A general theorem of existence of quasi absolutely minimal Lipschitz extensions

Abstract

In this paper we consider a wide class of generalized Lipschitz extension problems and the corresponding problem of finding absolutely minimal Lipschitz extensions. We prove that if a minimal Lipschitz extension exists, then under certain other mild conditions, a quasi absolutely minimal Lipschitz extension must exist as well. Here we use the qualifier “quasi” to indicate that the extending function in question nearly satisfies the conditions of being an absolutely minimal Lipschitz extension, up to several factors that can be made arbitrarily small.

Appendices

Appendix A: Equivalence of AMLE definitions

In this appendix we prove that the two definitions for an AMLE with a generalized functional $\varPhi$ are equivalent so long as the domain $(\mathbb {X},d_{\mathbb {X}})$ is path connected. First recall the two definitions:

Definition 7

Let $f \in \fancyscript{F}_{\varPhi }(\mathbb {X},Z)$ with $\mathrm {dom}(f)$ closed and let $U \in \fancyscript{F}_{\varPhi }(\mathbb {X},Z)$ be a minimal extension of $f$ with $\mathrm {dom}(U) = \mathbb {X}$. Then $U$ is an absolutely minimal Lipschitz extension of $f$ if for every open set $V \subset \mathbb {X}{\setminus } \mathrm {dom}(f)$ and every $\widetilde{U} \in \fancyscript{F}_{\varPhi }(\mathbb {X},Z)$ with $\mathrm {dom}(\widetilde{U}) = \mathbb {X}$ that coincides with $U$ on $\mathbb {X}{\setminus } V$,

$$\begin{aligned} \varPhi (U;V) \le \varPhi (\widetilde{U};V). \end{aligned}$$

Definition 8

Let $f \in \fancyscript{F}_{\varPhi }(\mathbb {X},Z)$ with $\mathrm {dom}(f)$ closed and let $U \in \fancyscript{F}_{\varPhi }(\mathbb {X},Z)$ be a minimal extension of $f$ with $\mathrm {dom}(U) = \mathbb {X}$. Then $U$ is an absolutely minimal Lipschitz extension of $f$ if

$$\begin{aligned} \varPhi (U; V) = \varPhi (U; \partial V), \quad \text {for all open } V \subset \mathbb {X}{\setminus } \mathrm {dom}(f). \end{aligned}$$

Proposition 2

Suppose that $(\mathbb {X},d_{\mathbb {X}})$ is path connected. Then Definition 7 is equivalent to Definition 8.

Proof

Since $(\mathbb {X},d_{\mathbb {X}})$ is path connected, the only sets that are both open and closed are $\emptyset$ and $\mathbb {X}$. Let $V \subset \mathbb {X}{\setminus } \mathrm {dom}(f)$. The case $V = \emptyset$ is vacuous for both definitions, and since $\mathrm {dom}(f) \ne \emptyset$, the case $V = \mathbb {X}$ is impossible. Thus every open set $V \subset \mathbb {X}{\setminus } \mathrm {dom}(f)$ is not also closed; in particular, $\partial V \ne \emptyset$.

We first prove that Definition 7 implies Definition 8. Let $U \in \fancyscript{F}_{\varPhi }(\mathbb {X},Z)$ be an AMLE for $f$ satisfying the condition of Definition 7, and suppose by contradiction that $U$ does not satisfy the condition of Definition 8. That would mean, in particular, that there exists an open set $V \subset \mathbb {X}{\setminus } \mathrm {dom}(f)$ such that $\varPhi (U; \partial V) < \varPhi (U; V)$. We can then define a new minimal extension $\widetilde{U} \in \fancyscript{F}_{\varPhi }(\mathbb {X},Z)$ as follows:

$$\begin{aligned} \widetilde{U}(x) \triangleq \left\{ \begin{array}{l@{\quad }l} H(U;V)(x), &{} \text {if } x \in V, \\ U(x), &{} \text {if } x \in \mathbb {X}{\setminus } V, \end{array} \right. \end{aligned}$$

where $H$ is the correction operator defined in Definition 6. But then $\widetilde{U}$ coincides with $U$ on $\mathbb {X}{\setminus } V$ and $\varPhi (\widetilde{U}; V) = \varPhi (U; \partial V) < \varPhi (U; V)$, which is a contradiction.

For the converse, suppose $U \in \fancyscript{F}_{\varPhi }(\mathbb {X},Z)$ satisfies Definition 8 but does not satisfy Definition 7. Then there exists and open set $V \subset \mathbb {X}{\setminus } \mathrm {dom}(f)$ and a function $\widetilde{U} \in \fancyscript{F}_{\varPhi }(\mathbb {X},Z)$ with $\mathrm {dom}(\widetilde{U}) = \mathbb {X}$ that coincides with $U$ on $\mathbb {X}{\setminus } V$ such that $\varPhi (\widetilde{U}; V) < \varPhi (U; V)$. Since $U$ and $\widetilde{U}$ coincide on $\mathbb {X}{\setminus } V$, $\varPhi (\widetilde{U}; \partial V) = \varPhi (U; \partial V)$. On the other hand, $\varPhi (\widetilde{U}; \partial V) \le \varPhi (\widetilde{U};V) < \varPhi (U;V) = \varPhi (U;\partial V)$. Thus we have a contradiction. $\square$

Appendix B: Proof that $(P_0)$–$(P_5)$ hold for 1-fields

In this appendix we consider the case of $1$-fields and the functional $\varPhi = \Gamma ^1$ first defined in Sect. 2.4.4. Recall that $(\mathbb {X},d_{\mathbb {X}}) = \mathbb {R}^d$ with $d_{\mathbb {X}}(x,y) = \Vert x-y\Vert$, where $\Vert \cdot \Vert$ is the Euclidean distance. The range $(Z,d_Z)$ is taken to be $\fancyscript{P}^1(\mathbb {R}^d,\mathbb {R})$, with elements $P \in \fancyscript{P}^1(\mathbb {R}^d,\mathbb {R})$ given by $P(a) = p_0 + D_0p \cdot a$, with $p_0 \in \mathbb {R}$, $D_0p \in \mathbb {R}^d$, and $a \in \mathbb {R}^d$. The distance $d_Z$ is defined as: $d_Z(P,Q) \triangleq |p_0 - q_0| + \Vert D_0p - D_0q\Vert$. For a function $f \in \fancyscript{F}(\mathbb {X},Z)$, we use the notation $x \in \mathrm {dom}(f) \mapsto f(x)(a) = f_x + D_xf \cdot (x-a)$, where $f_x \in \mathbb {R}$, $D_xf \in \mathbb {R}^d$, and once again $a \in \mathbb {R}^d$. Note that $f(x) \in \fancyscript{P}^1(\mathbb {R}^d,\mathbb {R})$. The functional $\varPhi$ is defined as:

$$\begin{aligned} \varPhi (f;x,y) = \Gamma ^1(f;x,y) \triangleq 2 \sup _{a \in \mathbb {R}^d} \frac{|f(x)(a) - f(y)(a)|}{\Vert x-a\Vert ^2 + \Vert y-a\Vert ^2}. \end{aligned}$$

(33)

Rather than $\varPhi$, we shall write $\Gamma ^1$ throughout the appendix. The goal is to show that the properties $(P_0)$–$(P_5)$ hold for $\Gamma ^1$ and the metric spaces $(\mathbb {X},d_{\mathbb {X}})$ and $(Z,d_Z)$.

1.1 B.1 $(P_0)$ and $(P_1)$ for $\Gamma ^1$

The property $(P_0)$ (symmetry and nonnegative) is clear from the definition of $\Gamma ^1$ in (33). The property $(P_1)$ (pointwise evaluation) is by definition.

1.2 B.2 $(P_2)$ for $\Gamma ^1$

The property $(P_2)$(existence of a minimal extension to $\mathbb {X}$ for each $f \in \fancyscript{F}_{\Gamma ^1}(\mathbb {X},Z)$) is the main result of [12]. We refer the reader to that paper for the details.

1.3 B.3 $(P_3)$ for $\Gamma ^1$

Showing property $P_3$, Chasles’ inequality, requires a detailed study of the domain of uniqueness for a biponctual 1-field (i.e., when $\mathrm {dom}(f)$ consists of two points). Let $\fancyscript{P}^m(\mathbb {R}^d,\mathbb {R})$ denote the space of polynomials of degree $m$ mapping $\mathbb {R}^d$ to $\mathbb {R}$.

For $f \in \fancyscript{F}_{\Gamma ^1}(\mathbb {X},Z)$ and $x,y \in \mathrm {dom}(f)$, $x \ne y$ we define

$$\begin{aligned} A(f;x,y) \triangleq \frac{2(f_x-f_y)+ (D_xf+D_yf) \cdot (y-x)}{\Vert x-y \Vert ^2} \end{aligned}$$

and

$$\begin{aligned} B(f;x,y) \triangleq \frac{\Vert D_xf-D_yf\Vert }{\Vert x-y \Vert }. \end{aligned}$$

(34)

Using [12], Proposition 2.2, we have for any $D \subset \mathrm {dom}(f)$,

$$\begin{aligned} \Gamma ^1(f;D) = \sup _{\begin{array}{c} x,y \in D \\ x \ne y \end{array}} \left( \sqrt{A(f;x,y)^2+B(f;x,y)^2} +|A(f;x,y)|\right) . \end{aligned}$$

(35)

For the remainder of this section, fix $f \in \fancyscript{F}_{\Gamma ^1}(\mathbb {X},Z)$, with $\mathrm {dom}(f) =\{x,y\}$, $x \ne y$, $f(x) = P_x$, $f(y) = P_y$, and set

$$\begin{aligned} M \triangleq \Gamma ^1 (f;\mathrm {dom}(f)). \end{aligned}$$

(36)

Also, for an arbitrary pair of points $a,b \in \mathbb {R}^d$, let $[a,b]$ denote the closed line segment with end points $a$ and $b$.

Proposition 3

Let $F$ be an extension of $f$ such that $\overline{B}_{1/2}(x,y) \subset \mathrm {dom}(F)$. Then there exists a point $c \in \overline{B}_{1/2}(x,y)$ that depends only on $f$ such that

$$\begin{aligned} \Gamma ^1(F;x,y) \le \max \{\Gamma ^1(F;x,a), \Gamma ^1(F;a,y)\}, \text { for all } a \in [x,c] \cup [c,y]. \end{aligned}$$

(37)

Remark 4

Proposition 3 implies that the operator $\Gamma ^1$ satisfies the Chasles’ inequality (property $(P_3)$). In particular, consider an arbitrary $1$-field $g \in \fancyscript{F}_{\Gamma ^1}(\mathbb {X},Z)$ with $x,y \in \mathrm {dom}(g)$ such that $\overline{B}_{1/2}(x,y) \subset \mathrm {dom}(g)$. Then $g$ is trivially an extension of the $1$-field $g|_{\{x,y\}}$, and so in particular satisfies (37). But this is the Chasles’ inequality with $\gamma = [x,c] \cup [c,y]$.

To prove proposition 3 we will use the following lemma.

Lemma 7

There exists $c \in \overline{B}_{1/2}(x,y)$ and $s \in \{-1,1\}$ such that

$$\begin{aligned} M = 2s\dfrac{ P_x(c)- P_y(c) }{\Vert x-c \Vert ^2 + \Vert y-c \Vert ^2}. \end{aligned}$$

Furthermore,

$$\begin{aligned} \begin{aligned}&c = \dfrac{x+y}{2} + s\dfrac{D_xf-D_yf}{2M}, \\&P_x(c) -s \dfrac{M}{2}\Vert x-c\Vert ^2 = P_y(c)+s\dfrac{M}{2}\Vert y-c\Vert ^2, \\&D_xf+sM (x-c) = D_yf-sM (y-c). \end{aligned} \end{aligned}$$

(38)

Moreover, all minimal extensions of $f$ coincide at $c$.

The proof of Lemma 7 uses [12], Propositions 2.2 and 2.13. The details are omitted. Throughout the remainder of this section, let $c$ denote the point which satisfies Proposition 7.

Lemma 8

Define $\widetilde{P}_c \in \fancyscript{P}^1(\mathbb {R}^d,\mathbb {R})$ as

$$\begin{aligned} \widetilde{P}_{c}(z) \triangleq \tilde{f}_c + D_c\tilde{f} \cdot (z-c), z \in \mathbb {R}^d, \end{aligned}$$

where

$$\begin{aligned} \tilde{f}_c \triangleq P_x(c)-s\dfrac{M}{2}\Vert x - c\Vert ^2 , \end{aligned}$$

and

$$\begin{aligned} D_c\tilde{f} \triangleq D_xf+sM(x-c). \end{aligned}$$

If $A(f;x,y)=0$, then the following polynomial

$$\begin{aligned} F(z) \triangleq \widetilde{P}_{c}(z) -s\dfrac{M}{2} \dfrac{[(z-c) \cdot (x-c)]^2}{\Vert x-c \Vert ^2} +s\dfrac{M}{2} \dfrac{[(z-c) \cdot (y-c)]^2}{\Vert y-c\Vert ^2}, z \in \mathbb {R}^d \end{aligned}$$

is a minimal extension of $f$.

If $A(f;x,y) \ne 0$, let $z \in \mathbb {R}^d$ and set $p(z) \triangleq (x-c) \cdot (z-c)$ and $q(z) \triangleq (y-c) \cdot (z-c)$. We define

$$\begin{aligned}&F(z) \nonumber \\&\quad \triangleq \left\{ \begin{array}{l@{\quad }l} \widetilde{P}_{c}(z) \!-\! s\dfrac{M}{2} \dfrac{[(z \!-\! c) \cdot (x \!-\! c)]^2}{\Vert x \!-\! c \Vert ^2}, &{} \text {if } p(z) \ge 0 \text { and } q(z) \!\le \! 0, \\ \widetilde{P}_{c}(z) \!+\! s\dfrac{M}{2} \dfrac{[(z \!-\! c) \cdot (y \!-\! c)]^2}{\Vert y \!-\! c \Vert ^2}, &{} \text {if } p(z) \le 0 \text { and } q(z) \!\ge \! 0, \\ \widetilde{P}_{c}(z), &{} \text {if } p(z) \!\le \! 0 \text { and } q(z) \!\le \! 0, \\ \widetilde{P}_{c}(z) \!-\! s\dfrac{M}{2} \dfrac{[(z \!-\! c) \cdot (x \!-\! c)]^2}{\Vert x \!-\! c \Vert ^2} \!+\! s\dfrac{M}{2} \dfrac{[(z \!-\! c) \cdot (y \!-\! c)]^2}{\Vert y \!-\! c\Vert ^2}, &{} \text {if } p(z) \!\ge \! 0 \text { and } q(z) \!\ge \! 0. \end{array} \right. \end{aligned}$$

Then $F$ is a minimal extension of $f$.

Remark 5

The function $F$ is an extension of the $1$-field $f$ in the following sense. $F$ defines a $1$-field via its first order Taylor polynomials; in particular, define the $1$-field $U$ with $\mathrm {dom}(U) = \mathrm {dom}(F)$ as:

$$\begin{aligned} U(a) \triangleq J_aF, a \in \mathrm {dom}(F), \end{aligned}$$

where $J_aF$ is the first order Taylor polynomial of $F$. We then have:

$$\begin{aligned}&U(x) = f(x) \quad \text {and} \quad U(y) = f(y), \\&\Gamma ^1(U;\mathrm {dom}(U)) = \Gamma ^1(f;\mathrm {dom}(f)). \end{aligned}$$

Proof

After showing that the equality $A(f;x,y)=0$ implies that $(x-c) \cdot (c-y) = 0$, the proof is easy to check. Suppose that $A(f;x,y)=0$. By (33) and (36) we have $M=B(f;x,y)$. By (38) we have

$$\begin{aligned} \Vert 2c-(x+y)\Vert = \dfrac{\Vert D_xf-D_yf\Vert }{M} = \Vert x-y\Vert . \end{aligned}$$

Therefore $(x-c) \cdot (c-y) = 0$. $\square$

The proof of the following lemma is also easy to check.

Lemma 9

Let $g \in \fancyscript{F}_{\Gamma ^1}(\mathbb {X},Z)$ such that for all $a \in \mathrm {dom}(g)$, $g(a) = Q_a \in \fancyscript{P}^1(\mathbb {R}^d,\mathbb {R})$, with $Q_a(z) = g_a + D_ag \cdot (z-a)$, where $g_a \in \mathbb {R}$, $D_ag \in \mathbb {R}^d$, and $z \in \mathbb {R}^d$. Suppose there exists $P \in \mathcal {P}^2(\mathbb {R}^d,\mathbb {R})$ such that

$$\begin{aligned} P(a) =g_a, \nabla P(a) = D_ag, \text { for all } a \in \mathrm {dom}(g). \end{aligned}$$

Then

$$\begin{aligned} A(g;a,b) = 0, \text { for all } a,b \in \mathrm {dom}(g). \end{aligned}$$

Proof

Omitted. $\square$

Lemma 10

All minimal extensions of $f$ coincide on the line segments $[x,c]$ and $[c, y]$.

Proof

First, let $F$ be the minimal extension of $f$ defined in Lemma 8, and let $U$ be the $1$-field corresponding to $F$ that was defined in remark 5. In particular, recall that we have:

$$\begin{aligned} U(a)(z) = J_aF(z) = F(a) + \nabla F(a) \cdot (z-a), a \in \mathrm {dom}(F). \end{aligned}$$

Now Let $W$ be an arbitrary minimal extension of $f$ such that for all $a \in \mathrm {dom}(W)$, $W(a) = Q_a \in \fancyscript{P}^1(\mathbb {R}^d,\mathbb {R})$, with $Q_a(z) = W_a + D_aW \cdot (z-a)$, where $W_a \in \mathbb {R}$, $D_aW \in \mathbb {R}^d$, and $z \in \mathbb {R}^d$. We now restrict our attention to $[x,c] \cup [c,y]$. For any $a \in [x,c] \cup [c,y]$, we write $W(a) = Q_a$ in the following form:

$$\begin{aligned} Q_a(z) = F(a) + \nabla F(a) \cdot (z-a) + \delta _a + \Delta _a \cdot (z-a), z \in \mathbb {R}^d, \end{aligned}$$

where $\delta _a \in \mathbb {R}$ and $\Delta _a \in \mathbb {R}^d$. In particular, we have

$$\begin{aligned} W_a&= F(a) + \delta _a, \\ D_aW&= \nabla F(a) + \Delta _a. \end{aligned}$$

Since $U$ is a minimal extension of $f$, it is enough to show that $\delta _a=0$ and $\varDelta _a = 0$ for $a \in [x,c] \cup [c,y]$. By symmetry, without lost generality let us suppose that $a \in [x,c]$. Since $W$ is a minimal extension of $f$, we have $W_x= F(x) = f_x$, and by Lemma 7, $W_c= F(c)$. Using (35) and (36), and once again since $W$ is a minimal extension of $f$, the following inequality must be satisfied:

$$\begin{aligned} |A(W;e,a)| + \dfrac{B(W;e,a)^2}{2M} \le \dfrac{M}{2}, \quad e \in \{x,c\}. \end{aligned}$$

(39)

Using Lemma 9 for $U$ restricted to $\{x,a,c\}$ we have

$$\begin{aligned} A(U;e,a)= 0, \quad e \in \{x,c\}. \end{aligned}$$

(40)

Therefore

$$\begin{aligned} A(W;e,a) = \dfrac{|-2\delta _a+ \Delta _a \cdot (e-a)|}{\Vert e-a\Vert ^2}, \quad e \in \{x,c\}. \end{aligned}$$

(41)

Since $a \in [x,c]$, we can write $a = c +\alpha (x-c)$ with $\alpha \in [0,1]$. Using (39) and (40), the definition of $U$, and after simplification, $\delta _a$ and $\varDelta _a$ must satisfy the following inequalities:

$$\begin{aligned} -2\delta _a + \alpha (1+s) \Delta _a \cdot (c-x) +\dfrac{\Vert \Delta _a\Vert ^2}{2M}&\le 0, \end{aligned}$$

(42)

$$\begin{aligned} 2\delta _a + \alpha (-1+s) \Delta _a \cdot (c-x) + \dfrac{\Vert \Delta _a\Vert ^2}{2M}&\le 0, \end{aligned}$$

(43)

$$\begin{aligned} -2\delta _a - (1- \alpha )(1+s) \Delta _a \cdot (c-x) + \dfrac{\Vert \Delta _a\Vert ^2}{2M}&\le 0, \end{aligned}$$

(44)

$$\begin{aligned} 2\delta _a- (1- \alpha )(-1+s) \Delta _a \cdot (c-x) + \dfrac{\Vert \Delta _a\Vert ^2}{2M}&\le 0. \end{aligned}$$

(45)

The inequality $(1-\alpha )((42)+ (43))+ \alpha ((44)+ (45))$ implies that $\varDelta _a =0$. Furthermore, the inequalities (42) and (43) imply that $\delta _a =0$. Now the proof is complete. $\square$

We finish this appendix by proving Proposition 3. Let us use the notations of Proposition 3 where $c$ satisfies Lemma 7. By Lemma 10, the extension $U$ (defined in Remark 5) of $f$ is the unique minimal extension of $f$ on the restriction to $[x,c] \cup [c,y]$. Moreover, we can check that

$$\begin{aligned} \Gamma ^1(f;x,y) = \max \{\Gamma ^1(U;x,a), \Gamma ^1(U;a,y)\}, \text { for all } a \in [x,c] \cup [c,y]. \end{aligned}$$

(46)

Let $W$ be an extension of $f$. By contradiction suppose that there exists $a \in [x,c] \cup [c,y]$ such that

$$\begin{aligned} \Gamma ^1(f;x,y) > \max \{\Gamma ^1(W;x,a), \Gamma ^1(W;a,y)\}. \end{aligned}$$

(47)

Using [12], Theorem 2.6, for the $1$-field $g \triangleq \{f(x),W(a),f(y)\}$ of domain $\{x,a,y\}$ there exists an extension $G$ of $g$ such that

$$\begin{aligned} \Gamma ^1(G;\mathrm {dom}(G)) \le \Gamma ^1(f;x,y). \end{aligned}$$

(48)

Therefore $G$ is a minimal extension of $f$. By Lemma (10) and the definition of $G$ we have $W(a) = G(a) = U(a)$. But then by (46),  (47), and (48) we obtain a contradiction. Now the proof of the Proposition 3 is complete.

1.4 B.4 $(P_4)$ for $\Gamma ^1$

Property $(P_4)$ (continuity of $\Gamma ^1$) can be shown using (35), and a series of elementary calculations. We omit the details.

1.5 B.5 $(P_5)$ for $\Gamma ^1$

To show property $(P_5)$ (continuity of $f \in \fancyscript{F}_{\Gamma ^1}(\mathbb {X},Z)$), we first recall the definition of $d_Z$. For $P \in \fancyscript{P}^1(\mathbb {R}^d,\mathbb {R})$ with $P(a) = p_0 + D_0p \cdot a$, $p_0 \in \mathbb {R}$, $D_0p \in \mathbb {R}^d$, we have

$$\begin{aligned} d_Z(P,Q) = |p_0 - q_0| + \Vert D_0p - D_0q\Vert . \end{aligned}$$

Recall also that for a $1$-field $f: E \rightarrow Z$, $E \subset \mathbb {X}$, we have:

$$\begin{aligned} x \in E \mapsto f(x)(a) = f_x + D_xf \cdot (a-x) = (f_x - D_xf \cdot x) + D_xf \cdot a, \quad a \in \mathbb {R}^d. \end{aligned}$$

To show continuity of $f \in \fancyscript{F}_{\Gamma ^1}(\mathbb {X},Z)$ at $x \in \mathbb {X}$, we need the following: for all $\varepsilon > 0$, there exists a $\delta > 0$ such that if $\Vert x-y\Vert < \delta$, then $d_Z(f(x),f(y)) < \varepsilon$. Consider the following:

$$\begin{aligned} d_Z(f(x),f(y))&= |f_x - D_xf \cdot x - f_y + D_yf \cdot y| + \Vert D_xf - D_yf\Vert \nonumber \\&\le |f_x - f_y| + |D_xf \cdot x - D_yf \cdot y| + \Vert D_xf - D_yf\Vert . \end{aligned}$$

(49)

We handle the three terms (49) separately and in reverse order.

For the third term, recall the definition of $B(f;x,y)$ in (34), and define $B(f;E)$ accordingly; we then have:

$$\begin{aligned} \Vert D_xf - D_yf\Vert \le B(f;E) \Vert x-y\Vert \le \Gamma ^1(f;E) \Vert x-y\Vert . \end{aligned}$$

(50)

Since $\Gamma ^1(f;E) < \infty$, that completes this term.

For the second term:

$$\begin{aligned} |D_xf \cdot x - D_yf \cdot y|&\le |D_xf \cdot (x-y)| + |(D_xf - D_yf) \cdot y| \\&\le \Vert D_xf\Vert \Vert x-y\Vert + \Vert D_xf - D_yf\Vert \Vert y\Vert \end{aligned}$$

Using (50), we see that this term can be made arbitrarily small using $\Vert x-y\Vert$ as well.

For the first term $|f_x - f_y|$, define $g: E \rightarrow \mathbb {R}$ as $g(x) = f_x$ for all $x \in E$. By Proposition 2.5 of [12], the function $g$ is continuous. This completes the proof.$\square$

Appendix C: Proof of Proposition 1

We prove Proposition 1, which we restate here:

Proposition 4

(Proposition 1) Let $f \in \fancyscript{F}_{\varPhi }(\mathbb {X},Z)$. For any open $V \subset \mathrm {dom}(f)$, $V \ne \mathbb {X}$, and $\alpha \ge 0$, let us define

$$\begin{aligned} V_{\alpha } \triangleq \{ x \in V \mid d_{\mathbb {X}}(x,\partial V) \ge \alpha \}. \end{aligned}$$

Then for all $\alpha > 0$,

$$\begin{aligned} \varPhi (f;V_{\alpha }) \le \max \{\varPsi (f;V;\alpha ), \varPhi (u;\partial V_{\alpha })\}, \end{aligned}$$

(51)

and

$$\begin{aligned} \varPhi (f;V) = \max \{\varPsi (f;V;0), \varPhi (f;\partial V)\}. \end{aligned}$$

(52)

Proof

For the first statement fix $\alpha >0$ and an open set $V \subset \mathbb {X}$, $V \ne \mathbb {X}$. For proving (51), it is sufficient to prove that for all $x \in \mathring{V}_{\alpha }$ and for all $y \in V_{\alpha }$ we have

$$\begin{aligned} \varPhi (f;x,y) \le \max \{\varPsi (f;V;\alpha ), \varPhi (f;\partial V_{\alpha })\}. \end{aligned}$$

(53)

Fix $x \in \mathring{V}_{\alpha }$. Let $B(x;r_{x}) \subset V$ be a ball such that $r_{x}$ is maximized and define

$$\begin{aligned} M(x) \triangleq \sup \left\{ \varPhi (f;x,y) \mid y \in \overline{V_{\alpha } {\setminus } B(x;r_{x})} \right\} , \end{aligned}$$

as well as

$$\begin{aligned} \varDelta (x) \triangleq \left\{ y \in \overline{V_{\alpha } {\setminus } B(x;r_{x})} \mid \varPhi (f;x,y) = M(x) \right\} , \end{aligned}$$

and

$$\begin{aligned} \delta (x) \triangleq \inf \{d_{\mathbb {X}}(x,y) \mid y\in \Delta (x) \}. \end{aligned}$$

We have three cases:

Case 1 Suppose $M(x) \le \sup \{\varPhi (f;x,y) \mid y \in B(x;r_{x}) \}$. Since $B(x;r_{x}) \subset V$ with $r_{x} \ge \alpha$ we have

$$\begin{aligned} \varPhi (f;x,y) \le \varPsi (f;V;\alpha ), \forall \, y \in B(x;r_{x}). \end{aligned}$$

Therefore $M(x) \le \varPsi (f;V;\alpha )$. That completes the first case.

For cases two and three, assume that $M(x) > \sup \{ \varPhi (f;x,y) \mid y \in B(x;r_x)\}$ and select $y \in \Delta (x)$ with $d_{\mathbb {X}}(x,y) = \delta (x)$.

Case 2 Suppose $y \in \mathrm {int}(V_{\alpha }{\setminus } B(x;r_{x}))$. Let $B(y;r_y)\subset V$ be a ball such that $r_y$ is maximal. Consider the curve $\gamma \in \Gamma (x,y)$ satisfying $(P_3)$. Let $m \in \gamma \cap B(y;r_y) \cap V_{\alpha }$, $m \ne x,y$. Using $(P_3)$, we have

$$\begin{aligned} \varPhi (f;x,y) \le \max \{\varPhi (f;x,m), \varPhi (f;m,y)\}. \end{aligned}$$

(54)

Using the monotonicity of $\gamma$ we have $d_{\mathbb {X}}(x,m) < d_{\mathbb {X}}(x,y)$. Using the minimality of the distance of $d_{\mathbb {X}}(x,y)$ and since $m \in V_{\alpha }$ we have $\varPhi (f;x,m) < \varPhi (f;x,y)$. Therefore

$$\begin{aligned} \varPhi (f;x,y) \le \varPhi (f;m,y). \end{aligned}$$

(55)

Since $m \in B(y;r_y)$ with $r_y \ge \alpha$, using the definition of $\varPsi$ we have $\varPhi (f;m,y) \le \varPsi (f;V;\alpha )$. Therefore $M(x) \le \varPsi (f;V;\alpha )$.

Case 3 Suppose $y \in \partial V_{\alpha }{\setminus } B(x;r_{x})$. As in case two, let $B(y;r_y)\subset V$ be a ball such that $r_y$ is maximal and consider the curve $\gamma \in \Gamma (x,y)$ satisfying $(P_3)$. Let $m \in \gamma \cap B(y;r_y) \cap V_{\alpha }$. If there exists $m \ne y$ in $V_{\alpha }$, we can apply the same reasoning as in case two and we have $M(x) \le \varPsi (f;V;\alpha )$.

If $m = y$ is the only element of $\gamma \cap B(y;r_y) \cap V_{\alpha }$, then there still exists $m' \in \gamma \cap \partial V_{\alpha }$ with $m' \ne y$. Using $(P_3)$ we have

$$\begin{aligned} \varPhi (f;x,y) \le \max \{\varPhi (f;x,m'), \varPhi (f;m',y)\}. \end{aligned}$$

(56)

Using the monotonicity of $\gamma$ we have $d_{\mathbb {X}}(x,m') < d_{\mathbb {X}}(x,y)$. Using the minimality of distance of $d_{\mathbb {X}}(x,y)$ and since $m' \in V_{\alpha }$ we have $\varPhi (f;x,m') < \varPhi (f;x,y)$. Therefore

$$\begin{aligned} \varPhi (f;x,y) \le \varPhi (f;m',y). \end{aligned}$$

(57)

Since $m',y \in \partial V_{\alpha }$, we obtain the following majoration

$$\begin{aligned} \varPhi (f;x,y) \le \varPhi (f;m',y) \le \varPhi (f;\partial V_{\alpha }), \end{aligned}$$

(58)

which in turn gives:

$$\begin{aligned} M(x) \le \varPhi (f;\partial V_{\alpha }). \end{aligned}$$

The inequality (51) is thus demonstrated.

For the second statement, we note that by the definition of $\varPsi$ we have

$$\begin{aligned} \max \{\varPsi (f;V;0), \varPhi (f;\partial V)\}\le \varPhi (f;V). \end{aligned}$$

Using (51), to show (52) it is sufficient to prove

$$\begin{aligned} \lim _{\alpha \rightarrow 0}\varPhi (f;V_{\alpha }) =\varPhi (f;V). \end{aligned}$$

(59)

Let $\varepsilon >0$. Then there exists $x_{\varepsilon } \in V$ and $y_{\varepsilon } \in \overline{V}$ such that

$$\begin{aligned} \varPhi (f;V) \le \varPhi (f;x_{\varepsilon },y_{\varepsilon })+ \varepsilon . \end{aligned}$$

Set $r_{\varepsilon }=d_{\mathbb {X}}(x_{\varepsilon },\partial V)$. If $y_{\varepsilon } \in V$, there exists $\tau _1$ with $0 < \tau _1 \le r_{\varepsilon }$ such that for all $\alpha$, $0 < \alpha \le \tau _1$, $(x_{\varepsilon },y_{\varepsilon }) \in V_{\alpha }\times V_{\alpha }$. Therefore

$$\begin{aligned} \varPhi (f;x_{\varepsilon },y_{\varepsilon }) \le \varPhi (f;V_{\alpha }), \quad \forall \, \alpha , 0 < \alpha \le \tau _1. \end{aligned}$$

If, on the other hand, $y_{\varepsilon } \in \partial V$, using $(P_4)$ there exists $\tau _2$ with $0 < \tau _2 \le \min \{r_{\varepsilon },\tau _1\}$, such that

$$\begin{aligned} |\varPhi (f;x_{\varepsilon },m)-\varPhi (f;x_{\varepsilon },y_{\varepsilon })| \le \varepsilon , \quad \forall \, m \in B(y_{\varepsilon }; \tau _2). \end{aligned}$$

By choosing $m \in B(y_{\varepsilon };\tau _2) \cap V_{\tau _2}$, we obtain

$$\begin{aligned} \varPhi (f;x_{\varepsilon },y_{\varepsilon })\le \varPhi (f;V_{\alpha })+\varepsilon , \quad \forall \, \alpha , 0< \alpha \le \tau _2. \end{aligned}$$

Therefore $\varPhi (f;V) \le \varPhi (f;V_{\alpha }) + 2\varepsilon$, for all $\alpha$ such that $0< \alpha \le \tau _2$ and for all $\varepsilon > 0$. Thus (59) is true. $\square$