Min–max formulas for nonlocal elliptic operators

Abstract

In this work, we give a characterization of Lipschitz operators on spaces of $C^2(M)$ functions (also $C^{1,1}$, $C^{1,\gamma }$, $C^1$, $C^\gamma$) that obey the global comparison property—i.e. those that preserve the global ordering of input functions at any points where their graphs may touch, often called “elliptic” operators. Here M is a complete Riemannian manifold. In particular, we show that all such operators can be written as a min–max over linear operators that are a combination of drift–diffusion and integro-differential parts. In the linear (and nonlocal) case, these operators had been characterized in the 1960s, and in the local, but nonlinear case—e.g. local Hamilton–Jacobi–Bellman operators—this characterization has also been known and used since approximately since 1960s or 1970s. Our main theorem contains both of these results as special cases. It also shows any nonlinear scalar elliptic equation can be represented as an Isaacs equation for an appropriate differential game. Our approach is to “project” the operator to one acting on functions on large finite graphs that approximate the manifold, use non-smooth analysis to derive a min–max formula on this finite dimensional level, and then pass to the limit in order to lift the formula to the original operator.

Appendices

Appendix A: Discretization of the gradient and the Hessian on M

First off, we shall construct proper discretizations for the covariant gradient and Hessian given M and ${\tilde{G}}_n$. Our point of view will be to think of a sufficiently smooth function $u:M\rightarrow {\mathbb {R}}$ as given. Then, the discrete gradient and Hessian of u will be defined at points in ${\tilde{G}}_n$ using only the values of u at points in ${\tilde{G}}_n$. We will see that the regularity of the original function u will control how far are these discrete operators from their continuum counterparts (Lemma A.14). Moreover, the regularity of u will control the regularity of discrete gradient and Hessian themselves, in a manner which is independent of the mesh size (Proposition A.15 and A.16).

Remark A.1

Before proceeding further, it is worthwhile to note that the discrete gradient and Hessian defined below are standard, and that this appendix has been made with the chief purpose of making the paper as self contained as possible. In fact, as with the discussion of Whitney extension, we failed to find a direct reference where the discretization of the gradient and Hessian is done in the context of a Riemannian manifold. Furthermore, for the purposes of this paper, we only need rather minimal properties of our discretization—essentially, their “consistency”. As such, the arguments and estimates here are far less optimal than what may be found in the numerical analysis literature where subtler issues are considered.

As we can only use the values of u at points of ${\tilde{G}}_n$, our first order of business is to single out admissible directions at $x\in {\tilde{G}}_n$ along which a (discrete) derivative may be computed. This is done in the following proposition.

Proposition A.2

Given $x \in {\tilde{G}}_n$ there are vectors

$$\begin{aligned} V_{n,1}(x),\ldots ,V_{n,d}(x) \in (TM)_x. \end{aligned}$$

Satisfying the following properties,

• (1) For each k,

  $$\begin{aligned} \exp _{x}(V_{n,k}(x)) \in {\tilde{G}}_n. \end{aligned}$$

• (2) Also for each k,

  $$\begin{aligned} 98 {\tilde{h}}_n \le |V_{n,k}(x)|_{g_x} \le 102 {\tilde{h}}_n. \end{aligned}$$

• (3) Finally, the family $\{V_{n,k}\}_{k=1}^{d}$ forms a basis which is “almost orthogonal”. To be concrete, for sufficiently large n, we have

  $$\begin{aligned} |({\hat{V}}_{n,l}(x),{\hat{V}}_{n,k}(x))_{g_x}|\le \frac{1}{20},\;\;\text { if } k\ne l. \end{aligned}$$

  Here, ${\hat{V}}$ denotes the unit vector in the direction of V, that is ${\hat{V}}:= V/|V|_{g_x}$.

Proof

Let us recall the constant $\delta \in (0,1)$ introduced in Remark 3.3, as well as ${\tilde{h}}_n$ (see (3.2)) which was given by

$$\begin{aligned} {\tilde{h}}_n := \sup \limits _{x\in M} d(x,{\tilde{G}}_n),\;\;\forall \;n, \end{aligned}$$

and which is such that $\lim \limits _n {\tilde{h}}_n = 0$. Next, recall that by (3.2), we have

$$\begin{aligned} 500 {\tilde{h}}_n < \delta . \end{aligned}$$

Fix $x \in {\tilde{G}}_n$ and let $e_1,\ldots ,e_d$ be an arbitrary orthonormal basis of $(TM)_x$. By definition of ${\tilde{h}}_n$,

$$\begin{aligned} d(\exp _{x}(100{\tilde{h}}_ne_k),{\tilde{G}}_n)\le {\tilde{h}}_n,\;\;k=1,\ldots ,d. \end{aligned}$$

In particular, for each x and each k, it is possible to pick a point $x_k$ such that

$$\begin{aligned} x_k \in {\tilde{G}}_n \text { and } d(\exp _{x}(100{\tilde{h}}_ne_k),x_k)\le {\tilde{h}}_n. \end{aligned}$$

Having made such a selection for each $x \in {\tilde{G}}_n$, we define

$$\begin{aligned} V_{n,k}(x) := (\exp _{x})^{-1}(x_k),\;\;\;k=1,\ldots ,d. \end{aligned}$$

Thus, the first property holds by construction. Next, observe that since $100 {\tilde{h}}_n<\delta$, both $x_k,x$ and $\exp _{x}(100 {\tilde{h}}_n e_k)$ all lie in a ball of radius $4\delta \sqrt{d}$. Therefore, using Remark 3.3 we can compare $|V_{n,k}(x)|_{g_x}$ and $|100{\tilde{h}}_n e_k|_{g_x}$. In particular, we have

$$\begin{aligned} |V_{n,k}-100 {\tilde{h}}_n e_k|_{g_x} \le \tfrac{101}{100} d(\exp _{x}(100{\tilde{h}}_ne_k),x_k) \le \tfrac{101}{100} {\tilde{h}}_n. \end{aligned}$$

(A.1)

Then, the triangle inequality yields,

$$\begin{aligned} |V_{n,k}(x)|_{g_x}&\le |100{\tilde{h}}_n e_k|_{g_x} + |V_{n,k}(x)-100{\tilde{h}}_n e_k|_{g_x} \le 100 {\tilde{h}}_n + \tfrac{101}{100} {\tilde{h}}_n, \le 102 {\tilde{h}}_n,\\ |V_{n,k}(x)|_{g_x}&\ge |100{\tilde{h}}_n e_k|_{g_x} -|V_{n,k}(x)-100{\tilde{h}}_n e_k|_{g_x} \ge 100 {\tilde{h}}_n - \tfrac{101}{100} {\tilde{h}}_n \ge 98 {\tilde{h}}_n. \end{aligned}$$

This proves the second property. It remains to prove the third one. For the sake of brevity, let us omit the x dependence in the computations below.

Let us express the inner product $(V_{n,l},V_{n,k})_{g_x}$ in terms of the orthonormal basis $e_k$,

$$\begin{aligned} (V_{n,l},V_{n,k})_{g_x}&= (V_{n,l}-100{\tilde{h}}_ne_l+100{\tilde{h}}_ne_l,V_{n,k} -100{\tilde{h}}_ne_k+100{\tilde{h}}_ne_k)_{g_x}\\&= (V_{n,l}-100{\tilde{h}}_ne_l,V_{n,k}-100{\tilde{h}}_ne_k +100{\tilde{h}}_ne_k)_{g_x}\\&\;\;\;\;+(100{\tilde{h}}_ne_l,V_{n,k}-100{\tilde{h}}_ne_k +100{\tilde{h}}_ne_k)_{g_x}\\&= (V_{n,l}-100{\tilde{h}}_ne_l,V_{n,k}-100{\tilde{h}}_ne_k)_{g_x} +(V_{n,l}-100{\tilde{h}}_ne_l,100{\tilde{h}}_ne_k)_{g_x}\\&\;\;\;\;+(100{\tilde{h}}_ne_l,V_{n,k}-100{\tilde{h}}_ne_k)_{g_x} +(100{\tilde{h}}_ne_l,100{\tilde{h}}_ne_k)_{g_x}. \end{aligned}$$

Since the $e_k$ are orthonormal, for $k\ne l$ it follows that

$$\begin{aligned} (V_{n,l},V_{n,k})_{g_x}&= (V_{n,l}-100{\tilde{h}}_ne_l,V_{n,k}-100{\tilde{h}}_ne_k)_{g_x} +(V_{n,l}-100{\tilde{h}}_ne_l,100{\tilde{h}}_ne_k)_{g_x}\\&\;\;\;\;+(100{\tilde{h}}_ne_l,V_{n,k}-100{\tilde{h}}_ne_k)_{g_x},\;\; k\ne l. \end{aligned}$$

We apply the estimate (A.1) to this last identity, it follows that

$$\begin{aligned} |(V_{n,l},V_{n,k})_{g_x}|&\le | V_{n,l}-100{\tilde{h}}_ne_l |_{g_x} |V_{n,k}-100{\tilde{h}}_ne_k|{g_x}+|V_{n,l}-100{\tilde{h}}_ne_l|_{g_x} |100{\tilde{h}}_ne_k|_{g_x}\\&\;\;\;\;+|100{\tilde{h}}_ne_l|_{g_x}|V_{n,k}-100{\tilde{h}}_ne_k|_{g_x} \\&\le \left( \tfrac{101}{100}\right) ^2{\tilde{h}}_n^2+ 2\left( \tfrac{101}{100}{\tilde{h}}_n\right) (100 {\tilde{h}}_n) \le 204{\tilde{h}}_n. \end{aligned}$$

Since $|V_{n,l}|_{g_x}^{-1}\ge 98 {\tilde{h}}_n$, it follows that

$$\begin{aligned} |({\hat{V}}_{n,l},{\hat{V}}_{n,k})_{g_x}|&\le 204 {\tilde{h}}_n^2 |V_{n,l}|_{g_x}^{-1}|V_{n,k}|_{g_x}^{-1} \le 204 (98)^{-2}\le \tfrac{1}{20}, \end{aligned}$$

and the third property is proved. $\square$

From here on, for each n and for every $x\in {\tilde{G}}_n$, we fix a selection of vectors $\{V_{n,1}(x),\ldots ,V_{n,d}(x)\} \in (TM)_x$ as in the previous proposition. Moreover, we fix $u \in C^\beta _b(M)$ for the rest of this section.

Definition A.3

(Discrete gradient) Given $x\in {\tilde{G}}_n$ and u, define $(\nabla _n)^1u(x) \in (TM)_x$ by solving the system of linear equations

$$\begin{aligned} (V_{n,k}(x),(\nabla _n)^1u(x))_{g_x} = u(\exp _{x}(V_{n,k}(x)))-u(x), \quad k=1,\ldots ,d. \end{aligned}$$

Note that, as the $V_{n,k}(x)$ are linearly independent, the above system always has a unique solution.

Remark A.4

Let us illustrate the above definition in a simple case. Let us take,

$$\begin{aligned} M={\mathbb {R}}^d,\;\;\; {\tilde{G}}_n = (2^{-n}{\mathbb {Z}}^d), \end{aligned}$$

and write ${\tilde{h}}_n = 2^{-n}$ and $V_{n,l}(x) = h_ne_l$, where $\{e_1,\ldots ,e_d\}$ denote the standard orthonormal basis of ${\mathbb {R}}^d$. Then,

$$\begin{aligned} u(x+{\tilde{h}}_ne_k)-u(x)&= u(\exp _{x}(V_{n,k}(x)))-u(x)\\&= \sum \limits _{l=1}^{d} (\nabla _n)_l^1u(x) (V_{n,k}(x),{\hat{V}}_{n,l}(x))_{g_x}\\&= (\nabla _n)^1_ku(x) {\tilde{h}}_n. \end{aligned}$$

Thus, in this case we have

$$\begin{aligned} (\nabla _n)^1_ku(x) = \frac{u(x+{\tilde{h}}_ne_k)-u(x)}{{\tilde{h}}_n} (\approx \partial _{x_k}u(x)), \end{aligned}$$

and the vector $(\nabla _n)^1u(x)$ is nothing but a discretization of the gradient.

Definition A.5

Let $x,y \in M$ be such that $d(x,y)<r_0$. Then let $\Gamma _{x,y}$ denote the linear map

$$\begin{aligned} \Gamma _{x,y}: (TM)_y \rightarrow (TM)_x, \end{aligned}$$

given by parallel transport along the unique minimal geodesic connecting x to y. We should recall this map is an isometry with respect to the inner products $g_x$ and $g_{y}$. If the point y is understood from context, we shall simply write $\Gamma _x$.

Definition A.6

Let V be a section of the tangent bundle TM. We say V is of class $C^\alpha$ if

$$\begin{aligned}{}[V]_{C^\alpha (M)} :=\sup \limits _{0<d(x,y)<r_0}\frac{|V(x)-\Gamma _{x,y}V(y)|_{g_x}}{d(x,y)^\alpha } < \infty \end{aligned}$$

Likewise, if $M:TM\rightarrow TM$, then

$$\begin{aligned}{}[M]_{C^\alpha (M)} := \sup \limits _{0<d(x,y)<r_0}\frac{|M(x)-M(y)\Gamma _{x,y}^{-1}|_{g_x}}{d(x,y)^\alpha }<\infty . \end{aligned}$$

These seminorms, when applied to $V=\nabla u$ and $M=\nabla ^2u$ allows to define the $C^\beta$ norm of u in the obvious manner.

Remark A.7

Let $\beta \in [0,3)$ be given. The following is a useful characterization of Hölder continuity that will be used later on. Let x(t) denote a geodesic and e(t) a parallel vector field along it with $|\dot{x}(t)|_{g_{x(t)}}=|e(t)|_{x(t)}=1$. Then,

$$\begin{aligned} |(\nabla u(x(t)),e(t))_{x(t)}-(\nabla u(x(s)),e(s))_{x(s)}|\le \Vert u\Vert _{C^\beta }|t-s|^{\min \{\beta -1,1\}},\;\;\text { if } \beta \ge 1, \end{aligned}$$

and

$$\begin{aligned} |(\nabla ^2 u(x(t))e(t),e(t))_{x(t)}-(\nabla ^2 u(x(s))e(s),e(s))_{x(s)}|\le \Vert u\Vert _{C^\beta }|t-s|^{\min \{\beta -2,1\}},\;\;\text { if } \beta \ge 2. \end{aligned}$$

Defining the discrete Hessian requires further preparation, we define first the following “second order difference”,

$$\begin{aligned} \delta u_{x}(V_1,V_2) := u(\exp _{\exp _x(V_1)}(V_2))-u(\exp _x(V_1))-u(\exp _{x}(\Gamma _{x}V_2))+u(x). \end{aligned}$$

Here $\Gamma _{x}$ denotes the operation of parallel transport, as introduced in Definition A.5.

Definition A.8

(Discrete Hessian) Given $x\in {\tilde{G}}_n$ and u, we will define a linear transformation

$$\begin{aligned} (\nabla _n)^2u(x): (TM)_x \rightarrow (TM)_x. \end{aligned}$$

Given $k=1,\ldots ,d$, define $(\nabla _n)^2u(x)V_{n,k}(x) \in (TM)_x$ as the solution V to the linear system

$$\begin{aligned} (V,\Gamma _{x}V_{n,l}(x_k))_{g_x} = \delta u_{x}(V_{n,k}(x),V_{n,l}(x_k)),\;\;l=1,\ldots ,d. \end{aligned}$$

Here, for the sake of brevity of notation, we have written

$$\begin{aligned} x_{k} = \exp _{x}(V_{n,k}(x)). \end{aligned}$$

Having indicated how $(\nabla _n)^2u(x)$ acts on the basis $\{V_{n,k}(x)\}_{k=1}^d$ of $(TM)_x$, the linear transformation is completely determined.

Let us elaborate on the linear algebra problem that was used to define $(\nabla _n)^2u$. Given a linear transformation $D:(TM)_x\rightarrow (TM)_x$, and a family of pairs of vectors $\{(V_k,W_k)\}_{k=1}^N$ for some N, we seek to recover the full matrix D from the values

$$\begin{aligned} (DV_k,W_k). \end{aligned}$$

We are given a basis $V_k$ ($k=1,\ldots ,d$), and for each k another basis $\{W_{k,l}\}$ ($l=1,\ldots ,d$). Then, we seek to completely determine a linear transformation M given the values

$$\begin{aligned} (DV_k,W_{k,l}), \;\text { for } k,l=1,\ldots ,d. \end{aligned}$$

Remark A.9

Let us again see what this definition says in a simple case. Let $M, {\tilde{G}}_n, {\tilde{h}}_n$ and $\{V_{n,k}(x)\}$ be as in Remark A.4. Then, given $x\in {\tilde{G}}_n$ and $k,l=1,\ldots ,d$ we have

$$\begin{aligned} \delta u_x(V_{n,k}(x),V_{n,l}(x_k))&= u(x+2^{-n}e_k+2^{-n}e_l)-u(x+2^{-n}e_k)-u(x+2^{-n}e_l)+u(x)\\&= 2^{-n}2^{-n}( (\nabla _n)^2u(x)e_k,e_l). \end{aligned}$$

It follows that the components of $(\nabla _n)^2u(x)$ are given by

$$\begin{aligned} (\nabla _n)_{kl}^2u(x) = \frac{u(x+2^{-n}e_k+2^{-n}e_l)-u(x+2^{-n}e_k)-u(x+2^{-n}e_l)+u(x)}{2^{-n}2^{-n}} (\approx \nabla ^2_{kl}u(x) ) \end{aligned}$$

and the matrix $(\nabla _n)_{kl}^2u(x)$ is nothing but a discretization of the standard Hessian.

Remark A.10

Let $x \in {\tilde{G}}_n$. Using the upper bound in part (2) of Proposition A.2, one notes that all the values of u taken in evaluating $\nabla _n^1 u(x)$ and $\nabla _n^2u(x)$ lie within a ball of radius $<250{\tilde{h}}_n$ centered at x. In particular, if $u\equiv 0$ in $B_{250h_n}(x)$, then

$$\begin{aligned} \nabla _n^1u(x) = 0,\;\;\nabla _n^2u(x) = 0. \end{aligned}$$

The previous remark guarantees that the extension operator is somewhat “local”, the locality becoming more and more exact as n becomes larger, this is made rigorous in the following proposition.

Proposition A.11

Let $u \in C^\beta$, and $x_0 \in M$. Then,

$$\begin{aligned} u \equiv 0 \text { in } B_{400{\tilde{h}}_n}(x_0) \Rightarrow E_n^\beta (u,\cdot ) \equiv 0 \text { in } B_{100{\tilde{h}}_n}(x_0). \end{aligned}$$

Proof

First, we claim that

$$\begin{aligned} x \in B_{100{\tilde{h}}_n}(x_0) \Rightarrow B_{250 {\tilde{h}}_n}({\hat{y}}_{n,k}) \subset B_{400 {\tilde{h}}_n}(x_0),\;\;\forall \; k\in K_x. \end{aligned}$$

(A.2)

Let us see how (A.2) implies the proposition. Fix $x \in B_{100 {\tilde{h}}_n}(x_0)$, with $x\in M{\setminus } {\tilde{G}}_n$, then

$$\begin{aligned} E_n^\beta (u,x)&= \sum \limits _{k} p_{(u,k)}^\beta (x)\phi _{n,k}(x) = \sum \limits _{k\in K_x} p_{(u,k)}^\beta (x)\phi _{n,k}(x). \end{aligned}$$

Then, thanks to (A.2), we have that

$$\begin{aligned} u \equiv 0 \text { in } B_{250 {\tilde{h}}_n}({\hat{y}}_{n,k}),\;\;\forall \;k\in K_x,\;\forall \; x\in B_{100{\tilde{h}}_n}(x_0). \end{aligned}$$

In this case, Remark A.10 guarantees that

$$\begin{aligned} p_{(u,k)}^\beta (x)\equiv 0,\;\;\forall \;k\in K_x,\;\forall \;x\in B_{100{\tilde{h}}_n}(x_0). \end{aligned}$$

In other words,

$$\begin{aligned} E_n^\beta (u,x) =0,\;\;\forall \;x\in B_{100{\tilde{h}}_n}(x_0). \end{aligned}$$

Which proves the proposition. It remains to prove (A.2). Fix $x \in B_{100 {\tilde{h}}_n}(x_0)$ and $k\in K_x$. By the triangle inequality, and the definition of ${\hat{y}}_{n,k}$, we have

$$\begin{aligned} d(x,{\hat{y}}_{n,k}) \le d(x,y_{n,k})+d({\hat{y}}_{n,k},y_{n,k})&= d(x,y_{n,k})+d(y_{n,k},{\tilde{G}}_n)\\&\le 2d(x,y_{n,k})+d(x,{\tilde{G}}_n)\\&\le 2\text {diam}(P_{n,k}^*)+d(x,{\tilde{G}}_n). \end{aligned}$$

Then, thanks to Remark 3.11,

$$\begin{aligned} d(x,{\hat{y}}_{n,k}) \le 15 d(x,{\tilde{G}}_n)\le 15 {\tilde{h}}_n,\;\;\forall \;k\in K_x. \end{aligned}$$

Furthermore,

$$\begin{aligned} d({\hat{y}}_{n,k},x_0)&\le d({\hat{y}}_{n,k},x)+d(x,x_0)\\&\le d({\hat{y}}_{n,k},{\hat{x}})+d(x,{\hat{x}})+d(x,x_0). \end{aligned}$$

We now recall that $d(x,{\hat{x}}) = d(x,{\tilde{G}}_n) \le {\tilde{h}}_n$, and $d(x,x_0)\le 100{\tilde{h}}_n$. Furthermore, as shown in (3.16) in the proof of Proposition we have $d({\hat{y}}_{n,k},{\hat{x}}) \le 16 d(x,{\tilde{G}}_n)$ for $k\in K_x$. Gathering these inequalities it follows that

$$\begin{aligned} d({\hat{y}}_{n,k},x_0) \le 117 {\tilde{h}}_n,\;\;\forall \;k\in K_x. \end{aligned}$$

From here, and the triangle inequality, we conclude that $B_{250 {\tilde{h}}_n}({\hat{y}}_{n,k})$ lies inside $B_{400{\tilde{h}}_n}(x_0)$, that is, (A.2). This proves the proposition. $\square$

In what follows, we will be using the functions l and q, introduced in Definition 3.14. In ${\mathbb {R}}^d$ this is a completely straightforward calculation using the Taylor polynomial. On a Riemannian manifold, we shall use the coordinates given by the exponential map. For the next proposition, we recall that the functions “linear” and “quadratic” functions l and q introduced in Definition 3.14 are defined in a ball of of radius $4\delta \sqrt{d}$ around their base point, where $\delta$ is as in Remark 3.3

Proposition A.12

Let $x_0,x \in M$ with $d(x,x_0)\le 4\delta \sqrt{d}$, and $u \in C^\beta _b(M)$. Then,

(1) If $C^{\beta }_b=C^1_b$, then

$$\begin{aligned} u(x)-u(x_0)-l(\nabla u(x_0),x_0;x) = o(d(x,x_0)), \end{aligned}$$

where the $o(d(x,x_0))$ term is controlled by the modulus of continuity of $\nabla u$.

(2) If $\beta \in [1,2]$, then

$$\begin{aligned} |u(x)-u(x_0)-l(\nabla u(x_0),x_0;x)|\le \Vert u\Vert _{C^\beta }d(x,x_0)^\beta . \end{aligned}$$

(3) If $C^\beta _b = C^2_b$, then

$$\begin{aligned} u(x)-u(x_0)-l(\nabla u(x_0),x_0;x)-q(\nabla ^2u(x_0),x_0;x) = o(d(x,x_0)^2), \end{aligned}$$

where the $o(d(x,x_0))$ term is controlled by the modulus of continuity of $\nabla ^2 u$.

(4) If $\beta \in [2,3]$, then

$$\begin{aligned} |u(x)-u(x_0)-l(\nabla u(x_0),x_0;x)-q(\nabla ^2u(x_0),x_0;x)|\le \Vert u\Vert _{C^\beta }d(x,x_0)^\beta . \end{aligned}$$

We omit the straightforward proof of Proposition A.12.

Remark A.13

From Definition 3.14 it is immediate that Proposition A.12 has the following equivalent formulation which will also be useful: given a unit vector $e \in (TM)_{x_0}$ and $h\le 4\delta \sqrt{d}$, we have

$$\begin{aligned}&u(\exp _{x_0}(he))-u(x_0)-h(\nabla u(x_0),e)_{g_{x_0}} = o(h), \text { if } C^\beta _b = C^1_b,\\&|u(\exp _{x_0}(he))-u(x_0)-h(\nabla u(x_0),e)_{g_{x_0}}|\le \Vert u\Vert _{C^\beta }h^{\beta }, \text { if } \beta \in (1,2],\\&u(\exp _{x_0}(he))-u(x_0)-h(\nabla u(x_0),e)_{g_{x_0}}-\frac{h^2}{2}(\nabla ^2 u(x_0)e,e)_{g_{x_0}} = o(h^2), \text { if } C^\beta _b = C^2_b,\\&|u(\exp _{x_0}(he))-u(x_0)-h(\nabla u(x_0),e)_{g_{x_0}}\!-\!\frac{h^2}{2}(\nabla ^2 u(x_0)e,e)_{g_{x_0}}|\!\le \! \Vert u\Vert _{C^\beta }h^{\beta }, \text { if } \beta \in [2,3]. \end{aligned}$$

Proof of Remark A.13

First estimate. Fix a unit vector $e\in (TM)_{x_0}$. For $h \in [0,r_0]$ let $x(h) := \exp _{x_0}(he)$, and let

$$\begin{aligned} \varepsilon (h) := u\left( x(h)\right) -u(x_0)-(\nabla u(x_0),he)_{g_{x_0}}. \end{aligned}$$

It is immediate that $\varepsilon (0)=0$, $d(x_0,x(h))=h$, and that

$$\begin{aligned} \varepsilon '(h)&= (\nabla u(x(h)),\dot{x}(h))_{g_{x(h)}}-(\nabla u(x_0),e)_{g_{x_0}}. \end{aligned}$$

Since $\dot{x}(0)= e$, we have $\varepsilon '(0) = 0$. Keeping in mind that $\dot{x}(h)$ is the parallel transport of e along x(h), the Hölder regularity of $\nabla u(x)$ yields

$$\begin{aligned} |\varepsilon '(h)| = |(\nabla u(x(h)),\dot{x}(h))_{x(h)}-(\nabla u(x_0),e)_{g_{x_0}}|&\le \Vert u\Vert _{C^\beta }d(x_0,x(h))^{\beta -1}\\&= \Vert u\Vert _{C^\beta }h^{\beta -1}. \end{aligned}$$

Integrating this last inequality from 0 to h, we obtain the first estimate, since

$$\begin{aligned} |\varepsilon (h)| = |\varepsilon (h)-\varepsilon (0)| \le \Vert u\Vert _{C^\beta }h^{\beta }. \end{aligned}$$

Second estimate. Let x(h) be as before, with $h \in [0,r_0]$. This time we consider the function

$$\begin{aligned} \varepsilon (h) := u\left( x(h) \right) -u(x_0)-h(\nabla u(x_0),e)_{g_{x_0}}-\frac{h^2}{2}( (\nabla ^2 u(x_0))e,e)_{g_{x_0}}. \end{aligned}$$

Then, as before it is clear that $\varepsilon (0)=\varepsilon '(0)=0$ and

$$\begin{aligned}&\varepsilon '(h) = (\nabla u(x(h)),\dot{x}(h))_{x(h)}-(\nabla u(x_0),e)_{g_{x_0}}-h( (\nabla ^2 u(x_0))e,e)_{g_{x_0}},\\&\varepsilon ''(h) = ( (\nabla ^2 u(x(h)))\dot{x}(h),\dot{x}(h))_{g_{x(h)}}-( (\nabla ^2 u(x_0))e,e)_{g_{x_0}}. \end{aligned}$$

As before, we make use of the fact that $\dot{x}(h)$ is a parallel vector along x(h), which leads to

$$\begin{aligned} |\varepsilon ''(h)| = | ( (\nabla ^2 u(x(h)))\dot{x}(h),\dot{x}(h))_{g_{x(h)}}-( (\nabla ^2 u(x_0))e,e)_{g_{x_0}}| \le \Vert u\Vert _{C^\beta }h^{\beta -2}. \end{aligned}$$

Integrating this inequality twice (and using that $\varepsilon (0)=\varepsilon '(0)=0$) it follows that

$$\begin{aligned} |\varepsilon (h)| = |\varepsilon (h)-\varepsilon (0)| = \left| \int _0^h \varepsilon '(s)\;ds \right| \le \Vert u\Vert _{C^\beta }h^\beta , \end{aligned}$$

which proves the second estimate.

$\square$

The next Lemma consists of a very important fact, namely, that the discrete difference operators $(\nabla _n)^1u$ and $(\nabla _n)^2u$ are “consistent” –i.e. they converge to the differential operators $\nabla u$ and $\nabla ^2 u$. Furthermore, we have that the error made when estimating the derivatives by the discrete operator is a quantity controlled by the $C^\beta$ norm of $u \in C^\beta _b(M)$.

Lemma A.14

Let $x \in {\tilde{G}}_n$ and $u\in C^\beta _b(M)$ then

$$\begin{aligned} |(\nabla _n)^1u(x)-\nabla u(x)|_{g_{x}} \le C\Vert u\Vert _{C^\beta }{\tilde{h}}_n^{\beta -1}, \text { if } \beta \in (1,2],\\ |(\nabla _n)^2 u(x)-\nabla ^2 u(x)|_{g_{x}} \le C\Vert u\Vert _{C^\beta }{\tilde{h}}_n^{\beta -2}, \text { if } \beta \in (2,3]. \end{aligned}$$

Furthermore, if $C^\beta _b = C^1_b$ or $C^\beta _b = C^2_b$ then, we have, respectively

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\sup \limits _{x \in K\cap {\tilde{G}}_n} |(\nabla _n)^1u(x)-\nabla u(x)|_{g_{x}} = 0,\\ \lim \limits _{n\rightarrow \infty }\sup \limits _{x \in K\cap {\tilde{G}}_n} |(\nabla _n)^2u(x)-\nabla ^2 u(x)|_{g_{x}} = 0, \end{aligned}$$

where K is an arbitrary compact subset of M.

Proof

First estimate. We may write

$$\begin{aligned} \nabla u(x) = \sum \limits _{l=1}^d \theta _l {\hat{V}}_{n,l}(x), \end{aligned}$$

where the numbers $\theta _1,\ldots ,\theta _d$ are determined from the system of equations

$$\begin{aligned} (\nabla u(x),{\hat{V}}_{n,k}(x))_{g_x} = \sum \limits _{l=d}^d \theta _l ({\hat{V}}_{n,k}(x),{\hat{V}}_{n,l}(x))_{g_x}. \end{aligned}$$

Now, Proposition A.12 says that

$$\begin{aligned} \left| \frac{u(\exp _{x}(V_{n,k}))-u(x)}{|V_{n,k}|_{g_x}} - \left( \nabla u(x),{\hat{V}}_{n,k} \right) _{g_x}\right| \le C\Vert u\Vert _{C^\beta }|V_{n,k}|_{g_x}^{\beta -1}, \end{aligned}$$

and, if $C^\beta _b = C^1_b$, it says that for any compact K,

$$\begin{aligned} \lim \limits _{n\rightarrow \infty } \sup \limits _{x\in K \cap {\tilde{G}}_n} \max \limits _{1\le k\le d}\left| \frac{u(\exp _{x}(V_{n,k}))-u(x)}{|V_{n,k}|_{g_x}} - \left( \nabla u(x),{\hat{V}}_{n,k} \right) _{g_x}\right| = 0, \end{aligned}$$

the convergence in the limit being determined by K, the continuity of $\nabla u$, and M. Then,

$$\begin{aligned} |(\nabla _n)_{l}^1u(x)-\theta _l|&\le C\Vert u\Vert _{C^\beta }{\tilde{h}}_n^{\beta -1} \;\;\; \forall \;x\in M,\text { if } \beta \in (0,1),\\ \lim \limits _{n\rightarrow \infty }\sup \limits _{x \in K \cap {\tilde{G}}_n}|(\nabla _n)_{l}^1u(x)-\theta _l|&= 0, \quad \quad \quad \quad \quad \ \forall \; K\subset \subset M, \text { if } C^\beta _b = C^1_b. \end{aligned}$$

The above holds for each $l=1,\ldots ,d$. Combining these inequalities it is immediate that

$$\begin{aligned} |(\nabla _n)^1u(x)-\nabla u(x)|\le C\Vert u\Vert _{C^\beta }{\tilde{h}}_n^{\beta -1}, \end{aligned}$$

and, for $C^\beta _b = C^1_b$,

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\sup \limits _{x \in K\cap {\tilde{G}}_n}|(\nabla _n)^1u(x)-\nabla u(x)| = 0. \end{aligned}$$

Second estimate. First, we need an elementary observation about geodesics. Observe that

$$\begin{aligned} \exp _{\exp _x(V_{n,k}(x))}(V_{n,l}(x_k)) = \exp _{x}(V_{n,k}(x)+\Gamma _xV_{n,l}(x_k)+ \text {(Error)}_0). \end{aligned}$$

(A.3)

Where the term $\text {(Error)}_0$ is term appearing due to possibly non-zero curvature. It turns out that this error term is at least a cubic error in terms of ${\tilde{h}}_n$, which is proved as follows: let J(t) be the Jacobi field along the geodesic $\gamma (t) = \exp _x(t {\hat{V}}_{n,k})$ determined by $J(0)=0$ and $J(|V_{n,k}|_{g_x}) = {\hat{V}}_{n,l}(x_k)$. Then, define $\sigma (t,s) \in (TM)_x$ by

$$\begin{aligned} \exp _{\gamma (t)}(s J(t) ) = \exp _{x}( \sigma (t,s)). \end{aligned}$$

Note that $\sigma \left( |V_{n,k}|_{g_x},|V_{n,l}(x_k)|_{g_x} \right)$ must be equal to the argument in the exponential on the right hand side of (A.3). Then, note that

$$\begin{aligned} \sigma (0,s) = 0,\;\;\forall \;s \Rightarrow \sigma (0,0)= \partial _{s}\sigma (0,0) =\partial _{ss}\sigma (0,0)=0. \end{aligned}$$

Furthermore, $\partial _t \sigma (0,0) = {\hat{V}}_{n,k}(x)$, so

$$\begin{aligned} \sigma (t,s) = t{\hat{V}}_{n,k}+st \partial _{ts}\sigma (0,0) + O((s^2+t^2)^{3/2}). \end{aligned}$$

Now, by contrasting the respective Jacobi and parallel transport equations, it can be shown that

$$\begin{aligned} |\partial _{ts}\sigma (0,0)-\Gamma _x {\hat{V}}_{n,k}(x_k)|\le C{\tilde{h}}_n. \end{aligned}$$

Given that $|V_{n,k}|_{g_x},|V_{n,l}(x_k)|_{g_x}\le h_n$, this leads to the bound

$$\begin{aligned} | \text {(Error)}_0|_{g_x} \le C{\tilde{h}}_n^3. \end{aligned}$$

(A.4)

The constant C depending only on the metric of M.

Let us analyze the first three terms appearing in the second order difference $\delta u_{x}(V_{n,k}(x),V_{n,l}(x_k))$. We consider the Taylor expansion and estimate the remainder via Proposition A.12. First of all, we have

$$\begin{aligned} u(\exp _{\exp _x(V_{n,k}(x))}(V_{n,l}(x_k))) = u(\exp _{x}(V_{n,k}(x)+\Gamma _xV_{n,l}(x_k)+ \text {(Error)}_0)). \end{aligned}$$

The estimate (A.4) guarantees in particular that $|V_{n,k}(x)+\Gamma _xV_{n,l}(x_k)+ \text {(Error)}_0|\le C{\tilde{h}}_n$. With this in mind, we apply Proposition A.12 in order to obtain the expansion

$$\begin{aligned}&u(\exp _{\exp _x(V_{n,k}(x))}(V_{n,l}(x_k)))\\&= u(x)+ (\nabla u(x),V_{n,k}(x)+\Gamma _xV_{n,l}(x_k)+ \text {(Error)}_0)_{g_x}\\&\;\;\;\; + \tfrac{1}{2} \left( \nabla ^2u(x)( V_{n,k}(x)+\Gamma _xV_{n,l}(x_k)+ \text {(Error)}_0) , V_{n,k}(x)+\Gamma _xV_{n,l}(x_k)+ \text {(Error)}_0 \right) _{g_x}\\&\;\;\;\; + \text {(Error)}, \end{aligned}$$

where $\text {(Error)}$, which denotes the remainder in the Taylor expansion, satisfies the bound

$$\begin{aligned} | \text {(Error)}| \le C\Vert u\Vert _{C^\beta }{\tilde{h}}_n^\beta . \end{aligned}$$

Expanding, we see that

$$\begin{aligned}&u(\exp _{\exp _x(V_{n,k}(x))}(V_{n,l}(x_k)))\\&= u(x)+ (\nabla u(x),V_{n,k}(x))_{g_x}+(\nabla u(x),\Gamma _xV_{n,l}(x_k))_{g_x}\\&\;\;\;\; + \tfrac{1}{2} \left( \nabla ^2u(x) V_{n,k}(x),V_{n,k}(x)\right) _{g_x}+\tfrac{1}{2} \left( \nabla ^2u(x)\Gamma _xV_{n,l}(x_k),\Gamma _xV_{n,l}(x_k)\right) _{g_x}\\&\;\;\;\; + \left( \nabla ^2u(x) V_{n,k}(x),\Gamma _xV_{n,l}(x_k)\right) _{g_x}\\&\;\;\;\;+(\nabla u(x),\text {(Error)}_0)_{g_x}+\frac{1}{2}\left( \nabla ^2u(x)\text {(Error)}_0,\text {(Error)}_0\right) _{g_x}\\&\;\;\;\;+\left( \nabla ^2u(x)\text {(Error)}_0,V_{n,k}(x)+\Gamma _xV_{n,l}(x_k)\right) _{g_x}+ \text {(Error)}. \end{aligned}$$

The terms involving a factor of $\text {(Error)}_0$ may be absorbed into $\text {(Error)}$. To see why, we use the estimate (A.4) and bound term by term

$$\begin{aligned} |(\nabla u(x),\text {(Error)}_0)_{g_x}|&\le C\Vert u\Vert _{C^1}h_n^3,\\ \left| \left( \nabla ^2u(x)\text {(Error)}_0,\text {(Error)}_0\right) _{g_x}\right|&\le C\Vert u\Vert _{C^2}{\tilde{h}}_n^{6},\\ \left| \left( \nabla ^2u(x)\text {(Error)}_0,V_{n,k}(x)\right) _{g_x}\right|&\le C\Vert u\Vert _{C^2}{\tilde{h}}_n^4,\\ \left| \left( \nabla ^2u(x)\text {(Error)}_0,\Gamma _xV_{n,l}(x_k)\right) _{g_x}\right|&\le C\Vert u\Vert _{C^2}{\tilde{h}}_n^4. \end{aligned}$$

Since $\beta \ge 2$, each of the above terms is bounded by $C\Vert u\Vert _{C^\beta }{\tilde{h}}_n^\beta$. Then, absorbing the terms involving $\text {(Error)}_0$ into $\text {(Error)}$ we obtain

$$\begin{aligned}&u(\exp _{\exp _x(V_{n,k}(x))}(V_{n,l}(x_k)))\\&\quad = u(x)+ (\nabla u(x),V_{n,k}(x))_{g_x}+(\nabla u(x),\Gamma _xV_{n,l}(x_k))_{g_x}\\&\quad \;\;\;\; + \tfrac{1}{2} \left( \nabla ^2u(x) V_{n,k}(x),V_{n,k}(x)\right) _{g_x}+\tfrac{1}{2} \left( \nabla ^2u(x)\Gamma _xV_{n,l}(x_k),\Gamma _xV_{n,l}(x_k)\right) _{g_x}\\&\quad \;\;\;\; + \left( \nabla ^2u(x) V_{n,k}(x),\Gamma _xV_{n,l}(x_k)\right) _{g_x}+ \text {(Error)}. \end{aligned}$$

As for the other two terms, we have

$$\begin{aligned} u(\exp _x(V_{n,k}(x)))&= u(x)+(\nabla u(x),V_{n,k}(x))_{g_x}\\&\;\;\;\;+\tfrac{1}{2} \left( \nabla ^2u(x)V_{n,k}(x),V_{n,k}(x) \right) _{g_x}+\text {(Error)}, \end{aligned}$$

and

$$\begin{aligned} u(\exp _{x}(\Gamma _{x}V_{n,l}(x_k)))&= u(x)+(\nabla u(x_0),\Gamma _x V_{n,l}(x_k))_{g_x}\\&\;\;\;\;+\tfrac{1}{2} \left( \nabla ^2u(x)\Gamma _x V_{n,l}(x_k),\Gamma _x V_{n,l}(x_k) \right) _{g_x}+\text {(Error)}. \end{aligned}$$

In each case, $|\text {(Error)}|$ is no larger than $C\Vert u\Vert _{C^\beta }{\tilde{h}}_n^\beta$, thanks to Proposition A.12.

Combining the last three formulas, it follows that

$$\begin{aligned} \delta u_{x}(V_{n,k}(x),V_{n,l}(x_k)) \end{aligned}$$

is equal to

$$\begin{aligned}&u(x)+(\nabla u(x),V_{n,k}(x))_{g_x}+(\nabla u(x),\Gamma _x(V_{n,l}(x_k))_{g_x}+\tfrac{1}{2}((\nabla ^2 u(x))V_{n,k}(x),V_{n,k}(x))_{g_x}\\&+((\nabla ^2u(x))V_{n,k}(x),\Gamma _xV_{n,l}(x_k))_{g_x} +\tfrac{1}{2}((\nabla ^2u(x))\Gamma _xV_{n,l}(x_k),\Gamma _xV_{n,l}(x_k))_{g_x}\\&-u(x)-(\nabla u(x),V_{n,k}(x))_{g_x}-\tfrac{1}{2}((\nabla ^2u(x))V_{n,k}(x),V_{n,k}(x))_{g_x} -u(x)\\&-(\nabla u(x),\Gamma _x(V_{n,l}(x_k))_{g_x}-\tfrac{1}{2}((\nabla ^2 u(x))\Gamma _x(V_{n,l}(x_k),\Gamma _x(V_{n,l}(x_k))_{g_x}+u(x)+\text {(Error)}. \end{aligned}$$

From the above, it is clear all but one of the terms in the first two lines above is cancelled out with a term in the last two lines. We then arrive at the formula

$$\begin{aligned} \delta u_{x}(V_{n,k}(x),V_{n,l}(x_k)) = ((\nabla ^2u(x))V_{n,k}(x),\Gamma _xV_{n,l}(x_k))_{g_x}+\text {(Error)}, \end{aligned}$$

where—thanks to Proposition A.12, as pointed out earlier—we have

$$\begin{aligned} |\text {(Error)}| \le C\Vert u\Vert _{C^\beta }{\tilde{h}}_n^\beta . \end{aligned}$$

Then, solving the linear problem corresponding to $(\nabla _n)^2u(x)$ and $\nabla ^2u(x)$ it follows that

$$\begin{aligned} |(\nabla _n)^2u(x)-\nabla ^2 u(x)|\le C\Vert u\Vert _{C^\beta }{\tilde{h}}_n^{\beta -2}. \end{aligned}$$

Finally, if $C^\beta _b = C^2_b$, the convergence of $(\nabla _n)^2u(x)$ to $\nabla ^2u(x)$ follows analogously to the convergence of $\nabla _n u(x)$ to $\nabla u(x)$ for $C^\beta _b = C^1_b$, we omit the details. $\square$

Given the proof of Lemma A.14 it should be clear that the $L^\infty ({\tilde{G}}_n)$ norm of $(\nabla _n)^iu$ ($i=1,2$) is controlled by the appropriate $C^\beta$ norm of u in a manner which is independent of n. This fact is the content of the next proposition.

Proposition A.15

Let $x \in {\tilde{G}}_n$, then we have the estimates

$$\begin{aligned} |(\nabla _n)^1u(x)|_{g_x}&\le C\Vert u\Vert _{C^1},\\ |(\nabla _n)^2u(x)|_{g_x}&\le C\Vert u\Vert _{C^2}. \end{aligned}$$

Proof of Proposition A.15

This is an immediate consequence of the previous proposition. Indeed, fix $u \in C^\beta _b(M)$ and $x\in {\tilde{G}}_n$. Then, we have

$$\begin{aligned} |(\nabla _n)^1u(x)|_{g_x}&\le |(\nabla _n)^1u(x)-\nabla u(x)|_{g_x} +|\nabla u(x)|_{g_x},\;\; \beta \ge 1.\\ |(\nabla _n)^2u(x)|_{g_x}&\le |(\nabla _n)^2u(x)-\nabla ^2 u(x)|_{g_x}+|\nabla ^2 u(x)|_{g_x},\;\;\beta \ge 2. \end{aligned}$$

Then, using the two estimates in Proposition A.12, we have

$$\begin{aligned} |(\nabla _n)^1u(x)|_{g_x}&\le C{\tilde{h}}_n^{\min \{\beta -1,1\} } \Vert u\Vert _{C^\beta }+\Vert u\Vert _{C^1} \le C\Vert u\Vert _{C^\beta },\;\;\beta \ge 1.\\ |(\nabla _n)^2u(x)|_{g_x}&\le C{\tilde{h}}_n^{\min \{\beta -2,1\}}\Vert u\Vert _{C^\beta }+\Vert u\Vert _{C^2} \le C\Vert u\Vert _{C^\beta },\;\;\beta \ge 2. \end{aligned}$$

$\square$

The next proposition yields a quantitative control on the “continuity” of $(\nabla _n)^{i}u$ in terms of the regularity of the original function u. As one may expect, if $\nabla u(x)$ and $\nabla ^2u(x)$ are Hölder continuous in M, then $(\nabla _n)^1u$ and $(\nabla _n)^2u$ enjoy a respective modulus of “continuity” on ${\tilde{G}}_n$, this being uniform in n.

Proposition A.16

Consider points $x,y \in M {\setminus } {\tilde{G}}_n$ and ${\hat{y}}$, ${\hat{x}}$ the corresponding points in ${\tilde{G}}_n$ with $d(x,{\tilde{G}}_n)=d(x,{\hat{x}})$, $d(y,{\tilde{G}}_n)=d(y,{\hat{y}})$, we have the following estimates with a universal C.

• (1) For $1\le \beta \le 2$,

  $$\begin{aligned}&|(\nabla _n)_a^1 u({\hat{x}})-(\nabla _n)_a^1 u({\hat{y}})| \le C\Vert u\Vert _{C^\beta }d({\hat{x}},{\hat{y}})^{\beta -1}. \end{aligned}$$

• (2) For $2\le \beta \le 3$,

  $$\begin{aligned}&| (\nabla _n)_{ab}^2 u({\hat{x}})-(\nabla _n)_{ab}^2u({\hat{y}})| \le C\Vert u\Vert _{C^\beta }d({\hat{x}},{\hat{y}})^{\beta -2}. \end{aligned}$$

Proof

If ${\hat{x}}={\hat{y}}$ both inequalities are trivial and there is nothing to prove, so let us assume ${\hat{x}},{\hat{y}}$ are two different points in ${\tilde{G}}_n$. In this case, and thanks to (3.3), we have

$$\begin{aligned} d({\hat{x}},{\hat{y}}) \ge \lambda {\tilde{h}}_n. \end{aligned}$$

(A.5)

First estimate. The triangle inequality yields,

$$\begin{aligned}&|(\nabla _n)_a^1 u({\hat{x}})-(\nabla _n)_a^1 u({\hat{y}})| \\&\le |(\nabla _n)_a^1 u({\hat{x}})-\nabla _a u({\hat{x}})|+ |\nabla _a u({\hat{x}})-\nabla _a u({\hat{y}})|+ |\nabla _a u({\hat{y}})-(\nabla _n)_a^1 u({\hat{y}})|. \end{aligned}$$

Let us estimate each of the three terms on the right. The middle term is straightforward,

$$\begin{aligned} |\nabla _a u({\hat{x}})-\nabla _a u({\hat{y}})|\le C\Vert u\Vert _{C^\beta }d({\hat{x}},{\hat{y}})^{\beta -1}. \end{aligned}$$

For the first and third term, we use the first part of Lemma A.14, which says that

$$\begin{aligned}&|(\nabla _n)_a^1 u({\hat{x}})-\nabla _a u({\hat{x}})| \le C\Vert u\Vert _{C^\beta }{\tilde{h}}_n^{\beta -1},\\&|(\nabla _n)_a^1 u({\hat{y}})-\nabla _a u({\hat{y}})| \le C\Vert u\Vert _{C^\beta }{\tilde{h}}_n^{\beta -1}. \end{aligned}$$

Using (A.5) it follows that

$$\begin{aligned}&|(\nabla _n)_a^1 u({\hat{x}})-\nabla _a u({\hat{x}})| \le C\Vert u\Vert _{C^\beta }d({\hat{x}},{\hat{y}})^{\beta -1},\\&|(\nabla _n)_a^1 u({\hat{y}})-\nabla _a u({\hat{y}})| \le C\Vert u\Vert _{C^\beta }d({\hat{x}},{\hat{y}})^{\beta -1}. \end{aligned}$$

Combining the bounds for the three terms the first estimate follows.

Second estimate. As before, we start by breaking the difference in three parts, so

$$\begin{aligned}&|(\nabla _n)_{ab}^2 u({\hat{x}})-(\nabla _n)_{ab}^2 u({\hat{y}})|\\&\quad \quad \le |(\nabla _n)_{ab}^2 u({\hat{x}})-\nabla _{ab}^2 u({\hat{x}})|+|\nabla _{ab}^2 u({\hat{x}})-\nabla _{ab}^2 u({\hat{y}})|+|\nabla _{ab}^2 u({\hat{y}})-(\nabla _n)_{ab}^2 u({\hat{y}})|. \end{aligned}$$

The middle term is bounded by

$$\begin{aligned} |\nabla _{ab}^2 u({\hat{x}})-\nabla _{ab}^2 u({\hat{y}})| \le C\Vert u\Vert _{C^\beta }d({\hat{x}},{\hat{y}})^{\beta -2}. \end{aligned}$$

Next, thanks to the second part of Lemma A.14,

$$\begin{aligned}&|(\nabla _n)_{ab}^2 u({\hat{x}})-\nabla _{ab}^2 u({\hat{x}})| \le C\Vert u\Vert _{C^\beta }{\tilde{h}}_n^{\beta -2},\\&|(\nabla _n)_{ab}^2 u({\hat{y}})-\nabla _{ab}^2 u({\hat{y}})| \le C\Vert u\Vert _{C^\beta }{\tilde{h}}_n^{\beta -2}. \end{aligned}$$

Using (A.5) again, we conclude that

$$\begin{aligned}&|(\nabla _n)_{ab}^2 u({\hat{x}})-\nabla _{ab}^2 u({\hat{x}})| \le C\Vert u\Vert _{C^\beta }d({\hat{x}},{\hat{y}})^{\beta -2},\\&|(\nabla _n)_{ab}^2 u({\hat{y}})-\nabla _{ab}^2 u({\hat{y}})| \le C\Vert u\Vert _{C^\beta }d({\hat{x}},{\hat{y}})^{\beta -2}. \end{aligned}$$

As in the previous case, the combined bounds for the three terms yields the estimate. $\square$

Appendix B: The Proof of Proposition 3.22

This section is dedicated to proving Proposition 3.22, which we re-record right here for the reader’s convenience.

Proposition

Let $x \in M{\setminus } {\tilde{G}}_n$ and $u\in C^\beta$. There is a universal constant C such that the following bounds hold. First, if $0\le \beta <1$,

$$\begin{aligned} |\nabla (E_n^\beta \circ T_n)u(x)|\le C\Vert u\Vert _{C^\beta }d(x,{\tilde{G}}_n)^{\beta -1}. \end{aligned}$$

If $1\le \beta <2$, we have

$$\begin{aligned} |\nabla ^2(E_n^\beta \circ T_n)u(x)|\le C\Vert u\Vert _{C^\beta }d(x,{\tilde{G}}_n)^{\beta -2}. \end{aligned}$$

Finally, if $2\le \beta <3$, we have

$$\begin{aligned} |\nabla ^3(E_n^\beta \circ T_n)u(x)|\le C\Vert u\Vert _{C^\beta }d(x,{\tilde{G}}_n)^{\beta -3}. \end{aligned}$$

Proof

As done throughout Sect. 3, for the sake of brevity we shall write $f=\pi _n^\beta u$.

The case $\beta \in [0,1)$. Since the sum defining f is locally finite, we may differentiate term by term, which yields

$$\begin{aligned} \nabla f(x)&= \sum \limits _{k} u({\hat{y}}_{n,k})\nabla \phi _{n,k}(x). \end{aligned}$$

Using (3.13) with $i=1$ we may rewrite the above as

$$\begin{aligned} \nabla f(x)&= \sum \limits _{k}(u({\hat{y}}_{n,k})-u({\hat{x}}))\nabla \phi _{n,k}(x),\;\;\forall \;x\in M {\setminus } {\tilde{G}}_n. \end{aligned}$$

Then, since the only non-zero terms are those with $k \in K_x$ ($K_x$ was introduced in Lemma 3.10),

$$\begin{aligned} |\nabla f(x)|_{g_x}&\le \sum \limits _{k}|u({\hat{y}}_{n,k})-u({\hat{x}})| |\nabla \phi _{n,k}(x)|_{g_x}\\&\le N \sup \limits _{k \in K_x}|u({\hat{y}}_{n,k})-u({\hat{x}})||\nabla \phi _{n,k}(x)|_{g_x}. \end{aligned}$$

For $k\in K_x$, using Remark 3.11, and the Hölder regularity of u one can check that

$$\begin{aligned} |u({\hat{y}}_{n,k})-u({\hat{x}})| |\nabla \phi _{n,k}(x)|_{g_x}&\le C\Vert u\Vert _{C^\beta }d(x,{\tilde{G}}_n)^\beta d(x,{\tilde{G}}_n)^{-1}. \end{aligned}$$

From here, it follows that

$$\begin{aligned} |\nabla f(x)|&\le C\Vert u\Vert _{C^\beta }d(x,{\tilde{G}}_n)^{\beta -1}. \end{aligned}$$

The case $\beta \in [1,2)$. This time, we shall compute the Hessian $\nabla ^2f$ using a local system of coordinates $\{x^{1},\ldots ,x^{d}\}$. Then, for any pair of indices a, b we have

$$\begin{aligned} \nabla ^2_{ab}\phi = \partial _{x_ax_b}^2\phi - \sum \limits _{k=1}^d\Gamma _{ab}^k\partial _{x_k}\phi . \end{aligned}$$

Then

$$\begin{aligned} \nabla ^2_{ab} f(x)&= \sum \limits _{k} \nabla ^2_{ab}\left( (u({\hat{y}}_{n,k})+l( \nabla ^1_n u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x)) \phi _{n,k}(x) \right) . \end{aligned}$$

We expand each term using the Leibniz rule, and conclude $\nabla ^2_{ab}f(x)$ is equal to

$$\begin{aligned} \text {I}(x)+\text {II}(x)+\text {III}(x), \end{aligned}$$

where, for the sake of brevity, we have written

$$\begin{aligned} \text {I}(x)&= \sum \limits _{k} (u({\hat{y}}_{n,k})+l( \nabla ^1_n u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x) )\nabla ^2_{ab}\phi _{n,k}(x),\\ \text {II}(x)&= \sum \limits _{k} \nabla _a l( \nabla ^1_n u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x) \nabla _b \phi _{n,k}(x)+\sum \limits _{k} \nabla _b \phi _{n,k}(x) \nabla _a l( \nabla ^1_n u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x),\\ \text {III}(x)&=\sum \limits _{k} \nabla ^2_{ab}l( \nabla ^1_n u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x) \phi _{n,k}(x). \end{aligned}$$

Since $x\in M {\setminus } {\tilde{G}}_n$, we can use (3.13) with $i=1,2$ to obtain

$$\begin{aligned}&\sum \limits _{k} (u({\hat{y}}_{n,k})+l( \nabla ^1_n u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x) )\nabla _{ab}^2\phi _{n,k}(x) \\&\quad \quad = \sum \limits _{k} (u({\hat{y}}_{n,k})+l( \nabla ^1_n u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x) - u({\hat{x}}) )\nabla _{ab}^2\phi _{n,k}(x), \end{aligned}$$

and

$$\begin{aligned}&\sum \limits _{k} \nabla _a l( \nabla ^1_n u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x) \nabla _b \phi _{n,k}(x) \\&\quad \quad = \sum \limits _{k} \left( \nabla _a l( \nabla ^1_n u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x) -\nabla _a l( \nabla ^1_n u ({\hat{x}}),{\hat{x}};x) \right) \nabla _b \phi _{n,k}(x). \end{aligned}$$

Let us bound each of these. The triangle inequality says

$$\begin{aligned}&|u({\hat{y}}_{n,k})+l( (\nabla _n)^1u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x)-u(x)|\le |u({\hat{y}}_{n,k})+l( \nabla u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x)\\&\quad -u(x)|+|l((\nabla _n)^1u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x)-l( \nabla u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x)|. \end{aligned}$$

By Proposition A.12 the first term on the right is no larger than $C\Vert u\Vert _{C^\beta }d(x,{\hat{y}}_{n,k})^\beta$. On the other hand, from the definition of $l(\cdot ,\cdot ;\cdot )$, it is immediate that the second term is no larger than

$$\begin{aligned}&|(\nabla _n)^1u ({\hat{y}}_{n,k})- \nabla u ({\hat{y}}_{n,k})|_{g_{{\hat{y}}_{n,k}}}d(x,{\hat{y}}_{n,k}). \end{aligned}$$

Now, Lemma A.14 says that $|(\nabla _n)^1u ({\hat{y}}_{n,k})- \nabla u ({\hat{y}}_{n,k})|_{g_{{\hat{y}}_{n,k}}} \le C\Vert u\Vert _{C^\beta }{\tilde{h}}_n^{\beta -1}$. Noting that $d(x,{\hat{y}}_{n,k})$ is no larger than $C{\tilde{h}}_n$ for $x\in P_{n,k}^*$, we obtain the estimate

$$\begin{aligned} |(\nabla _n)^1u ({\hat{y}}_{n,k})- \nabla u ({\hat{y}}_{n,k})|_{g_{{\hat{y}}_{n,k}}} \le C\Vert u\Vert _{C^\beta }d(x,{\hat{y}}_{n,k})^{\beta -1}. \end{aligned}$$

Combining the last three estimates, we conclude that

$$\begin{aligned}&|u({\hat{y}}_{n,k})+l( (\nabla _n)^1u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x)-u(x)|\le C\Vert u\Vert _{C^\beta }d(x, {\tilde{G}}_n)^\beta ,\;\;\forall \;x\in P_{n,k}^*. \end{aligned}$$

Using the estimates for the size of $\nabla ^2\phi _{n,k}$, the above implies that

$$\begin{aligned}&|u({\hat{y}}_{n,k})+l( (\nabla _n)^1u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x)-u(x)||\nabla ^2\phi _{n,k}(x)|\\&\quad \le C\Vert u\Vert _{C^\beta }d(x,{\tilde{G}}_n)^{\beta -2},\;\;\forall \;x\in P_{n,k}^*. \end{aligned}$$

Finally, let us recall that the only nonzero terms appearing in the sum $\text {I}(x)$ are those with $k\in K_x$ (i.e. $x \in P_{n,k}^*$), and that there at most N of these terms. Then, we conclude that

$$\begin{aligned} \text {I}(x) \le C\Vert u\Vert _{C^\beta }d(x,{\tilde{G}}_n)^{\beta -2}. \end{aligned}$$

Let us now bound $\text {II}(x)$, observe that

$$\begin{aligned} \left| \nabla _a l( (\nabla _n)u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x)-\nabla _a l( \nabla ^1_n u ({\hat{x}}),{\hat{x}};x)\right| \le C\Vert u\Vert _{C^\beta }d({\hat{x}},{\hat{y}}_{n,k})^{\beta -1},\;\;\forall \;x\in P_{n,k}^*. \end{aligned}$$

Therefore

$$\begin{aligned} \sup \limits _{x\in P_{n,k}^*}\left| \nabla _a l( \nabla ^1_n u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x)-\nabla _a l( \nabla ^1_n u ({\hat{x}}),{\hat{x}};x)\right|&\le C\Vert u\Vert _{C^\beta }d(x,{\tilde{G}}_n)^{\beta -1}. \end{aligned}$$

As before, the only nonzero terms adding up to $\text {II}(x)$ are those with $x \in P_{n,k}^*$, therefore, the above bound implies that

$$\begin{aligned} \left| \sum \limits _{k} \nabla _a l( \nabla ^1_n u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x) \nabla _b \phi _{n,k}(x) \right|&\le \sum \limits _{k\in K_x}\Vert u\Vert _{C^\beta }d(x,{\tilde{G}}_n)^{\beta -1}Cd(x,{\tilde{G}}_n)^{-1}\\&\le CN \Vert u\Vert _{C^{\beta }}d(x,{\tilde{G}}_n)^{\beta -2}. \end{aligned}$$

Therefore,

$$\begin{aligned} \text {II}(x) \le C \Vert u\Vert _{C^{\beta }}d(x,{\tilde{G}}_n)^{\beta -2}. \end{aligned}$$

It remains to bound $\text {III}(x)$. According to Proposition 3.17 and Proposition A.15,

$$\begin{aligned} |\nabla _{ab}^2l( \nabla ^1_n u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x)| \le C\Vert u\Vert _{C^\beta (M)}. \end{aligned}$$

Therefore, using (3.9) (from Lemma 3.10) it follows that

$$\begin{aligned} \left| \sum \limits _{k} \nabla _{ab}^2l( \nabla ^1_n u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x) \phi _{n,k}(x) \right|&\le C\sum \limits _{k\in K_x}\Vert u\Vert _{C^\beta }\phi _{n,k}(x)\\&\le CN \Vert u\Vert _{C^\beta }. \end{aligned}$$

Gathering the last three estimates, we conclude that

$$\begin{aligned} |\nabla ^2_{ab}f(x)|\le C\Vert u\Vert _{C^\beta }(d(x,{\tilde{G}}_n)^{\beta -2}+1). \end{aligned}$$

Moreover, since the indices a, b were arbitrary, and since $d(x,{\tilde{G}}_n)$ is bounded from above for $x \in M {\setminus } {\tilde{G}}_n$ by a constant C, we conclude that

$$\begin{aligned} |\nabla ^2f(x)|\le C\Vert u\Vert _{C^\beta }d(x,{\tilde{G}}_n)^{\beta -2}. \end{aligned}$$

The case $\beta \in [2,3)$. The proof is entirely analogous to the previous case, and we only highlight the overall steps of the proof: as before, we pick a local system of coordinates $\{x_1,\ldots ,x_d\}$ and use the identity

$$\begin{aligned} \nabla ^3_{abc}f(x)&= \sum \limits _{k} \nabla _{abc}^3\left[ \left( u({\hat{y}}_{n,k})+l( \nabla ^1_n u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};x) + q(\nabla ^2_nu ({\hat{y}}_{n,k} ),{\hat{y}}_{n,k};x) \right) \phi _{n,k}(x) \right] , \end{aligned}$$

which holds for any three indices a, b,  and c. The expression on the right may be expanded via Leibniz rule, resulting in terms mixing various derivatives of $\phi _{n,k}$, $l( \nabla ^1_n u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};\cdot )$, and $q(\nabla ^2_n u ({\hat{y}}_{n,k}),{\hat{y}}_{n,k};\cdot )$.

It can then be checked that $\nabla ^{3}_{abc}f(x)$ is given by a sum in k of terms involving $\phi _{n,k}$ and values of u on ${\tilde{G}}_n$ –in a manner analogue to the case $\beta \in [1,2)$. Now, to bound each of the resulting terms we will use (3.13) with $i=1,2$ as before, but this time also with $i=3$. The bounds will follow by applying at difference instances Propositions A.12 and A.15, as well as Lemma A.14. All throughout, we will make us of the fact that the only non-zero terms appearing in the sums are those with $k\in P_{n,k}^*$. At the end, we arrive at the bound,

$$\begin{aligned} |\nabla ^3_{abc}f(x)|\le C\Vert u\Vert _{C^\beta }(d(x,{\tilde{G}}_n)^{\beta -3}+1), \end{aligned}$$

which holds for any choice of the indices a, b and c. This means that

$$\begin{aligned} |\nabla ^3 f(x)|\le C\Vert u\Vert _{C^\beta }d(x,{\tilde{G}}_n)^{\beta -3}, \end{aligned}$$

where we have used again that $d(x,{\tilde{G}}_n)$ is bounded from above for $x \in M {\setminus } {\tilde{G}}_n$. $\square$