# Tietze type extensions for absolutely continuous functions in the plane

Ian Doust
Alan Stoneham
School of Mathematics and Statistics
University of New South Wales
UNSW Sydney 2052
Australia

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###### Abstract

It is an open problem whether one can always extend an absolutely continuous
function (in the sense of Ashton and Doust) on a compact subset of the plane
to a larger compact set. In this paper we show that this can be done for a
large family of initial domains whose components consist of polygons and
convex curves. An application is given to the spectral theory of $AC(\sigma)$
operators.

††2020 _Mathematics Subject Classification_ : Primary 26B30; Secondary 47B40,
54C20.†† _Key words and phrases_ : Functions of bounded variation, Absolutely
continuous functions, $AC(\sigma)$ operators.

##  1 Introduction

A Hilbert space operator $T$ is normal if and only if it admits a
$C(\sigma(T))$ functional calculus. Indeed, if there is any nonempty compact
set $\sigma\subseteq\mathbb{C}$ and a $C^{*}$-algebra isomorphism
$\Phi:C(\sigma)\to B(H)$ such that $\Psi$ maps the identity function on
$\sigma$ to $T$, then $T$ is normal. In this case $\sigma(T)\subseteq\sigma$
and the kernel of $\Phi$ is just the set of continuous functions which vanish
on $\sigma(T)$.

In [4] Doust and Ashton introduced a class of Banach space operators called
${AC}(\sigma)$ operators which generalize the classes of well-bounded
operators and trigonometrically well-bounded operators. In the setting of
Hilbert spaces, all self-adjoint operators are well-bounded and all unitary
operators are trigonometrically well-bounded, and in this context $AC(\sigma)$
operators may be seen as an analogue of normal operators (see, for example,
[12] and [5, 6, 7] for more details of these classes of operators). An
operator is an $AC(\sigma)$ operator if it admits a functional calculus for an
algebra $AC(\sigma)$ of ‘absolutely continuous’ functions on $\sigma$ which
was introduced in [3]. (Details of the spaces $AC(\sigma)$ for compact sets
$\sigma\subseteq\mathbb{C}$ will be given in Section 2.)

If $T$ is an $AC(\sigma)$ operator, then certainly $\sigma(T)\subseteq\sigma$,
but at present it remains an open problem as to whether $T$ must in fact be an
$AC(\sigma(T))$ operator. Such questions are closely related to classical
extension theorems. Suppose that $\sigma_{0}\subseteq\sigma$ are two compact
subsets of the plane. The Tietze Extension Theorem ensures that if $f\in
C(\sigma_{0})$ then there exists ${\hat{f}}\in C(\sigma)$ such that
$f={\hat{f}}|\sigma_{0}$ and such that $\left\lVert
f\right\rVert_{\infty}=\lVert{\hat{f}}\rVert_{\infty}$. The main focus of this
paper concerns the open question as to whether a corresponding extension
result holds for absolutely continuous functions. In both settings, the
restriction map $g\mapsto g|\sigma_{0}$ is a norm 1 algebra homomorphism. The
extension question is asking whether this map is onto.

The above fact about the restriction map means that if one can extend a
function on $\sigma_{0}$ to any suitably large square or disk, then one can
extend it to any compact superset. This means that the primary issue in such
questions is the nature of the original domain $\sigma_{0}$, rather than the
nature of $\sigma$.

For certain classes of sets, an extension of an absolutely continuous function
is always possible. This is easily seen to be the case if, for example,
$\sigma_{0}$ is a compact subset of the real line. On the other hand, as the
following example shows, not every ‘natural’ extension of an absolutely
continuous function is absolutely continuous.

######  Example 1.1.

Let $C:[0,1]\to[0,1]$ denote the Cantor function. Let
$\sigma_{0}=\\{(x,C(x))\,:\,x\in[0,1]\\}$, let $\sigma=[0,1]\times[0,1]$ and
let $f:\sigma_{0}\to\mathbb{C}$ be defined by $f(x,y)=y$,
$(x,y)\in\sigma_{0}$. Then (by the definitions in the next section) $f\in
AC(\sigma_{0})$. One might consider both the functions ${\hat{f}}_{1}(x,y)=y$
and ${\hat{f}}_{2}(x,y)=C(x)$ as natural continuous extensions of $f$ to
$\sigma$, but only the first of these is absolutely continuous.

A few particular extension results for absolutely continuous functions have
appeared in the literature. For example, a formula which extends an absolutely
continuous function on the boundary of a square to its interior was given in
[4, Theorem 3.5].

Figure 1: An example of a plane set $\sigma_{0}$ which will satisfy the
hypotheses of the main extension result, Theorem 10.1.

The main result of this paper, Theorem 10.1, gives sufficient conditions on
$\sigma_{0}$ which ensure that absolutely continuous extensions to supersets
always exist. The main theorem covers sets such as the one in Figure 1 which
are made up of polygonal regions with polygonal holes and convex curves
satisfying certain hypotheses. Since the sets may be made up of pieces of a
somewhat different nature, the proof essentially progresses by dealing with
different parts of the set separately and then patching the pieces together.
This will involve proving a number of ‘joining theorems’.

These theorems, which are of independent interest, examine when one can say
that a function which is absolutely continuous on each of two sets is then
absolutely continuous on their union. We give some examples to show that even
when the two sets are quite simple, such a conclusion may not hold.

In the simplest situation, where
$\sigma_{0}\subseteq\sigma\subseteq\mathbb{R}$, one can always extend an
absolutely continuous function on $\sigma_{0}$ to one on $\sigma$ without an
increase in the variation norm. For the plane sets considered in this paper,
it is not known whether one can always find an extension with the same norm.
Certainly our methods do not produce such an extension. On the other hand we
are able to at least show that the norm of the extension is controlled by some
constant (depending on $\sigma_{0}$) times the norm of the function on
$\sigma_{0}$. In order to keep the proof as simple as possible we have
generally made no attempt to optimize the bounds, although in a few places we
shall note that where our methods require that the constants that appear are
greater than one.

There are a number of relatively simple sets for which it is not known whether
extensions always exist. These include closed disks and the graph of the
Cantor function considered above.

In the next section we shall recall the main definitions and properties of
absolutely continuous functions. The following section will introduce the
classes of compact sets which appear in the main result, Theorem 10.1.

Although the great majority of the paper is purely concerned with the algebras
of absolutely continuous functions, we shall return in the final section to
revisit the original motivating question concerning $AC(\sigma)$ operators
referred to above. Due to this, we shall work throughout with complex-valued
functions. However, all the results remain true for real-valued functions.

##  2 Definitions and properties of $AC(\sigma)$

Throughout this paper, $\sigma_{0}$ and $\sigma$ will denote nonempty compact
subsets of the plane. We shall consider the plane to be either
$\mathbb{R}^{2}$ or $\mathbb{C}$ as is notationally convenient, and identify
the real line with the $x$-axis in the plane.

For the moment then, fix $\sigma\subseteq\mathbb{R}^{2}$. We shall now briefly
recall the definition of the Banach algebra of functions of bounded variation
on $\sigma$ and of its subalgebra, the set of absolutely continuous functions
on $\sigma$. Further details and proofs may be found in [10].

Suppose that $f:\sigma\to\mathbb{C}$. Given an ordered finite list of (not
necessarily distinct) elements
$S=[{\boldsymbol{x}}_{0},\dots,{\boldsymbol{x}}_{n}]$ in $\sigma$, we set

$\operatorname*{\rm
cvar}(f,S)=\sum_{k=1}^{n}|f({\boldsymbol{x}}_{k})-f({\boldsymbol{x}}_{k-1})|.$

The variation factor of the list $S$, denoted $\operatorname{vf}(S)$, is a
positive integer which roughly measures the maximum number of times that the
piecewise linear path joining the points in $S$ in order crosses any line.
More formally, let $\gamma_{S}:[0,1]\to\mathbb{C}$ be a parameterization of
the piecewise linear curve joining the elements in $S$ in order. Given a line
$\ell$ in the plane, let $L(S,\ell)=\gamma_{S}^{-1}(\ell)\subseteq[0,1]$ and
let $\operatorname{vf}(S,\ell)$ denote the number of connected components of
$L(S,\ell)$. The variation factor of $S$ is then
$\operatorname{vf}(S)=\sup_{\ell}\operatorname{vf}(S,\ell)$. The main fact
that we shall need in the later sections is that if $P$ is an simple $m$-gon
and $S=[{\boldsymbol{x}}_{0},\dots,{\boldsymbol{x}}_{n}]$ has $k$ line
segments $\overline{\vphantom{\vbox
to5.16663pt{}}{\boldsymbol{x}}_{j-1}\,{\boldsymbol{x}}_{j}}$ with one endpoint
inside $P$ and one endpoint outside $P$, then $\operatorname{vf}(S)\geq\lceil
k/m\rceil$. We refer the reader to [10, Section 2.1] and the appendix to [2]
for further details of the properties of the variation factor.

The variation of $f$ over $\sigma$ is found by taking a supremum over all such
finite lists of points:

$\operatorname*{var}(f,\sigma)=\sup_{S}\frac{\operatorname*{\rm
cvar}(f,S)}{\operatorname{vf}(S)}.$

The space of functions of bounded variation on $\sigma$, denoted $BV(\sigma)$,
consists of all functions $f:\sigma\to\mathbb{C}$ for which

$\left\lVert
f\right\rVert_{BV(\sigma)}=\sup_{{\boldsymbol{x}}\in\sigma}|f({\boldsymbol{x}})|+\operatorname*{var}(f,\sigma)$

is finite.

Calculating $\operatorname*{var}(f,\sigma)$ precisely is often challenging, so
we shall mainly be using the properties of variation in order to obtain good
bounds. We recall here the most important of these properties. Further details
can be found in [10].

######  Theorem 2.1.

Suppose that $\sigma$ is a nonempty compact subset of $\mathbb{R}^{2}$. Then

  1. (i)

$BV(\sigma)$ is a Banach algebra under the norm
$\left\lVert\cdot\right\rVert_{BV(\sigma)}$.

  2. (ii)

$BV(\sigma)$ contains the (restrictions of) complex polynomials in two real
variables.

  3. (iii)

If $\sigma=[a,b]\subseteq\mathbb{R}$ then $BV(\sigma)$ is the usual classical
space of functions of bounded variation, and $\operatorname*{var}(f,[a,b])$ is
just the usual measure of the variation of a function on the interval $[a,b]$.

  4. (iv)

If $\phi$ is an invertible affine transformation of the plane and
$\sigma^{\prime}=\phi(\sigma)$, then $\Phi(f)=f\circ\phi^{-1}$ is an isometric
isomorphism from $BV(\sigma)$ to $BV(\sigma^{\prime})$.

######  Definition 2.2.

The space $AC(\sigma)$ of absolutely continuous functions on $\sigma$ is the
closure of the complex polynomials in two real variables in $BV(\sigma)$.

We record some of the most important properties of $AC(\sigma)$ spaces.

######  Theorem 2.3.

Suppose that $\sigma$ is a nonempty compact subset of $\mathbb{R}^{2}$.

  1. (i)

If $\sigma=[a,b]\subseteq\mathbb{R}$ then $AC(\sigma)$ is the usual classical
space of absolutely continuous functions.

  2. (ii)

(Affine invariance) If $\phi$ is an invertible affine transformation of the
plane and $\sigma^{\prime}=\phi(\sigma)$, then $\Phi(f)=f\circ\phi^{-1}$ is an
isometric isomorphism from $AC(\sigma)$ to $AC(\sigma^{\prime})$.

  3. (iii)

If $\sigma_{0}$ is a nonempty compact subset of $\sigma$ and $f\in
AC(\sigma)$, then $f|\sigma_{0}\in AC(\sigma_{0})$ with $\left\lVert
f|\sigma_{0}\right\rVert_{BV(\sigma_{0})}\leq\left\lVert
f\right\rVert_{BV(\sigma)}$.

  4. (iv)

If $\sigma_{x}$ denotes the projection of $\sigma$ onto the $x$-axis, and
$f\in AC(\sigma_{x})$, then $\hat{f}:\sigma\to\mathbb{C}$ defined by
$\hat{f}(x,y)=f(x)$ is in $AC(\sigma)$ and satisfies
$\|\hat{f}\|_{BV(\sigma)}\leq\|f\|_{BV(\sigma_{x})}$.

When there is no risk of confusion, when $f:\sigma\to\mathbb{C}$ and
$\sigma_{0}\subseteq\sigma$ we shall often write $\left\lVert
f\right\rVert_{BV(\sigma_{0})}$ for the norm of the restriction of $f$ to the
smaller set.

An important fact about this sense of absolutely continuity is that it is a
‘local property’. We shall say that a set $U$ is a compact neighbourhood of a
point ${\boldsymbol{x}}\in\sigma$ if there exists a bounded open set
$V\subseteq\mathbb{R}^{2}$ containing ${\boldsymbol{x}}$ such that
$U=\sigma\cap\overline{V}$.

######  Theorem 2.4 (Patching Lemma [10, Theorem 4.1]).

Suppose that $\sigma$ is a nonempty compact subset of $\mathbb{R}^{2}$ and
that $f:\sigma\to\mathbb{C}$. Then $f\in AC(\sigma)$ if and only if for every
point ${\boldsymbol{x}}\in\sigma$ there exists a compact neighbourhood
$U_{\boldsymbol{x}}$ of ${\boldsymbol{x}}$ such that $f|U_{\boldsymbol{x}}\in
AC(U_{\boldsymbol{x}})$.

We note that $AC(\sigma)$ always contains a relatively rich collection of
functions. It is shown in [11, Section 5] that the space of functions which
admit a $C^{1}$ extension to an open neighbourhood of $\sigma$ is always dense
in $AC(\sigma)$, as is the space $\mathrm{CTPP}(\sigma)$ of functions which
are continuous and triangularly piecewise planar. On the other hand, while
Lipschitz functions on a real interval are always absolutely continuous, this
result does not extend to arbitrary compact subsets of the plane [10, Example
3.13].

##  3 Classes of sets

The aim of this paper is to identify the largest class of compact sets in the
plane for which we can prove that any absolutely continuous function on such a
set can always be extended to be absolutely continuous on any larger compact
set. This class contains all finite unions of polygonal regions, as well as
most sets which are finite unions of convex curves.

Given a set $A\subseteq\mathbb{R}^{2}$, we shall denote its closure by
$\mathop{\mathrm{cl}}(A)$, its interior by ${\rm int}(A)$ and its boundary by
$\partial A$.

###  3.1 Polygonal regions

By a polygonal region we mean a connected subset of the plane with (simple)
polygonal boundary. A set $W$ is a (polygonal) window in a polygonal region
$P$ if it is the interior of a polygon $P^{\prime}$ which lies in $P$. Let
$PW^{+}$ denote the collection of all compact subsets of the plane which are
finite unions of polygonal regions with finitely many windows removed.

We allow here that a window may comprise the entire interior of a polygonal
region. Note that for a set in $PW^{+}$, one may have windows inside polygons
inside windows inside polygons, and so forth (see Figure 2). Every set in
$PW^{+}$ is a union of a finite number of polygonal regions and line segments.
While not all such sets are in $PW^{+}$ (for example a single line segment),
our main theorem will cover these cases.

$R$$\sigma$ Figure 2: A set $\sigma\in PW^{+}$ inside a rectangle $R$. The
blue region $R\setminus(\sigma\cup\partial R)$ can always be triangulated.

Let $R$ be a rectangular region containing $\sigma_{0}\in PW^{+}$ in its
interior. Then ${\hat{\sigma}}_{0}=\sigma_{0}\cup\partial R$ is also in
$PW^{+}$. The region $R\setminus{\hat{\sigma}}_{0}$ can always be
triangulated. Given $f\in AC(\sigma_{0})$ our aim will be to inductively
extend $f$ to be absolutely continuous, first on the boundary of each of the
triangles, and then on the whole triangular region. This will require that we
first show that we can extend from a subset of a convex curve to the whole
curve, and then (in Section 6) from the boundary of a polygon to its interior.
We shall then show that the final function obtained is absolutely continuous
on the whole of $R$. This will require us to show (in Section 7) that we can
join absolutely continuous functions defined on polygonal regions.

###  3.2 Almost polygonally inscribed curves

The class of polygonally inscribed curves, or $\mathrm{PIC}$ sets, was
introduced in [1]. These are compact connected sets which can be written as a
finite union of convex curves with certain constraints on how the curves can
meet. For our main theorem we need to relax this condition slightly to deal
with the union of a convex curve and a polygonal region. We cannot however
completely eliminate all restrictions of how the curves may meet since, as we
shall note in Section 8, a function can be absolutely continuous on each of
two convex curves, but not on their union.

######  Definition 3.1.

A (convex) polygonal mosaic in the plane is a finite collection $\mathcal{M}$
of convex polygonal regions such that

  1. 1.

$\bigcup_{P\in\mathcal{M}}P$ is connected;

  2. 2.

if $P$ and $Q$ are distinct elements of $\mathcal{M}$, then $P\cap Q$ is
either

     * •

empty

     * •

a single point which is a vertex of both $P$ and $Q$, or

     * •

a line segment.

######  Definition 3.2.

Let $P$ be a convex polygonal region. A set $c\subseteq P$ is a convex curve
in $P$ if it is a differentiable convex curve joining two vertices of $P$
which only touches the boundary of $P$ at those points.

######  Definition 3.3.

A nonempty compact connected set $\sigma$ is a polygonally inscribed curve if
there exists a polygonal mosaic $\mathcal{M}=\\{P_{i}\\}_{i=1}^{M}$ whose
union contains $\sigma$ and such that for each $i$, $c_{i}=\sigma\cap P_{i}$
is a convex curve in $P_{i}$. The curves $c_{i}$ will be called the components
of $\sigma$. The collection of all polygonally inscribed curves in the plane
will be denoted by $\mathrm{PIC}$.

######  Definition 3.4.

We shall say that a connected compact set $\sigma\subseteq\mathbb{R}^{2}$ is
an almost polygonally inscribed curve, written $\sigma\in\mathrm{APIC}$, if
there exists a polygonal mosaic $\mathcal{M}=\\{P_{i}\\}_{i=1}^{M}$ whose
union contains $\sigma$ and such that for each $i$, $C_{i}=\sigma\cap P_{i}$
is nonempty and is the the union of at most one convex curve in $P_{i}$ and a
(possibly empty) subset of the sides of $P_{i}$. Given
$\sigma\in\mathrm{APIC}$ with a polygonal mosaic $\mathcal{M}$, the components
of $\sigma$ determined by $\mathcal{M}$ are the convex curves and line
segments determined by $\mathcal{M}$ whose union is $\sigma$.

Once split into components, every $\mathrm{PIC}$ or $\mathrm{APIC}$ set can be
considered as the drawing of a planar graph, with the components giving the
edges of the graph, and the intersections of the components the vertices. The
graph is not uniquely determined by the set, since such a set can be
decomposed into components in many different ways.

It is worth noting the differences between $\mathrm{PIC}$ sets and
$\mathrm{APIC}$ sets. Two components of a $\mathrm{PIC}$ set can meet
tangentially, but only with strictly opposite convexity. The new behaviour
permitted for an $\mathrm{APIC}$ set is that a curve and a line may meet
tangentially at a vertex. Such a vertex will be called a line tangential
vertex.

######  Definition 3.5.

Let $\mathrm{SPIC}$ denote the class of all nonempty compact subsets of
$\mathbb{R}^{2}$ which are subsets of some almost polygonally inscribed curve.

$\sigma_{1}$$\sigma_{2}$$\sigma_{3}$$\sigma_{4}$ Figure 3:
$\sigma_{1}\in\mathrm{PIC}$, with a polygonal mosaic shown. The sets
$\sigma_{2},\sigma_{3}$ are in $\mathrm{APIC}\setminus\mathrm{PIC}$ since they
contain line tangential vertices. The set $\sigma_{4}$ is not in
$\mathrm{APIC}$ due to the curves meeting at a vertex with the same convexity.

The components of sets in $\mathrm{PIC}$ or $\mathrm{APIC}$ are differentiable
compact convex curves. A differentiable convex curve $c$ joining
${\boldsymbol{x}}$ to ${\boldsymbol{y}}$ is said to be projectable if the
orthogonal projection of $c$ onto the line through ${\boldsymbol{x}}$ and
${\boldsymbol{y}}$ is just the line segment joining those points. As is shown
in [1], any set in $\mathrm{PIC}$ can be broken into projectable component
curves. This result can be extended to sets in $\mathrm{APIC}$.

######  Lemma 3.6.

Suppose that $\sigma\in\mathrm{APIC}$. Then there exists a polygonal mosaic
$\mathcal{M}=\\{P_{i}\\}_{i=1}^{M}$ for $\sigma$ such that for all $i$, if
$\sigma$ contains a convex curve in $P_{i}$ this curve is projectable.

###### Proof.

Let $\mathcal{M}_{0}=\\{Q_{i}\\}_{i=1}^{M_{0}}$ be a polygonal mosaic for
$\sigma$. Suppose that $1\leq i\leq M_{0}$ and that $\sigma$ contains a
nonprojectable convex curve $c$ in $Q_{i}$. As is shown in [1, Section 5] we
can recursively subdivide $Q_{i}$ into (finitely many) smaller convex polygons
so that whenever $c$ passes through ones of the subpolygons, this part of $c$
is projectable (see Figure 4). We can then replace $Q_{i}$ in the mosaic with
the subpolygons which contain part of $c$, as well as any subpolygons (such as
$P_{4}$ in Figure 4) with sides containing parts of $\sigma$. Doing this for
each such nonprojectable curve will give the required mosaic. ∎

$c$$\ell_{1}$$\ell_{2}$$P_{1}$$P_{2}$$P_{3}$$P_{4}$$Q_{i}$ Figure 4: Lemma
3.6: Subdividing a polygon $Q_{i}$ to produce a finer mosaic containing only
projectable convex curves and sides of polygons. Here $\sigma\cap
Q_{i}=c\cup\ell_{1}\cup\ell_{2}$. The polygon $Q_{i}$ would be replaced here
with $P_{1},P_{2},P_{3},P_{4}$.

Note that any finite union of disjoint $\mathrm{APIC}$ sets is in
$\mathrm{SPIC}$ since we can connect the components with line segments. On the
other hand, the union of two $\mathrm{PIC}$ sets need not be in
$\mathrm{SPIC}$ since one may get two convex curves meeting tangentially with
the same convexity (as in $\sigma_{4}$ in Figure 3). The following lemma will
at least allow us to add line segments to an $\mathrm{SPIC}$ set.

######  Lemma 3.7.

Suppose that $\sigma_{0}\in\mathrm{SPIC}$ and that $\ell$ is a line segment in
$\mathbb{R}^{2}$. Then $\sigma=\sigma_{0}\cup\ell\in\mathrm{SPIC}$.

###### Proof.

By definition, there exists $\tau_{0}\in\mathrm{APIC}$ with
$\sigma_{0}\subseteq\tau_{0}$. Let $\mathcal{M}=\\{P_{i}\\}_{i=1}^{n}$ be a
polygonal mosaic for $\tau_{0}$ with $c_{i}$ denoting the convex curve in
$P_{i}$. Without loss we may assume that $\tau_{0}$ contains all the sides of
all the polygons in $\mathcal{M}$ since $\mathcal{M}$ is also a polygonal
mosaic for this larger set. Let $M=\cup_{i=1}^{n}P_{i}$ and let $\hat{\ell}$
denote the full line in $\mathbb{R}^{2}$ containing $\ell$.

If $\hat{\ell}$ does not intersect $M$, then one can certainly connect $\ell$
to $M$ with a line segment $\ell^{\prime}$, and then find suitable polygons
containing $\ell$ and $\ell^{\prime}$ so that adding these polygons to
$\mathcal{M}$ gives a polygonal mosaic for
$\tau=\tau_{0}\cup\ell\cup\ell^{\prime}$. Thus $\tau$ is an $\mathrm{APIC}$
set containing $\sigma$ and hence $\sigma\in\mathrm{SPIC}$.

Otherwise $\hat{\ell}$ intersects at least one polygon in $\mathcal{M}$. Let
$\ell^{\prime}$ denote the smallest line segment containing $\ell$ for which
$\hat{\ell}\cap M=\ell^{\prime}\cap M$, and let
$\tau=\tau_{0}\cup\ell^{\prime}$. Our aim now is to adapt $\mathcal{M}$ so
that it is a polygonal mosaic for $\tau$.

First choose, if necessary, convex polygons $Q_{1},\dots,Q_{m}$ around any
parts of $\ell^{\prime}$ which are disjoint from $M$.

Next, suppose that $1\leq i\leq n$ and that $\ell_{i}=\ell^{\prime}\cap
P_{i}\neq\emptyset$. If $\ell_{i}$ is a subset of the boundary of $P_{i}$
there will be nothing to be done as $\ell_{i}\subset\tau_{0}$. We may suppose
then that $\ell_{i}$ intersects the interior of $P_{i}$. Let
$c_{i}^{\circ}=c_{i}\cap{\rm int}(P_{i})$. There are three cases to consider.

$\ell_{i}$$c_{i}$$P_{i,1}$$P_{i,2}$$\ell_{i}$$c_{i}$$P_{i,1}$$P_{i,2}$ Figure
5: The cases where $\ell_{i}$ meets $c_{i}$ once.
$\ell_{i}$$c_{i}$$\ell_{i}$$c_{i}$ Figure 6: The cases where $\ell_{i}$ meets
$c_{i}$ more than once. One may also have $\ell_{i}$ intersecting $c_{i}$ on a
line segment whose endpoints are both in the interior of $P_{i}$.

  1. 1.

$\ell_{i}\cap c_{i}^{\circ}=\emptyset$. In this case, $\ell_{i}$ splits
$P_{i}$ into two convex polygons, and in this case we will replace $P_{i}$ in
$\mathcal{M}$ with the smaller polygon containing $c_{i}$.

  2. 2.

$\ell_{i}$ intersects $c_{i}^{\circ}$ at a single point ${\boldsymbol{z}}$. In
this case, as in the proof of the Partition Lemma [1, Lemma 1], we may replace
$P_{i}$ with two smaller convex polygons $P_{i,1}$ and $P_{i,2}$ whose union
contains $c_{i}$, and so that $\ell_{i}$ lies in the union of the boundaries
of these polygons. (See Figure 5.)

  3. 3.

$\ell_{i}$ intersects $c_{i}^{\circ}$ in more than one point. In this case we
can partition $c_{i}$ into two or three parts as in Figure 6.

By making these replacements for each polygon in $\mathcal{M}$ which meets
$\hat{\ell}$, and by adding in $Q_{1},\dots,Q_{m}$ we will produce a polygonal
mosaic for $\tau$, which shows that $\sigma\in\mathrm{SPIC}$. (See Figure 7.)
∎

$\ell$$\sigma_{0}$$\sigma$ Figure 7: Lemma 3.7: adapting a mosaic for the
addition of an extra line segment. The top diagram shows a line segment $\ell$
(in green) and a $\mathrm{SPIC}$ set $\sigma_{0}$ (in red). A polygonal mosaic
for $\sigma_{0}$ is shown by the dashed polygons. The lower diagram shows the
addition of three new polygons (in green) and splitting of two polygons into
smaller pieces to form a polygonal mosaic for $\sigma=\sigma_{0}\cup\ell$.

##  4 Extending from the real line

The first, and rather elementary, step in the proof of the main theorem is to
show that if $\sigma_{0}$ is any compact subset of the real line, and $f\in
AC(\sigma_{0})$, then $f$ can be extended to any larger compact subset.

Suppose then that $\sigma_{0}\subseteq\sigma\subseteq\mathbb{R}$. Let
$a_{0}=\min\sigma_{0}$ and let $b_{0}=\max\sigma_{0}$. Let $J=[a,b]$ denote a
compact interval containing $\sigma$ in its interior. Given $f\in
AC(\sigma_{0})$ define the extension $\iota(f):J\to\mathbb{C}$ by making it
constant on $[a,a_{0}]$ and $[b_{0},b]$ and piecewise linear on the open
intervals of $[a_{0},b_{0}]\setminus\sigma_{0}$.

######  Lemma 4.1.

[3, Corollary 2.14] The map $\iota$ is a linear isometry from $AC(\sigma_{0})$
to $AC(J)$.

Extending from a subset of the line to a larger subset of the plane is
straightforward. By affine invariance, the following result can be applied to
any function which is constant on a family of parallel lines.

######  Theorem 4.2.

Suppose that $\sigma_{0}$ is a nonempty compact subset of $\mathbb{R}$ and
that $\sigma\subseteq\mathbb{R}^{2}$ is a compact superset of $\sigma_{0}$. If
$f\in AC(\sigma_{0})$ then there exists an extension ${\hat{f}}\in AC(\sigma)$
with $\lVert{\hat{f}}\rVert_{BV(\sigma)}=\left\lVert
f\right\rVert_{BV(\sigma_{0})}$.

###### Proof.

Combining Lemma 4.1 and Theorem 2.3(iv) gives an extension ${\hat{f}}$ to
$\sigma$ with norm at most $\left\lVert f\right\rVert_{BV(\sigma_{0})}$. That
the norms must be equal follows from Theorem 2.3(iii). ∎

##  5 Convex curves

The results for subsets of the real line can be extended to subsets of curves
as long as these curves are well enough behaved. The following theorem
contains the content of Section 8 of [1]. (We remind the reader that a
projectable curve is differentiable except at its endpoints.)

######  Theorem 5.1.

Suppose that $c$ is a projectable convex curve in the plane. Then there exists
a Banach algebra isomorphism $\Phi:AC(c)\to AC[0,1]$ with
$\left\lVert\Phi\right\rVert\leq 2$ and $\left\lVert\Phi^{-1}\right\rVert=1$.

More specifically, if $h:[0,1]\to c$ is a differentiable parameterization of
$c$, then we can take the isomorphism to be $\Phi_{h}(f)=f\circ h$.

######  Lemma 5.2.

Suppose that $c$ is a projectable convex curve parameterized by a
differentiable function $h:[0,1]\to c$, and that
$\emptyset\neq\sigma_{0}\subseteq c$. Let
$\tau_{0}=h^{-1}(\sigma_{0})\subseteq[0,1]$. Then $\Phi_{0}(f)=f\circ h$
defines an isomorphism between $AC(\sigma_{0})$ and $AC(\tau_{0})$.

###### Proof.

First note that since $\Phi_{h}(f)=f\circ h$ defines an isomorphism between
$AC(c)$ and $AC[0,1]$, [2, Theorem 4.7] implies that $h\in AC[0,1]$ and
$h^{-1}\in AC(c)$. But by Theorem 2.3(iii), this means that $h|\tau_{0}\in
AC(\tau_{0})$ and $h^{-1}|\sigma_{0}\in AC(\sigma_{0})$. It is an easy
consequence of [2, Theorem 3.11] that $\Phi_{0}$ is a Banach algebra
isomorphism from $BV(\sigma_{0})$ to $BV(\tau_{0})$. Applying [2, Theorem 4.7]
again then, we can deduce that $\Phi_{0}$ acts as an isomorphism from
$AC(\sigma_{0})$ to $AC(\tau_{0})$. ∎

######  Theorem 5.3.

Suppose that $c$ is a projectable convex curve, and that
$\emptyset\neq\sigma_{0}\subseteq c$. If $f\in AC(\sigma_{0})$ then there
exists an extension $\hat{f}\in AC(c)$ with $\lVert\hat{f}\rVert_{BV(c)}\leq
2\left\lVert f\right\rVert_{BV(\sigma_{0})}$

###### Proof.

Suppose that $f\in AC(\sigma_{0})$. Let $\Phi_{0}$ and $h$ be as in Lemma 5.2.
Let $g=\Phi_{0}(f)\in AC(h^{-1}(\sigma_{0}))$. Lemma 4.1 allows us to extend
$g$ to ${\hat{g}}\in AC[0,1]$, and so ${\hat{f}}=\Phi_{h}^{-1}(\hat{g})\in
AC(c)$. The norm bounds follow from the properties of $\Phi$. ∎

One can extend Theorem 4.2 to cover the case where $\sigma_{0}$ is a nonempty
compact subset of a projectable curve. For the moment however, we shall just
give a special case which we shall need in the later sections. In this result,
the curve $c$ need not be projectable.

######  Lemma 5.4.

Suppose that $T$ is a triangle in $\mathbb{R}^{2}$ with vertices at
${\boldsymbol{x}}$, ${\boldsymbol{y}}$ and ${\boldsymbol{z}}$. Let $c$ be a
differentiable convex curve joining ${\boldsymbol{x}}$ to ${\boldsymbol{y}}$.
Then every $f\in AC(c)$ admits an extension $\hat{f}\in AC(T)$ with
$\lVert\hat{f}\rVert_{BV(T)}\leq 2\left\lVert f\right\rVert_{BV(c)}$.

###### Proof.

Let ${\boldsymbol{x}}^{\prime}=(0,0)$, ${\boldsymbol{y}}^{\prime}=(1,0)$ and
${\boldsymbol{z}}^{\prime}=(\frac{1}{2},1)$ and let $h$ be the affine
transformation which maps ${\boldsymbol{x}}$ to ${\boldsymbol{x}}^{\prime}$,
${\boldsymbol{y}}$ to ${\boldsymbol{y}}^{\prime}$ and ${\boldsymbol{z}}$ to
${\boldsymbol{z}}^{\prime}$. Let $T^{\prime}=h(T)$ and let $c^{\prime}=h(c)$.
Then $c^{\prime}$ is a projectable convex curve which we may assume is the
graph of the convex function $k:[0,1]\to\mathbb{R}$. As above, we may define
$g\in AC[0,1]$ by $g(t)=f(h^{-1}(t,k(t)))$, and then setting
${\hat{g}}(x,y)=g(x)$ gives an extension of $g$ to all of $T^{\prime}$. Now
setting ${\hat{f}}={\hat{g}}\circ h$ gives an absolutely continuous extension
of $f$ to $T$. The bounds follow from Theorem 2.3(ii) and Theorem 4.2. ∎

##  6 Filling in polygons

The other main building block of the sets to which the main theorem will apply
are polygonal regions of the plane. The first step is to show that one may
always extend an absolutely continuous function defined on the boundary of
such a set into its interior.

In the case where the polygonal region is the unit square
$S=[0,1]\times[0,1]$, one can write an explicit formula for an extension. This
result, with a slightly different bound, is [4, Theorem 3.5].

######  Theorem 6.1.

Suppose that $\sigma_{0}$ is the boundary of the unit square $\sigma=S$ and
that $f\in AC(\sigma_{0})$. Then $f$ admits an extension ${\hat{f}}\in
AC(\sigma)$ with

$\lVert{\hat{f}}\rVert_{BV(\sigma)}\leq 10\left\lVert
f\right\rVert_{AC(\sigma_{0})}.$

###### Proof.

The restrictions of $f$ to the four sides of $S$ are absolutely continuous:

$b(x)=f(x,0),\qquad t(x)=f(x,1),\qquad\ell(y)=f(0,y),\qquad r(y)=f(1,y).$

For $(x,y)\in S$ let

$\displaystyle{\hat{f}}(x,y)$
$\displaystyle=b(x)+\ell(y)-b(0)+(b(0)-b(1)+r(y)-\ell(y))x$
$\displaystyle\hphantom{b(x)b(x)+\ell(y)-b(0}+(\ell(0)-\ell(1)+t(x)-b(x))y$
(1) $\displaystyle\hphantom{b(x)b(x)+\ell(y)-b(0}+(b(1)-b(0)+t(0)-t(1))xy.$

It is readily verified that ${\hat{f}}$ is indeed an extension of $f$.
Further, using Theorem 4.2, it is clear that ${\hat{f}}\in AC(\sigma)$.

Note that ${\hat{f}}$ can also be written as

$\displaystyle{\hat{f}}(x,y)$
$\displaystyle=b(0)(x-1)(1-y)+b(1)x(1-y)+t(0)(x-1)y+t(1)xy$ (2)
$\displaystyle\hphantom{b(x)b(x)+\ell(y)-b(0}+b(x)(1-y)+\ell(y)(1-x)+r(y)x+t(x)y$

and so $\lVert{\hat{f}}\rVert_{\infty}\leq 8\left\lVert
f\right\rVert_{\infty}$. Using Lemma 2.7 and Theorem 2.16 from [10],

$\displaystyle\operatorname*{var}((b(0)-b(1))x,\sigma)$
$\displaystyle\leq\operatorname*{var}(b,[0,1])\operatorname*{var}(x,\sigma)=\operatorname*{var}(b,[0,1])$
$\displaystyle\operatorname*{var}(r(y)x,\sigma)$
$\displaystyle\leq\operatorname*{var}(r(y),\sigma)\left\lVert
x\right\rVert_{\infty}+\left\lVert
r\right\rVert_{\infty}\operatorname*{var}(x,\sigma)=\left\lVert
r\right\rVert_{BV[0,1]}$

while

$\displaystyle\operatorname*{var}((b(1)-b(0)$
$\displaystyle+t(0)-t(1))xy,\sigma)\leq(\operatorname*{var}(b,[0,1])+\operatorname*{var}(t,[0,1])\operatorname*{var}(xy,\sigma)$
$\displaystyle\leq(\operatorname*{var}(b,[0,1])+\operatorname*{var}(t,[0,1])(\left\lVert
x\right\rVert_{\infty}\operatorname*{var}(y,\sigma)+\operatorname*{var}(x,\sigma)\left\lVert
y\right\rVert_{\infty}$
$\displaystyle=2(\operatorname*{var}(b,[0,1])+\operatorname*{var}(t,[0,1]).$

Using similar estimates for the other terms in (6), this shows that

$\displaystyle\operatorname*{var}({\hat{f}},\sigma)$
$\displaystyle\leq\operatorname*{var}(b,[0,1])+\operatorname*{var}(\ell,[0,1])+\left\lVert
r\right\rVert_{BV[0,1]}+\left\lVert\ell\right\rVert_{BV[0,1]}$
$\displaystyle\qquad+\operatorname*{var}(\ell,[0,1])+\left\lVert
t\right\rVert_{BV[0,1]}+\left\lVert b\right\rVert_{BV[0,1]}$
$\displaystyle\qquad+2(\operatorname*{var}(b,[0,1])+\operatorname*{var}(t,[0,1])$
$\displaystyle\leq 4\operatorname*{var}(f,\sigma_{0})+\left\lVert
b\right\rVert_{BV[0,1]}+\left\lVert
t\right\rVert_{BV[0,1]}+\left\lVert\ell\right\rVert_{BV[0,1]}+\left\lVert
r\right\rVert_{BV[0,1]}.$

In the proof of Theorem 3.4 of [4] it was shown that

$\left\lVert b\right\rVert_{BV[0,1]}+\left\lVert
t\right\rVert_{BV[0,1]}+\left\lVert\ell\right\rVert_{BV[0,1]}+\left\lVert
r\right\rVert_{BV[0,1]}\leq 2\left\lVert f\right\rVert_{BV(\sigma_{0})}.$

Combining all of this gives

$\displaystyle\lVert{\hat{f}}\rVert_{BV(\sigma)}$
$\displaystyle=\lVert{\hat{f}}\rVert_{\infty}+\operatorname*{var}({\hat{f}},\sigma)$
$\displaystyle\leq 8\left\lVert
f\right\rVert_{\infty}+4\operatorname*{var}(f,\sigma_{0})+2\left\lVert
f\right\rVert_{BV(\sigma_{0})}\leq 10\left\lVert
f\right\rVert_{BV(\sigma_{0})}.\qed$

The second author [13] obtained a slightly better bound for an extension from
the boundary of a triangle to its interior. As the construction is a little
more complicated than the one for a square, we just record the result here.

######  Theorem 6.2.

[13, Theorem 6.3.6] Suppose that $\sigma_{0}$ is the boundary of a triangular
region $\sigma$ in the plane. If $f\in AC(\sigma_{0})$ then $f$ admits an
extension ${\hat{f}}\in AC(\sigma)$ with

$\lVert{\hat{f}}\rVert_{BV(\sigma)}\leq 7\left\lVert
f\right\rVert_{AC(\sigma_{0})}.$

It is likely that the constants in both cases for the bounds on the extensions
are not sharp, even for the explicit extensions constructed. It is known that
the construction in the proof of Theorem 6.1 may produce an extension with a
larger norm than the initial function.

######  Example 6.3.

Consider the following example of the construction from Theorem 6.1. If
$b(x)=t(x)=x(1-x)$ and $\ell(y)=r(y)=y(1-y)$, then
${\hat{f}}(x,y)=x(1-x)+y(1-y)$. It is easy to check that
$\lVert{\hat{f}}\rVert_{BV(\sigma)}=6$. Now, as in the proof of Section 2.6 of
[10], $\operatorname*{var}(f,\sigma_{0})$ is at most twice the parameterized
variation of $f$ around the edges of the square, so $\left\lVert
f\right\rVert_{BV(\sigma_{0})}\leq 1+2=3$.

Using the results of [9] (see Theorem 6.3), one can obtain extensions to the
region bounded by any polygon.

######  Theorem 6.4.

Suppose that $\sigma$ is a polygonal region in $\mathbb{R}^{2}$ with boundary
$\sigma_{0}$. Then there exists a homeomorphism
$h:\mathbb{R}^{2}\to\mathbb{R}^{2}$, made up of the composition of locally
piecewise affine maps, such that

  * •

$h(\sigma)$ is the square $S=[0,1]\times[0,1]$,

  * •

$AC(\sigma_{0})$ is isomorphic to $AC(\partial S)$,

  * •

$AC(\sigma)$ is isomorphic to $AC(S)$.

In each case the algebra isomorphism is given by $\Phi(g)=g\circ h^{-1}$.

######  Theorem 6.5.

Suppose that $\sigma$ is a polygonal region in $\mathbb{R}^{2}$ with boundary
$\sigma_{0}$, and that $f\in AC(\sigma_{0})$. Then $f$ admits an extension
${\hat{f}}\in AC(\sigma)$ with

$\lVert{\hat{f}}\rVert_{BV(\sigma)}\leq K_{\sigma}\left\lVert
f\right\rVert_{BV(\sigma_{0})}$

where $K_{\sigma}$ only depends on $\sigma$.

###### Proof.

Let $h$ and $\phi$ be the maps guaranteed by Theorem 6.4. Then $g=\Phi(f)\in
AC(\partial S)$ and so by Theorem 6.1, $g$ admits an extension ${\hat{g}}\in
AC(S)$. The form of $\Phi$ then ensures that $\Phi^{-1}({\hat{g}})$ is an
extension of $f$.

Finally
$\lVert{\hat{f}}\rVert\leq\left\lVert\Phi^{-1}\right\rVert\left\lVert\hat{g}\right\rVert\leq
18\left\lVert\Phi^{-1}\right\rVert\left\lVert\Phi\right\rVert\left\lVert
f\right\rVert$, so we can take
$K_{\sigma}=18\left\lVert\Phi^{-1}\right\rVert\left\lVert\Phi\right\rVert$
which only depends on $\sigma$. ∎

Examining the proof of Theorem 6.4 in [9] shows that the norm $K_{\sigma}$ in
the above proof actually only depends on the number of sides of $\sigma$.

##  7 Joining theorems

Many of the later constructions will require that an extension be separately
constructed on different parts of the larger set $\sigma$. The challenge then
is to show that the function is absolutely continuous on the whole of the set.
In general, if $\sigma=\sigma_{1}\cup\sigma_{2}$ then knowing that
$f|\sigma_{1}\in AC(\sigma_{1})$ and $f|\sigma_{2}\in AC(\sigma_{2})$ is not
enough to deduce that $f\in AC(\sigma)$.

######  Example 7.1.

Suppose that $\\{x_{k}\\}_{k=1}^{\infty}$ is a strictly decreasing sequence of
positive numbers with limit $0$. Let
$\sigma=\\{x_{k}\,:\,k=1,2,3,\dots\bigr{\\}}\cup\\{0\\}$,
$\sigma_{1}=\\{x_{2k}\,:\,k=1,2,3,\dots\\}\cup\\{0\\}$ and
$\sigma_{2}=\\{x_{2k-1}\,:\,k=1,2,3,\dots\\}\cup\\{0\\}$. Define
$f:\sigma\to\mathbb{C}$ by $f(x_{k})=\frac{(-1)^{k}}{k}$, with $f(0)=0$. Then
the restrictions of $f$ to $\sigma_{1}$ and to $\sigma_{2}$ are absolutely
continuous, but $f$ is not even of bounded variation on $\sigma$.

If $f_{1}:\sigma_{1}\to\mathbb{C}$ and $f_{2}:\sigma_{2}\to\mathbb{C}$ agree
on $\sigma_{1}\cap\sigma_{2}$, we shall call the function determined by
$f({\boldsymbol{x}})=f_{i}({\boldsymbol{x}})$ for
${\boldsymbol{x}}\in\sigma_{i}$ the joined function of $f_{1}$ and $f_{2}$. If
$\sigma_{1}$ and $\sigma_{2}$ are disjoint, then the Patching Lemma (Theorem
2.4) implies that $f\in AC(\sigma)$. The difficulty then occurs in the case
that $\sigma_{1}$ and $\sigma_{2}$ overlap. Under suitable conditions on
$\sigma_{1}$ and $\sigma_{2}$ one can at least rule out the possibility that
the joined function fails to be of bounded variation.

We shall say that $\sigma_{1}$ and $\sigma_{2}$ join convexly if for given any
${\boldsymbol{x}}\in\sigma_{1}\setminus\sigma_{2}$ and
${\boldsymbol{y}}\in\sigma_{2}\setminus\sigma_{1}$ the line segment joining
${\boldsymbol{x}}$ and ${\boldsymbol{y}}$ contains a point
${\boldsymbol{w}}\in\sigma_{1}\cap\sigma_{2}$.

######  Theorem 7.2.

[10, Theorem 3.8] Suppose that $\sigma_{1},\sigma_{2}\subseteq\mathbb{R}^{2}$
are nonempty compact sets which are disjoint except at their boundaries, and
that $\sigma_{1}$ and $\sigma_{2}$ join convexly. Let
$\sigma=\sigma_{1}\cup\sigma_{2}$. If $f_{1}\in BV(\sigma_{1})$ and $f_{2}\in
BV(\sigma_{2})$ agree on $\sigma_{1}\cap\sigma_{2}$, then their joined
function $f$ lies in $BV(\sigma)$ and $\left\lVert
f\right\rVert_{BV(\sigma)}\leq\left\lVert
f_{1}\right\rVert_{BV(\sigma_{1})}+\left\lVert
f_{2}\right\rVert_{BV(\sigma_{2})}$.

If $\sigma_{1}$ and $\sigma_{2}$ lie on either side of a straight line then
one can, at the cost of an extra factor, remove the convexity hypothesis. Let
$H^{+}$ and $H^{-}$ denote the closed upper and lower half-planes in
$\mathbb{R}^{2}$ and let $L$ be the $x$-axis.

######  Theorem 7.3.

Suppose that $\sigma_{1}\subseteq H^{+}$ and $\sigma_{2}\subseteq H^{-}$ are
nonempty compact sets, that $\sigma_{2}\cap L\subseteq\sigma_{1}\cap L$, and
that $\sigma=\sigma_{1}\cup\sigma_{2}$. If $f:\sigma\to\mathbb{C}$ then

$\left\lVert f\right\rVert_{BV(\sigma)}\leq 2\bigl{(}\left\lVert
f|\sigma_{1}\right\rVert_{BV(\sigma_{1})}+\left\lVert
f|\sigma_{2}\right\rVert_{BV(\sigma_{2})}\bigr{)}.$

###### Proof.

Suppose that
$S=[{\boldsymbol{x}}_{0},{\boldsymbol{x}}_{1},\dots,{\boldsymbol{x}}_{n}]$ is
a list of elements of $\sigma$ and let $J=\\{1,\dots,n\\}$. Let

$\displaystyle J_{1}$ $\displaystyle=\\{j\in
J\,:\,{\boldsymbol{x}}_{j-1},{\boldsymbol{x}}_{j}\in\sigma_{1}\\},$
$\displaystyle J_{2}$ $\displaystyle=\\{j\in
J\,:\,{\boldsymbol{x}}_{j-1},{\boldsymbol{x}}_{j}\in\sigma_{2}\\},$
$\displaystyle J_{3}$ $\displaystyle=J\setminus(J_{1}\cup J_{2}).$

Then, noting that $J_{1}$ and $J_{2}$ need not be disjoint, and considering
empty sums as zero,

$\operatorname*{\rm
cvar}(f,S)=\sum_{j=1}^{n}|f({\boldsymbol{x}}_{j})-f({\boldsymbol{x}}_{j-1})|\\\
\leq\sum_{k=1}^{3}\sum_{j\in
J_{k}}|f({\boldsymbol{x}}_{j})-f({\boldsymbol{x}}_{j-1})|.$

As in the proof of [10, Lemma 4.2], $\sum_{j\in
J_{1}}|f({\boldsymbol{x}}_{j})-f({\boldsymbol{x}}_{j-1})|\leq\operatorname*{var}(f,\sigma_{1})\operatorname{vf}(S)$
and $\sum_{j\in
J_{2}}|f({\boldsymbol{x}}_{j})-f({\boldsymbol{x}}_{j-1})|\leq\operatorname*{var}(f,\sigma_{2})\operatorname{vf}(S)$.
Note that if $j\in J_{3}$ then one end of the line segment
$s_{j}=\overline{\vphantom{\vbox
to5.16663pt{}}{\boldsymbol{x}}_{j-1}\,{\boldsymbol{x}}_{j}}$ lies $H^{+}$
while the other end, which we shall denote
${\boldsymbol{u}}_{j}=(u_{j},v_{j})$, lies in the open lower half-plane. Thus
$v=\sup_{j\in J_{3}}v_{j}$ is strictly negative and each of these line
segments crosses the horizontal line $y=v/2$. This implies that
$\operatorname{vf}(S)$ is at least the number of elements in $J_{3}$. Thus

$\sum_{j\in
J_{3}}|f({\boldsymbol{x}}_{j})-f({\boldsymbol{x}}_{j-1})|\leq\bigl{(}\left\lVert
f|\sigma_{1}\right\rVert_{\infty}+\left\lVert
f|\sigma_{2}\right\rVert_{\infty}\bigr{)}\operatorname{vf}(S).$

It follows that

$\frac{\operatorname*{\rm cvar}(f,S)}{\operatorname{vf}(S)}\leq\left\lVert
f|\sigma_{1}\right\rVert_{BV(\sigma_{1})}+\left\lVert
f|\sigma_{2}\right\rVert_{BV(\sigma_{2})},$

and so, on taking the supremum over all such lists $S$,

$\left\lVert f\right\rVert_{BV(\sigma)}=\left\lVert
f\right\rVert_{\infty}+\operatorname*{var}(f,\sigma)\leq 2\bigl{(}\left\lVert
f|\sigma_{1}\right\rVert_{BV(\sigma_{1})}+\left\lVert
f|\sigma_{2}\right\rVert_{BV(\sigma_{2})}\bigr{)}.\qed$

By considering the case where $\sigma_{1}$ and $\sigma_{2}$ are disjoint and
$f$ is the characteristic function of $\sigma_{1}$, one can observe that the
factor 2 in the above theorem is necessary. Of course by affine invariance,
the same result holds with the $x$-axis replaced by any other line in the
plane.

An important special case is when a line and a convex curve meet at a line
tangential vertex.

######  Corollary 7.4.

Suppose that the line segment $\ell=\overline{\vphantom{\vbox
to5.16663pt{}}{\boldsymbol{x}}\,{\boldsymbol{z}}}$ meets the convex curve $c$
tangentially at ${\boldsymbol{x}}$, and let $\sigma=\ell\cup c$. If
$f:\sigma\to\mathbb{C}$ then

$\left\lVert f\right\rVert_{BV(\sigma)}\leq 2\bigl{(}\left\lVert
f\right\rVert_{BV(\ell)}+\left\lVert f\right\rVert_{BV(c)}\bigr{)}.$

###### Proof.

After applying an appropriate affine transformation one need consider the case
where $\ell=[0,1]$ and $c$ lies in the closed lower half-plane. The result
then follows from Theorem 7.3 with $\sigma_{1}=\ell$ and $\sigma_{2}=c$. ∎

For the later results we shall piece together functions defined on polygonal
pieces. It is not hard to adapt Example 7.1 to show that even in the case of
two polygonal sets $\sigma_{1}$ and $\sigma_{2}$, there is no constant $C$
such that $\left\lVert f\right\rVert_{BV(\sigma)}\leq C(\left\lVert
f\right\rVert_{BV(\sigma_{1})}+\left\lVert f\right\rVert_{BV(\sigma_{2})})$.
The next result will however allow us some control over the norms if we add
triangular pieces one at a time.

Let $R_{1}$ and $R_{2}$ be two distinct rays in $\mathbb{R}^{2}$ starting at
the origin and let $S_{1}$ and $S_{2}$ be the two closed sectors of the plane
with boundary $B=R_{1}\cup R_{2}$. A half-plane affine map will refer to a
homeomorphism of the plane which is affine on each of a pair of complementary
half-planes. (Further details can be found in [9, Section 4].)

######  Theorem 7.5.

Suppose that $\sigma_{1}\subseteq S_{1}$ and $\sigma_{2}\subseteq S_{2}$ are
nonempty compact subsets of complementary sectors with boundary $B$,
$\sigma_{1}\cap B=\sigma_{2}\cap B$, and $\sigma=\sigma_{1}\cup\sigma_{2}$. If
$f:\sigma\to\mathbb{C}$, then

$\left\lVert f\right\rVert_{BV(\sigma)}\leq 8\bigl{(}\left\lVert
f|\sigma_{1}\right\rVert_{BV(\sigma_{1})}+\left\lVert
f|\sigma_{2}\right\rVert_{BV(\sigma_{2})}\bigr{)}.$

###### Proof.

There exists a half-plane affine map $\alpha$ which maps $B$ onto a straight
line through the origin. Let ${\hat{\sigma}}_{1}$, ${\hat{\sigma}}_{2}$ and
$\hat{\sigma}$ be the images of $\sigma_{1}$, $\sigma_{2}$ and $\sigma$ under
$\alpha$, and let $g:{\hat{\sigma}}\to\mathbb{C}$ be $g=f\circ\alpha^{-1}$.
Then by [9, Theorem 4.5] and Theorem 7.3

$\displaystyle\left\lVert f\right\rVert_{BV(\sigma)}$ $\displaystyle\leq
2\lVert g\rVert_{BV({\hat{\sigma}})}$ $\displaystyle\leq 4\bigl{(}\lVert
g\rVert_{BV({\hat{\sigma}}_{1})}+\lVert
g\rVert_{BV({\hat{\sigma}}_{2})}\bigr{)}$ $\displaystyle\leq
8\bigl{(}\left\lVert f\right\rVert_{BV({\sigma}_{1})}+\left\lVert
f\right\rVert_{BV({\sigma}_{2})}\bigr{)}.\qed$

The factor $8$ here is unlikely to be sharp, but again by considering
characteristic functions, one can see that one needs at least a factor of $3$
in the bound.

######  Theorem 7.6.

Suppose that $\sigma_{1}\subseteq H^{+}$ and $\sigma_{2}\subseteq H^{-}$ are
nonempty compact sets, that $\sigma_{2}\cap L\subseteq\sigma_{1}\cap L$, and
that $\sigma=\sigma_{1}\cup\sigma_{2}$. Suppose that $f:\sigma\to\mathbb{C}$.
If $f|\sigma_{1}\in{AC}(\sigma_{1})$ and $f|\sigma_{2}\in{AC}(\sigma_{2})$,
then $f\in{AC}(\sigma)$ and

$\left\lVert f\right\rVert_{{BV}(\sigma)}\leq 2\bigl{(}\left\lVert
f|\sigma_{1}\right\rVert_{{BV}(\sigma_{1})}+\left\lVert
f|\sigma_{2}\right\rVert_{{BV}(\sigma_{2})}\bigr{)}.$

$\sigma_{1}$$\sigma_{2}$$P$ Figure 8: Diagram for Theorem 7.6.

###### Proof.

The bound on the variation norm comes from the previous theorem, so the task
is to show that $f$ is absolutely continuous. If
$\sigma_{1}\cap\sigma_{2}=\emptyset$, then the absolutely continuity follows
immediately from Theorem 2.4

Suppose then that $\sigma_{1}\cap\sigma_{2}\neq\emptyset$ so that
$\tau=\sigma_{1}\cap L$ is nonempty. Choose $a_{0},b_{0}\in\mathbb{R}$ such
that $a_{0}\leq x\leq b_{0}$ for all $(x,y)\in\sigma$ and let
$\tau_{0}=[a_{0},b_{0}]\times\\{0\\}$. Let
$P=[a_{0},b_{0}]\times[c_{0},d_{0}]$ be a closed rectangle containing
$\sigma$. Let $P_{1}=[a_{0},b_{0}]\times[0,d_{0}]$ and let
$P_{2}=[a_{0},b_{0}]\times[c_{0},d_{0}]$, so that $\sigma_{1}\subseteq P_{1}$
and $\sigma_{2}\subseteq P_{2}$.

Now $f|\tau\in AC(\tau)$ so by Lemma 4.1 and Theorem 2.3(iv) there is a
function $g\in AC(\tau_{0})$ such that $g|\tau=f|\tau$. By setting
$g(x,y)=g(x,0)$ for all $y$, we can then extend to $g\in AC(P)$ with
$\left\lVert g\right\rVert_{BV(P)}=\left\lVert f\right\rVert_{BV(\tau)}$.

Fix $\epsilon>0$. Let $f_{1}=f-g|\sigma$. Clearly $f\equiv 0$ on $\tau$,
$f_{1}|\sigma_{1}\in AC(\sigma_{1})$ and $f_{1}|\sigma_{2}\in AC(\sigma_{2})$.

As $f_{1}|\sigma_{1}\in AC(\sigma_{1})$, by Lemma 3.2 and Theorem 6.1 of [11]
there exists $h_{1}\in\mathrm{CTPP}(P_{1})$ with $\left\lVert
f_{1}-h_{1}\right\rVert_{BV(\sigma_{1})}\leq\epsilon/8$.

Now $h_{1}|\tau_{0}$ is piecewise linear, and in particular it is absolutely
continuous. We can therefore extend $h_{1}$ to $P_{2}$ by setting
$h_{1}(x,y)=h_{1}(x,0)$ for all $(x,y)\in P_{2}$. Note that
$h_{1}\in\mathrm{CTPP}(P)$. Now

$\displaystyle\left\lVert h_{1}\right\rVert_{BV(\sigma_{2})}=\left\lVert
h_{1}\right\rVert_{BV(\tau_{0})}$ $\displaystyle=\left\lVert
f_{1}-h_{1}\right\rVert_{BV(\tau_{0})}$ $\displaystyle=\left\lVert
f_{1}-h_{1}\right\rVert_{BV(\tau)}\leq\left\lVert
f_{1}-h_{1}\right\rVert_{BV(\sigma_{1})}<\epsilon/8.$

Similarly, construct $h_{2}\in\mathrm{CTPP}(P_{2})$ with $\left\lVert
f_{1}-h_{2}\right\rVert_{BV(\sigma_{2})}<\epsilon/8$ and extend this to all of
$P$ as above, so that $\left\lVert
h_{2}\right\rVert_{BV(\sigma_{1})}<\epsilon/8$.

Let $h=h_{1}+h_{2}\in\mathrm{CTPP}(P)$. Then

$\left\lVert f_{1}-h\right\rVert_{BV(\sigma_{1})}\leq\left\lVert
f_{1}-h_{1}\right\rVert_{BV(\sigma_{1})}+\left\lVert
h_{2}\right\rVert_{BV(\sigma_{1})}<\epsilon/4.$

Similarly, $\left\lVert f_{1}-h\right\rVert_{BV(\sigma_{2})}<\epsilon/4$ and
so, by Theorem 7.3, $\left\lVert f_{1}-h\right\rVert_{BV(\sigma)}<\epsilon$.
Thus, $f_{1}\in AC(\sigma)$. Since $g|\sigma\in AC(\sigma)$, this implies that
$f\in AC(\sigma)$. ∎

The next theorem says that we can join functions which are absolutely
continuous on polygonal regions.

######  Theorem 7.7.

Let $Q_{1},\dots,Q_{n}$ be disjoint bounded open sets whose boundaries consist
of a finite number of line segments. Let $P_{i}=\mathop{\mathrm{cl}}(Q_{i})$,
$1\leq i\leq n$, and let $\sigma=\cup_{i=1}^{n}P_{i}$. Suppose that
$f:\sigma\to\mathbb{C}$. Then $f\in AC(\sigma)$ if and only if $f|P_{i}\in
AC(P_{i})$ for each $i$.

###### Proof.

The forward implication is immediate.

Suppose that $f\in AC(P_{i})$ for each $i$. We shall use the Patching Lemma to
show that $f\in AC(\sigma)$.

Suppose that ${\boldsymbol{x}}\in\sigma$. There are three mutually exclusive
cases to consider.

  1. 1.

${\boldsymbol{x}}$ lies in a single set $P_{i}$. In this case there is
certainly compact neighbourhood of ${\boldsymbol{x}}$ in $\sigma$ on which $f$
is absolutely continuous.

  2. 2.

${\boldsymbol{x}}$ lies in the boundary of two regions, say $P_{1}$ and
$P_{2}$, but not at a corner of either of the region. In this case one may
choose a small closed disk $D\subseteq P_{1}\cup P_{2}$ centred at
${\boldsymbol{x}}$ such that $f$ is absolutely continuous on $D\cap P_{1}$ and
on $D\cap P_{2}$. By Theorem 7.6, $f$ is then absolutely continuous on $D$,
which is a compact neighbourhood of ${\boldsymbol{x}}$ in $\sigma$.

  3. 3.

${\boldsymbol{x}}$ lies in $m\geq 2$ regions, say $P_{1},\dots,P_{m}$, and it
lies at a corner of at least one of these. Since we are only interested in the
local behaviour near ${\boldsymbol{x}}$, by intersecting with a small square
around ${\boldsymbol{x}}$ we may assume that each region is in fact a polygon.
If ${\boldsymbol{x}}$ lies in $P_{i}$ but not at a corner, one can split
$P_{i}$ into two smaller polygons each with a corner at ${\boldsymbol{x}}$,
and of course $f$ is absolutely continuous on each of the smaller polygons. We
may therefore also assume that ${\boldsymbol{x}}$ lies at a corner of each of
$P_{1},\dots,P_{m}$, and by again splitting a polygon if necessary, we can
assume that the polygons can be labelled so that

     * •

each $P_{i}$ is convex,

     * •

$P_{1},\dots,P_{r}$ are in one closed half-plane, and that
$P_{r+1},\dots,P_{m}$ are in the complementary closed half-plane (see Figure
9). (Note that it may be that all the polygons lie in a single half-plane.)

One may now inductively apply Theorem 7.6 to show that the restriction of $f$
to $\cup_{i=1}^{r}P_{i}$ and $\cup_{i=r+1}^{m}P_{i}$ are both absolutely
continuous. Applying this theorem one more time shows that
$\cup_{i=1}^{m}P_{i}$ is a compact neighbourhood of ${\boldsymbol{x}}$ on
which $f$ is absolutely continuous.

In all cases, there is a compact neighbourhood of ${\boldsymbol{x}}$ on which
$f$ is absolutely continuous, and so $f\in AC(\sigma)$ by the Patching Lemma.
∎

$P_{1}$$P_{2}$$P_{3}$$P_{4}$${\boldsymbol{x}}$ Figure 9: Diagram for Theorem
7.7. Here the bottom polygon is split in two so that $P_{1},P_{2},P_{3}$ lie
in one half-plane, and $P_{4}$ lies in the complementary half-plane.

As noted earlier, in the above theorem, it is not possible to bound the norm
of $f$ by some absolute constant times the sum of the norms of the
restrictions of $f$ to the components $P_{i}$. On the other hand, one can give
a bound which will depend on the geometry of the components.

######  Theorem 7.8.

Let $P\in PW^{+}$ and let $T$ be a closed triangular whose interior is
disjoint from $P$. Let $\sigma=P\cup T$. Then there is a constant $K$ such
that if $f:\sigma\to\mathbb{C}$ then

$\left\lVert f\right\rVert_{BV(\sigma)}\leq K\bigl{(}\left\lVert
f\right\rVert_{BV(P)}+\left\lVert f\right\rVert_{BV(T)}\bigr{)}.$

###### Proof.

Extending the sides of $T$ allows us to split the exterior of $T$ into 3
sectors $R_{1},R_{2},R_{3}$ whose vertices lie at vertices of $T$, as in
Figure 10. Applying Theorem 7.5 shows that

$\left\lVert f\right\rVert_{BV(T\cap R_{1})}\leq 8\bigl{(}\left\lVert
f\right\rVert_{BV(T)}+\left\lVert f\right\rVert_{BV(R_{1}\cap P)}\bigr{)}\leq
8\bigl{(}\left\lVert f\right\rVert_{BV(T)}+\left\lVert
f\right\rVert_{BV(P)}\bigr{)}.$

Applying Theorem 7.5 twice more shows that

$\left\lVert f\right\rVert_{BV(\sigma)}\leq 8^{3}\bigl{(}\left\lVert
f\right\rVert_{BV(T)}+\left\lVert f\right\rVert_{BV(P)}\bigr{)}$

as required. ∎

$T$$P$$R_{1}$$R_{2}$$R_{3}$ Figure 10: Joining a triangle to a polygonal
region.

######  Corollary 7.9.

Suppose that $P_{1}\in PW^{+}$ and that $P_{2}$ is a closed polygonal region
in the plane whose interior is disjoint from $P_{1}$, Let $\sigma=P_{1}\cup
P_{2}$. Then there exists a constant $K(P_{1},P_{2})$ such that if
$f:\sigma\to\mathbb{C}$ then

$\left\lVert f\right\rVert_{BV(\sigma)}\leq K(P_{1},P_{2})\bigl{(}\left\lVert
f\right\rVert_{BV(P_{1})}+\left\lVert f\right\rVert_{BV(P_{2})}\bigr{)}.$

###### Proof.

Triangulate $P_{2}$ as $\cup_{k=1}^{n}T_{k}$. Applying the previous theorem
repeatedly then shows that

$\displaystyle\left\lVert f\right\rVert_{BV(\sigma)}$ $\displaystyle\leq
K^{n}\bigl{(}\left\lVert f\right\rVert_{BV(P_{1})}+\left\lVert
f\right\rVert_{BV(T_{1})}+\dots+\left\lVert f\right\rVert_{BV(T_{n})}\bigr{)}$
$\displaystyle\leq K^{n}\bigl{(}\left\lVert
f\right\rVert_{BV(P_{1})}+n\left\lVert f\right\rVert_{BV(P_{2})}\bigr{)}.\qed$

One can of course extend this result to deal with any finite number of
polygons. As can be seen from the proof, the constant $K(P_{1},P_{2})$ can be
chosen so that it only depends on the minimum number of triangles needed to
triangulate one of the the polygons. (The constant needs to depend at least
linearly in this number, but it seems unlikely that it needs the dependence
given in the proof.)

The next result says that if we can extend a function to be absolutely
continuous on the ‘holes’ of a $PW^{+}$ set, then the function will be
absolutely continuous on the filled in set.

######  Theorem 7.10.

Suppose that $\sigma_{0}\in PW^{+}$. Let $W_{1},\dots,W_{d}$ denote the
bounded components of the complement of $\sigma_{0}$, and let
$\sigma=\sigma_{0}\cup\bigl{(}\cup_{i=1}^{d}\mathop{\mathrm{cl}}(W_{i})\bigr{)}$.
Then $f\in AC(\sigma)$ if and only if $f|\sigma_{0}\in AC(\sigma_{0})$ and
$f|\mathop{\mathrm{cl}}(W_{i})\in AC(\mathop{\mathrm{cl}}(W_{i}))$ for $1\leq
i\leq d$.

###### Proof.

One only needs to prove the converse direction. Suppose then that
$f|\sigma_{0}\in AC(\sigma_{0})$ and $f|\mathop{\mathrm{cl}}(W_{i})\in
AC(\mathop{\mathrm{cl}}(W_{i}))$ for $1\leq i\leq d$. One can triangulate
$\sigma_{0}$ and each of the sets $W_{i}$. The set $\sigma$ is then a union of
triangles and the restriction of $f$ to each of these is absolutely continuous
so the result follows from Theorem 7.7. ∎

At this point we have enough to show that if $\sigma_{0}$ is a polygonal
region with finitely many polygonal windows, then, using Theorem 6.5, we can
extend the function to those windows and hence produce an absolutely
continuous function on the ‘filled-in’ set. An alternative proof of this
result may be found in [13, Theorem 6.4.2], where an explicit bound on the
norm of the extension is given in terms of the total number of edges on the
polygonal region and its windows.

It is worth noting that in all the known examples where a joined function
fails to be absolutely continuous, the issue has been a failure to control the
variation norm. This leads to the following open question.

######  Question 7.11.

If $f_{1}\in AC(\sigma_{1})$ and $f_{2}\in AC(\sigma_{2})$ and $f\in
BV(\sigma)$, must we have $f\in AC(\sigma)$?

##  8 An $\mathrm{APIC}$ joining theorem

Theorem 7.6 is strong enough to prove that a function $f$ defined on a
$\mathrm{PIC}$ set $\sigma=\cup_{k=1}^{n}c_{k}$ is absolutely continuous if
and only if it is absolutely continuous on each of its component curves. Our
next task is to extend this result to the class of $\mathrm{APIC}$ sets. Note
that as is shown in Theorem 6.2.1 of [13], if $c_{1}=\\{(x,x^{2})\,:\,0\leq
x\leq 1\\}$ and $c_{2}=\\{(x,2x^{2})\,:\,0\leq x\leq 1\\}$, and
$\sigma=c_{1}\cup c_{2}$ then one can find a function which is absolutely
continuous on each of the components, but which is not absolutely continuous
on $\sigma$, so one cannot extend this result to arbitrary unions of convex
curves. (Of course, this doesn’t imply that one can’t extend absolutely
continuous functions from such sets.)

We begin by giving showing that one can join functions of bounded variation on
different components of an $\mathrm{APIC}$ set.

######  Lemma 8.1.

Let $P$ be a convex $N$-gon with sides $\ell_{1},\dots,\ell_{N}$ and let $c$
be a convex curve in $P$. Suppose that $\mathcal{C}$ is a nonempty subset of
$\\{c,\ell_{1},\dots,\ell_{N}\\}$ and let
$\sigma=\cup\\{\tau\,:\,\tau\in\mathcal{C}\\}$. If $f:\sigma\to\mathbb{C}$
then

$\max\bigl{\\{}\left\lVert
f\right\rVert_{BV(\tau)}\,:\,\tau\in\mathcal{C}\bigr{\\}}\leq\left\lVert
f\right\rVert_{BV(\sigma)}\leq
K_{N}\Bigl{(}\sum_{\tau\in\mathcal{C}}\left\lVert
f\right\rVert_{BV(\tau)}\Bigr{)}.$

###### Proof.

The left hand inequality is clear, so it remains to establish the equality on
the right. Label the sets in $\mathcal{C}$ as $\tau_{0},\dots,\tau_{r}$,
where, if $c\in\mathcal{C}$ we choose $\tau_{0}=c$.

Let $\sigma_{0}=\tau_{0}$ and for $j=1,\dots,r$ let
$\sigma_{j}=\sigma_{j-1}\cup\tau_{j}$. Since $P$ is convex, one may apply
Theorem 7.3 to show that for each $j$

$\left\lVert f\right\rVert_{BV(\sigma_{j})}\leq 2\bigl{(}\left\lVert
f\right\rVert_{BV(\sigma_{j-1})}+\left\lVert
f\right\rVert_{BV(\tau_{j})}\bigr{)}$

from which the result follows easily. ∎

######  Theorem 8.2.

Suppose that $\sigma\in\mathrm{APIC}$. Let $\mathcal{C}$ denote the components
of $\sigma$ with respect to a polygonal mosaic $\mathcal{M}$. Then there
exists $K_{\sigma,\mathcal{M}}$ such that if $f:\sigma\to\mathbb{C}$ then

$\left\lVert f\right\rVert_{BV(\sigma)}\leq
K_{\sigma,\mathcal{M}}\sum_{\tau\in\mathcal{C}}\left\lVert
f\right\rVert_{BV(\tau)}.$

###### Proof.

Let $\mathcal{M}=\\{P_{i}\\}_{i=1}^{n}$ be the polygonal mosaic. As above, for
$j=1,\dots,n$, let $\sigma_{j}=\cup_{i=1}^{j}(P_{i}\cap\sigma)$ and let
$\mathcal{C}_{j}$ denote the components of $\sigma$ which lie in $\sigma_{j}$.
The previous lemma implies that $\left\lVert f\right\rVert_{BV(\sigma_{1})}$
is bounded by a constant $K_{1}$ (depending on the number of sides of $P_{1}$)
times the sum of the norms of $f$ on the components of $\sigma$ in
$\mathcal{C}_{1}$.

Suppose now that $1\leq j<n$ and that

(3) $\left\lVert f\right\rVert_{BV(\sigma_{j})}\leq
K_{j}\sum_{s\in\mathcal{C}_{j}}\left\lVert f\right\rVert_{BV(s)}.$

Let $r$ denote the number of sides of $P_{j+1}$. Extending the sides of
$P_{j+1}$ one can split the exterior of $P_{j+1}$ into $r$ regions
$R_{1},\dots,R_{r}$, each contained in a sector whose vertex is one of the
vertices of the polygon, as in Figure 11.

As in the proof of Theorem 7.8, one can now repeatedly apply Theorem 7.5 to
show that

$\left\lVert f\right\rVert_{BV(\sigma_{k+1})}\leq 8^{r}\big{(}\left\lVert
f\right\rVert_{BV(\sigma_{k})}+\left\lVert f\right\rVert_{BV(\sigma\cap
P_{k+1})}\bigr{)}.$

Applying Lemma 8.1 and using (3) gives the result. ∎

$P_{k+1}$$R_{1}$$R_{2}$$R_{3}$$R_{4}$$R_{5}$ Figure 11: Splitting the exterior
of a convex polygon into sectors $R_{1},\dots,R_{r}$.

Our next step is to ensure that a function $f$ defined on an $\mathrm{APIC}$
set is absolutely continuous if and only if it is absolutely continuous on
each component curve. Recall from Example 7.1 that this result fails if one
considers more general finite unions of convex curves.

######  Lemma 8.3.

Suppose that $T$ is a triangle in $\mathbb{R}^{2}$ with vertices at
${\boldsymbol{x}}$, ${\boldsymbol{y}}$ and ${\boldsymbol{z}}$. Let $c$ be a
differentiable convex curve joining ${\boldsymbol{x}}$ to ${\boldsymbol{y}}$
through the interior of $T$ which meets the side
$\ell=\overline{\vphantom{\vbox
to5.16663pt{}}{\boldsymbol{x}}\,{\boldsymbol{z}}}$ tangentially at
${\boldsymbol{x}}$. Let $\sigma=c\cup\ell$ and suppose that
$f:\sigma\to\mathbb{C}$. If $f|c\in AC(c)$ and $f|\ell\in AC(\ell)$ then $f\in
AC(\sigma)$. Furthermore $f$ has an extension $\hat{f}\in AC(T)$ with

$\lVert\hat{f}\rVert_{BV(T)}\leq 3\left\lVert f\right\rVert_{BV(\sigma)}.$

###### Proof.

By applying a suitable affine transformation we may assume that
${\boldsymbol{x}}=(0,0)$, that $\ell=\\{(0,y)\,:\,0\leq y\leq 1\\}$, and that
$c$ joins $(0,0)$ to $(1,1)$. We may assume that $c$ is the graph of the
convex function $k:[0,1]\to\mathbb{R}$. Let $R=[0,1]\times[0,1]$.

Let $g(x,y)=f(0,y)$, $(x,y)\in R$. Then, as $f|\ell\in AC(\ell)$, Theorem
2.3(iv) implies that $g\in AC(R)$. Define $f_{1}:c\to\mathbb{C}$ by
$f_{1}(x,y)=f(x,y)-g(x,y)$, $(x,y)\in c$. Now $f_{1}\in AC(c)$, so as $c$ is a
convex curve, the function $h:[0,1]\to\mathbb{C}$, $h(x)=f_{1}(x,k(x))$ is
also absolutely continuous. Indeed ${\hat{f}}_{1}(x,y)=h(x)$ is then
absolutely continuous on all of $R$. It follows that

${\hat{f}}(x,y)={\hat{f}}_{1}(x,y)+g(x,y),\qquad(x,y)\in R$

is also absolutely continuous on $R$. But of course ${\hat{f}}$ is an
extension of $f$ and so $f={\hat{f}}|\sigma\in AC(\sigma)$.

The estimate for $\|\hat{f}\|_{BV(T)}$ simply follows from the triangle
inequality and the facts that $\|\hat{f}_{1}\|_{BV(T)}\leq 2\|f\|_{\sigma}$
(by Lemma 5.4) and $\|g\|_{BV(T)}\leq\|f\|_{BV(\sigma)}$ (by Theorem 2.3(iv)).
∎

The following theorem is a generalization of [1, Theorem 6] for $\mathrm{PIC}$
sets.

######  Theorem 8.4.

Suppose that $\sigma\in\mathrm{APIC}$ is represented as above as a union of
convex curves and lines meeting these curves tangentially, and that
$f:\sigma\to\mathbb{C}$. Then $f\in AC(\sigma)$ if and only if

  1. 1.

$f|c_{k}\in AC(c_{k})$ for $k=1,\dots,n$, and

  2. 2.

$f|\ell_{j}\in AC(\ell_{j})$ for $j=1,\dots,m$.

###### Proof.

The forward direction follows from the general results about restrictions of
absolutely continuous functions.

Suppose then that (1) and (2) hold. By the Patching Lemma, it suffices to show
that $f$ is absolutely continuous on a compact neighbourhood of each point in
$\sigma$. Note that if ${\boldsymbol{x}}\in\sigma$ is not a line tangential
vertex, then, by the $\mathrm{PIC}$ joining theorem [1, Corollary 3], $f$ is
absolutely continuous on a compact neighbourhood of ${\boldsymbol{x}}$. So
suppose that ${\boldsymbol{x}}$ is a line tangential vertex.

Fix a small polygonal region $P$ centred at ${\boldsymbol{x}}$ and split this
into triangles $T_{1},\dots,T_{r}$ so that (the closure of) each triangle
contains at most one curve $c_{k}$ (and necessarily any line which meets
$c_{k}$ tangentially at ${\boldsymbol{x}}$). (See Figure 12.) Using Lemma 8.3,
one can deduce that $f$ is absolutely continuous on $T_{j}\cap\sigma$ for all
$j$.

Again, as in the proof of Theorem 7.7, one may inductively use Theorem 7.6 to
show that $f$ is absolutely continuous on the compact neighbourhood
$P\cap\sigma$ of ${\boldsymbol{x}}$. Thus $f$ is absolutely continuous on all
of $\sigma$. ∎

${\boldsymbol{x}}$ Figure 12: Triangulating around a line tangential vertex.

A consequence of the preceding results is that $\mathrm{APIC}$ is a
‘Gelfand–Kolmogorov family’ of compact sets, which generalizes [1, Theorem 7].

######  Corollary 8.5.

Suppose that $\sigma,\tau\in\mathrm{APIC}$. Then $AC(\sigma)$ is isomorphic
(as a Banach algebra) to $AC(\tau)$ if and only if $\sigma$ is homeomorphic to
$\tau$.

A final result which we will need is to be able to extend an absolutely
continuous function defined on an $\mathrm{APIC}$ set on each polygon of its
polygonal mosaic.

######  Theorem 8.6.

Suppose that $c$ is a projectable convex curve joining two vertices
${\boldsymbol{x}},{\boldsymbol{y}}$ of a convex polygon $P$ to each other
through the interior of $P$. Let $\sigma_{0}$ be a compact set which is the
union of $c$ and a (possibly empty) subset of the boundary of $P$, and suppose
that $f\in AC(\sigma_{0})$. Then $f$ admits an extension $\hat{f}\in AC(P)$.

###### Proof.

Note that $f$ is initially specified at at least two points of the boundary of
$P$, namely the endpoints of $c$. First extend $f$ to all of the sides of $P$
by making it continuous and affine on any parts of $\partial P$ which were not
part of $\sigma_{0}$ (and hence, by Lemma 4.1 and Theorem 8.4, it is in
$AC(\partial P)$). If $c$ is a straight line, then the conclusion follows by
applying Theorems 6.5 and 7.7.

Suppose then that $c$ is not a straight line. Let $P_{1}\subseteq P$ be the
convex polygon containing $c$ which has the diagonal
$\overline{\vphantom{\vbox to5.16663pt{}}{\boldsymbol{x}}\,{\boldsymbol{y}}}$
as a side. (If $\overline{\vphantom{\vbox
to5.16663pt{}}{\boldsymbol{x}}\,{\boldsymbol{y}}}$ was already a side, then
$P_{1}=P$.) Since $c$ is convex, one can find a line segment $\ell_{1}$ in
$P_{1}$ which is tangential to $c$ at ${\boldsymbol{x}}$. Choose a point
${\boldsymbol{z}}$ on $c$ so that the orthogonal projection ${\boldsymbol{w}}$
of ${\boldsymbol{z}}$ lies on $\ell_{1}$, and consider the triangle $T_{1}$
with vertices ${\boldsymbol{x}},{\boldsymbol{z}},{\boldsymbol{w}}$. If
$\ell_{1}$ forms part of a side of $P$, and $f$ has been specified on that
side, then extend $f$ to all of $T_{1}$ using Lemma 8.3. Otherwise extend $f$
to $T_{1}$ using Lemma 5.4. In either case one has an absolutely continuous
extension of $f$ to this triangle.

The same procedure can be repeated choosing a small right-angle triangle
$T_{2}$ with vertices
${\boldsymbol{y}},{\boldsymbol{z}}^{\prime},{\boldsymbol{w}}^{\prime}$ to
define an extension of $f$ on $T_{2}$. The points
${\boldsymbol{z}},{\boldsymbol{z}}^{\prime}\in c$ now lie in the interior of
$P$. Let $R$ be a rectangle with side $\overline{\vphantom{\vbox
to5.16663pt{}}{\boldsymbol{z}}\,{\boldsymbol{z}}^{\prime}}$ which contains
$c^{\prime}$, the part of $c$ between ${\boldsymbol{z}}$ and
${\boldsymbol{z}}^{\prime}$. Choose a polygon $Q\subseteq R\cap P_{1}$ which
contains $c^{\prime}$. Since $c$ is projectable we may use Theorem 2.3(iv) to
extend $f$ to all of $Q$ by making it constant on lines which are orthogonal
to $\overline{\vphantom{\vbox
to5.16663pt{}}{\boldsymbol{z}}\,{\boldsymbol{z}}^{\prime}}$.

We have now specified the extension $\hat{f}$ on the green regions in Figure
13, as well as on the blue boundary of $P$. This leaves a finite number of
polygonal regions on which $\hat{f}$ given on the boundary, but not in the
interiors of each of the regions. Using Theorem 6.5 one can now extend
$\hat{f}$ so that it is absolutely continuous on each of these regions.
Finally, we can apply Theorem 7.7 to deduce that $\hat{f}$ is absolutely
continuous on the whole set $P$. ∎

${\boldsymbol{x}}$${\boldsymbol{y}}$${\boldsymbol{z}}$${\boldsymbol{z}}^{\prime}$${\boldsymbol{w}}$${\boldsymbol{w}}^{\prime}$$T_{1}$$T_{2}$$Q$$P_{1}$$c$$R$
Figure 13: The construction in the proof of Theorem 8.6. The value of $f$ is
initially given on the red curves.

##  9 The $\mathrm{SPIC}$ extension theorem

Suppose that $\sigma_{0}\in\mathrm{SPIC}$. Then there exists a set
$\sigma_{1}\in\mathrm{APIC}$ with $\sigma_{0}\subseteq\sigma_{1}$. By Lemma
3.6, we may choose a polygonal mosaic so that each of the components of
$\sigma_{1}$ is projectable, and we shall assume that that is the case
throughout this section.

We will see below that we can always construct extensions of $AC$ functions
defined on APIC sets, and then extend this result to the case of SPIC sets.

######  Lemma 9.1.

Suppose that $\emptyset\neq\sigma_{0}\subseteq\sigma_{1}$ with
$\sigma_{1}\in\mathrm{APIC}$. If $f\in AC(\sigma_{0})$, then there exists an
extension $\hat{f}\in AC(\sigma_{1})$ with
$\lVert\hat{f}\rVert_{BV(\sigma_{1})}\leq C_{\sigma_{1}}\left\lVert
f\right\rVert_{BV(\sigma_{0})}$ for some $C_{\sigma_{1}}>0$.

###### Proof.

Let $\mathcal{C}$ denote the set of components of $\sigma_{1}$ with respect to
a suitable polygonal mosaic. Suppose that $f\in AC(\sigma_{0})$. Let
$\sigma_{0}^{\prime}$ denote the union of $\sigma_{0}$ and all of the
(finitely many) endpoints of the components of $\sigma_{1}$. Any endpoint
which is not in $\sigma_{0}$ is an isolated point of $\sigma_{0}^{\prime}$.
Thus we can extend $f$ to $\sigma_{0}^{\prime}$ by setting it to be zero at
any such point and this extension is in $AC(\sigma_{0}^{\prime})$. The norm of
the extension may increase by a factor of 3 on each component.

Let $\tau\in\mathcal{C}$ be a component of $\sigma_{1}$. By Theorem 5.3 we can
extend $f$ from $\sigma_{0}^{\prime}\cap\tau$ to ${\hat{f}}\in AC(\tau)$ (with
a possible doubling of the norm). Since $\sigma_{0}^{\prime}$ contains all the
endpoints of the components of $\sigma_{1}$, doing this for each
$\tau\in\mathcal{C}$ produces a well-defined function
$\hat{f}:\sigma_{1}\to\mathbb{C}$. By Theorem 8.4, $\hat{f}\in
AC(\sigma_{1})$.

The existence of the constant $C_{\sigma_{1}}$ follows from Theorem 8.2. ∎

######  Theorem 9.2.

Suppose that $\sigma_{0}\in\mathrm{SPIC}$ and that $\sigma$ is a compact
superset of $\sigma_{0}$. If $f\in AC(\sigma_{0})$ then there exists an
extension $\hat{f}\in AC(\sigma)$ with $\lVert\hat{f}\rVert_{BV(\sigma)}\leq
K_{\sigma_{0}}\left\lVert f\right\rVert_{BV(\sigma_{0})}$ for some
$K_{\sigma_{0}}>0$.

###### Proof.

Suppose that $f\in AC(\sigma_{0})$. As $\sigma_{0}\in\mathrm{SPIC}$, there
exists $\sigma_{1}\in\mathrm{APIC}$ with $\sigma_{0}\subseteq\sigma_{1}$. By
Lemma 9.1 we can find an absolutely continuous extension of $f$ to
$\sigma_{1}$.

Let $\mathcal{M}=\\{P_{i}\\}_{i=1}^{n}$ be a polygonal mosaic for $\sigma_{1}$
chosen so that each component convex curve is projectable. By Theorem 8.6 we
can extend $f$ from $\sigma_{1}\cap P_{1}$ to be absolutely continuous on
$P_{1}$. We may now inductively further extend $f$ to each polygon $P_{i}$ in
turn, taking into account the values which have already been determined on the
boundary of the polygon at each stage. By Theorem 7.7, the extension $\hat{f}$
is absolutely continuous on $M=\cup_{i=1}^{n}P_{i}$ (the blue region in Figure
14).

Let $R_{0}$ be a large rectangle which contains $M$ in its interior. Let
$\sigma_{2}=M\cup\partial R_{0}$ and extend ${\hat{f}}$ to $\sigma_{2}$ by
making it zero on the boundary of $\sigma_{2}$. As in [10, Lemma 4.2],
${\hat{f}}\in AC(\sigma_{2})$ with $\lVert{\hat{f}}\rVert_{BV(\sigma_{2})}\leq
5\left\lVert f\right\rVert_{BV(M)}$.

The region $R_{0}\setminus M$ can be triangulated by $\\{T_{i}\\}_{i=1}^{m}$
with each triangle having vertices in $\sigma_{2}$. If necessary, use Theorem
9.1 to extend ${\hat{f}}$ to the whole boundary of $T_{1}$. One can then use
Theorem 6.5 to extend ${\hat{f}}$ to an absolutely continuous function on
$T_{1}$. As above, one can repeat this procedure with each triangle in turn,
taking into account the values of ${\hat{f}}$ already determined.

If necessary, one now chooses an even larger rectangle $R$ containing
$R_{0}\cup\sigma$ and extends ${\hat{f}}$ further by setting ${\hat{f}}=0$ on
$R\setminus R_{0}$. Applying Theorem 7.7 again shows that ${\hat{f}}\in
AC(R)$. At each step we have a control over the norm of the extension. This
provides a bound on $\lVert{\hat{f}}\rVert_{BV(R)}$ which depends only on the
choice of $R_{0}$, and not on $\sigma$.

Finally we can restrict $\hat{f}$ to $\sigma$ to give the conclusion of the
theorem. ∎

$R_{0}$$\sigma$$\sigma_{1}$$M$ Figure 14: Theorem 9.2. The function $f$ is
first extended to $\sigma_{1}\in\mathrm{APIC}$, then to the blue polygons
whose union is $M$, and then, one triangle at a time, to the rectangle
$R_{0}$. The final step is to restrict to the set $\sigma$.

##  10 The main theorem

######  Theorem 10.1.

Suppose that $\emptyset\neq\sigma_{0}=\sigma_{P}\cup\sigma_{S}$ with
$\sigma_{P}\in PW^{+}$ and $\sigma_{S}\in\mathrm{SPIC}$ and that $\sigma$ is a
compact superset of $\sigma_{0}$. Then there exists a constant
$C_{\sigma_{0}}$ such that for all $f\in AC(\sigma_{0})$ there exists an
extension $\hat{f}\in AC(\sigma)$ with $\|\hat{f}\|_{BV(\sigma)}\leq
C_{\sigma_{0}}\|f\|_{BV(\sigma_{0})}$.

###### Proof.

Choose a large rectangle $R$ which contains $\sigma$ in its interior. Let
${\hat{\sigma}}_{P}=\sigma_{P}\cup\partial R\in PW^{+}$. As in the proof of
Theorem 9.2, at the cost of a factor 5 in the norm, we can extend $f$ by
setting it to be zero on $\partial R$. Let
$\mathcal{W}=\\{W_{1},\dots,W_{d}\\}$ denote the connected components of
$R\setminus{\hat{\sigma}}_{P}$.

Fix $W\in\mathcal{W}$. Since the boundary of $W$ consists of a finite number
of line segments, we can use Lemma 3.7 to show that $\sigma_{W}=\partial
W\cup(\sigma_{S}\cap W)\in\mathrm{SPIC}$. Thus, by Theorem 9.2, there is a
constant $K_{W}$ and an extension $f_{W}$ of $f|\sigma_{W}$ to an absolutely
continuous function on all of $\mathop{\mathrm{cl}}(W)$ with $\left\lVert
f_{W}\right\rVert_{BV(W)}\leq K_{W}\left\lVert
f\right\rVert_{BV(\sigma_{0})}$.

Having done this for each $W\in\mathcal{W}$, setting

$\hat{f}({\boldsymbol{x}})=\begin{cases}f({\boldsymbol{x}}),&{\boldsymbol{x}}\in\sigma_{0},\\\
f_{W}({\boldsymbol{x}}),&{\boldsymbol{x}}\notin\sigma_{0},\
{\boldsymbol{x}}\in W\in\mathcal{W}\end{cases}$

gives a function which is absolutely continuous on all of $R$ by Theorem 7.10.
Finally we can restrict to $\sigma$ to obtain the required extension in
$AC(\sigma)$. The bound follows from applying Corollary 7.9 repeatedly. ∎

Theorem 10.1 covers almost all the sets $\sigma_{0}$ for which it it is known
that one can always find an absolutely continuous extension to a larger
compact set. It is likely that the differentiability restriction on the convex
curves considered can be relaxed. It is not too difficult, for example, to use
the results of this paper to prove an extension result for a single convex
curve consisting of an infinite collection of line segments.

It would of course be of interest to know whether one can always extend an
absolutely continuous function on a disk to larger sets.

##  11 An application to operator theory

The motivation for this work is to be able to show that if a Banach space
operator $T$ is an $AC(\sigma)$ operator, then it is in fact an
$AC(\sigma(T))$ operator. That is, it has an $AC(\sigma(T))$ functional
calculus.

Suppose then that $T$ is bounded operator on a Banach space $X$ with
$\sigma(T)=\sigma_{P}\cup\sigma_{S}$ with $\sigma_{P}\in PW^{+}$ and
$\sigma_{S}\in\mathrm{SPIC}$. Suppose further than $T$ has an $AC(\sigma)$
functional calculus, which implies that $\sigma(T)\subseteq\sigma$. That is,
there is a Banach algebra homomorphism $\Psi:AC(\sigma)\to B(X)$ which extends
the natural polynomial functional calculus.

By Theorem 10.1 there is a map $e:AC(\sigma(T))\to AC(\sigma)$ such that
$e(f)$ is an extension of $f$ to $\sigma$. Thus the map $\Phi:AC(\sigma(T))\to
B(X)$ given by $\Phi(f)=\Phi(e(f))$ is well-defined. It is of course not
immediately clear that this map is a functional calculus map for $T$ since the
map $e$ has not been shown to have any algebraic properties.

Suppose then that $f,g\in AC(\sigma(T))$. Then $e(f+g)-(e(f)+e(g))$ is
identically zero on $\sigma(T)$. Theorem 3.1.6 of [8] ensures that the support
of the functional calculus map $\Psi$ is $\sigma(T)$, which implies that

$0=\Psi\big{(}e(f+g)-(e(f)+e(g))\big{)}=\Psi(e(f+g))-\Psi(e(f))-\Psi(e(g))$

or $\Phi(f+g)=\Phi(f)+\Phi(g)$. A similar proof shows that $\Phi$ is
multiplicative. Since
$\left\lVert\Phi(f)\right\rVert\leq\left\lVert\Psi\right\rVert\left\lVert
e(f)\right\rVert_{BV(\sigma)}\leq
C_{\sigma(T)}\left\lVert\Psi\right\rVert\left\lVert
f\right\rVert_{{BV}(\sigma(T))}$ we now have that $\Phi$ is a Banach algebra
homomorphism. Finally, let $\lambda(z)=z$ be the identity function on the
complex plane. Since $e(\lambda)-\lambda$ is identically zero on $\sigma(T)$
we must have that $\Phi(\lambda)=\Psi(e(\lambda))=\Psi(\lambda)=T$, and so
$\Phi$ is an $AC(\sigma(T))$ functional calculus for $T$.

We note in particular that this conclusion holds whenever
$\sigma(T)\subset\mathbb{R}$ or $\sigma(T)\subseteq\mathbb{T}$, which covers
the cases of well-bounded and trigonometrically well-bounded operators.

The remaining challenge is to be able to remove the above hypotheses on the
spectrum of $T$.

### Acknowledgements

The work of the second author was supported by the Research Training Program
of the Department of Education and Training of the Australian Government.

### Rights Retention Statement

This research was produced in whole or part by UNSW Sydney researchers and is
subject to the UNSW Intellectual property policy. For the purposes of Open
Access, the authors have applied a Creative Commons Attribution CC-BY licence
to any Author Accepted Manuscript(AAM) version arising from this submission.

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