16118 LABEL:LastPageApr. 24, 2018Feb. 17, 2020

*This paper is a major extension of [22].

# McShane-Whitney extensions in constructive analysis

Iosif Petrakis  Mathematics Institute, Ludwig-Maximilians-Universität München
<EMAIL_ADDRESS>
###### Abstract.

Within Bishop-style constructive mathematics we study the classical McShane-
Whitney theorem on the extendability of real-valued Lipschitz functions
defined on a subset of a metric space. Using a formulation similar to the
formulation of McShane-Whitney theorem, we show that the Lipschitz real-valued
functions on a totally bounded space are uniformly dense in the set of
uniformly continuous functions. Through the introduced notion of a McShane-
Whitney pair we describe the constructive content of the original McShane-
Whitney extension and we examine how the properties of a Lipschitz function
defined on the subspace of the pair extend to its McShane-Whitney extensions
on the space of the pair. Similar McShane-Whitney pairs and extensions are
established for Hölder functions and $\nu$-continuous functions, where $\nu$
is a modulus of continuity. A Lipschitz version of a fundamental corollary of
the Hahn-Banach theorem, and the approximate McShane-Whitney theorem are
shown.

###### Key words and phrases:

constructive analysis, Lipschitz analysis, McShane-Whitney extension

## Introduction

A central theme in mathematical analysis, classical or constructive, is the
extension of a continuous function $f$, defined on a subspace $A$ of a space
$X$ and satisfying some property $P$, to a continuous function $f^{*}$ defined
on $X$ and satisfying the property $P$. A “general extension theorem” is a
statement providing conditions on $X$ and $A$ such that such an extension of
$f$ to $f^{*}$ is possible111Note that the general pattern of an extension
theorem is not to bestow extra properties upon $f$.. As it is noted in [11],
p. 149, “general extension theorems are rare and usually have interesting
topological consequences”.

If $X$ is a normal (and Hausdorff) topological space, $A$ is a closed subspace
of $X$, the continuous function on $A$ takes values in ${\mathbb{R}}$, and
$P(f)$ is “$f$ is bounded”, the corresponding general extension theorem is
Tietze’s extension theorem, which is actually a characterisation of normality
(see [11], pp. 149-151) and its proof rests on Urysohn’s lemma. If $X$ is a
metric space, hence normal, we get the restriction of the classical Tietze
extension theorem to metric spaces. If $X$ is a complete and separable metric
space, Urysohn’s lemma and Tietze’s extension theorem are provable in
$\mathrm{RCA}_{0}$ (see [23], pp. 90-91). Within Bishop-style constructivism,
which is the framework of constructive mathematics we work within, a Tietze
extension theorem for metric spaces is proved in [4], p. 120. According to it,
if $X$ is a metric space, $A$ is a locally compact222In Bishop-style
constructive analysis a locally compact metric space $X$ is a metric space
such that every bounded subset of $X$ is included in a compact subset of $X$.
subset of $X$, and $f:A\to[a,b]$ is continuous i.e., it is uniformly
continuous on every bounded subset of $A$ (see [4], p. 110), there is an
extension function $f^{*}:X\to[a,b]$ of $f$ on $X$, which is also uniformly
continuous on every bounded subset of $X$. As it is customary in the
constructive treatment of a classical extension theorem, the constructive
theorem is less general, and sometimes the notions involved have to be
understood differently. The benefit of the constructive approach though, is
that the, usually hidden, computational content of the classical theorem is
revealed.

If $X$ is a normed space, $A$ is a linear subspace of $X$ and the property
$P(f)$ for the continuous function $f:A\to{\mathbb{R}}$ is “$f$ is linear and
$||f||=a$”, where $a\in[0,+\infty)$, the corresponding extension theorem is
(one version of) the Hahn-Banach theorem. Its proof is non-constructive, as it
rests on the axiom of choice. Constructively, one needs to add some extra
properties on $X$, in order to avoid the axiom of choice. E.g., in [13] the
normed space $X$ is also uniformly convex, complete and its norm is Gâteaux
differentiable. Since constructively the existence of the norm of the
continuous functional is equivalent to the locatedness of its kernel (see [7],
pp. 53-54), in the formulation of the constructive Hahn-Banach threorem we
need to suppose that the norm $||f||$ of $f$ exists. Then the norm $||f^{*}||$
of the extension function $f^{*}$ exists too, and $||f^{*}||=||f||$. The
computational content of the classical theorem is revealed in the
constructively provable, approximate version of the Hahn-Banach theorem (see
[6], pp. 39-40). According to it, if $X$ is a separable normed space and
$||f||$ exists, then for every $\epsilon>0$ there is a continuous and linear
extension $f^{*}$ of $f$ with a norm $||f^{*}||$ satisfying

$||f^{*}||\leq||f||+\epsilon.$

Hence, in the approximate version of the Hahn-Banach theorem the numerical
part of $P(f)$ is satisfied only approximately.

If $X$ is a metric space, $A$ is any subspace of $X$ and the property $P(f)$
for the (necessarily continuous) function $f:A\to{\mathbb{R}}$ is “$f$ is
Lipschitz”, the corresponding extension theorem is the McShane-Whitney
theorem, which appeared first333To determine metric spaces $X$ and $Y$, such
that a similar extension theorem for $Y$-valued Lipschitz functions defined on
a subset $A$ of $X$ holds, is a non-trivial problem under active current study
(see [2], [5] and [24]). in [19] and [26]. This theorem admits a classical
proof that uses the axiom of choice, and it is very similar to the proof of
the analytic version of the Hahn-Banach theorem (see [9], pp. 224-5).

Following [25], pp.16-17, if $\sigma>0$ and $f$ is $\sigma$-Lipschitz i.e.,
$\forall_{a,a^{\prime}\in A}(|f(a)-f(a^{\prime})|\leq\sigma d(a,a^{\prime}))$,
and if $x\in X\setminus A$, a $\sigma$-Lipschitz extension $g$ of $f$ on
$A\cup\\{x\\}$ must satisfy

$\forall_{a\in A}\big{(}|g(x)-g(a)|=|g(x)-f(a)|\leq\sigma d(x,a)\big{)},$

which is equivalent to

$\forall_{a\in A}\big{(}f(a)-\sigma d(x,a)\leq g(x)\leq f(a)+\sigma
d(x,a)\big{)}.$

Since $\forall_{a,a^{\prime}\in A}\big{(}|f(a)-f(a^{\prime})|\leq\sigma
d(a,a^{\prime})\leq\sigma(d(a,x)+d(x,a^{\prime}))\big{)}$, we get

$\forall_{a,a^{\prime}\in A}\big{(}f(a)-\sigma d(a,x)\leq f(a^{\prime})+\sigma
d(x,a^{\prime})\big{)}.$

Hence, if we fix some $a\in A$ (the case $A=\emptyset$ is trivial), then for
all $a^{\prime}\in A$ we get

$f(a)-\sigma d(a,x)\leq f(a^{\prime})+\sigma d(x,a^{\prime}),$

therefore, using the classical completeness property of real numbers, we get

$f(a)-\sigma d(x,a)\leq\inf\\{f(a^{\prime})+\sigma d(x,a^{\prime})\mid
a^{\prime}\in A\\}.$

Since $a$ is an arbitrary element of $A$, we conclude that

$s_{x}\equiv\sup\\{f(a)-\sigma d(x,a)\mid a\in A\\}\leq\inf\\{f(a)+\sigma
d(x,a)\mid a\in A\\}\equiv i_{x}.$

If we define $g(x)$ to be any value in the interval $[s_{x},i_{x}]$, we get
the required extension of $f$ on $A\cup\\{x\\}$. If we apply Zorn’s lemma on
the non-empty poset

$\\{(B,h)\mid B\supseteq A\ \&\ h:B\to{\mathbb{R}}\ \mbox{is}\
\sigma\mbox{-Lipschitz}\ \&\ h_{|A}=f\\},$

where

$(B_{1},h_{1})\leq(B_{2},h_{2}):\Leftrightarrow B_{1}\subseteq B_{2}\ \&\
{h_{2}}_{|B_{1}}=h_{1},$

its maximal element is the required $\sigma$-Lipschitz extension of $f$.

What McShane and Whitney independently observed though, was that if one
defines $g^{*}(x)=s_{x}$ and ${}^{*}g(x)=i_{x}$, for all $x\in X$, the
resulting functions $g^{*}$ and ${}^{*}g$ on $X$ are $\sigma$-Lipschitz
extensions of $f$, and, as expected by the above possible choice of $g(x)$ in
$[s_{x},i_{x}]$, every other $\sigma$-Lipschitz extension of $f$ is between
$g^{*}$ and ${}^{*}g$. In this way, the McShane-Whitney explicit construction
of the extensions $g^{*}$ and ${}^{*}g$ of $f$ avoids Zorn’s lemma.

If $X$ is a metric space and $A$ is a subset of $X$, we say that $(X,A)$ is a
McShane-Whitney pair, if the functions $g^{*}$ and ${}^{*}g$ are well-defined
(Definition 2.1). Classically, all pairs $(X,A)$, where $A\neq\emptyset$, are
McShane-Whitney pairs. The definitions of $g^{*}$ and ${}^{*}g$ though, depend
on the existence of the supremum and infimum of non-empty, bounded above and
bounded below subsets of reals, respectively. Since the classical completeness
property of reals implies constructively the principle of the excluded middle
$(\mathrm{PEM})$ (see [7], p. 32), it is not a surprise that with a similar
argument we can show that the statement “every pair $(X,A)$ is a McShane-
Whitney pair” implies constructively the principle $\mathrm{PEM}$ (Remark
2.1).

In this paper we study the McShane-Whitney extension within Bishop’s informal
system of constructive mathematics $\mathrm{BISH}$ (see [3], [4], [6] and
[7]). A formal system for $\mathrm{BISH}$ is Myhill’s system $\mathrm{CST}$
with dependent choice (see [20]), or Aczel’s system of constructive set theory
$\mathrm{CZF}$ with dependent choice (see [1]).

The main results included here reveal, in our view, the deep interactions
between the McShane-Whitney theorem and the Hahn-Banach theorem. We can
summarise our results as follows:

  * •

If $X$ is a totally bounded metric space, we prove the uniform density of the
Lipschitz functions on $X$ in the set of uniformly continuous functions on $X$
(Corollary 1.1). For that we prove Theorem 1 that has a formulation similar to
the formulation of the McShane-Whitney theorem (Theorem 2.1).

  * •

We introduce the notion of a McShane-Whitney pair that reveals the
constructive content of the classical McShane-Whitney extension in Theorem
2.1.

  * •

We describe how properties of $f$, like boundedness, existence of the
Lipschitz constant $L(f)$ of $f$, linearity, and convexity, extend to $f^{*}$
and ${}^{*}f$ (subsections 2.1 and 3.1).

  * •

We provide conditions on $X$ and/or $Y$ so that $(X,A)$ is a McShane-Whitney
pair, from the constructive point of view (Proposition 2.1).

  * •

We define a local version of a McShane-Whitney pair (subsection 2.2).

  * •

We prove a Lipschitz version of a fundamental corollary of the Hahn-Banach
theorem (Theorem 3.1).

  * •

We show how the McShane-Whitney pairs and the McShane-Whitney construction can
be applied to Hölder functions and $\nu$-continuous functions, where $\nu$ is
a modulus of continuity (subsection 2.3).

  * •

We prove the approximate McShane-Whitney theorem for Lipschitz functions
defined on a finite-dimensional subspace of a normed space (Theorem 3.2). This
is, in our view, the most technically interesting result of this paper.

First we prove some fundamental facts in the constructive theory of Lipschitz
functions, which is quite different from its classical counterpart.

##  1\. Lipschitz functions, constructively

Constructive Lipschitz analysis is quite underdeveloped. Some related results
are found scattered e.g., in [10], [15], [17], [8]. For the sake of
completeness, we include the definitions of some fundamental concepts.

######  Definition .

Let $A\subseteq{\mathbb{R}}$ and $b\in{\mathbb{R}}$. If $A$ is bounded above,
we define

$b\geq A:\Leftrightarrow\forall_{a\in A}(b\geq a),$
$[A):=\\{c\in{\mathbb{R}}\mid c\geq A\\},$ $l:=\sup A:\Leftrightarrow l\geq A\
\&\ \forall_{\epsilon>0}\exists_{a\in A}(a>l-\epsilon),$
$\lambda:=\mathrm{lub}A:\Leftrightarrow\lambda\geq A\ \&\
\forall_{b\in[A)}(b\geq\lambda).$

If $A$ is bounded below, $b\leq A$, $(A]$, $m:=\inf A$, and
$\mu:=\mathrm{glb}A$ are defined in a dual way.

According to the classical completeness of ${\mathbb{R}}$, if
$A\subseteq{\mathbb{R}}$ is inhabited and bounded above, then $\sup A$ exists.
Constructively, this statement implies the principle of the excluded middle
$\mathrm{PEM}$ (i.e., the scheme $P\vee\neg P$) (see [7], p. 32). A similar
argument444This argument goes as follows. If
$A:=\\{0\\}\cup\\{x\in{\mathbb{R}}\mid x=1\ \&\ P\\}$ and
$\lambda:=\mathrm{lub}A$ exists, then by the constructive version of the
classical trichotomy of reals (Corollary (2.17) in [4], p. 26) we have that
$\lambda>0$ or $\lambda<1$. If $\lambda>0$, we suppose $\neg P$, hence
$A=\\{0\\}$, and $\lambda$ cannot be the least upper bound of $\\{0\\}$. Hence
we showed $\neg\neg P$. If $\lambda<1$, we suppose $P$, hence $A=\\{0,1\\}$
and we get the contradiction $1\leq\lambda<1$. Hence we showed $\neg P$. shows
that the corresponding existence of $\mathrm{lub}A$ implies the weak principle
of the excluded middle $\mathrm{WPEM}$ (i.e., the scheme $\neg P\vee\neg\neg
P$). It is immediate to show that if $\sup A$ exists, then $\mathrm{lub}A$
exists and they are equal, for every $A\subseteq{\mathbb{R}}$. The converse is
not generally true (see [18], p. 27).

Note that Bishop used the terms $\mathrm{lub}A$ and $\sup A$ (and similarly
the terms $\mathrm{glb}A$ and $\inf A$) without distinction. In [18]
Mandelkern gave reasons to keep both notions at work. In [18], pp. 24-25, he
also gave a necessary and sufficient condition for the existence of
$\mathrm{lub}A$ and $\mathrm{glb}A$, similar to the constructive version of
completeness given by Bishop and Bridges in [4], p. 37, and he proved the
following remark: If $A\subseteq{\mathbb{R}}$ is bounded and $\mathrm{glb}A$
exists, then $\sup(A]$ exists and $\sup(A]=\mathrm{glb}A$, while if
$\mathrm{lub}A$ exists, then $\inf[A)$ exists and $\inf[A)=\mathrm{lub}A$.

######  Definition .

If $X,Y$ are sets, we denote by ${\mathbb{F}}(X,Y)$ the set of functions of
type $X\rightarrow Y$ and by ${\mathbb{F}}(X)$ the set of functions of type
$X\rightarrow{\mathbb{R}}$. If $a\in{\mathbb{R}}$, then $\overline{a}_{X}$
denotes the constant map in ${\mathbb{F}}(X)$ with value $a$, and
$\mathrm{Const}(X)$ is the set of constant maps on $X$. If $(X,d),(Y,\rho)$
are metric spaces, then $C_{u}(X,Y)$ denotes the set of uniformly continuous
functions from $X$ to $Y$, where $f\in C_{u}(X,Y)$, if there is a function
$\omega_{f}:{\mathbb{R}}^{+}\to{\mathbb{R}}^{+}$, a modulus of uniform
continuity of $f$, such that for every $\epsilon>0$

$\forall_{x,y\in
X}\big{(}d(x,y)\leq\omega_{f}(\epsilon)\Rightarrow\rho(f(x),f(y))\leq\epsilon\big{)}.$

We denote by $C_{u}(X)$ the set $C_{u}(X,{\mathbb{R}})$, where ${\mathbb{R}}$
is equipped with its standard metric. The metric $d_{x_{0}}$ at the point
$x_{0}\in X$ is defined by $d_{x_{0}}(x):=d(x_{0},x),$ for every $x\in X$, and
$U_{0}(X):=\\{d_{x_{0}}\mid x_{0}\in X\\}.$ The set $X_{0}$ of the
$d$-distinct pairs of $X$ is defined by

$X_{0}:=\\{(x,y)\in X\times X\mid d(x,y)>0\\}.$

The Lipschitz functions are defined constructively as in the classical
setting. Throughout this section, $X$ is a set equipped with a metric $d$, and
$Y$ is a set equipped with a metric $\rho$.

######  Definition .

The set of Lipschitz functions from $X$ to $Y$ is defined by

$\mathrm{Lip}(X,Y):=\bigcup_{\sigma\geq 0}\mathrm{Lip}(X,Y,\sigma),$
$\mathrm{Lip}(X,Y,\sigma):=\\{f\in\mathbb{F}(X,Y)\mid\forall_{x,y\in
X}(\rho(f(x),f(y))\leq\sigma d(x,y))\\}.$

If $Y={\mathbb{R}}$, we write $\mathrm{Lip}(X)$ and $\mathrm{Lip}(X,\sigma)$,
respectively.

Clearly, $\mathrm{Lip}(X,Y)\subseteq C_{u}(X,Y)$. An element of
$\mathrm{Lip}(X,Y)$ sends a bounded subset of $X$ to a bounded subset of $Y$,
which is not generally the case for elements of $C_{u}(X,Y)$; the identity
function $\mathrm{id}:{\mathbb{N}}\rightarrow{\mathbb{R}}$, where
${\mathbb{N}}$ is equipped with the discrete metric, is in
$\mathrm{C_{u}({\mathbb{N}})}\setminus\mathrm{Lip}({\mathbb{N}})$ and
$\mathrm{id}({\mathbb{N}})={\mathbb{N}}$ is unbounded in ${\mathbb{R}}$. The
following proposition is easy to show.

######  Proposition .

The set $\mathrm{Lip}(X)$ includes the sets $U_{0}(X)$, $\mathrm{Const}(X)$,
and it is closed under addition and multiplication by reals. If every element
of $C_{u}(X)$ is a bounded function, then $\mathrm{Lip}(X)$ is closed under
multiplication.

Next we show555This proof appeared first in [21]. constructively the uniform
density of $\mathrm{Lip}(X)$ in $C_{u}(X)$, in case $X$ is totally bounded
i.e., if there is a finite $\epsilon$-approximation to $X$, for every
$\epsilon>0$ (see [4], p. 94). The formulation and the proof of the following
theorem are analogous to the formulation and the proof of Theorem 2.1.

######  Theorem .

Let $(X,d)$ be totally bounded. If $f\in C_{u}(X)$ and $\epsilon>0$, there are
$\sigma>0$, $g^{*}\in\mathrm{Lip}(X,\sigma)$ and
${{}^{*}}g\in\mathrm{Lip}(X,\sigma)$ such that the following hold:
$(i)$ $f-\epsilon\leq g^{*}\leq f\leq{{}^{*}}g\leq f+\epsilon$.
$(ii)$ For every $e\in\mathrm{Lip}(X,\sigma)$, $e\leq f\Rightarrow e\leq
g^{*}$.
$(iii)$ For every $e\in\mathrm{Lip}(X,\sigma)$, $f\leq
e\Rightarrow{{}^{*}}g\leq e$.
$(iv)$ The pair $(g^{*}$, ${{}^{*}}g)$ is the unique pair of functions in
$\mathrm{Lip}(X,\sigma)$ satisfying conditions $(i)$-$(iii)$.

###### Proof.

(i) Let $\omega_{f}$ be a modulus of uniform continuity of $f$, and $M_{f}>0$
a bound of $f$. We define the functions $h_{x}:X\rightarrow{\mathbb{R}}$ and
$g^{*}:X\rightarrow{\mathbb{R}}$ by

$h_{x}:=f+\sigma d_{x},$ $\sigma:=\frac{2M_{f}}{\omega_{f}(\epsilon)}>0,$
$g^{*}(x):=\inf\\{h_{x}(y)\mid y\in X\\}=\inf\\{f(y)+\sigma d(x,y)\mid y\in
X\\},$

for every $x\in X$. Note that $g^{*}(x)$ is well-defined, since $h_{x}\in
C_{u}(X)$ and the infimum of $h_{x}$ exists, since $X$ is totally bounded (see
[4], p.94 and p.38). First we show that $g\in\mathrm{Lip}(X,\sigma)$. If
$x_{1},x_{2},y\in X$, the inequality $d(x_{1},y)\leq
d(x_{2},y)+d(x_{1},x_{2})$ implies that $f(y)+\sigma
d(x_{1},y)\leq(f(y)+\sigma d(x_{2},y))+\sigma d(x_{1},x_{2})$, hence
$g^{*}(x_{1})\leq(f(y)+\sigma d(x_{2},y))+\sigma d(x_{1},x_{2})$, therefore
$g^{*}(x_{1})\leq g^{*}(x_{2})+\sigma d(x_{1},x_{2})$, or
$g^{*}(x_{1})-g^{*}(x_{2})\leq\sigma d(x_{1},x_{2})$. Starting with the
inequality $d(x_{2},y)\leq d(x_{1},y)+d(x_{1},x_{2})$ and working similarly we
get that $g^{*}(x_{2})-g^{*}(x_{1})\leq\sigma d(x_{1},x_{2})$, therefore
$|g^{*}(x_{1})-g^{*}(x_{2})|\leq\sigma d(x_{1},x_{2})$. Next we show that

$\forall_{x\in X}\big{(}f(x)-\epsilon\leq g^{*}(x)\leq f(x)\big{)}.$

Since $f(x)=f(x)+\sigma d(x,x)=h_{x}(x)\geq\inf\\{h_{x}(y)\mid y\in
X\\}=g^{*}(x)$, for every $x\in X$, we have that $g^{*}\leq f$. Next we show
that $\forall_{x\in X}(f(x)-\epsilon\leq g^{*}(x))$. For that we fix $x\in X$
and we show that $\neg{(f(x)-\epsilon>g^{*}(x))}$. Note that if
$A\subseteq{\mathbb{R}},b\in{\mathbb{R}}$, then666By the definition of $\inf
A$ in [4], p.37, we have that $\forall_{\epsilon>0}\exists_{a\in A}(a<\inf
A+\epsilon)$, therefore if $b>\inf A$ and $\epsilon=b-\inf A>0$ we get that
$\exists_{a\in A}(a<\inf A+(b-\inf A)=b)$. $b>\inf A\Rightarrow\exists_{a\in
A}(a<b)$. Therefore,

$\displaystyle f(x)-\epsilon>g^{*}(x)$ $\displaystyle\Leftrightarrow
f(x)-\epsilon>\inf\\{f(y)+\sigma d(x,y)\mid y\in X\\}$
$\displaystyle\Rightarrow\exists_{y\in X}(f(x)-\epsilon>f(y)+\sigma d(x,y))$
$\displaystyle\Leftrightarrow\exists_{y\in X}(f(x)-f(y)>\epsilon+\sigma
d(x,y)).$

For this $y$ we show that $d(x,y)\leq\omega_{f}(\epsilon)$. If
$d(x,y)>\omega_{f}(\epsilon)$, we have that

$2M_{f}\geq f(x)+M_{f}\geq
f(x)-f(y)>\epsilon+2M_{f}\frac{d(x,y)}{\omega_{f}(\epsilon)}>\epsilon+2M_{f}>2M_{f},$

which is a contradiction. Hence, by the uniform continuity of $f$ we get that
$|f(x)-f(y)|\leq\epsilon$, therefore the contradiction $\epsilon>\epsilon$ is
reached, since

$\epsilon\geq|f(x)-f(y)|\geq f(x)-f(y)>\epsilon+\sigma d(x,y)\geq\epsilon.$

Next we define the functions $h_{x}^{*}:X\rightarrow{\mathbb{R}}$ and
${{}^{*}}g:X\rightarrow{\mathbb{R}}$ by

$h_{x}^{*}:=f-\sigma d_{x},$ ${{}^{*}}g(x):=\sup\\{h_{x}^{*}(y)\mid y\in
X\\}=\sup\\{f(y)-\sigma d(x,y)\mid y\in X\\},$

for every $x\in X$, and $\sigma=\frac{2M_{f}}{\omega_{f}(\epsilon)}$.
Similarly777To show that $\neg{(g^{*}(x)>f(x)+\epsilon)}$ we just use the fact
that if $A\subseteq{\mathbb{R}},b\in{\mathbb{R}}$, then $\sup
A>b\Rightarrow\exists_{a\in A}(a>b)$. The function ${{}^{*}}g$ is mentioned in
[27], where non-constructive properties of the classical $({\mathbb{R}},<)$
are used. to the proof for $g^{*}$, we get that
${{}^{*}}g\in\mathrm{Lip}(X,\sigma)$ and

$\forall_{x\in X}(f(x)\leq{{}^{*}}g(x)\leq f(x)+\epsilon).$

(ii) Let $e\in\mathrm{Lip}(X,\sigma)$ such that $e\leq f$. If we fix some
$x\in X$, then for every $y\in X$ we have that
$e(x)-e(y)\leq|e(x)-e(y)|\leq\sigma d(x,y)$, hence $e(x)\leq e(y)+\sigma
d(x,y)\leq f(y)+\sigma d(x,y)$, therefore $e(x)\leq g^{*}(x)$.
(iii) Let $e\in\mathrm{Lip}(X,\sigma)$ such that $f\leq e$. If we fix some
$x\in X$, then for every $y\in X$ we have that
$e(y)-e(x)\leq|e(y)-e(x)|\leq\sigma d(x,y)$, hence $f(y)-\sigma d(x,y)\leq
e(y)-\sigma d(x,y)\leq e(x)$, therefore ${{}^{*}}g(x)\leq e(x)$.
(iv) Let $h^{*},{{}^{*}}h\in\mathrm{Lip}(X,\sigma)$ such that the following
hold:
(i)${}^{{}^{\prime}}$ $f-\epsilon\leq h^{*}\leq f\leq{{}^{*}}h\leq
f+\epsilon$.
(ii)${}^{{}^{\prime}}$ For every $e\in\mathrm{Lip}(X,\sigma)$, $e\leq
f\Rightarrow e\leq h^{*}$.
(iii)${}^{{}^{\prime}}$ For every $e\in\mathrm{Lip}(X,\sigma)$, $f\leq
e\Rightarrow{{}^{*}}h\leq e$.
Since $h^{*}\leq f$, we get by (ii) that $h^{*}\leq g^{*}$. Since $g^{*}\leq
f$, we get by (ii)${}^{{}^{\prime}}$ that $g^{*}\leq h^{*}$. Using similarly
(iii) and (iii)${}^{{}^{\prime}}$, we get ${{}^{*}}h={{}^{*}}g$. ∎

From Theorem 1(i) we get immediately the following corollary.

######  Corollary .

If $X$ is totally bounded, then $\mathrm{Lip}(X)$ is uniformly dense in
$C_{u}(X)$.

The constructive behavior of Lipschitz functions is different from the
classical one. For example, Lebesgue’s theorem on the almost everywhere
differentiability of a Lipschitz function $f:(a,b)\to{\mathbb{R}}$, and hence
Rademacher’s generalization (see [12], p. 18), do not hold constructively;
Demuth has shown in [10] that in the recursive variety RUSS of BISH (see [6])
there is a Lipschitz function $f:[0,1]\to{\mathbb{R}}$, which is nowhere
differentiable. As we explain in the rest of this paper, there are important
differences between the classical and the constructive theory of Lipschitz
functions at an even more elementary level.

######  Definition .

If $f\in{\mathbb{F}}(X,Y)$, we define the following sets:

$\Lambda(f):=\\{\sigma\geq 0\mid\forall_{x,y\in X}(\rho(f(x),f(y))\leq\sigma
d(x,y))\\},$ $\Xi(f):=\\{\sigma\geq 0\mid\forall_{x,y\in
X}(\rho(f(x),f(y))\geq\sigma d(x,y))\\},$
$M_{0}(f):=\\{\sigma_{x,y}(f)\mid(x,y)\in X_{0}\\},$
$\sigma_{x,y}(f):=\frac{\rho(f(x),f(y))}{d(x,y)}.$

######  Proposition .

If $f\in{\mathbb{F}}(X,Y)$, then $\Lambda(f)=[M_{0}(f))$ and
$\Xi(f)=(M_{0}(f)]$.

###### Proof.

First we show that $\Lambda(f)=[M_{0}(f))$. If $b\geq M_{0}(f)$, we show that
$b\in\Lambda(f)$ i.e., $\forall_{x,y\in X}(\rho(f(x),f(y))\leq bd(x,y)).$
Since $b\geq 0$, let $x,y\in X$ such that $\rho(f(x),f(y))>bd(x,y)$. Suppose
next that $d(x,y)>0$, hence $\sigma_{x,y}(f)\in M_{0}(f)$, and since $b\geq
M_{0}(f)$, we get that $\sigma_{x,y}(f)\leq
b\Leftrightarrow\rho(f(x),f(y))\leq bd(x,y)$ which together with our initial
hypothesis $bd(x,y)<\rho(f(x),f(y))$ leads to a contradiction. Hence,
$d(x,y)\leq 0\Leftrightarrow d(x,y)=0\Leftrightarrow x=y$, and our initial
hypothesis is reduced to $0>0$. The converse implication,
$\sigma\in\Lambda(f)\Rightarrow\sigma\geq M_{0}(f)$ follows immediately. The
proof of the equality $\Xi(f)=(M_{0}(f)]$ is similar. ∎

By the classical completeness of ${\mathbb{R}}$, if $f\in\mathrm{Lip}(X,Y)$,
then $\inf\Lambda(f)$ exists. Constructively, we don’t have in general the
existence of $\inf\Lambda(f)$. Actually, the statement “$\inf\Lambda(f)$
exists, for every $f\in\mathrm{Lip}(A)$ and every $A\subseteq{\mathbb{R}}$”
implies $\mathrm{WPEM}$. To see this, we work as follows. If $P$ is a
syntactically well-formed proposition, let
$A:=\\{0\\}\cup\\{x\in{\mathbb{R}}\mid x=1\ \&\ P\\}$. If $f:A\to{\mathbb{R}}$
is defined by $f(a):=a$, for every $a\in A$, then $f\in\mathrm{Lip}(A,1)$ and
$1\in\Lambda(f)$. If $m:=\inf\Lambda(f)$, then by the constructive version of
the classical trichotomy of reals (Corollary (2.17) in [4], p. 26) we have
that $m>0$ or $m<1$. If $m>0$, we suppose $\neg P$, hence $A=\\{0\\}$. In this
case $0\in\Lambda(f)$, and we have that $m\leq 0<m$, which is a contradiction,
hence we showed $\neg\neg P$. If $m<1$, we suppose $P$, hence $A=\\{0,1\\}$.
By the definition of $\inf\Lambda(f)$ we can find $\epsilon>0$ and
$\sigma\in\Lambda(f)$ such that $\sigma<m+\epsilon<1$. Hence,
$1=|0-1|\leq\sigma|0-1|=\sigma<1$, which is a contradiction. Hence we showed
$\neg P$.

Classically, if $\inf\Lambda(f)$ exists, then $\sup M_{0}(f)$ exists and it is
equal to $\inf\Lambda(f)$. Constructively, from the existence of
$\inf\Lambda(f)$ we only get the existence of $\mathrm{lub}M_{0}(f)$. If $\sup
M_{0}(f)$ exists though, we can infer constructively the existence of
$\inf\Lambda(f)$.

######  Proposition .

Let $f\in\mathrm{Lip}(X,Y)$.
$(i)$ If $\sup M_{0}(f)$ exists, then $\inf\Lambda(f)$ exists and
$\inf\Lambda(f)=\min\Lambda(f)=\sup M_{0}(f)$.
$(ii)$ If $\inf\Lambda(f)$ exists, then $\mathrm{lub}M_{0}(f)$ exists and
$\mathrm{lub}M_{0}(f)=\inf\Lambda(f)$.
$(iii)$ If $\mathrm{lub}M_{0}(f)$ exists, then $\inf\Lambda(f)$ exists and
$\inf\Lambda(f)=\mathrm{lub}M_{0}(f)$.

###### Proof.

(i) Since for $l=\sup M_{0}(f)$ we have that $l\geq M_{0}(f)$, by Proposition
1 we get that $l\in\Lambda(f)$. Next we show that $l\leq\Lambda(f)$. If
$\sigma\in\Lambda(f)$, then $\sigma\geq M_{0}(f)$, therefore $\sigma\geq l$,
since the supremum of a set is its least upper bound. Since $l\in\Lambda(f)$,
if $\epsilon>0$, then $l<l+\epsilon$, hence $l=\inf\Lambda(f)$.
(ii) If $v=\inf\Lambda(f)$, then $v\geq M_{0}(f)$, as we have explained in
(i). Suppose next that $b\geq M_{0}(f)$. We show that $b\geq v$, by supposing
$b<v$ and reaching a contradiction. Since $b\geq M_{0}(f)$, by Proposition 1
we have that $b\in\Lambda(f)$, and consequently $v\leq b<v$.
(iii) By Proposition 1 we have that $\Lambda(f)=[M_{0}(f))$, hence by
Mandelkern’s remark after Definition 1 we get that
$\mathrm{lub}M_{0}(f)=\inf[M_{0}(f))=\inf\Lambda(f)$. ∎

######  Definition .

Let $f\in\mathrm{Lip}(X,Y)$. If $\sup M_{0}(f)$ exists, we call $L(f):=\sup
M_{0}(f)$ the Lipschitz constant of $f$. If $\mathrm{lub}M_{0}(f)$ exists, we
call $L^{*}(f):=\mathrm{lub}M_{0}(f)$ the weak Lipschitz constant of $f$.

Constructively, $L(f)$ does not generally exist. Actually, the statement
“$L(f)$ exists, for every $f\in\mathrm{Lip}(l_{2})$” implies the limited
principle of omniscience (LPO), where LPO is a weak form of PEM, and
$l_{2}:=\\{(x_{n})\in\mathbb{F}(\mathbb{N},{\mathbb{R}})\mid\sum_{n\in\mathbb{N}}|x_{n}|^{2}\in{\mathbb{R}}\\}$
is equipped with the $2$-norm. This is explained as follows. A linear
functional on a normed space $X$ is bounded, if there is $\sigma>0$ such that
$|f(x)|\leq\sigma||x||$, for every $x\in X$ (see [7], p. 51). By the linearity
of $f$ we then get $f\in\mathrm{Lip}(X,\sigma)$. Since in this case it is easy
to show that $M_{0}(f)=\\{|f(x)|\mid||x||=1\\}$, the existence of $\sup
M_{0}(f)$ is equivalent to the existence of $||f||$. Hence, the statement
“$L(f)$ exists, for every $f\in\mathrm{Lip}(l_{2})$” implies the statement
“every bounded linear functional on $l_{2}$ is normable”. The last statement
implies LPO (see [14]).

If $L(f)$ exists, and since $L(f)\geq M_{0}(f)$, by Proposition 1 we get
$\forall_{x,y\in X}(\rho(f(x),f(y))\leq L(f)d(x,y)),$ i.e.,
$f\in\mathrm{Lip}(X,Y,L(f)).$ If $L^{*}(f)$ exists, then proceeding similarly
we get $\forall_{x,y\in X}(\rho(f(x),f(y))\leq L^{*}(f)d(x,y))$ i.e.,
$f\in\mathrm{Lip}(X,Y,L^{*}(f)).$

######  Proposition .

Let $A\subseteq X$, $f\in\mathrm{Lip}(A,Y)$ and $g\in{\mathbb{F}}(X,Y)$ such
that

$g_{|A}=f\ \&\ \forall_{\sigma\geq 0}(f\in\mathrm{Lip}(A,Y,\sigma)\Rightarrow
g\in\mathrm{Lip}(X,Y,\sigma)).$

$(i)$ If $L(f)$ exists, then $L(g)$ exists and $L(g)=L(f)$.
$(ii)$ If $L^{*}(f)$ exists, then $L^{*}(g)$ exists and $L^{*}(g)=L^{*}(f)$.

###### Proof.

(i) If $L(f)=\sup M_{0}(f)$ we show that $L(f)=\sup M_{0}(g)$, where
$M_{0}(f)=\\{\sigma_{a,b}(f)\mid(a,b)\in
A_{0}\\},M_{0}(g)=\\{\sigma_{x,y}(g)\mid(x,y)\in X_{0}\\},$ and
$M_{0}(f)\subseteq M_{0}(g)$. First we show that $L(f)\geq M_{0}(g)$. Since
$L(f)\geq 0$ and $f\in\mathrm{Lip}(A,Y,L(f))$, we get by hypothesis that $g$
is also in $\mathrm{Lip}(X,Y,L(f))$, therefore $L(f)\geq M_{0}(g)$. If
$\epsilon>0$, there exists $\sigma_{a,b}(f)\in M_{0}(f)$ such that
$\sigma_{a,b}(f)>L(f)-\epsilon$, which implies that
$\sigma_{a,b}(f)=\sigma_{a,b}(g)>L(f)-\epsilon$, and $\sigma_{a,b}(g)\in
M_{0}(g)$, since $A_{0}\subseteq X_{0}$.
(ii) If $L^{*}(f)=\mathrm{lub}M_{0}(f)$ we show that
$L^{*}(f)=\mathrm{lub}M_{0}(g)$. First we show that $L^{*}(f)\geq M_{0}(g)$.
As in (i), since $L^{*}(f)\geq 0$ and $f\in\mathrm{Lip}(A,Y,L^{*}(f))$, we get
by hypothesis that $g$ is also in $\mathrm{Lip}(X,Y,L^{*}(f))$, therefore
$L^{*}(f)\geq M_{0}(g)$. If $m\geq M_{0}(g)$, then $m\geq M_{0}(f)$, since
$M_{0}(g)\subseteq M_{0}(g)$, therefore $m\geq L^{*}(f)$. ∎

Note that if $f\in\mathrm{Lip}(A,Y),g\in\mathrm{Lip}(X,Y)$ such that
$L(f),L(g)$ exist and $L(f)=L(g)$, then it is immediate to see that
$\forall_{\sigma\geq 0}(f\in\mathrm{Lip}(A,Y,\sigma)\Rightarrow
g\in\mathrm{Lip}(X,Y,\sigma))$. Next follows the Lipschitz-version of the
extendability of a uniformly continuous function defined on a dense subset of
a metric space with values in a complete metric space. Its expected proof,
which is based on the proof of Lemma 3.7 in [4], pp. 91-2, is omitted.

######  Proposition .

Let $D\subseteq X$ be dense in $X$, $Y$ complete, and $f\in\mathrm{Lip}(D,Y)$.
$(i)$ $\exists_{!g\in{\mathbb{F}}(X,Y)}\big{(}g_{|A}=f\ \&\
\forall_{\sigma\geq 0}(f\in\mathrm{Lip}(D,Y,\sigma)\Rightarrow
g\in\mathrm{Lip}(X,Y,\sigma))\big{)}$.
$(ii)$ If $L(f)$ exists and $g$ is as in $($i$)$, then $L(g)$ exists and
$L(g)=L(f)$.
$(iii)$ If $L^{*}(f)$ exists and $g$ is as in $($i$)$, then $L^{*}(g)$ exists
and $L^{*}(g)=L^{*}(f)$.

##  2\. McShane-Whitney pairs

###  2.1. Metric spaces

Throughout this section $X$ is a set equipped with a metric $d$, and $Y$ is a
set equipped with a metric $\rho$.

######  Definition .

Let $A\subseteq X$. We call $(X,A)$ a McShane-Whitney pair, or simply an MW-
pair, if for every $\sigma>0$ and $g\in\mathrm{Lip}(A,\sigma)$ the functions
$g^{*},{{}^{*}}g:X\rightarrow{\mathbb{R}}$ are well-defined, where for every
$x\in X$

$g^{*}(x):=\sup\\{g(a)-\sigma d(x,a)\mid a\in A\\},$
${{}^{*}}g(x):=\inf\\{g(a)+\sigma d(x,a)\mid a\in A\\}.$

Note that $(X,A)$ is a McShane-Whitney pair if and only if for every
$g\in\mathrm{Lip}(A,1)$ the function $g^{*}$ is well-defined, since we can
work with $\frac{1}{\sigma}h$, for an arbitrary $\sigma>0$ and
$h\in\mathrm{Lip}(A,\sigma)$, and use the equality ${{}^{*}}g=-[(-g)^{*}]$ for
the extension ${{}^{*}}g$. The classical statement “every pair $(X,A)$ is a
McShane-Whitney pair” cannot be accepted constructively.

######  Remark .

The statement “$({\mathbb{R}},A)$ is a McShane-Whitney pair, for every non-
empty $A\subseteq{\mathbb{R}}$” implies $\mathrm{PEM}$.

###### Proof.

If $P$ is a syntactically well-formed proposition, let
$A:=\\{0\\}\cup\\{x\in{\mathbb{R}}\mid x=1\ \&\ P\\}$. If $g:A\to{\mathbb{R}}$
is defined by $g(a):=a$, for every $a\in A$, then $g\in\mathrm{Lip}(A,1)$. If
$x:=2$, we suppose that

$g^{*}(2):=\sup\\{a-|2-a|\mid a\in A\\}\in{\mathbb{R}}.$

By the constructive version of the classical trichotomy of reals (Corollary
(2.17) in [4], p. 26) we have that $g^{*}(2)>-\frac{3}{2}$ or $g^{*}(2)<-1$.
In the first case, by the definition of $\sup A$ there is $a\in A$ such that
$a-|2-a|>-\frac{3}{2}$. If $a=0$, then $0-|2-0|=-2<-\frac{3}{2}$, hence there
must be $a\in\\{x\in{\mathbb{R}}\mid x=1\ \&\ P\\}$ that satisfies this
property. This implies $P$ and $g^{*}(2)=0>-\frac{3}{2}$. If $g^{*}(2)<-1$, we
suppose $P$, hence $A=\\{0,1\\}$. Since then $g^{*}(2)=0$, we get the
contradiction $0=g^{*}(2)<-1$ i.e., we showed $\neg P$. ∎

The next theorem reveals the computational content of the classical McShane-
Whitney theorem.

######  Theorem  (McShane-Whitney).

If $(X,A)$ is an MW-pair and $g\in\mathrm{Lip}(A,\sigma)$, the following hold:
$(i)$ $g^{*},{{}^{*}}g\in\mathrm{Lip}(X,\sigma)$.
$(ii)$ ${g^{*}}_{|A}=({{}^{*}}g)_{|A}=g$.
$(iii)$ $\forall_{f\in\mathrm{Lip}(X,\sigma)}\big{(}f_{|A}=g\Rightarrow
g^{*}\leq f\leq{{}^{*}}g\big{)}$.
$(iv)$ The pair $(g^{*}$, ${{}^{*}}g)$ is the unique pair of functions
satisfying conditions $(i)$-$(iii)$.

###### Proof.

(i) Let $x_{1},x_{2}\in X$ and $a\in A$. Then $d(x_{1},a)\leq
d(x_{2},a)+d(x_{2},x_{1})$ and $\sigma d(x_{1},a)\leq\sigma d(x_{2},a)+\sigma
d(x_{1},x_{2})$, therefore

$\displaystyle g(a)+\sigma d(x_{1},a)\leq(g(a)+\sigma d(x_{2},a))+\sigma
d(x_{1},x_{2})\Rightarrow$ $\displaystyle{{}^{*}g}(x_{1})\leq(g(a)+\sigma
d(x_{2},a))+\sigma d(x_{1},x_{2})\Rightarrow$
$\displaystyle{{}^{*}g}(x_{1})-\sigma d(x_{1},x_{2})\leq g(a)+\sigma
d(x_{2},a)\Rightarrow$ $\displaystyle{{}^{*}g}(x_{1})-\sigma
d(x_{1},x_{2})\leq{{}^{*}g}(x_{2})\Rightarrow$
$\displaystyle{{}^{*}g}(x_{1})-{{}^{*}g}(x_{2})\leq\sigma d(x_{1},x_{2}).$

If we start from the inequality $d(x_{2},a)\leq d(x_{1},a)+d(x_{2},x_{1})$ and
work as above, we get ${{}^{*}}g(x_{2})-{{}^{*}g}(x_{1})\leq\sigma
d(x_{1},x_{2})$, therefore $|{{}^{*}g}(x_{1})-{{}^{*}g}(x_{2})|\leq\sigma
d(x_{1},x_{2})$. Working similarly we get that $g^{*}$ is an extension of $g$
which is in $\mathrm{Lip}(X,\sigma)$.
(ii) We show that ${{}^{*}g}$ extends $g$. If $a_{0}\in A$, then
${{}^{*}g}(a_{0})=\inf\\{g(a)+\sigma d(a_{0},a)\mid a\in A\\}\leq
g(a_{0})+\sigma d(a_{0},a_{0})=g(a_{0})$. If $a\in A$, then
$g(a_{0})-g(a)\leq|g(a_{0})-g(a)|\leq\sigma d(a_{0},a)$, hence $g(a)+\sigma
d(a_{0},a)\geq g(a_{0})$. Since $a$ is arbitrary, ${{}^{*}g}(a_{0})\geq
g(a_{0})$.
(iii) If $f\in\mathrm{Lip}(X,\sigma)$ such that $f_{|A}=g$, $x\in X$ and $a\in
A$ we have that

$\displaystyle f(x)-g(a)=f(x)-f(a)\leq|f(x)-f(a)|\leq\sigma d(x,a)\Rightarrow$
$\displaystyle f(x)\leq g(a)+\sigma d(a,x)\Rightarrow$ $\displaystyle
f(x)\leq{{}^{*}g}(x),$ $\displaystyle
g(a)-f(x)=f(a)-f(x)\leq|f(a)-f(x)|\leq\sigma d(x,a)\Rightarrow$ $\displaystyle
g(a)-\sigma d(a,x)\leq f(x)\Rightarrow$ $\displaystyle g^{*}(x)\leq f(x).$

(iv) Let $h^{*},{{}^{*}}h$ satisfy conditions (i)-(iii). Since
${h^{*}}_{|A}=({{}^{*}}h)_{|A}=g$, we have that $g^{*}\leq h^{*}\leq{{}^{*}}g$
and $g^{*}\leq{{}^{*}}h\leq{{}^{*}}g$. Since
${g^{*}}_{|A}=({{}^{*}}g)_{|A}=g$, we have that $h^{*}\leq g^{*}\leq{{}^{*}}h$
and $h^{*}\leq{{}^{*}}g\leq{{}^{*}}h$, hence $h^{*}=g^{*}$ and
${{}^{*}}h={{}^{*}}g$. ∎

Since in the previous proof we only used the “least upper bound” property of
$\sup$ and the “greatest lower bound” property of $\inf$, we could have
defined $g^{*}$ and ${{}^{*}}g$ through $\mathrm{lub}$ and $\mathrm{glb}$,
respectively. Next we recall some fundamental definitions in the constructive
theory of metric spaces.

######  Definition .

A space $X$ is compact, if it is complete and totally bounded, while it is
locally compact (totally bounded), if every bounded subset of $X$ is included
in a compact (totally bounded) subset of $X$. A subset $A$ of $X$ is located,
if $d(x,A):=\inf\Delta(x,A)$, where $\Delta(x,A):=\\{d(x,a)\mid a\in A\\}$,
exists, for every $x\in X$.

######  Proposition .

The following sets $X$ and $A$ form MW-pairs.
$(i)$ $A$ is a totally bounded subset of $X$.
$(ii)$ $X$ is totally bounded and $A$ is located.
$(iii)$ $X$ is locally compact $($totally bounded$)$ and $A$ is bounded and
located.

###### Proof.

(i) If $\sigma>0$, $g\in\mathrm{Lip}(A)$ and $x\in X$, then $g+\sigma
d_{x},g-\sigma d_{x}\in\mathrm{Lip}(A)\subseteq C_{u}(A)$, hence $g^{*}(x)$
and ${{}^{*}}g(x)$ are well defined, since $A$ is totally bounded (see [4],
Corollary 4.3, p.94).
(ii) A located subset of $X$ is also totally bounded (see [4], p.95), and we
use (i).
(iii) If $A$ is bounded and located, there is compact (totally bounded)
$K\subseteq X$ with $A\subseteq K$. Since $A$ is located in $X$, it is located
in $K$, hence $A$ is totally bounded, and we use (i).
∎

######  Proposition .

Let $(X,A)$ be an MW-pair and $g\in\mathrm{Lip}(A,\sigma)$. The following
hold.
$(i)$ The set $A$ is located.
$(ii)$ If $M\in{\mathbb{R}}$ such that $\forall_{a\in A}(g(a)\leq M)$, then
$\forall_{x\in X}(g^{*}(x)\leq M)$.
$(iii)$ If $m\in{\mathbb{R}}$ such that $\forall_{a\in A}(g(a)\geq m)$, then
$\forall_{x\in X}({{}^{*}}g(x)\geq m)$.
$(iv)$ If $\inf g$ and $\sup g$ exist, then $\inf{{}^{*}g}$, $\sup g^{*}$
exist and the following equalities hold

$\inf_{x\in X}{{}^{*}g}=\inf_{a\in A}g\ \ \&\ \ \sup_{x\in
X}{g^{*}}=\sup_{a\in A}g.$

###### Proof.

(i) Let $r\in{\mathbb{R}}$ and $\sigma>0$. Since
$\overline{r}_{A}\in\mathrm{Lip}(A,\sigma)$, by hypothesis
${{}^{*}}\overline{r}_{A}$ is well-defined, where
${{}^{*}}\overline{r}_{A}(x)=\inf\\{r+\sigma d(x,a)\mid a\in A\\}$, for every
$x\in X$. If $x\in X$ and $a\in A$, then $d(x,a)=\frac{1}{\sigma}(r+\sigma
d(x,a)-r)$, and $\Delta(x,A)=\\{\frac{1}{\sigma}(r+\sigma d(x,a)-r)\mid a\in
A\\}$. Hence

$\displaystyle d(x,A)$ $\displaystyle=\inf\\{\frac{1}{\sigma}(r+\sigma
d(x,a)-r)\mid a\in A\\}$ $\displaystyle=\frac{1}{\sigma}(\inf\\{r+\sigma
d(x,a)\mid a\in A\\}-r)$
$\displaystyle=\frac{1}{\sigma}({{}^{*}}\overline{r}_{A}(x)-r).$

(ii) Let $x\in X$. Since $g(a)\leq M$, $g(a)-\sigma d(x,a)\leq M-\sigma
d(x,a)\leq M$. Hence $g^{*}(x)\leq M$.
(iii) Let $x\in X$. Since $g(a)\geq M$, $g(a)+\sigma d(x,a)\geq M$. Hence
${{}^{*}}g\leq M$.
(iv) We show that $m:=\inf\\{g(a)\mid a\in A\\}$ satisfies the properties of
$\inf\\{{{}^{*}}g(x)\mid x\in X\\}$. It suffices to show that
$m\leq{{}^{*}}g(X)$, since the other definitional condition of $\inf$ follows
immediately; if $\epsilon>0$, then there exists $a\in A\subseteq X$ such that
$g(a)={{}^{*}}g(a)<m+\epsilon$. If $x\in A$, then $m\leq g(x)={{}^{*}}g(x)$,
since $m=\inf g$. Since $A$ is located, the set

$-A:=\\{x\in X\mid d(x,A)>0\\}$

is well-defined. If $x\in-A$, then $d(x,a)\geq d(x,A)>0$, for every $a\in A$.
Hence

$\displaystyle g(a)+\sigma d(x,a)>g(a)\geq\inf_{a\in A}g$
$\displaystyle\Rightarrow\inf_{a\in A}(g(a)+\sigma d(x,a))\geq\inf_{a\in A}g$
$\displaystyle\Leftrightarrow{{}^{*}g}(x)\geq m.$

Since $A$ is located, the set $A\cup(-A)$ is dense in $X$ (see [4], p.88). If
$x\in X$, there is some sequence $(d_{n})_{n\in{\mathbb{N}}}\subseteq
A\cup(-A)$ such that $d_{n}\stackrel{{\scriptstyle n}}{{\rightarrow}}x$. By
the continuity of ${{}^{*}}g$ we have that
${{}^{*}}g(d_{n})\stackrel{{\scriptstyle n}}{{\rightarrow}}{{}^{*}}g(x)$.
Suppose that ${{}^{*}}g(x)<m$. Since ${{}^{*}}g(d_{n})\geq m$, for every
$n\in{\mathbb{N}}$, if $\epsilon:=(m-{{}^{*}}g(x))>0$, there is some $n_{0}$
such that for every $n\geq n_{0}$ we have that

$|{{}^{*}}g(d_{n})-{{}^{*}}g(x)|={{}^{*}}g(d_{n})-{{}^{*}}g(x)<m-{{}^{*}}g(x)\Leftrightarrow{{}^{*}}g(d_{n})<m,$

which is a contradiction. Hence ${{}^{*}}g(x)\geq m$. For the existence of
$\sup g^{*}$ and the equality $\sup_{x\in X}{g^{*}}=\sup_{a\in A}g$ we work
similarly. ∎

If $X$ is a normed space and $x_{0}\in X$, then it is not generally the case
that the set ${\mathbb{R}}x_{0}:=\\{\lambda
x_{0}\mid\lambda\in{\mathbb{R}}\\}$ is a located subset of $X$. If
$X={\mathbb{R}}$, the locatedness of ${\mathbb{R}}x_{0}$, for every
$x_{0}\in{\mathbb{R}}$, is equivalent to LPO (see [4], p. 122). If for every
$x_{0}\in{\mathbb{R}}$ we have that $({\mathbb{R}},{\mathbb{R}}x_{0})$ is a
McShane-Whitney pair, then by Proposition 2.1(i) every set of the form
${\mathbb{R}}x_{0}$ is located, which implies LPO.

The next proposition expresses the “step-invariance” of the McShane-Whitney
extension. If $A\subseteq B\subseteq X$ such that $(X,A),(X,B)$ and $(B,A)$
are McShane-Whitney pairs and $g\in\mathrm{Lip}(A)$, then $g^{*_{X}}$ is the
$(A$-$X)$ extension of $g$, $g^{*_{B}*_{X}}$ is the $(B$-$X)$ extension of the
$(A$-$B)$ extension $g^{*_{B}}$ of $g$, and similarly for ${}^{*_{X}}g$ and
${{}^{*_{X}*_{B}}}g$.

######  Proposition .

If $A\subseteq B\subseteq X$ such that $(X,A),(X,B)$, $(B,A)$ are MW-pairs and
$g\in\mathrm{Lip}(A,\sigma)$, for some $\sigma>0$, then

$g^{*_{X}}=g^{*_{B}*_{X}},\ \ \ ^{*_{X}}g={{}^{*_{X}*_{B}}}g.$

###### Proof.

We show only the first equality and for the second we work similarly. By
definition $g^{*_{B}}:B\rightarrow{\mathbb{R}}\in\mathrm{Lip}(B,\sigma)$ and
$g^{*_{B}}(b)=\sup\\{g(a)-\sigma d(b,a)\mid a\in A\\}$, for every $b\in B$.
Moreover, $g^{*_{B}*_{X}}:X\rightarrow{\mathbb{R}}\in\mathrm{Lip}(X,\sigma)$
and

$g^{*_{B}*_{X}}(x)=\sup\\{g^{*_{B}}(b)-\sigma d(x,b)\mid b\in B\\},$

for every $x\in X$. For the $(A$-$X)$ extension of $g$ we have that
$g^{*_{X}}:X\rightarrow{\mathbb{R}}\in\mathrm{Lip}(X,\sigma)$ and
$g^{*_{X}}(x)=\sup\\{g(a)-\sigma d(x,a)\mid a\in A\\}$, for every $x\in X$.
Since $(g^{*_{B}*_{X}})_{|B}=g^{*_{B}}$, we have that
$(g^{*_{B}*_{X}})_{|A}=(g^{*_{B}})_{|A}=g$. Therefore $g^{*_{X}}\leq
g^{*_{B}*_{X}}\leq{{}^{*_{X}}}g,$ and
$(g^{*_{X}})_{|B}\leq(g^{*_{B}*_{X}})_{|B}=g^{*_{B}}\leq({{}^{*_{X}}}g)_{|B}.$
Since $(g^{*_{X}})_{|A}=g$, we get that $((g^{*_{X}})_{|B})_{|A}=g$, therefore
$g^{*_{B}}\leq((g^{*_{X}})_{|B})\leq{{}^{*_{B}}}g.$ Since
$(g^{*_{X}})_{|B}\leq g^{*_{B}}$ and $g^{*_{B}}\leq(g^{*_{X}})_{|B}$, we get
$(g^{*_{X}})_{|B}=g^{*_{B}}$. Hence $g^{*_{B}*_{X}}\leq
g^{*_{X}}\leq{{}^{*_{X}*_{B}}}g$ i.e., we have shown both $g^{*_{X}}\leq
g^{*_{B}*_{X}}$ and $g^{*_{B}*_{X}}\leq g^{*_{X}}$. ∎

######  Proposition .

If $(X,A)$ is an MW-pair and $g\in\mathrm{Lip}(A)$ such that $L(g)$ exists,
the following hold.
$(i)$ $g\in\mathrm{Lip}(A,L(g))$.
$(ii)$ If $f$ is an $L(g)$-Lipschitz extension of $g$, then $L(f)$ exists and
$L(f)=L(g)$.
$(iii)$ $L({{}^{*}}g),L(g^{*})$ exist and $L({{}^{*}}g)=L(g)=L(g^{*}).$

###### Proof.

(i) See our comment before the Proposition 1.
(ii) Since $f\in\mathrm{Lip}(X,L(g))$, we get $L(g)\in\Lambda(f)$ and
$L(g)\geq M_{0}(f)$. Let $\epsilon>0$. Since $L(g)=\sup M_{0}(g)$, there
exists $(a,b)\in A_{0}\subseteq X_{0}$ such that
$\sigma_{a,b}(g)>L(g)-\epsilon$. Since $f$ extends $g$,
$\sigma_{a,b}(g)=\sigma_{a,b}(f)$.
(iii) By definition ${{}^{*}}g,g^{*}\in\mathrm{Lip}(X,L(g))$ and they extend
$g$. Hence we use (ii). ∎

######  Proposition .

Let $(X,A)$ be an MW-pair,
$g_{1}\in\mathrm{Lip}(A,\sigma_{1}),g_{2}\in\mathrm{Lip}(A,\sigma_{2})$ and
$g\in\mathrm{Lip}(A,\sigma)$, for some $\sigma_{1},\sigma_{2},\sigma>0$. The
following hold.
$(i)$ $(g_{1}+g_{2})^{*}\leq g_{1}^{*}+g_{2}^{*}$ and
${{}^{*}}(g_{1}+g_{2})\geq{{}^{*}}g_{1}+{{}^{*}}g_{2}$.
$(ii)$ If $\lambda>0$, then $(\lambda g)^{*}=\lambda g^{*}$ and
${{}^{*}}(\lambda g)=\lambda{{}^{*}}g$.
$(iii)$ If $\lambda<0$, then $(\lambda g)^{*}=\lambda{{}^{*}g}$ and
${{}^{*}}(\lambda g)=\lambda g^{*}$.

###### Proof.

In each case we show only one of the two facts.
(i) We have that $g_{1}+g_{2}\in\mathrm{Lip}(A,\sigma_{1}+\sigma_{2})$ and

$\displaystyle(g_{1}+g_{2})^{*}(x)$
$\displaystyle=\sup\\{g_{1}(a)+g_{2}(a)-(\sigma_{1}+\sigma_{2})d(x,a)\mid a\in
A\\}$
$\displaystyle=\sup\\{(g_{1}(a)-\sigma_{1}d(x,a))+(g_{2}(a)-\sigma_{2}d(x,a))\mid
a\in A\\}$ $\displaystyle\leq\sup\\{g_{1}(a)-\sigma_{1}d(x,a)\mid a\in
A\\}+\sup\\{g_{2}(a)-\sigma_{2}d(x,a)\mid a\in A\\}$
$\displaystyle=g_{1}^{*}(x)+g_{2}^{*}(x).$

(ii) If $\lambda\in{\mathbb{R}}$, then $\lambda
g\in\mathrm{Lip}(A,|\lambda|\sigma)$ and if $\lambda>0$, then

$\displaystyle(\lambda g)^{*}(x)$ $\displaystyle=\sup\\{\lambda
g(a)-|\lambda|\sigma d(x,a)\mid a\in A\\}$
$\displaystyle=\lambda\sup\\{g(a)-\sigma d(x,a)\mid a\in A\\}$
$\displaystyle=\lambda g^{*}(x).$

(iii) If $\lambda<0$, then

$\displaystyle(\lambda g)^{*}(x)$ $\displaystyle=\sup\\{\lambda
g(a)-|\lambda|\sigma d(x,a)\mid a\in A\\}$ $\displaystyle=\sup\\{\lambda
g(a)-(-\lambda)\sigma d(x,a)\mid a\in A\\}$
$\displaystyle=\sup\\{\lambda(g(a)+\sigma d(x,a))\mid a\in A\\}$
$\displaystyle=\lambda\inf\\{g(a)+\sigma d(x,a)\mid a\in A\\}$
$\displaystyle=\lambda{{}^{*}}g(x).$ ∎

###  2.2. Local McShane-Whitney pairs

Next we define a local notion of an MW-pair.

######  Definition .

The pair $(X,A)$ is a local McShane-Whitney pair, or simply a local MW-pair,
if for every bounded $B\subseteq A$ there is $A{{}^{\prime}}\subseteq A$ such
that $B\subseteq A{{}^{\prime}}$ and $(X,A{{}^{\prime}})$ is a MW-pair.

Clearly, an MW-pair is a local MW-pair. The next lemma is shown in [6], p. 33
and in [7], p. 46. If $A,B\subseteq X$, we use Sambin’s notation $A\between B$
to denote that the intersection $A\cap B$ is inhabited.

######  Lemma .

If $A$ is a located subset of a metric space $X$ and $T$ is a totally bounded
subset of $X$ such that $T\between A$, there is totally bounded $S\subseteq X$
such that $T\cap A\subseteq S\subseteq A$.

The next proposition is the “local” version of Proposition 2.1(i)-(ii).

######  Proposition .

If $A\subseteq X$, the following hold.
$(i)$ If $X$ is locally totally bounded and $A$ is located, then $(X,A)$ is a
local MW-pair.
$(ii)$ If $A$ is a locally totally bounded subset of $X$, then $(X,A)$ is a
local MW-pair.

###### Proof.

(i) Let $B\subseteq A$ bounded. By the definition of local total boundedness
there is a totally bounded $T\subseteq X$ such that $B\subseteq T$. Since
$B\subseteq T\cap A$, and since by the definition of a bounded set, $B$ is
inhabited, we get $A\between T$. By Lemma 2.2 there is a totally bounded
$S\subseteq X$ such that $B\subseteq T\cap A\subseteq S\subseteq A$. Since $S$
is totally bounded, by Proposition 2.1(i) we have that $(X,S)$ is an MW-pair.
(ii) Let $B\subseteq A$ be bounded. By the definition of local total
boundedness there is a totally bounded $T\subseteq A$ such that $B\subseteq
T$. As in case (i), we get that $(X,T)$ is an MW-pair. ∎

###  2.3. Hölder continuous functions of order $\alpha$ and $\nu$-continuous
functions

As expected, the McShane-Whitney extension works for the standard
generalizations of Lipschitz functions.

######  Definition .

If $\sigma\geq 0$ and $\alpha\in(0,1]$, the Hölder continuous functions of
order $\alpha$ are defined as follows:

$\mathrm{{H{\mathrm{\text{\"{o}}}}l}}(X,Y,\alpha,\sigma):=\\{f\in\mathbb{F}(X,Y)\mid\forall_{x,y\in
X}(\rho(f(x),f(y))\leq\sigma d(x,y)^{\alpha})\\},$
$\mathrm{{H{\mathrm{\text{\"{o}}}}l}}(X,Y,\alpha):=\bigcup_{\sigma\geq
0}\mathrm{{H{\mathrm{\text{\"{o}}}}l}}(X,Y,\alpha,\sigma).$

If $Y={\mathbb{R}}$ is equipped with its standard metric, we write
$\mathrm{{H{\mathrm{\text{\"{o}}}}l}}(X,\alpha,\sigma)$ and
$\mathrm{{H{\mathrm{\text{\"{o}}}}l}}(X,\alpha)$, respectively.

Clearly,
$\mathrm{{H{\mathrm{\text{\"{o}}}}l}}(X,Y,1,\sigma)=\mathrm{Lip}(X,Y,\sigma)$
and $\mathrm{{H{\mathrm{\text{\"{o}}}}l}}(X,Y,1)=\mathrm{Lip}(X,Y)$. One can
define MW-pairs with respect to Hölder continuous functions of order $\alpha$,
and show a McShane-Whitney extension theorem for them. We give some more
details of these two steps only for the following generalization of Hölder
continuity. The next definition is a reformulation of the definition found in
the formulation of Theorem 5.6 in [4], p. 102.

######  Definition .

A modulus of continuity is a function $\nu:[0,+\infty)\rightarrow[0,+\infty)$
satisfying the following conditions:
(i) $\nu(0)=0$.
(ii) $\forall_{x,y\in[0,+\infty)}\big{(}\nu(x+y)\leq\nu(x)+\nu(y)\big{)}$.
(iii) It is strictly increasing i.e.,
$\forall_{s,t\in[0,+\infty)}\big{(}s<t\Rightarrow\nu(s)<\nu(t)\big{)}$.
(iv) It is uniformly continuous on every bounded subset of $[0,+\infty)$.
If $\nu$ is a modulus of continuity the set of $\nu$-continuous functions from
$X$ to $Y$, and from $X$ to ${\mathbb{R}}$, are defined, respectively, as
follows:

$S(X,Y,\nu):=\\{f\in C_{u}(X,Y)\mid\forall_{x,y\in
X}(\rho(f(x),f(y))\leq\nu(d(x,y))\\},$ $S(X,\nu):=\\{f\in
C_{u}(X)\mid\forall_{x,y\in X}(|f(x)-f(y)|\leq\nu(d(x,y))\\},$

Clearly, if $\nu_{1}(t)=\sigma t$, $\nu_{2}(t)=\sigma t^{\alpha}$, for every
$t\in[0,+\infty)$, then

$S(X,\nu_{1})=\mathrm{Lip}(X,\sigma),$
$S(X,\nu_{2})=\mathrm{{H{\mathrm{\text{\"{o}}}}l}}(X,\alpha,\sigma).$

Notice that since $a\in(0,1]$, we have that $\nu_{2}$ is concave, hence
subadditive. It is immediate to show that if $\nu$ is a modulus of continuity,
then

$\forall_{s,t\in[0,+\infty)}\big{(}s\leq t\Rightarrow\nu(s)\leq\nu(t)\big{)}.$

######  Definition .

Let $A$ be a subset of $X$. We call $(X,A)$ a $\nu$-McShane-Whitney pair, or
simply a $\nu$-MW-pair, if for every $g\in S(A,\nu)$ the functions
$g^{*}_{\nu},{{}^{*}}g_{\nu}:X\rightarrow{\mathbb{R}}$ are well-defined, where
for every $x\in X$

$g^{*}_{\nu}(x):=\sup\\{g(a)-\nu(d(x,a))\mid a\in A\\},$
${{}^{*}}g_{\nu}(x):=\inf\\{g(a)+\nu(d(x,a))\mid a\in A\\}.$

In order to get examples of $\nu$-MW-pairs, one proceeds as in Proposition
2.1(i)-(iii). If, for example, $A$ is totally bounded, then, since $d_{x}$ is
uniformly continuous, $d_{x}(A)$ is a bounded subset of $[0,+\infty)$ and
$\nu\circ d_{x}=\nu_{|d_{x}(A)}\circ d_{x}$ is a uniformly continuous function
as a composition of uniformly continuous functions. Hence ${{}^{*}g}_{\nu}(x)$
and $g^{*}_{\nu}(x)$ are well-defined.

######  Theorem  (McShane-Whitney theorem for $\nu$-continuous functions).

If $(X,A)$ is a $\nu$-MW-pair and $g\in S(A,\nu)$, the following hold.
$(i)$ $g^{*}_{\nu},{{}^{*}}g_{\nu}\in S(X,\lambda)$.
$(ii)$ ${g^{*}_{\nu}}_{|A}=({{}^{*}}g_{\nu})_{|A}=g$.
$(iii)$ $\forall_{f\in S(X,\nu)}(f_{|A}=g\Rightarrow g^{*}_{\nu}\leq
f\leq{{}^{*}}g_{\nu})$.
$(iv)$ The pair $(g^{*}_{\nu}$, ${{}^{*}}g_{\nu})$ is the unique pair of
functions satisfying conditions $(i)$-$(iii)$.

###### Proof.

Based on the properties of Definition 2.3, all steps of the proof of Theorem
2.1 are repeated for the functions ${{}^{*}g_{\nu}},g^{*}_{\nu}$. To show the
implication $d(x_{1},a)\leq
d(x_{2},a)+d(x_{1},x_{2})\Rightarrow\nu(d(x_{1},a))\leq\nu(d(x_{2},a)+d(x_{1},x_{2}))\leq\nu(d(x_{2},a))+\nu(d(x_{1},x_{2}))$
we use the fact that $\forall_{s,t\in[0,+\infty)}\big{(}s\leq
t\Rightarrow\nu(s)\leq\nu(t)\big{)}.$ ∎

##  3\. Normed spaces

###  3.1. McShane-Whitney pairs in normed spaces

Throughout this section $X$ is a normed space with norm $||.||$.

######  Definition .

A subset $C$ of $X$ is called convex, if $\forall_{x,y\in
C}\forall_{\lambda\in(0,1)}(\lambda x+(1-\lambda)y\in C)$. If $C\subseteq X$
is convex, a function $g:C\rightarrow{\mathbb{R}}$ is called convex, if
$\forall_{x,y\in C}\forall_{\lambda\in[0,1]}(g(\lambda
x+(1-\lambda)y)\leq\lambda g(x)+(1-\lambda)g(y)),$ and $g$ is called concave,
if $\forall_{x,y\in C}\forall_{\lambda\in[0,1]}(g(\lambda
x+(1-\lambda)y)\geq\lambda g(x)+(1-\lambda)g(y))$. A function
$f:X\rightarrow{\mathbb{R}}$ is called sublinear if it is subadditive and
positively homogeneous i.e., if $f(x+y)\leq f(x)+f(y)$, and $f(\lambda
x)=\lambda f(x)$, for every $x,y\in X$ and $\lambda>0$, respectively.
Similarly, $f$ is called superlinear, if it is superadditive i.e., if
$f(x+y)\geq f(x)+f(y)$, for every $x,y\in X$, and positively homogeneous.

######  Proposition .

If $C\subseteq X$ is convex, $(X,C)$ is an MW-pair, and
$g\in\mathrm{Lip}(C,\sigma)$, the following hold.
$(i)$ If $g$ is convex, then ${{}^{*}}g$ is convex.
$(ii)$ If $g$ is concave, then $g^{*}$ is concave.

###### Proof.

We show only (i), and for (ii) we work similarly. Let $x,y\in X$, and
$\lambda\in[0,1]$. If

$C_{x}=\\{\lambda(g(c)+\sigma||x-c||)\mid c\in C\\},$ $\ \ \ \ \ \ \
D_{y}=\\{(1-\lambda)(g(c)+\sigma||y-c||)\mid c\in C\\},$

an element of $C_{x}+D_{y}$ has the form
$\lambda(g(c)+\sigma||x-c||)+(1-\lambda)(g(d)+\sigma||y-d||)$, for some
$c,d\in C$. We show that

${{}^{*}}g(\lambda
x+(1-\lambda)y)\leq\lambda(g(c)+\sigma||x-c||)+(1-\lambda)(g(d)+\sigma||y-d||),$

where $c,d\in C$. Since $C$ is convex, $c{{}^{\prime}}:=\lambda
c+(1-\lambda)d\in C$, and by the convexity of $g$ we get

$\displaystyle{{}^{*}}g(\lambda x+(1-\lambda)y)$ $\displaystyle\leq
g(c{{}^{\prime}})+\sigma||\lambda x+(1-\lambda)y-c{{}^{\prime}}||$
$\displaystyle\leq\lambda
g(c)+(1-\lambda)g(d)+\lambda\sigma||x-c||+(1-\lambda)\sigma||y-d||$
$\displaystyle=\lambda(g(c)+\sigma||x-c||)+(1-\lambda)(g(d)+\sigma||y-d||).$

Since the element of $C_{x}+D_{y}$ considered is arbitrary, we get that
${{}^{*}}g(\lambda x+(1-\lambda)y)\leq\inf(C_{x}+D_{y})=\inf C_{x}+\inf
D_{y}=\lambda{{}^{*}}g(x)+(1-\lambda){{}^{*}}g(y).$ ∎

The next proposition says that if $g$ is linear, then ${{}^{*}}g$ is sublinear
and $g^{*}$ is superlinear.

######  Proposition .

If $A$ is a non-trivial subspace of $X$ such that $(X,A)$ is an MW-pair, and
if $g\in\mathrm{Lip}(A,\sigma)$ is linear, the following hold.
$(i)$ $g^{*}(x_{1}+x_{2})\geq g^{*}(x_{1})+g^{*}(x_{2})$, and
${{}^{*}}g(x_{1}+x_{2})\leq{{}^{*}}g(x_{1})+{{}^{*}}g(x_{2})$.
$(ii)$ If $\lambda>0$, then $g^{*}(\lambda x)=\lambda g^{*}(x)$ and
${{}^{*}}g(\lambda x)=\lambda{{}^{*}}g(x)$.
$(iii)$ If $\lambda<0$, then $g^{*}(\lambda x)=\lambda{{}^{*}}g(x)$ and
${{}^{*}}g(\lambda x)=\lambda g^{*}(x)$.

###### Proof.

In each case we show only one of the two facts.
(i) If $a_{1},a_{2}\in A$, then

$\displaystyle{{}^{*}}g(x_{1}+x_{2})$
$\displaystyle=\inf\\{g(a)+\sigma||(x_{1}+x_{2})-a||\mid a\in A\\}$
$\displaystyle\leq g(a_{1}+a_{2})+\sigma||x_{1}+x_{2}-(a_{1}+a_{2})||$
$\displaystyle\leq
g(a_{1})+\sigma||x_{1}-a_{1}||+g(a_{2})+\sigma||x_{2}-a_{2}||,$

therefore

$\displaystyle{{}^{*}}g(x_{1}+x_{2})$
$\displaystyle\leq\inf\\{g(a_{1})+\sigma||x_{1}-a_{1}||+g(a_{2})+\sigma||x_{2}-a_{2}||\mid
a_{1},a_{2}\in A\\}$ $\displaystyle=\inf\\{g(a_{1})+\sigma||x_{1}-a_{1}||\mid
a_{1}\in A\\}+\inf\\{g(a_{2})+\sigma||x_{2}-a_{2}||\mid a_{2}\in A\\}$
$\displaystyle={{}^{*}}g(x_{1})+{{}^{*}}g(x_{2}).$

(ii) First we show that ${{}^{*}}g(\lambda x)\leq\lambda{{}^{*}}g(x)$. If
$a\in A$, then

$\displaystyle{{}^{*}}g(\lambda x)$ $\displaystyle=\inf\\{g(a)+\sigma||\lambda
x-a||\mid a\in A\\}$ $\displaystyle\leq g(\lambda a)+\sigma||\lambda x-\lambda
a||$ $\displaystyle=\lambda g(a)+|\lambda|\sigma||x-a||$
$\displaystyle=\lambda(g(a)+\sigma||x-a||),$

therefore

$\displaystyle{{}^{*}}g(\lambda x)$
$\displaystyle\leq\inf\\{\lambda(g(a)+\sigma||x-a||)\mid a\in A\\}$
$\displaystyle=\lambda\inf\\{g(a)+\sigma||x-a||\mid a\in A\\}$
$\displaystyle=\lambda{{}^{*}}g(x).$

For the inequality ${{}^{*}}g(\lambda x)\geq\lambda{{}^{*}}g(x)$ we work as
follows.

$\displaystyle\lambda{{}^{*}}g(x)$
$\displaystyle=\lambda\inf\\{g(a)+\sigma||x-a||\mid a\in A\\}$
$\displaystyle\leq\lambda(g(\frac{1}{\lambda}a)+\sigma||x-\frac{1}{\lambda}a||)$
$\displaystyle=g(a)+\sigma|\lambda|||x-\frac{1}{\lambda}a||$
$\displaystyle=g(a)+\sigma||\lambda(x-\frac{1}{\lambda}a)||$
$\displaystyle=g(a)+\sigma||\lambda x-a||,$

therefore

$\displaystyle\lambda{{}^{*}}g(x)$
$\displaystyle\leq\inf\\{g(a)+\sigma||\lambda x-a||\mid a\in
A\\}={{}^{*}}g(\lambda x).$

(iii) First we show that ${{}^{*}}g(\lambda x)\leq\lambda g^{*}(x)$. If $a\in
A$, then

$\displaystyle{{}^{*}}g(\lambda x)$ $\displaystyle=\inf\\{g(a)+\sigma||\lambda
x-a||\mid a\in A\\}$ $\displaystyle\leq g(\lambda a)+\sigma||\lambda x-\lambda
a||$ $\displaystyle=\lambda g(a)+|\lambda|\sigma||x-a||$
$\displaystyle=\lambda(g(a)-\sigma||x-a||),$

therefore

$\displaystyle{{}^{*}}g(\lambda x)$
$\displaystyle\leq\inf\\{\lambda(g(a)-\sigma||x-a||)\mid a\in A\\}$
$\displaystyle=\lambda\sup\\{g(a)-\sigma||x-a||\mid a\in A\\}$
$\displaystyle=\lambda g^{*}(x).$

For the inequality ${{}^{*}}g(\lambda x)\geq\lambda g^{*}(x)$ we work as
follows. Since

$\displaystyle g^{*}(x)$ $\displaystyle=\sup\\{g(a)-\sigma||x-a||\mid a\in
A\\}$ $\displaystyle\geq g(\frac{1}{\lambda}a)-\sigma||x-\frac{1}{\lambda}a||$

and $\lambda<0$, we get

$\displaystyle\lambda g^{*}(x)$
$\displaystyle\leq\lambda(g(\frac{1}{\lambda}a)-\sigma||x-\frac{1}{\lambda}a||)$
$\displaystyle=g(a)-\lambda\sigma||x-\frac{1}{\lambda}a||$
$\displaystyle=g(a)+\sigma|\lambda|||x-\frac{1}{\lambda}a||$
$\displaystyle=g(a)+\sigma||\lambda(x-\frac{1}{\lambda}a)||$
$\displaystyle=g(a)+\sigma||\lambda x-a||,$

therefore

$\displaystyle\lambda g^{*}(x)$ $\displaystyle\leq\inf\\{g(a)+\sigma||\lambda
x-a||\mid a\in A\\}={{}^{*}}g(\lambda x).$ ∎

As we have already said in subsection 2.1, if $x_{0}\in X$, it is not
generally the case that ${\mathbb{R}}x_{0}:=\\{\lambda
x_{0}\mid\lambda\in{\mathbb{R}}\\}$ is a located subset of $X$. If
$||x_{0}||>0$ though, ${\mathbb{R}}x_{0}$ is a $1$-dimensional subspace of $X$
i.e., a closed and located linear subset of $X$ of dimension one (see [4], p.
307). Of course, ${\mathbb{R}}x_{0}$ is a convex subset of $X$.

A standard corollary of the classical Hahn-Banach theorem is that if
$x_{0}\neq 0$, there is a bounded linear functional $u$ on $X$ such that
$||u||=1$ and $u(x_{0})=||x_{0}||$. Its proof is based on the extension of the
obvious linear map on ${\mathbb{R}}x_{0}$ to $X$ through the Hahn-Banach
theorem. Next follows a first approach to the translation of this corollary in
Lipschitz analysis. Of course, if $x_{0}\in X$, then $||.||\in\mathrm{Lip}(X)$
and $L(||.||)=1$. To find a Lipschitz function on $X$, which is also the
McShane-Whitney extension of some Lipschitz function defined on the set of a
non-trivial McShane-Whitney pair, we work as follows. It is immediate to see
that if $x_{0}\in X$ such that $||x_{0}||>0$, then
${\mathbb{I}}x_{0}:=\\{\lambda x_{0}\mid\lambda\in[-1,1]\\}$ is a compact
subset of $X$. If we define $g:{\mathbb{I}}x_{0}\rightarrow{\mathbb{R}}$ by
$g(\lambda x_{0})=\lambda||x_{0}||,$ for every $\lambda\in[-1,1]$, then
$g\in\mathrm{Lip}({\mathbb{I}}x_{0})$ and $L(g)=1$; if $\lambda,\mu\in[-1,1]$,
then $|g(\lambda x_{0})-g(\mu
x_{0})|=|\lambda||x_{0}||-\mu||x_{0}|||=|\lambda-\mu|||x_{0}||=||(\lambda-\mu)x_{0}||=||\lambda
x_{0}-\mu x_{0}||,$ and since

$M_{0}(g)=\bigg{\\{}\sigma_{\lambda x_{0},\mu x_{0}}(g)=\frac{|g(\lambda
x_{0})-g(\mu x_{0})|}{||\lambda x_{0}-\mu
x_{0}||}=1\mid(\lambda,\mu)\in[-1,1]_{0}\bigg{\\}},$

we get that $L(g)=\sup M_{0}(g)=1$. Since ${\mathbb{I}}x_{0}$ is totally
bounded, by Proposition 2.1(i) we have that $(X,{\mathbb{I}}x_{0})$ is an MW-
pair, hence the extension ${{}^{*}}g$ of $g$ is in $\mathrm{Lip}(X)$. By
Proposition 2.1 we have that $L({{}^{*}}g)=L(g)=1$.

######  Theorem .

Let $x_{0}\in X$ such that $||x_{0}||>0$. If $(X,{\mathbb{R}}x_{0})$ is a MW-
pair, there exists a linear function $g\in\mathrm{Lip}({\mathbb{R}}x_{0})$
with $L(g)=1$, such that ${{}^{*}}g$ is a sublinear Lipschitz function on $X$
with ${{}^{*}}g(x_{0})=||x_{0}||$ and $L({{}^{*}}g)=1$, and $g^{*}$ is a
superlinear Lipschitz function on $X$ with $g^{*}(x_{0})=||x_{0}||$ and
$L(g^{*})=1$.

###### Proof.

As in the proof of the previous remark, the function
$g:{\mathbb{R}}x_{0}\rightarrow{\mathbb{R}}$, defined by $g(\lambda
x_{0})=\lambda||x_{0}||,$ for every $\lambda\in{\mathbb{R}}$, is in
$\mathrm{Lip}({\mathbb{R}}x_{0})$ and $L(g)=1$. Since $(X,{\mathbb{R}}x_{0})$
is an MW-pair, the extension ${{}^{*}}g$ of $g$ is a Lipschitz function, and
by Proposition 2.1 we get $L({{}^{*}}g)=L(g)=1$. Since $g$ is linear, by
Proposition 3.1 we have that ${{}^{*}}g$ is sublinear. Similarly, the
extension $g^{*}$ of $g$ is a Lipschitz function, and by Proposition 2.1 we
get $L(g^{*})=L(g)=1$. Since $g$ is linear, by Proposition 3.1 we have that
$g^{*}$ is superlinear. ∎

###  3.2. Approximate McShane-Whitney theorem for normed spaces

Next we show the approximate version of the McShane-Whitney theorem for normed
spaces. The technique of our proof888The author would like to thank Hajime
Ishihara for bringing this technique to our attention. is found in the proof
of the lemma that is necessary to prove the approximate version of the
constructive Hahn-Banach theorem (Lemma 5.9 in [6], p. 39). What we get in the
conclusion of the approximate McShane-Whitney theorem is not a
$\sigma$-Lipschitz extension of a given $\sigma$-Lipschitz function $g$, but a
$(1+\epsilon)\sigma$-Lipschitz extension of $g$, for every $\epsilon>0$, where
the Lipschitz function $g$ is defined on a finite-dimensional subspace of a
normed space. Of course, if we consider the trivial subspace $\\{0\\}$ of $X$,
the McShane-Whitney theorem holds trivially.

######  Theorem  (Approximate McShane-Whitney theorem for normed spaces).

Let $A$ be a non-trivial, finite dimensional, linear subspace of $X$ and
$g\in\mathrm{Lip}(A,\sigma)$, for some $\sigma>0$. If $\epsilon>0$, there
exist functions $g^{*}_{\epsilon},{{}^{*}}g_{\epsilon}:X\to{\mathbb{R}}$ such
that:
$(i)$
$g^{*}_{\epsilon},{{}^{*}}g_{\epsilon}\in\mathrm{Lip}(X,(1+\epsilon)\sigma)$.
$(ii)$ $({g^{*}_{\epsilon}})_{|A}=({{}^{*}}g_{\epsilon})_{|A}=g$.
$(iii)$ $\forall_{f\in\mathrm{Lip}(X,(1+\epsilon)\sigma)}(f_{|A}=g\Rightarrow
g^{*}_{\epsilon}\leq f\leq{{}^{*}}g_{\epsilon})$.
$(iv)$ The pair $(g^{*}_{\epsilon}$, ${{}^{*}}g_{\epsilon})$ is the unique
pair of functions satisfying conditions $(i)$-$(iii)$.

###### Proof.

Suppose first that $g(0)=0$.
Let $a,b\in A$ and $x\in X$. Since
$g(a)-g(b)\leq\sigma||a-b||\leq\sigma||a-x||+\sigma||x-b||$, we have that

$g(a)-\sigma||a-x||\leq g(b)+\sigma||x-b||,$

hence

$\phi_{\epsilon,x}(a):=g(a)-(1+\epsilon)\sigma||a-x||\leq
g(b)+(1+\epsilon)\sigma||x-b||=:\theta_{\epsilon,x}(b).$

We find $r>0$, such that for every $a\in A$

$||a||\geq
r\Rightarrow\phi_{\epsilon,x}(a)\leq-(1+\epsilon)\sigma||x||=\phi_{\epsilon,x}(0).$
(1)

Let

$r:=\frac{2(1+\epsilon)}{\epsilon}||x||.$

We show first that

$||a||\geq r\Rightarrow||a||-(1+\epsilon)||a-x||\leq-(1+\epsilon)||x||.$ (2)

We have that

$\displaystyle||a-x||\geq||a||-||x||$
$\displaystyle\Rightarrow-||a-x||\leq||x||-||a||$
$\displaystyle\stackrel{{\scriptstyle\times(1+\epsilon)}}{{\Rightarrow}}-(1+\epsilon)||a-x||\leq(1+\epsilon)||x||-(1+\epsilon)||a||$
$\displaystyle\stackrel{{\scriptstyle+||a||}}{{\Rightarrow}}||a||-(1+\epsilon)||a-x||\leq(1+\epsilon)||x||-(1+\epsilon)||a||+||a||$
$\displaystyle\Leftrightarrow||a||-(1+\epsilon)||a-x||\leq(1+\epsilon)||x||-\epsilon||a||.$

By the definition of $r$ we have that

$\displaystyle||a||\geq\frac{2(1+\epsilon)}{\epsilon}||x||$
$\displaystyle\Rightarrow-||a||\leq-\frac{2(1+\epsilon)}{\epsilon}||x||$
$\displaystyle\stackrel{{\scriptstyle\times\epsilon}}{{\Rightarrow}}-\epsilon||a||\leq-2(1+\epsilon)||x||,$

hence

$||a||-(1+\epsilon)||a-x||\leq(1+\epsilon)||x||-\epsilon||a||\leq(1+\epsilon)||x||-2(1+\epsilon)||x||=-(1+\epsilon)||x||.$

Hence, if $||a||\geq r$, multiplying the conclusion of (2) by $\sigma>0$ we
get

$\sigma||a||-(1+\epsilon)\sigma||a-x||\leq-(1+\epsilon)\sigma||x||.$

Since

$g(a)=g(a)-g(0)\leq|g(a)-g(0)|\leq\sigma||a-0||=\sigma||a||,$

we get

$g(a)-(1+\epsilon)\sigma||a-x||\leq-(1+\epsilon)\sigma||x||,$

which is the required conclusion of (1). Since $A$ is finite dimensional
space, the set $K_{r}:=\\{a\in A\mid||a||\leq r\\}$ is compact (see [6], p.
34), and

$s_{\epsilon,x}:=\sup\\{\phi_{\epsilon,x}(a)\mid a\in K_{r}\\}$

is well defined, since $\phi$ is uniformly continuous and $K_{r}$ is totally
bounded. Since $0\in K_{r}$ and since the set $\\{\phi_{\epsilon,x}(a)\mid
a\in A\ \&\ ||a||\geq r\\}$ is bounded above by $\phi_{\epsilon,x}(0)$, we
get999Here we use the simple fact that if $D\subseteq B\subseteq{\mathbb{R}}$
such that $D$ is dense in $B$ and $\sup D$ exists, then $\sup B$ exists and
$\sup B=\sup D$. The set $E=K_{r}\cup\\{a\in A\mid||a||\geq r\\}$ is dense in
$A$ and $\phi_{\epsilon,x}(E)$ is dense in $\phi_{\epsilon,x}(A)$.

$s_{\epsilon,x}=\sup\\{\phi_{\epsilon,x}(a)\mid a\in A\\}.$

Hence we define

$g^{*}_{\epsilon}(x):=s_{\epsilon,x}=\sup\\{g(a)-(1+\epsilon)\sigma||x-a||\mid
a\in A\\}.$

Next we show that $g^{*}_{\epsilon}\in\mathrm{Lip}(X,(1+\epsilon)\sigma)$.
Since $||x-a||\leq||x-y||+||y-a||$, we get $g(a)-(1+\epsilon)\sigma||x-a||\geq
g(a)-(1+\epsilon)\sigma||y-a||-(1+\epsilon)\sigma||x-y||$, and
$(1+\epsilon)\sigma||x-y||+g(a)-(1+\epsilon)\sigma||x-a||\geq
g(a)-(1+\epsilon)\sigma||y-a||$, hence
$(1+\epsilon)\sigma||x-y||+g(a)-(1+\epsilon)\sigma||x-a||\geq
g^{*}_{\epsilon}(y)$, and $(1+\epsilon)\sigma||x-y||+g^{*}_{\epsilon}(x)\geq
g^{*}_{\epsilon}(y)$. Consequently,

$g^{*}_{\epsilon}(y)-g^{*}_{\epsilon}(x)\leq(1+\epsilon)\sigma||x-y||.$

Working similarly, we also get
$g^{*}_{\epsilon}(x)-g^{*}_{\epsilon}(y)\leq(1+\epsilon)\sigma||x-y||.$
Suppose next that $a_{0}\in A$. By definition we have that
$g^{*}_{\epsilon}(a_{0})=\sup\\{g(a)-(1+\epsilon)||a_{0}-a||\mid a\in A\\}.$
Clearly, $g(a_{0})=g(a_{0})-(1+\epsilon)\sigma||a_{0}-a_{0}||\leq
g^{*}_{\epsilon}(a_{0})$. If $a\in A$, then
$g(a)-g(a_{0})\leq\sigma||a-a_{0}||\leq(1+\epsilon)\sigma||a-a_{0}||$, hence
$-g(a_{0})\leq-g(a)+(1+\epsilon)\sigma||a-a_{0}||$ and $g(a_{0})\geq
g(a)-(1+\epsilon)\sigma||a-a_{0}||$. Since $a\in A$ is arbitrary, we get
$g(a_{0})\geq g^{*}_{\epsilon}(a_{0})$. If
$f\in\mathrm{Lip}(X,(1+\epsilon)\sigma)$ such that $f_{|A}=g$, then
$g(a)-f(x)=f(a)-f(x)\leq(1+\epsilon)\sigma||x-a||$, hence
$g(a)-(1+\epsilon)\sigma||x-a||\leq f(x)$, and since $a\in A$ is arbitrary, we
get $g^{*}_{\epsilon}(x)\leq f(x)$.
For the construction of ${{}^{*}}g_{\epsilon}$ we work similarly, as follows.
The same $r>0$ that we chose earlier satisfies, for every $a\in A$, also the
property

$||a||\geq
r\Rightarrow\theta_{\epsilon,x}(a)=g(a)+(1+\epsilon)\sigma||x-a||\geq(1+\epsilon)\sigma||x||=\theta_{\epsilon,x}(0).$

The reason for this is that since we can choose $||a||\geq r$ such that
$||a||\leq(1+\epsilon)(||a-x||-||x||)$, we have that

$\displaystyle-||a||\geq(1+\epsilon)(||x||-||a-x||)\Rightarrow$
$\displaystyle-||a||+(1+\epsilon)(||a-x||)\geq(1+\epsilon)||x||\Rightarrow$
$\displaystyle-\sigma||a||+(1+\epsilon)\sigma(||a-x||)\geq(1+\epsilon)\sigma||x||\Rightarrow$
$\displaystyle g(a)+(1+\epsilon)\sigma(||a-x||)\geq(1+\epsilon)\sigma||x||,$

since from the inequality $-g(a)\leq|g(a)|\leq\sigma||a||$, we get
$g(a)\geq-\sigma||a||$. Similarly,

$\iota_{\epsilon,x}:=\inf\\{\theta_{\epsilon,x}(a)\mid a\in K_{r}\\}$

is well-defined. Since the set $\\{\theta_{\epsilon,x}(a)\mid a\in A\ \&\
||a||\geq r\\}$ is bounded below by $\theta_{\epsilon,x}(0)$, we get

$\iota_{\epsilon,x}=\inf\\{\theta_{\epsilon,x}(a)\mid a\in A\\}.$

Hence, we define

${{}^{*}}g_{\epsilon}(x):=\iota_{\epsilon,x}=\inf\\{g(a)+(1+\epsilon)\sigma||x-a||\mid
a\in A\\}.$

To show the properties of ${{}^{*}}g_{\epsilon}$ in (i)-(iii), we proceed as
we did in the proof of the properties of $g^{*}_{\epsilon}$. The uniqueness of
the pair $(g^{*}_{\epsilon},{{}^{*}}g_{\epsilon})$ follows immediately.
In the general case, where we do not assume that $g(0)=0$, we work as follows.
If $h:A\to{\mathbb{R}}$ is defined by

$h(a):=g(a)-g(0),$

for every $a\in A$, then $h(0)=0$. Clearly, $h\in\mathrm{Lip}(A,\sigma)$,
hence by our proof above there are functions
$h^{*}_{\epsilon},{{}^{*}}h_{\epsilon}:X\to{\mathbb{R}}$ such that:
(i)${}^{{}^{\prime}}$
$h^{*}_{\epsilon},{{}^{*}}h_{\epsilon}\in\mathrm{Lip}(X,(1+\epsilon)\sigma)$.
(ii)${}^{{}^{\prime}}$
$({h^{*}_{\epsilon}})_{|A}=({{}^{*}}h_{\epsilon})_{|A}=h$.
(iii)${}^{{}^{\prime}}$
$\forall_{F\in\mathrm{Lip}(X,(1+\epsilon)\sigma)}(F_{|A}=h\Rightarrow
h^{*}_{\epsilon}\leq F\leq{{}^{*}}h_{\epsilon})$.
(iv)${}^{{}^{\prime}}$ The pair $(h^{*}_{\epsilon}$, ${{}^{*}}h_{\epsilon})$
is the unique pair of functions satisfying conditions
(i)${}^{{}^{\prime}}{-}$(iii)${}^{{}^{\prime}}$.
We define $g^{*}_{\epsilon},{{}^{*}}g_{\epsilon}:X\to{\mathbb{R}}$ by

$g^{*}_{\epsilon}(x):=h^{*}_{\epsilon}(x)+g(0),$
${{}^{*}}g_{\epsilon}(x):={{}^{*}}h_{\epsilon}(x)+g(0),$

for every $x\in X$. Properties (i)-(iv) for $g^{*}_{\epsilon}$ and
${{}^{*}}g_{\epsilon}$ follow immediately from properties
(i)${}^{{}^{\prime}}{-}$(iv)${}^{{}^{\prime}}$ for $h^{*}_{\epsilon}$ and
${{}^{*}}h_{\epsilon}$. ∎

##  4\. Concluding remarks and open questions

In this paper we examined the McShane-Whitney extension of Lipschitz functions
from the Bishop-style, constructive point of view. As in the case of Tietze’s
theorem, or the Hahn-Banach theorem, the generality of the classical McShane-
Whitney theorem is lost. In reward though, the computational content of this
extension is constructively studied through the notion of a McShane-Whitney
pair.

Moreover, the analogies between the Hahn-Banach theorem and the McShane-
Whitney theorem made possible to transfer methods and results from one to the
other. A significant example of the interplay between the two theorems is the
approximate McShane-Whitney theorem, the proof of which uses the proof-
technique of the approximate Hahn-Banach theorem.

Another example that we plan to elaborate in future work is the following. If
$p$ is a seminorm on a vector space $X$ i.e., $p:X\to{\mathbb{R}}$ such that
(i) $p(x)\geq 0$, (ii) $p(x+y)\leq p(x)+p(y),$ and (iii) $p(\lambda
x)=|\lambda|p(x)$, for every $x,y\in X$ and $\lambda\in{\mathbb{R}}$, the set
of $p$-Lipschitz functions on $X$ is defined by

$p{-}\mathrm{Lip}(X):=\bigcup_{\sigma\geq 0}p{-}\mathrm{Lip}(X,\sigma),$
$p{-}\mathrm{Lip}(X,\sigma):=\\{f\in\mathbb{F}(X,Y)\mid\forall_{x,y\in
X}\big{(}|f(x)-f(y)|\leq\sigma p(x-y)\big{)}\\}.$

One can define a $p$-MW-pair in the obvious way. In order to get examples of
$p$-MW-pairs, one can define $p$-versions of locatedness, total boundedness
and uniform continuity, and prove, similarly, the $p$-versions of the
propositions in Bishop and Bridges pp. 94-95 and the $p$-version of
Proposition 2.1. The proof of the expected McShane-Whitney theorem for
$p$-Lipschitz functions is similar to the proof of Theorem 2.1. The following
MW-version of the analytic Hahn-Banach theorem for seminorms101010According to
it, if $p$ is a seminorm on $X$, $A$ a subspace of $X$ and
$g:A\to{\mathbb{R}}$ a linear function such that $\forall_{a\in
A}\big{(}g(a)\leq p(a)\big{)}$, there is linear function $f:X\to{\mathbb{R}}$
that extends $g$ and $\forall_{x\in X}\big{(}f(x)\leq p(x)\big{)}$. In its
general formulation, the analytic Hahn-Banach theorem refers to sublinear
functionals $p$, where the condition (iii) of a seminorm is replaced by the
condition $p(\lambda x)=\lambda p(x)$, for $\lambda\geq 0$ (see [16], p. 120).
Its proof uses Zorn’s lemma. can be shown.
If $p$ is a seminorm on $X$, $(X,A)$ is a $p$-MW-pair, $A$ is a subspace of
$X$, and $g:A\to{\mathbb{R}}$ is linear, then if $\forall_{a\in A}(g(a)\leq
p(a))$, there is a superlinear extension $g^{*}$ of $g$ such that
$g^{*}\in\mathrm{p{\text{-}}Lip}(X,1)$ and $\forall_{x\in X}(g^{*}(x)\leq
p(x))$, while if $\forall_{a\in A}(g(a)\geq p(a))$, there is a sublinear
extension ${{}^{*}}g$ of $g$ such that
${{}^{*}}g\in\mathrm{p{\text{-}}Lip}(X,1)$ and $\forall_{x\in
X}({{}^{*}}g(x)\geq p(x))$.
With respect to the analytic Hahn-Banach theorem for seminorms we lose
linearity of the extension function, since in the case (i) of the previous
result $g^{*}$ is only superlinear, but instead we have that
$g^{*}\in\mathrm{p{\text{-}}Lip}(X,1)$, and we avoid Zorn’s lemma. We expect
to find good applications of the previous result in convex analysis, where
linearity is not as important as sublinearity, or superlinearity.

Moreover, the following problems and questions arise naturally from our work.

  1. (1)

To find necessary and sufficient conditions on metric spaces $X,Y$ and on
function $f\in\mathrm{Lip}(X,Y)$ such that $L(f)$ and/or $L^{*}(f)$ exist.
This problem is similar to the normability of a bounded linear functional $T$,
which, constructively is equivalent to the locatedness of the kernel of $T$.

  2. (2)

To characterise, for given metric space $X$, its MW-pairs $(X,A)$.

  3. (3)

To find necessary and/or sufficient conditions on a normed space $(X,||.||)$
such that $(X,{\mathbb{R}}x_{0})$ is an MW-pair for some given $x_{0}\in X$
with $||x_{0}||>0$. We expect that the properties added by Ishihara in [13] to
the hypotheses of the Hahn-Banach theorem i.e., the uniform convexity and
separability of $X$ and the Gâteaux differentiability of its norm, are
interesting properties to be tested for this question too.

  4. (4)

To connect the separation version of the Hahn-Banach theorem to the McShane-
Whitney theorem.

We hope to address these questions and problems in future work.

The first version of this paper was completed during our research visit to the
Japan Advanced Institute of Science and Technology (JAIST) that was funded by
the European-Union project “Computing with Infinite Data”. We would like to
thank Hajime Ishihara for hosting us in JAIST and discussing with us the
material of this work. The second version of this paper was completed during
our research visit to the Philosophy Department of Carnegie Mellon University
that was funded by the same European-Union project. We would like to thank
Wilfried Sieg for hosting us in CMU.

We would also like to thank the anonymous referees for their very helpful
comments and suggestions.

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