Trace operator on von Koch's snowflake
Krystian Kazaniecki, Micha{\l} Wojciechowski

TL;DR
This paper investigates the boundary trace operator on Sobolev spaces over the von Koch snowflake domain, providing a combinatorial proof of its properties, identifying the trace space as , and explaining the non-existence of right inverses in certain domains.
Contribution
It offers a new combinatorial proof of the trace operator's properties on the von Koch snowflake and characterizes the trace space as , also explaining the non-existence of right inverses in regular boundary domains.
Findings
The trace space is isomorphic to .
A combinatorial proof of the trace operator properties is provided.
The non-existence of right inverses for domains with regular boundaries is explained.
Abstract
We study properties of the boundary trace operator on the Sobolev space . Using the density result by Koskela and Zhang, we define a surjective operator \mbox{}, where is von Koch's snowflake and is a trace space with the quotient norm. Since is a uniform domain whose boundary is Ahlfors-regular with an exponent strictly bigger than one, it was shown by L. Mal\'y that there exists a right inverse to , i.e. a linear operator such that . In this paper we provide a different, purely combinatorial proof based on geometrical structure of von Koch's snowflake. Moreover we identify the isomorphism class of the trace space as . As an additional consequence of our approach we obtain a simple proof of the Peetre's…
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Taxonomy
TopicsAdvanced Mathematical Modeling in Engineering · Holomorphic and Operator Theory · Advanced Harmonic Analysis Research
Trace operator on von Koch’s snowflake
Krystian Kazaniecki*†,*
Institute of Analysis, Johannes Kepler University Linz
Institute of Mathematics, University of Warsaw
Michał Wojciechowski
Institute of Mathematics, Polish Academy of Sciences
Abstract
We study properties of the boundary trace operator on the Sobolev space . Using the density result by Koskela and Zhang [13], we define a surjective operator , where is von Koch’s snowflake and is a trace space with the quotient norm. Since is a uniform domain whose boundary is Ahlfors-regular with an exponent strictly bigger than one, it was shown by L. Malý [16] that there exists a right inverse to , i.e. a linear operator such that . In this paper we provide a different, purely combinatorial proof based on geometrical structure of von Koch’s snowflake. Moreover we identify the isomorphism class of the trace space as . As an additional consequence of our approach we obtain a simple proof of the Peetre’s theorem [18] about non-existence of the right inverse for domain with regular boundary, which explains Banach space geometry cause for this phenomenon.
††footnotetext: Corresponding author
Keywords: Sobolev Spaces, Trace operator, von Koch’s Snowflake
** MSC 2020** : 46E35, 46B25, 46B03, 46B45
It was shown by Gagliardo ([8]) that the trace operator maps the space onto for domains with Lipschitz boundary. From this theorem immediately arises a question whether there exists a right inverse operator to the trace, i.e. a continuous, linear operator such that . It turns out that in general such operator does not exist. This was proved by Peetre ([18]). In his paper he has shown the non-existence of right inverse to the trace operator for a half plane. From that by straightening out the boundary one can deduce the non-existence of the right inverse for with a Lipschitz boundary. More recent proofs can be found in [20], [2]. Note that the trace was studied also in the context of spaces given by more general differential constraints [9]. In this article we present an exceptionally simple proof based on geometry of a Whitney covering and basic properties of classical Banach spaces.
Theorem 1**.**
Let be an open domain with Lipschitz boundary and be a Jordan curve. Let be a trace operator. Then there is no continuous, linear operator such that .
Actually the proof of Theorem 1 presented here works with weaker assumptions. Whenever
one can define reasonable trace operator i.e. functions which are continuous up to the boundary are dense in and the Trace is just the restriction operator on such functions; when is a Jordan curve this is provided by Koskela, Zhang theorem ( [13]), 2. 2.
trace space contains isomorphic copy of .
In [10] Hajlasz and Martio studied the existence of a right inverse to the trace operator in the case of Sobolev spaces for and they characterize trace space as a generalized Sobolev space. In paper [1] the trace of Hajłasz-Sobolev spaces to porous Ahlfors regular closed subspace is studied for sufficiently large exponent.
In the case of the behavior of the trace space ( the smallest Banach space for which the trace operator is bounded) changes dramatically for the domains with fractal boundary. L. Malý [16] characterized the trace spaces for Ahlfors regular, uniform domains in terms of Besov spaces on fractal sets and he constructed the linear extension operator from the trace space.
The second goal of this paper is to study the trace operator on the Sobolev space on von Koch’s snowflake . This is a very particular domain and the proofs provided in this article depend heavily on its combinatorial structure. However they are quite different than Malý approach and does not use Besov norm at all. The Besov space obtained by Malý as a trace space in the case of von Koch’s snowflake turns out to be isomorphic to . Indeed, it follows from Theorem 1 and Theorem 3 of Chapter VI of [11] that the trace space is complemented subspace of . By Proposition 7 p. 200 of [17] the space is isomorphic to . By Pełczyński’s theorem complemented subspaces of are isomorphic to [19]. Our proof identifies the trace space directly as an Arens-Eells space (otherwise known as Lipschitz free space) for a suitable metric on the boundary by a combinatorial argument. It follows by an old Ciesielski’s argument [5] that this space is isomorphic to . The existence of a right inverse operator is just a property of . In the proof we use the structure of a specific Whitney covering of described in the Appendix, which is quite interesting in itself. To summarize:
Theorem 2**.**
Let be a trace operator, where is a trace space defined by (2). Then and there exists a continuous, linear operator such that .
We want to stress that the novelty of this article lies in proposing new combinatorial method of proof rather than the result itself, which is already known. To keep the clarity of the presentation we decided to focus on the von Koch’s snowflake. Our method could be applied to more general class of domains (some of them could be obtained e.g. by Carleson’s construction [4]). However we do not know how wide the actual range of possible applications of this approach is.
In the following section we define the trace operator, trace space and auxiliary properties of needed in the proof.
1 Properties of and trace operator
From now on we assume that , is a Jordan curve. Our approach to Theorem 1 up to technical differences works in higher dimensions. However in the proof of the Theorem 2 the properties of two dimensional euclidean space are crucial. We define the trace operator and the trace space for . Let us recall a notion of (slightly generalized) Whitney covering of .
Definition 3**.**
We call the family of polygons a Whitney decomposition of an open set if it satisfies:
*For the boundaries are uniformly bi-lipschitz *images of . 2. 2.
* and elements of have pairwise disjoint interiors.* 3. 3.
** 4. 4.
If ha a positive one dimensional Hausdorff measure then
- (a)
. 2. (b)
** 3. (c)
,
where denotes length of a curve, and denotes the area of the polygon. 5. 5.
For a given polygon there exists at most N polygons s.t. .
For the purpose of this article we will also assume that polygons of are uniformly star shaped in the following sense
For every there exists a point and positive numbers , s.t. , where is fixed and the polygon is star shaped with respect to . We call such point a center of .
Let be such covering then we can define a graph describing it’s geometry.
Definition 4**.**
Let be a Whitney decomposition. We call a graph a graph of if and only if boundaries of and have intersection of positive one dimensional Hausdorff measure.
We denote by a space of measures of bounded variation i.e.
[TABLE]
with a norm
[TABLE]
We introduce some special subspaces of .
Definition 5**.**
Let be a Whitney decomposition of . We define the following subspaces of
[TABLE]
and
[TABLE]
It is a known fact that for a given Whitney decomposition the space is a complemented subspace of . A proof of this fact can be found in ([21],[6]).
Lemma 6**.**
For any domain :
[TABLE]
Let us observe that we can easily calculate the norm of function .
[TABLE]
Definition 7**.**
Let be a simply connected planar domain with Poincare inequality for and be its Whitney decomposition. We will call a spanning tree of the graph a Whitney tree of if it satisfies the following conditions:
for every
[TABLE] 2. 2.
*for every point on the boundary there is *an infinite branch of s.t. and as , where . For a sequence of real numbers we call a limit a limit along the branch .
In their unpublished preprint Derezinski, Nazarov, Wojciechowski [7] have proven that for any bounded simply connected planar domain there exists a Whitney tree of the graph . However in the present paper we will not use this result. An explicit construction of a Whitney tree for von Koch’s snowflake is given in Section 3 and in the Appendix. Moreover the obtained tree has very regular structure.
It follows immediately that , where is a set with the norm . Using the above notation we define trace of . Since is a domain with a Jordan curve as boundary it follows from theorem of Koskela and Zhang ( [13], see [12] for the case d>2) that restrictions of Lipschitz function are dense in . For we define the trace operator as a restriction of to the boundary. We define a trace space as completion of a space with respect to the norm , where
[TABLE]
Since Lipschitz functions on are dense in we can define the trace operator on the whole space . It is obvious that is a continuous linear operator and it is surjective. We want to extend the trace operator to (see also [14]).
Lemma 8**.**
There exists a continuous, linear operator s.t. for every
[TABLE]
Proof.
We begin with a construction of an operator which smooths out the function from inside of a fixed ball . We formulate it in two-dimensional case, however it is valid in domains of arbitrary dimension with the analogous proof.
Lemma 9**.**
For every ball there exists a bounded linear operator such that and for . Moreover
[TABLE]
with a constant independent of the radius of a ball. Moreover for we have
[TABLE]
Proof of Lemma 9.
We will use operator from Proposition 4.2 in [20]. It is a continuous operator from to satisfying . This operator is given by a convolution like formula:
[TABLE]
Therefore for we have
[TABLE]
For every we have and diffeomorphism such that There exists such that for we have
[TABLE]
We can choose a finite cover of by and with . There exists a decomposition of unity corresponding to this covering. We denote it by . We define our operator by the formula
[TABLE]
where is a mollifier with , . Clearly
[TABLE]
From properties of the operator we have
[TABLE]
for every . Thus
[TABLE]
Moreover from the properties of the operator and convolution for we get
[TABLE]
Let be a homothety with a center at zero and scale we define
[TABLE]
First we estimate the norm of the gradient
[TABLE]
Note that is a bounded linear operator on . Thus is bounded linear operator on . We estimate the norm of the function
[TABLE]
It is clear satisfies estimates. Now we define
[TABLE]
Note that from the properties of follows that satisfies all of the properties from the statement of the Lemma 9. ∎
We return to the proof of Lemma 8. We a define family of balls where and . We put , where . This collection is a Besicovitch covering of . There are at most families of disjoint balls, which cover . We define operators
[TABLE]
and
[TABLE]
It is clear, that we smooth out the function on neighbourhoods of every point . Hence for we get a function . Moreover
[TABLE]
Functions are constant on . Thus
[TABLE]
where depends on constants from Definition 3.3. Balls from only intersect the neighbouring cubes. By Definition 3.3-4 we know that on the neighbouring cubes the right hand sides of (5) are comparable. Therefore by (4) we get
[TABLE]
The estimates above remain comparable on the neighbouring cubes. Repeating this argument times we get
[TABLE]
The function is modified only on the set . The radii of balls intersecting are comparable to . Thus for depending on the bi-lipschitz constant (Def 3.1) we have
[TABLE]
Therefore
[TABLE]
∎
Let be a projection from onto . We define by the formula
[TABLE]
If then the function is continuous on . Therefore its trace is a restriction of to the boundary. However the value of the restriction at point for the function from is equal to the limit of \mathchoice{{\vbox{\hbox{\textstyle- }}\kern-7.83337pt}}{{\vbox{\hbox{\scriptstyle- }}\kern-5.90005pt}}{{\vbox{\hbox{\scriptscriptstyle- }}\kern-4.75003pt}}{{\vbox{\hbox{\scriptscriptstyle- }}\kern-4.25003pt}}\!\int_{A}\Phi(Pf(y))dy along the branch . From (3) and the definition of the space
[TABLE]
Since are dense in and the operator is an extension of the trace operator to . We will abuse the notation and from now on we will denote by . From the definition of the trace operator it follows that
[TABLE]
2 Proof of Peetre’s theorem
In this section we will give a proof of Theorem 1.
Proof.
Since has Lipschitz boundary by theorem of Gagliardo - the space of functions integrable with respect to the -dimensional Hausdorff measure. Let us denote by the projection onto . Assume there exists such that . Then the following diagram is commutative
{L^{1}(\partial\Omega)}$${BV(\Omega)}$${L^{1}(\partial\Omega)}$${BV_{G}}SPTr$$Tr
From (6) and the theorem of Gagliardo we conclude that Tr|_{BV_{{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}G}}} is onto . On the other hand, . Hence is isomorphic to a subspace of . The definition of implies that is isomorphic to a subspace of . Since the measure on the boundary is non atomic, . However, it is well known that could not be embedded in . (To see this, note that by Khintchine inequality, Radamacher functions span in space. The space could not be embedded in because every subspace of contains a copy ([15], Proposition 1.a.11). ∎
3 Trace operator on von Koch’s snowflake
Let be a domain bounded by von Koch’s curve. Since is simply connected and von Koch’s curve is a Jordan curve, we can use all the properties from the first section. It is enough to show that there exists a right inverse to the trace on because then and , where is an operator from Lemma 8.
It is a well known fact that satisfies Poincaré inequality (eg. [3]). Therefore (here and further in this section the constants are absolute and specific for von Koch’s snowflake and it’s concrete Whitney’s covering; in possible generalizations for other domains given by Carleson construction they will depend on specific geometry, bi-lipschitz constants of Whitney’s covering and corresponding scaling factor).
[TABLE]
where is a total variation of a measure on . This inequality implies
[TABLE]
where
[TABLE]
where is the space of constant functions on . In this case the norm is equal to
[TABLE]
Similarly for the quotient space . We have already established that for all we have
[TABLE]
We reduce the problem to finding a right inverse operator to the trace . We will show it’s existence for a carefully chosen Whitney covering. Construction of this covering is described in the Appendix. We introduce the following notation
Definition 10**.**
For a given tree by we will denote the root of . For a vertex by we denote descendants of of order exactly and we put . For a vertex by we denote its unique father. We will denote by the set of all descendants of i.e. .
We take a covering as shown on the Figure 1.
This covering of von Koch’s snowflake is easy to describe if we look at its Whitney tree . The root of is a six pointed star with six "pants" shaped descendants. We denote it by . In this tree there are three types of polygons/vertices. The aforementioned root, "pants" shaped polygons and "palace" shaped polygons. The type of a vertex describes direct descendants of this vertex (Figure 2).
Polygons in are similar to polygons from with a scale . The tree is the tree from Definition 7. Indeed let G be a Whitney graph of this decomposition. We denote by . We see that
[TABLE]
However for we have from the triangle inequality
[TABLE]
Since we use edge in the estimate for at most two edges from we get
[TABLE]
We get by induction
[TABLE]
The geometric sequence with quotient is convergent. Therefore for such Whitney covering the norm of satisfies
[TABLE]
Further we will use the above formula as a norm on . We want to study the norm on . To be precise, we want to define and calculate the norm of .
Definition 11**.**
Let us denote by a cylinder of , i.e. . We call an arc rational if there exists a finite sequence s.t. and we say that points x,y are rational points.
For a given arc there exists a sequence of vertices s.t. and sets are pairwise disjoint. Moreover this sequence can be taken maximal in the sense that if a vertex is in the sequence then there exists , which is not in . Such a sequence is unique for . Let be a natural number such that . Let
[TABLE]
We introduce an auxiliary metric on the boundary
[TABLE]
It is easy to check that is a metric which is greater than two dimensional euclidean metric. We prefer this metric over the euclidean one because it is a monotone function on an arc with respect to the natural order on the arc. In the lemma below we show that for every rational arc and every monotone right continuous function on this arc there exists a "good" extension of this function to . We call an arc [x,y] a short arc iff
[TABLE]
We will say that a function is monotone on an arc if it is monotone with respect to the natural order on the arc.
Lemma 12**.**
Let and be a short arc. Let function be a monotone and continuous function on the arc and . There exists such that
** 2. 2.
F(z)=\underset{\begin{subarray}{c}A\in\operatorname{br}(z)\\ A\rightarrow z\end{subarray}}{\lim}\mathchoice{{\vbox{\hbox{\textstyle- }}\kern-7.83337pt}}{{\vbox{\hbox{\scriptstyle- }}\kern-5.90005pt}}{{\vbox{\hbox{\scriptscriptstyle- }}\kern-4.75003pt}}{{\vbox{\hbox{\scriptscriptstyle- }}\kern-4.25003pt}}\!\int h(y)dy\qquad\forall\,z\in\partial\Omega_{K}.
Proof.
First we prove the existence of the good extension for characteristics functions on arcs . Since arc is a short arc then an arc is a short arc and can be written as a countable sum in a unique way mentioned in the definition of . From this assumption it is clear that . Let us put
[TABLE]
Clearly along every infinite branch the limit of exists and it is equal to . We need to estimate the total variation of . From the definition of it follows that
[TABLE]
Let us assume that is an increasing function. For let be its Lebesgue-Stieltjes measure i.e. . From the definition of the measure and the assumptions on we get
[TABLE]
We define by the formula
[TABLE]
It follows from the definition of that
[TABLE]
Since it follows from Lebesgue’s dominated convergence theorem that
[TABLE]
∎
Lemma 13**.**
For every and there is a monotone Lipschitz function , with respect to the euclidean metric, on the arc such that and .
Proof.
In order to construct such function we proceed inductively. If we have defined values in the arc only at a points and we choose point such that and (since arc is a one dimensional curve there exists such a point). We put . This procedure allows us to define on a dense subset. We extend to the whole arc. The function has desired properties. ∎
In the lemma below we prove the existence of a class of functions in , which have desirable properties and every function from this class provides a good approximation of the norm of its trace on the boundary.
Lemma 14**.**
Let . There are sequences of functions , , and such that:
, 2. 2.
For every , \mathds{1}_{[x,y]}(z)=\underset{\begin{subarray}{c}A\in\operatorname{br}(z)\\ A\rightarrow z\end{subarray}}{\lim}\mathchoice{{\vbox{\hbox{\textstyle- }}\kern-7.83337pt}}{{\vbox{\hbox{\scriptstyle- }}\kern-5.90005pt}}{{\vbox{\hbox{\scriptscriptstyle- }}\kern-4.75003pt}}{{\vbox{\hbox{\scriptscriptstyle- }}\kern-4.25003pt}}\!\int_{A}f_{n}(y)dy, 3. 3.
. 4. 4.
* is a Cauchy sequence in * 5. 5.
. 6. 6.
.
Proof.
We use Lemma 13. For every and every rational arc , the characteristic function of can be written as sum of a Lipschitz function and a two monotone Lipschitz functions , with supports in arcs respectively. Moreover are rational, and the monotone functions are bounded uniformly by one. Hence from the Lemma 12 for every function there exists a function such that and for every
[TABLE]
Any Lipschitz extension of to is in ). Hence is in the trace space. From the definition of the trace space there exists a such that
[TABLE]
Since is in we have \underset{\begin{subarray}{c}A\in\operatorname{br}(x)\\ A\rightarrow x\end{subarray}}{\lim}\mathchoice{{\vbox{\hbox{\textstyle- }}\kern-7.83337pt}}{{\vbox{\hbox{\scriptstyle- }}\kern-5.90005pt}}{{\vbox{\hbox{\scriptscriptstyle- }}\kern-4.75003pt}}{{\vbox{\hbox{\scriptscriptstyle- }}\kern-4.25003pt}}\!\int_{A}g_{\varepsilon}(y)dy=g(x). Therefore the function has desired properties. The limits along of \mathchoice{{\vbox{\hbox{\textstyle- }}\kern-7.83337pt}}{{\vbox{\hbox{\scriptstyle- }}\kern-5.90005pt}}{{\vbox{\hbox{\scriptscriptstyle- }}\kern-4.75003pt}}{{\vbox{\hbox{\scriptscriptstyle- }}\kern-4.25003pt}}\!\int_{A}g_{\varepsilon}(y)dy exist and are equal to for every and
[TABLE]
where the term is the estimate on the norms of the functions . For every we choose suitable and we get the desired properties. The sequence is Cauchy sequence. Indeed for a given function and there exists a continuous piecewise monotone function with support on a small set on the boundary such that
[TABLE]
From Lemma 12 there exists a function with a small norm such that
[TABLE]
The size of the support of depends only on . Therefore
[TABLE]
for sufficiently large . ∎
The Cauchy sequence defines an element . From the analogous argument as in the above Lemma if satisfies \mathds{1}_{[x,y]}(z)=\underset{\begin{subarray}{c}A\in\operatorname{br}(z)\\ A\rightarrow z\end{subarray}}{\lim}\mathchoice{{\vbox{\hbox{\textstyle- }}\kern-7.83337pt}}{{\vbox{\hbox{\scriptstyle- }}\kern-5.90005pt}}{{\vbox{\hbox{\scriptscriptstyle- }}\kern-4.75003pt}}{{\vbox{\hbox{\scriptscriptstyle- }}\kern-4.25003pt}}\!\int_{A}f(y)dy for every on the boundary then . To simplify the notation we denote . From the point 6. of the Lemma 14 it follows
[TABLE]
Since the projection from onto preserves the trace, we may assume that functions are from . Therefore the function , whose arcs are rational, satisfies
[TABLE]
where consists of such that the limit exists for every and it is equal to .
Remark 15**.**
In the above lemmas we abuse the notation a bit. For rational points there are two branches . If we look at a finite linear combination of characteristic functions of arcs, the are finitely many points (endpoints of segments) on which the limits over this two the branches are different. However they are equal to the value of the trace either on left or right side of that endpoint. Further in the article we are only interested in branches which contain some specific vertex . Hence we are interested only in one of the problematic branches and it is clear what we mean by the limit.
We want to characterize the space . We introduce, a metric on von Koch’s curve by the formula
[TABLE]
where is a characteristic function of an arc on the von Koch’s curve which connects and . It does not matter which one of the two arcs we take because the difference between their characteristic functions is constant. Further in the proof it will be clear which arc is considered. Since is a norm, is a metric on the boundary. For a given metric space we define the Arens-Eells space ([22]).
Definition 16**.**
Let be a metric space. We call a function a molecule if it has finite support and . Let . We define special type of a molecule - an atom : , where is a characteristic of a set {\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\{a\}}. Let be a molecule, i.e. , then the Arens-Eells norm of is
[TABLE]
where the infimum is taken over all possible representations of as a sum of . The Aerens-Eells space is the completion of molecules with respect to the norm .
We want to show that is isomorphic to the Arens-Eells space with the metric . We will denote by the linear space of molecules. Clearly it is a non-complete norm space. By the definition it is dense in . We define the candidate for the isomorphism on the a linearly dense subsets of both spaces. We set by the formula
[TABLE]
Lemma 17**.**
* is an isomorphism between Banach spaces.*
Proof.
By the triangle inequality and the definitions of and Arens-Eells space, it follows that is continuous
[TABLE]
Proving the estimate from below is more involved. In the trace space we have following density result.
Lemma 18**.**
* is dense in .*
Proof.
From [13] we know that the restrictions of Lipschitz functions on are dense in . Therefore Lipschitz functions are dense in . Hence for any there exists a sequence of Lipschitz functions such that
[TABLE]
So it is enough to approximate Lipschitz functions with piecewise constant functions. Let be a Lipschitz function. We define a piecewise constant function , where are rational points of order i.e. such that . We define a function by the formula
[TABLE]
The function satisfies
[TABLE]
We will estimate the norm of the function . The function is also Lipschitz with respect to the metric . Let be the Lipschitz constant of with respect to . Observe that due to the Lipschitz continuity of the function there are positive numbers such that for every pants shaped polygon and we have
[TABLE]
Similarly for palace shaped polygon
[TABLE]
Let . We can prove inductively that . Let we have the following estimate on the variation on the sub-tree , starting with
[TABLE]
We sum the above inequalities over all and we get
[TABLE]
The left hand side tends to zero with . Hence is dense in . ∎
To show that is an isomorphism we need to prove the estimate from below on the norm of . The next auxiliary lemma reduces our problem to a finite tree.
Lemma 19**.**
Let and for every . Function given by the formula
[TABLE]
satisfies
[TABLE]
Proof.
Fix such that . Without loss of generality we assume that and . If is a descendant of it follows from the definition that .
We can assume that for the value does not exceed one. Indeed if is such that and then we define an auxiliary function
[TABLE]
The function has the same trace as and differs from only on . Since
[TABLE]
and is constant on it follows that
[TABLE]
We can assume that is monotone (non-decreasing) on with respect to the descendancy relation i.e. if and is a descendant of then . Indeed suppose that for some . Since for functions in the value of trace is defined as the limit along , but for the limit is one. Therefore on every branch such that there exists a vertex such that and . We denote by the set of all such vertices. Let be a tree with a root and the set of leafs is equal to . We define an auxiliary function by the formula
[TABLE]
On the tree the variation of is equal to the weighted sum of differences on leafs. However for every
[TABLE]
Therefore
[TABLE]
We have reduced our problem to the set of functions such that iff it is a non-decreasing function on with respect to the descendancy relation, for every and for . We introduce a partial order on . For
[TABLE]
If is a chain with respect to the relation then it has an upper bound in . Indeed the function defined by the formula
[TABLE]
is an upper bound. Function is a supremum of non-decreasing functions hence it is non-decreasing. If every non-decreasing sequence is convergent to one as then converges to one. Therefore has the same trace as functions in . In particular for . By the definition if then for every and the total variation . Hence for every we can choose a sequence such that
[TABLE]
and for every . Therefore the following estimate is satisfied
[TABLE]
Taking the limit with we get
[TABLE]
Since every chain in has an upper bound in by the Kuratowski-Zorn Lemma, there exists an element of maximal with respect to . Let be the maximal element. By the monotonicity of , it follows that for every . Since for every the set of direct descendants has at least three elements,
[TABLE]
The function is maximal with respect to , hence for every . Therefore there is an infinite branch such that and is constant on . However for the limit over any branch is equal to one. Hence for every . We have proven that changing the values of to one on the descendants of does not increase the total variation. It remains to consider the value at the point . By the triangle inequality and the fact that for every vertex , we have
[TABLE]
Therefore changing the value of on and its descendants to one, will not increase the total variation. Since the only assumption on was that we have desired estimate
[TABLE]
∎
Lemma 20**.**
Let and then
[TABLE]
Proof.
For any such that we define
[TABLE]
From the Lemma 19 it follows that
[TABLE]
However
[TABLE]
The right hand side of the inequality is the total variation of a function given by the formula
[TABLE]
∎
Let us observe that the set of functions , where are rational, is dense in . Indeed for every irrational arc there exists a sequence of points such that
[TABLE]
Similarly we observe that molecules , where are rational, are dense in Arens-Eells space.
We fix , where arcs are rational and pairwise disjoint. Let be any function such that . There exists such that for either there exists an arc such that or and are disjoint. We define the function by
[TABLE]
It is easy to observe that . Moreover from Lemma 19 it follows that
[TABLE]
Therefore
[TABLE]
Since we minimize the total variation over the set , the values are fixed for , . Therefore the total variation on this set is a function of finitely many variables. Moreover it is a piecewise linear function with finitely many pieces. Therefore the minimum is attained. We denote the total variation minimizer by . We define by
[TABLE]
Therefore by Abel’s summation formula
[TABLE]
A simple calculation gives us
[TABLE]
The function minimize the variation for a given trace, hence
[TABLE]
Therefore from ,
[TABLE]
Therefore is an isomorphism of Banach spaces. ∎
We have proven that the trace space is isomorphic to the Arens-Eells space.
We will characterize further.
Lemma 21**.**
* is isomorphic to *
Proof.
In order to characterize we introduce another metric on the von Koch’s curve. The von Koch’s curve is constructed inductively. The induction starts with a triangle and every segment of the triangle is replaced with a piecewise linear curve . This curve is made of from 4 segments. In the next step every old segment is replaced with a rescaled copy of . Every segment is indexed in the following way. The segment is replaced with segments .
xx,1x,2x,3x,4
is a set of all infinite indices of segments in the von Koch’s curve construction. For every point there is a corresponding index such that segments as . We define a bijection between set of indices and a one dimensional Torus with the euclidean metric
[TABLE]
Every has a unique index in . Abusing notation we denote it by . We can define a metric on by
[TABLE]
As is easily on a Figure 1, if is a "pants" shaped polygon then , where . It is so because its descendants cover two segments of -th generation. Similarly if is a "palace" shaped polygon, . In any of the above cases we have
[TABLE]
For rational points we define
[TABLE]
Obviously . Since are rational, there exists unique finite sequence of , such that . Let . From the definition of we deduce that have disjoint supports, and for every there are at most polygons in . Therefore
[TABLE]
and we have an analogous estimate for . Hence
[TABLE]
Therefore . Since the claim of the lemma follows from the theorem below,
Theorem 22**.**
Let and is isometric to an infinite compact subset of . If , are metrics on s.t for then the space is isomorphic to .
The case N=1 was proven by Z. Ciesielski [5] and for the above Theorem follows from Theorem 3.5.5 and Theorem 3.3.3 in [22]. ∎
Therefore is isomorphic to . Let . From the definition of the trace space for every there exists such that and . Hence the given by the formula
[TABLE]
is the desired right inverse operator with . Indeed
[TABLE]
This concludes the proof of Theorem 2.
Appendix
In this section we show the steps of the construction of the Whitney Covering of the von Koch’s snowflake used in our proof. We divide the von Koch’s snowflake into six identical parts. We focus our attention on one of them.
We start our construction with the third step of the iterative construction of von Koch’s snowflake. In most steps of the construction we limit the description of it to showing pictures. In our construction all the lines we use are parallel to the sides of the equilateral triangle (the starting point for von Koch’s snowflake construction). We denote the vectors pointing in those three directions respectively by and by we denote the line parrallel to a vector , which contains point . In the first step we construct vertices of the first generation of polygons - "pants" polygon.
The green points are intersections of following lines:
[TABLE]
Now we take next generation of the approximation of von Koch’s snowflake. We observe that we can cover the boundary using few blue and lime regions.
In blue regions we repeat the first step of the construction. In the lime part we do the following (Figure 6).
[TABLE]
and , , where are points from the previous step of construction. In the end we have constructed five new polygons. What is important vertices on the boundary of given region coincide with the corresponding vertices from the neighboring regions (Figure 7).
We take the next iterative step of the approximation of von Koch’s snowflake and we observe that again we can cover the neighborhood of the boundary by the lime and blue regions.
In every region we repeat the construction according to the color of the region. Let us observe that in the next generation every lime region has 3 subregions (lime,blue,lime) and every blue region has 5 subregions (lime,blue,blue,blue,lime). Since the vertices of neighboring polygons coincide we can repeat the construction inductively.
Let be -th approximation of von Koch’s snowflake. By we denote the set covered by polygons from -th generation of the construction, where is just the six pointed star in middle of von Koch’s snowflake.
The polygons on the -th step of the construction almost cover the set (except a narrow strip next to the boundary). Observe that we always perform the same construction on lime and blue regions. However the regions on n-th step are the scaled copies of regions from the second step with a scale for . Therefore there exists a constant such that
[TABLE]
Since for we have and for any the sequence is non-increasing. We get
[TABLE]
Therefore
[TABLE]
where is von Koch’s snowflake. Obviously
[TABLE]
Hence the family of polygons we constructed covers von Koch’s snowflake. Other properties of the Whitney covering follow easily from the construction.
Acknowledgments We would like to thank Anna Kamont for valuable comments and suggestions. We would like to thank anonymous reviewers for stimulating comments. During the work on the article, a by-product, a handmade fractal carpet, was created. As an artistic object, it was exhibited at the Bridges Linz 2019 conference. Its manufacturing was supported by the Copernicus Science Center in Warsaw. We thank them for their support of our artistic endeavors.
Funding This research was partially supported by the National Science Centre, Poland, and Austrian Science Foundation FWF joint CEUS programme. National Science Centre project no. 2020/02/Y/ST1/00072 and FWF project no. I5231.
The reference list from the paper itself. Each links out to its DOI / PubMed record.
- 1[1] Mario Bonk, Eero Saksman, and Tomás Soto. Triebel-Lizorkin spaces on metric spaces via hyperbolic fillings. Indiana Univ. Math. J. , 67(4):1625–1663, 2018.
- 2[2] Alexander Brudnyi and Yuri Brudnyi. Methods of geometric analysis in extension and trace problems. Volume 1 , volume 102 of Monographs in Mathematics . Birkhäuser/Springer Basel AG, Basel, 2012.
- 3[3] S. Buckley and P. Koskela. Sobolev-Poincaré implies John. Mathematical Research Letters , 2(5):577–593, 1995.
- 4[4] Lennart Carleson. On the support of harmonic measure for sets of Cantor type. Ann. Acad. Sci. Fenn. Ser. A I Math. , 10:113–123, 1985.
- 5[5] Z. Ciesielski. On the isomorphisms of the spaces H α subscript 𝐻 𝛼 H_{\alpha} and m 𝑚 m . Bulletin de l’Académie Polonaise des Sciences. Série des Sciences Mathématiques, Astronomiques et Physiques , 8:217–222, 1960.
- 6[6] Michał Derezinski. Isomorphic properties of function space bv on simply connected planar sets. Master’s thesis, Uniwersity of Warsaw, 2013.
- 7[7] Michał Derezinski, Fedor Nazarov, and Michał Wojciechowski. Isomorphic properties of function space bv on simply connected planar sets. in preparation.
- 8[8] Emilio Gagliardo. Caratterizzazioni delle tracce sulla frontiera relative ad alcune classi di funzioni in n 𝑛 n variabili. Rendiconti del Seminario Matematico della Università di Padova. The Mathematical Journal of the University of Padova , 27:284–305, 1957.
