n$ exists and satisfies

(5.19)\begin{equation} \Delta^{\|\cdot\|}_{\mu}u(0_n)=\frac{1}{2} \textrm{Tr} (M_\nu \nabla^2u)(0_n). \end{equation}

Proof. From the Taylor theorem, there exist a neighbourhood $\Omega ' \subset \Omega$ of $0_n$ and functions $h_{ij}: \Omega ' \to \mathbb {R}$ such that $\lim \limits _{x \to 0_n} h_{ij}(x)=0$ and

\[ u(x) - u(0_n) = \sum_{i=1}^n \frac{\partial u}{ \partial x_i}(0_n) x_i + \frac{1}{2}\sum_{i,j=1}^n \frac{\partial^2 u}{ \partial x_i \partial x_j}(0_n)x_ix_j + \sum_{i,j=1}^n h_{ij}(x)x_ix_j \]

for any $x \in \Omega '$. Then, for any $r>0$ such that $B_r(0_n) \subset \Omega '$,

(5.20)\begin{align} \Delta^{\|\cdot\|}_{\mu,r}u(0_n) & = \sum_{i=1}^n \frac{\partial u}{ \partial x_i}(0_n) \frac{1}{r^2} {\unicode{x2A0F}}_{B_r(0_n)} x_i \mathop{}\!\mathrm{d} \mu(x)\nonumber\\ & \quad + \frac{1}{2} \sum_{i,j=1}^n \frac{\partial^2 u}{ \partial x_i \partial x_j}(0_n)\frac{1}{r^2} {\unicode{x2A0F}}_{B_r(0_n)} x_i x_j\mathop{}\!\mathrm{d} \mu(x) \nonumber\\ & + \sum_{i,j=1}^n \frac{1}{r^2} {\unicode{x2A0F}}_{B_r(0_n)} h_{ij}(x)x_ix_j \mathop{}\!\mathrm{d} \mu(x). \end{align}

Step 1. We fix $i,\,j \in \{1,\,\ldots,\,n\}$ and show that

(5.21)\begin{equation} \lim\limits_{r \downarrow 0} \frac{1}{r^2} {\unicode{x2A0F}}_{B_r(0_n)} h_{ij}(x)x_ix_j \mathop{}\!\mathrm{d} \mu(x) = 0. \end{equation}

Fix $\varepsilon >0$. Then, there exists $\delta \in (0,\,r_0)$ such that $|h_{ij}(x)|<\varepsilon$ for any $x \in B_\delta (0_n)$. For any $r \in (0,\,\delta )$,

\[ \left| \frac{1}{r^2} {\unicode{x2A0F}}_{B_r(0_n)} h_{ij}(x)x_ix_j \mathop{}\!\mathrm{d} \mu(x) \right| \le \frac{1}{r^2 } {\unicode{x2A0F}}_{B_r(0_n)} |h_{ij}(x)| |x_i| |x_j| \mathop{}\!\mathrm{d} \mu(x) < \varepsilon, \]

where we have used that $|x_i| \le r$ and $|x_j|\le r$ to get the last inequality. This yields (5.21).

Step 2. We fix $i \in \{1,\,\ldots,\,n\}$ and show that

(5.22)\begin{equation} \lim\limits_{r \downarrow 0} \frac{1}{r^2}{\unicode{x2A0F}}_{B_r(0_n)} x_i \mathop{}\!\mathrm{d} \mu(x) = 0. \end{equation}

Observe that

(5.23)\begin{equation} {\unicode{x2A0F}}_{B_r(0_n)} x_i \mathop{}\!\mathrm{d} \mu(x) = \frac{1}{\int_{B_r(0_n)} w(x)\mathop{}\!\mathrm{d} x}\int_{B_r(0_n)} x_i w(x) \mathop{}\!\mathrm{d} x \end{equation}

and

(5.24)\begin{align} \int_{B_r(0_n)} x_i w(x) \mathop{}\!\mathrm{d} x & = \frac{1}{2} \int_{B_r(0_n)} x_i w(x) \mathop{}\!\mathrm{d} x + \frac{1}{2} \int_{B_r(0_n)} x_i w(x) \mathop{}\!\mathrm{d} x \nonumber\\ & = \frac{1}{2} \int_{B_r(0_n)} x_i w(x) \mathop{}\!\mathrm{d} x + \frac{1}{2} \int_{B_r(0_n)} -x_i' w({-}x') \mathop{}\!\mathrm{d} x' \nonumber\\ & = \frac{1}{2}\int_{B_r(0_n)} x_i [w(x)-w({-}x)] \mathop{}\!\mathrm{d} x, \end{align}

where we have used the change of variable $x'=-x$ in the second integral to pass from the first line to the second. Set $S_w(r):= \mathscr {L}^n-\mathop{\textrm{ess sup}}_{x \in B_r(0_n)} |w(x)-w(-x)|.$ Then,

\begin{align*} \left|\frac{1}{r^2}{\unicode{x2A0F}}_{B_r(0_n)} x_i \mathop{}\!\mathrm{d} \mu(x)\right| & = \frac{1}{2r^2|\int_{B_r(0_n)} w(x)\mathop{}\!\mathrm{d} x|} \left| \int_{B_r(0_n)} x_i [w(x)-w({-}x)] \mathop{}\!\mathrm{d} x \right| \\ & \le \frac{1}{2r^2|\int_{B_r(0_n)} w(x)\mathop{}\!\mathrm{d} x|} \left( \int_{B_r(0_n)} |x_i| |w(x)-w({-}x)| \mathop{}\!\mathrm{d} x \right) \\ & \le \frac{S_w(r)}{2r|\int_{B_r(0_n)} w(y)\mathop{}\!\mathrm{d} y|} \int_{B_r(0_n)} 1 \mathop{}\!\mathrm{d} x \\ & = \frac{S_w(r)}{2r|{\unicode{x2A0F}}_{B_r(0_n)} w(y)\mathop{}\!\mathrm{d} y|} \end{align*}

where we have used $|x_i|\le r$ and $|w(x)-w(-x)|\le S_w(r)$ to get the second inequality. Then, (5.22) follows from the infinitesimal evenness assumption on $w$.

Step 3. We fix $i,\,j \in \{1,\,\ldots,\,n\}$ and show that

(5.25)\begin{equation} \lim\limits_{r \downarrow 0} \frac{1}{r^2} {\unicode{x2A0F}}_{B_r(0_n)} x_ix_j \mathop{}\!\mathrm{d} \mu(x) = \int_{B_1(0_n)} y_iy_j \mathop{}\!\mathrm{d} \nu(y). \end{equation}

Performing the change of variable $y=x/r$, we have

\begin{align*} \frac{1}{r^2} {\unicode{x2A0F}}_{B_r(0_n)} x_ix_j \mathop{}\!\mathrm{d} \mu(x) & = \frac{1}{r^2 \mu(B_r(0_n))} \int_{B_1(0_n)} (ry_i)(ry_j)w(ry) r^n \mathop{}\!\mathrm{d} y\\ & = \int_{B_1(0_n)} y_i y_j \mathop{}\!\mathrm{d} \nu_r(y) \to\int_{B_1(0_n)} y_i y_j \mathop{}\!\mathrm{d} \nu(y) \qquad \text{as }r\downarrow 0. \end{align*}

Remark 5.9 By the Prokhorov theorem, the probability measures $\{\nu _r\}$ always admit accumulation points in the weak topology of measures as $r \downarrow 0$. The assumption made in the previous proposition demands that there is a unique limit. If this is not satisfied, then $\Delta ^{\|\cdot \|}_{\mu }u(0_n)$ may be understood as multivalued, namely we get

\[ \Delta^{\|\cdot\|}_{\mu}u(0_n)=\left\{ \frac{1}{2} \textrm{Tr} (M_{\bar{\nu}_\alpha} \nabla^2u)(0_n)\right\} \]

where $\{\bar {\nu }_\alpha \}$ are the limit points of $\{\nu _r\}$ as $r \downarrow 0$.

A version of proposition 5.8 also holds in the case of the $\operatorname {SAMV}$ Laplacian, but under slightly different assumptions.

Proposition 5.10 Consider $(\Omega,\,\|\cdot \|,\,w \mathscr {L}^n)$ where $w$ is a vanishing weight at $0_n$ of domain $\Omega$ which is infinitesimally even at $0_n$. Assume that $\mu := w \mathscr {L}^n$ satisfies a comparability condition at $0_n$, that the weight

(5.26)\begin{equation} \tilde{w}({\cdot}) := \frac{w({\cdot})}{2}\left( 1 + \frac{\mu(B_r(0_n))}{\mu(B_r({\cdot}))} \right) \end{equation}

is infinitesimally even at $0_n$, and that there exists a Radon measure $\tilde {\nu }$ on $B_1(0_n)$ such that

(5.27)\begin{equation} \left( \frac{\tilde{w}(r\, \cdot)r^n}{\mu(B_r(0_n))} \right) \mathscr{L}^n {\unicode{x02EA}} B_1(0_n) =: \tilde{\nu}_r \rightharpoonup \tilde{\nu} \qquad \text{as $r \downarrow 0$.} \end{equation}

Then, the pointwise $\operatorname {SAMV}$ Laplacian at $0_n$ of any function $u : \Omega \to \mathbb {R}$ two times differentiable at $0_n$ exists and satisfies

(5.28)\begin{equation} \tilde{\Delta}^{\|\cdot\|}_{\mu}u(0_n)=\frac{1}{2} \textrm{Tr} (M_{\tilde{\nu}} \nabla^2u)(0_n). \end{equation}

Proof. Let $u : \Omega \to \mathbb {R}$ be two times differentiable at $0_n$. Using the Taylor theorem like in the proof of proposition 5.8, we get that for any small enough $r$, it holds that

(5.29)\begin{align} \tilde{\Delta}_{\mu,r}^{\lVert\cdot\rVert}u(0_n) & = \sum_{i=1}^n \frac{\partial u}{ \partial x_i}(0_n) \frac{1}{r^2 \mu(B_r(0_n))} \int_{B_r(0_n)} x_i \tilde{w}(x)\mathop{}\!\mathrm{d} x \nonumber\\ & \quad+ \sum_{i,j=1}^n \frac{\partial^2 u}{ \partial x_i \partial x_j}(0_n)\frac{1}{r^2\mu(B_r(0_n))} \int_{B_r(0_n)} x_i x_j \tilde{w}(x) \mathop{}\!\mathrm{d} x \nonumber\\ & \quad+ \sum_{i,j=1}^n \frac{1}{r^2\mu(B_r(0_n))} \int_{B_r(0_n)} h_{ij}(x)x_ix_j \tilde{w}(x) \mathop{}\!\mathrm{d} x. \end{align}

Fix $i,\,j \in \{1,\,\ldots,\,n\}$. For any $\varepsilon >0$ there exists $\delta \in (0,\,r_0)$ such that $|h_{ij}(x)|<\varepsilon$ for any $x \in B_\delta (0_n)$. For any $r \in (0,\,\delta )$,

\begin{align*} & \left| \frac{1}{r^2\mu(B_r(0_n))} \int_{B_r(0_n)} h_{ij}(x)x_ix_j \tilde{w}(x) \mathop{}\!\mathrm{d} x \right| \\ & \quad \le \frac{1}{r^2 \mu(B_r(0_n))} \int_{B_r(0_n)} |h_{ij}(x)| |x_i| |x_j| \tilde{w}(x) \mathop{}\!\mathrm{d} x\\ & \quad \le \frac{\varepsilon}{\mu(B_r(0_n))} \int_{B_r(0_n)} \tilde{w}(x) \mathop{}\!\mathrm{d} x \le (1+C) \varepsilon/2, \end{align*}

where we have used the local comparability condition at $0_n$ to get the last inequality. This yields

(5.30)\begin{equation} \lim\limits_{r \downarrow 0} \frac{1}{r^2\mu(B_r(0_n))} \int_{B_r(0_n)} h_{ij}(x)x_ix_j \tilde{w}(x) \mathop{}\!\mathrm{d} x = 0. \end{equation}

Moreover, we have that

(5.31)\begin{equation} \lim\limits_{r \downarrow 0}\frac{1}{r^2\mu(B_r(0_n))} \int_{B_r(0_n)}x_i \tilde{w}(x) \mathop{}\!\mathrm{d} x = 0. \end{equation}

Indeed, acting like in (5.24) we get

\[ \int_{B_r(0_n)} x_i \tilde{w}(x) \mathop{}\!\mathrm{d} x = \frac{1}{2}\int_{B_r(0_n)} x_i [ \tilde{w}(x)- \tilde{w}({-}x)] \mathop{}\!\mathrm{d} x. \]

Set $S_ {\tilde {w}}(r):= \mathscr {L}^n-\mathop{\textrm{ess sup}}_{x \in B_r(0_n)} | \tilde {w}(x)- \tilde {w}(-x)|.$ Like in the previous proof, we obtain

\begin{align*} \left|\frac{1}{r^2\mu(B_r(0_n))}\int_{B_r(0_n)} x_i \tilde{w}(x)\mathop{}\!\mathrm{d} x\right| & \le \frac{S_{ \tilde{w}}(r)}{2r|{\unicode{x2A0F}}_{B_r(0_n)} w(x)\mathop{}\!\mathrm{d} x|} \le \frac{(1+C)S_{ \tilde{w}}(r)}{4r|{\unicode{x2A0F}}_{B_r(0_n)} \tilde{w}(x)\mathop{}\!\mathrm{d} x|} \end{align*}

where we have used the local comparability condition at $0_n$ to get the last inequality. Then, (5.31) follows from the infinitesimal evenness assumption on $\tilde {w}$.

Lastly, acting as in the previous proof and using the definition of $\tilde {\nu }_r$, we obtain that

(5.32)\begin{equation} \lim\limits_{r \downarrow 0} \frac{1}{r^2\mu(B_r(0_n))} \int_{B_r(0_n)} x_ix_j \tilde{w}(x)\mathop{}\!\mathrm{d} x = \int_{B_1(0_n)} y_iy_j \mathop{}\!\mathrm{d} \tilde{\nu}(y) \end{equation}

for any $i,\,j \in \{1,\,\ldots,\,n\}$.

The conclusion follows by letting $r \downarrow 0$ in (5.29) and using (5.30), (5.31) and (5.32).

Remark 5.11 A simple computation shows that

(5.33)\begin{equation} \tilde{\nu}_r = \frac{1}{2} \left(\nu_r + \frac{r^n}{\mu(B_r({\cdot}))}\mu {\unicode{x02EA}} B_1(0_n)\right). \end{equation}

In this way, it clearly appears that there exists $r_0>0$ such that the measures $\{\tilde {\nu }_r\}_{0< r< r_0}$ are uniformly bounded, thanks to the local comparability condition at $0_n$ of $\mu$. Therefore, like in the non-symmetrized case (see remark 5.9), the Prokhorov theorem implies that the measures $\{\tilde {\nu }_r\}_{r>0}$ admit limit points in the weak topology of measures as $r \downarrow 0$. If these limit points are not unique then $\tilde {\Delta }^{\|\cdot \|}_{\mu }u(0)$ may be understood as multivalued, namely

\[ \tilde{\Delta}^{\|\cdot\|}_{\mu}u(0_n)=\left\{ \frac{1}{2} \textrm{Tr} (M_{\hat{\nu}_\alpha} \nabla^2u)(0_n)\right\} \]

where $\{\hat {\nu }_\alpha \}$ are the limit points of $\{\tilde {\nu }_r\}$ as $r \downarrow 0$.

Remark 5.12 If $w$ is even, then so is $\tilde {w}$. Indeed, in this case, obvious changes of variable show that

\begin{align*} \mu(B_r({-}x)) & = \int_{B_r(0_n)}w({-}x+y)\mathop{}\!\mathrm{d} y = \int_{B_r(0_n)}w({-}x-y)\mathop{}\!\mathrm{d} y\\ & = \int_{B_r(0_n)}w(x+y)\mathop{}\!\mathrm{d} y = \mu(B_r(x)) \end{align*}

from which evenness of $\tilde {w}$ follows from the definition in (5.26).

5.2.2. Weights $w(\cdot )=|\cdot |^{\alpha }$

Proposition 5.8 may be applied to the case of weights $w(\cdot )=|\cdot |^{\alpha }$ where $|\cdot |$ is the Euclidean norm and $\alpha >0$.

Corollary 5.13 Take $\alpha > 0$ and set $w(\cdot ):= |\cdot |^{\alpha }$ where $|\cdot |$ denotes the Euclidean norm. Consider $(\mathbb {R}^n,\,\mathsf {d}_e,\,\mu )$ where $\mu := w\mathscr {L}^n$. Then, for any $u : \mathbb {R}^n \to \mathbb {R}$ two times differentiable at $0_n$ the pointwise $\operatorname {AMV}$ and $\operatorname {SAMV}$ Laplacians of $u$ in $0_n$ exist and are given by

(5.34)\begin{align} \Delta^{\mathsf{d}_e}_{\mu}u(0_n) & = \frac{n+\alpha}{2n(n+\alpha+2)}\Delta u (0_n), \end{align}

(5.35)\begin{align} \tilde{\Delta}^{\mathsf{d}_e}_{\mu}u(0_n) & = \frac{n^2 + (2 + \alpha)n+ \alpha}{2n(n+\alpha+2)(n+2)}\Delta u (0_n), \end{align}

where $\Delta := \sum _{i} \partial _{ii}$ is the classical Laplacian.

Proof. The weight $w$ is continuous, hence $0_n$ is a Lebesgue point with $w(0_n)=0$. Moreover, the infinitesimal evenness of $w$ is trivially satisfied because $w$ is even. Finally the $\alpha$-homogeneity of $w$ implies that the measures $\{\nu _r\}_{r>0}$ are constantly equal to the measure

\[ \nu = \frac{n+\alpha}{\sigma_{n-1}}|\cdot|^\alpha \mathscr{L}^n {\unicode{x02EA}} B_1(0_n), \]

hence proposition 5.8 applies and yields

(5.36)\begin{equation} \Delta^{\mathsf{d}_e}_{\mu}u(0_n)=\frac{1}{2} \textrm{Tr} (M_\nu \nabla^2u)(0_n) \end{equation}

where we recall that $M_\nu$ is defined in (5.18). Let us compute $M_\nu$. If $n = 1$,

\[ \int_{{-}1}^1 y^2 |y|^{\alpha} \mathop{}\!\mathrm{d} \nu(y) = \frac{1+\alpha}{\sigma_{0}} \int_{{-}1}^1 y^2 |y|^{\alpha} \mathop{}\!\mathrm{d} \nu(y) = \frac{1+\alpha}{2 + \alpha +1}\, \cdot \]

Assume now $n>1$. Consider $i,\,j \in \{1,\,\ldots,\, n\}$ such that $i \neq j$. Then, the map $y \mapsto y_i y_j |y|^{\alpha }$ is odd, so

\[ \int_{B_1(0_n)} y_i y_j |y|^{\alpha} \mathop{}\!\mathrm{d} y =0. \]

Moreover, the invariance of the Lebesgue measure under exchanging coordinates $\theta _i$ and $\theta _{n}$ yields

\begin{align*} \int_{B_1(0_n)} y_i^2 |y|^{\alpha} \mathop{}\!\mathrm{d} y & = \int_{B_1(0_n)} y_{n}^2 |y|^{\alpha} \mathop{}\!\mathrm{d} y = \int_0^1 \int_{\mathbb{S}^{n-1}} r^{\alpha + n + 1} \theta_n^2 \mathop{}\!\mathrm{d} \sigma(\theta)\mathop{}\!\mathrm{d} r\\ & =\frac{1}{n+\alpha+2} \int_{\mathbb{S}^{n-1}} \theta_{n}^2 \mathop{}\!\mathrm{d} \sigma(\theta) \end{align*}

where $\sigma$ is the usual surface measure on $\mathbb {S}^{n-1}$. Since

\[ \sum_{1\le \ell \le n} \int_{\mathbb{S}^{n-1}} \theta_{\ell}^2 \mathop{}\!\mathrm{d} \sigma(\theta) = \int_{\mathbb{S}^{n-1}} |\theta|^2 \mathop{}\!\mathrm{d} \sigma(\theta) = \sigma_{n-1} \]

we get

\[ \int_{\mathbb{S}^{n-1}} \theta_{n}^2 \mathop{}\!\mathrm{d} \sigma(\theta) = \frac{\sigma_{n-1}}{n} \, \cdot \]

In the end, for any $n \ge 1$,

\[ M_\nu = \frac{n+\alpha}{n(n+\alpha+2)} I_n \]

and the result follows.

For the $\operatorname {SAMV}$ Laplacian, consider $\tilde {w}$ as defined in (5.26). We want to apply proposition 5.10. The weight $\tilde {w}$ is even by remark 5.12. Thanks to (5.33), in order to prove (5.27), one is left with computing the limit as $r \downarrow 0$ of

\[ \frac{r^n}{\mu(B_r(x))} \]

for any $x \neq 0$. This follows from the Lebesgue differentiation theorem:

\[ \frac{r^n}{\mu(B_r(x))} = \left( \omega_n {\unicode{x2A0F}}_{B_r(x)} w(y) \mathop{}\!\mathrm{d} y \right)^{{-}1} \to \frac{1}{|x|^\alpha \omega_n} \, \cdot \]

Consequently,

\[ \tilde{\nu} = \frac{1}{2} \left( \nu + \frac{\mathscr{L}^n}{\omega_n}\right). \]

Finally, computing the second order moments w.r.t. to $\tilde {\nu }$, simplifying and putting back into (5.28) gives (5.35).

5.2.3. Separable weights

After proposition 5.8 it is natural to consider the class of separable weights, defined as follows.

Definition 5.14 We say that a vanishing weight $w$ of domain $\Omega$ is separable in a neighbourhood of $0_n$ if there exist $r>0$ such that $B_r(0_n) \subset \Omega$ and $f \in L^1([0,\,r],\,\mathscr {L}^1)$, $g \in L^1(\mathbb {S}^{n-1},\,\sigma )$, such that $w(x)=f(|x|)g(x/|x|)$ for $\mathscr {L}^n$-a.e. $x \in \Omega \backslash \{0_n\}$. We say that $f$ (resp. $g$) is the radial (resp. angular) part of $w$.

Proposition 5.15 Consider $(\Omega,\,\mathsf {d}_e,\,w \mathscr {L}^n)$ where $w$ is a vanishing weight of domain $\Omega$ which is separable in a neighbourhood of $0_n$. Let $f$ and $g$ be the radial and angular parts of $w$, respectively. Set $\mu =w \mathscr {L}^n$ and

\[ \nu := (c_f \, g) \,\sigma \]

where $\sigma$ is the normalized surface measure on the sphere $\mathbb {S}^{n-1}$ and

\[ c_f := \int_0^1 f(\rho) \rho^{n+1} \mathop{}\!\mathrm{d} \rho \in (0,+\infty). \]

Then, the following hold.

1. (1) For any function $u$ two-times differentiable at $0_n$, the $\operatorname {AMV}$ Laplacian $\Delta _{\mu }^{\mathsf {d}_e} u(0_n)$ exists and satisfies

   (5.37)\begin{equation} \Delta_{\mu}^{\mathsf{d}_e} u(0_n) = \frac{1}{2}\textrm{Tr} (M_\nu \nabla^2 u)(0_n) \end{equation}
   if and only if
   (5.38)\begin{equation} \bigg\langle \nabla u (0_n), \int_{\mathbb{S}^{n-1}} \theta g(\theta) \mathop{}\!\mathrm{d} \sigma(\theta) \bigg\rangle = 0. \end{equation}

2. (2) For any function $u$ two-times differentiable at $0_n$ such that $\nabla u (0_n)=0$, the $\operatorname {AMV}$ Laplacian $\Delta _{\mu }^{\mathsf {d}_e} u(0_n)$ exists and satisfies (5.37).

3. (3) If $\int _{\mathbb {S}^{n-1}} \theta g(\theta ) \mathop {}\!\mathrm {d} \sigma (\theta )=0_n$, then for any function $u$ two-times differentiable at $0_n$, the $\operatorname {AMV}$ Laplacian $\Delta _{\mu }^{\mathsf {d}_e} u(0_n)$ exists and satisfies (5.37).

Proof. We only prove the first assertion since the others are direct consequences. Acting as in the proof of proposition 5.8, we get that $\Delta _{\mu }^{\mathsf {d}_e}u(0_n)$ exists and satisfies (5.37) if and only if

\[ \lim\limits_{r \downarrow 0} \sum_{i=1}^n \frac{\partial u}{ \partial x_i}(0_n) \frac{1}{r^2} {\unicode{x2A0F}}_{B_r(0_n)} x_i \mathop{}\!\mathrm{d} \mu(x) = 0. \]

Thanks to (5.23), we know that for any $i \in \{1,\,\ldots,\,n\}$,

\[ \frac{1}{r^2 }{\unicode{x2A0F}}_{B_r(0_n)} x_i \mathop{}\!\mathrm{d} \mu(x) = \frac{1}{ r \int_{B_1(0_n)} w(ry)\mathop{}\!\mathrm{d} y}\left( \int_{B_1(0_n)} y_i w(ry) \mathop{}\!\mathrm{d} y\right) \]

from which the separation assumption yields

\[ \frac{1}{r^2 }{\unicode{x2A0F}}_{B_r(0_n)} x_i \mathop{}\!\mathrm{d} \mu(x) =\frac{\int_0^1 f(r \rho) \rho^{n+1} \mathop{}\!\mathrm{d} \rho}{r \int_0^1 f(r \rho) \rho \mathop{}\!\mathrm{d} \rho} \frac{\int_{\mathbb{S}^{n-1}} \theta_i [g(\theta)-g(-\theta)] \mathop{}\!\mathrm{d} \sigma(\theta)}{\int_{\mathbb{S}^{n-1}} g(\theta) \mathop{}\!\mathrm{d} \sigma(\theta)} \, \cdot \]

Now

\begin{align*} \int_{\mathbb{S}^{n-1}} \theta_i [g(\theta)-g(-\theta)] \mathop{}\!\mathrm{d} \sigma(\theta) & = \int_{\mathbb{S}^{n-1}} \theta_i g(\theta)\mathop{}\!\mathrm{d} \sigma(\theta) - \int_{\mathbb{S}^{n-1}} \theta_i g(-\theta)\mathop{}\!\mathrm{d} \sigma(\theta)\\ & = 2 \int_{\mathbb{S}^{n-1}} \theta_i g(\theta)\mathop{}\!\mathrm{d} \sigma(\theta) \end{align*}

where we have used the change of variable $\theta '=-\theta$ in the second integral to get the last term. Thus,

(5.39)\begin{align} \sum_{i=1}^n \frac{\partial u}{ \partial x_i}(0_n) \frac{1}{r^2} {\unicode{x2A0F}}_{B_r(0_n)} x_i \mathop{}\!\mathrm{d} \mu(x) & = \frac{\int_0^1 f(r \rho) \rho^{n+1} \mathop{}\!\mathrm{d} \rho}{r \int_0^1 f(r \rho) \rho \mathop{}\!\mathrm{d} \rho} \sum_{i=1}^n \frac{\partial u}{ \partial x_i}(0_n) \frac{2\int_{\mathbb{S}^{n-1}} \theta_i g(\theta) \mathop{}\!\mathrm{d} \sigma(\theta)}{\int_{\mathbb{S}^{n-1}} g(\theta) \mathop{}\!\mathrm{d} \sigma(\theta)} \nonumber\\ & = \frac{\int_0^1 f(r \rho) \rho^{n+1} \mathop{}\!\mathrm{d} \rho}{r \int_0^1 f(r \rho) \rho \mathop{}\!\mathrm{d} \rho} \frac{\bigg\langle \nabla u (0_n), 2\int_{\mathbb{S}^{n-1}} \theta g(\theta)\mathop{}\!\mathrm{d} \sigma(\theta) \bigg\rangle}{\int_{\mathbb{S}^{n-1}} g(\theta) \mathop{}\!\mathrm{d} \sigma(\theta)} \, \cdot \end{align}

This immediately shows that if (5.38) holds, then $\Delta _\mu ^{\mathsf {d}_e}u(0_n)$ exists and is given by (5.19).

Assume now that (5.38) does not hold. Then,

\[ c:=\left| \bigg\langle \nabla u (0_n), 2\int_{\mathbb{S}^{n-1}} \theta g(\theta) \mathop{}\!\mathrm{d} \sigma(\theta)\bigg\rangle \right|>0. \]

Let $m \in (0,\,1)$ be the median of the measure $f(r\rho )\mathop {}\!\mathrm {d} \rho$ on $[0,\,1]$, so that

\[ \int_0^m f(r \rho) \mathop{}\!\mathrm{d} \rho = \int_m^1 f(r \rho) \mathop{}\!\mathrm{d} \rho= \frac{1}{2}\int_0^1 f(r \rho) \mathop{}\!\mathrm{d} \rho. \]

Since $f$ is non-negative and $\int _m^1 f(r \rho )\rho ^{n+1} \mathop {}\!\mathrm {d} \rho \ge m^{n+1} \int _m^1 f(r \rho ) \mathop {}\!\mathrm {d} \rho$, we get

\[ \frac{\int_0^1 f(r \rho) \rho^{n+1} \mathop{}\!\mathrm{d} \rho}{r \int_0^1 f(r \rho) \rho \mathop{}\!\mathrm{d} \rho} \ge \frac{m^{n+1} \int_m^1 f(r \rho) \mathop{}\!\mathrm{d} \rho}{r \int_0^1 f(r \rho) \rho \mathop{}\!\mathrm{d} \rho} = \frac{m^{n+1} \int_m^1 f(r \rho) \mathop{}\!\mathrm{d} \rho}{ 2 r \int_m^1 f(r \rho) \rho \mathop{}\!\mathrm{d} \rho} \ge \frac{m^{n+1}}{2r} , \]

where we have used $\int _m^1 f(r \rho ) \rho \mathop {}\!\mathrm {d} \rho \le \int _m^1 f(r \rho ) \mathop {}\!\mathrm {d} \rho$ to get the last inequality. Then, from (5.39) we get

\[ \left| \sum_{i=1}^n \frac{\partial u}{ \partial x_i}(0_n) \frac{1}{r^2 }{\unicode{x2A0F}}_{B_r(0_n)} x_i \mathop{}\!\mathrm{d} \mu(x) \right| \ge \frac{m^{n+1}}{2r}c \to +\infty \]

as $r \downarrow 0$, so $\Delta _\mu ^{\mathsf {d}_e}u(0_n)$ does not converge.

When $n=2$, the Fourier theory easily yields a concrete equivalent form of the sufficient condition given in (3).

Corollary 5.16 Assume $n = 2$ and consider $(\Omega,\,\mathsf {d}_e,\,w \mathscr {L}^n)$ and $u$ as in the previous proposition. If there exist $c \ge 1$ and $\varphi (\cdot ) \in \mathrm {Span}(\{\cos (m\, \cdot ),\, \sin (m \, \cdot )\}_{m \ge 2})$ such that

\[ g(e^{\bf{i} \, \cdot}) = c + \varphi({\cdot}), \]

then for any function $u$ two times differentiable at $0_2$, the $\operatorname {AMV}$ Laplacian $\Delta _\mu ^{\mathsf {d}_e}u(0_2)$ exists and is given by (5.37).

Proof. Let $h \in L^1(\mathbb {R})$ be $2 \pi$-periodic and such that $g(e^{\bf {i} t})=h(t)$ for any $t \in \mathbb {R}$. Since

\[ \int_{\mathbb{S}^{1}} \theta g(\theta) \mathop{}\!\mathrm{d} \sigma(\theta) = \int_{0}^{2\pi} e^{\bf{i} t} g(e^{\bf{i} t}) \mathop{}\!\mathrm{d} t = \int_{0}^{2\pi} e^{\bf{i} t} h(t) \mathop{}\!\mathrm{d} t \]

then

\begin{align*} \int_{\mathbb{S}^{1}} \theta g(\theta) \mathop{}\!\mathrm{d} \sigma(\theta) = 0_2 \quad \Longleftrightarrow \quad (*) \,:\, \begin{cases} \displaystyle \int_0^{2\pi} \cos(t)h(t) \mathop{}\!\mathrm{d} t=0,\\ \displaystyle\int_0^{2\pi} \sin(t) h(t) \mathop{}\!\mathrm{d} t=0. \end{cases} \end{align*}

Since $g$ is non-negative, so is $h$, hence $(*)$ is equivalent to

\[ h=c + \varphi \]

where $c\ge 1$ and $\varphi \in \operatorname {Span}(\{\cos (m \cdot ),\, \sin (m\cdot )\}_{m \ge 2})$, thanks to the Fourier theory.