Trace operator on von Koch's snowflake

We study properties of the boundary trace operator on the Sobolev space $W^1_1(\Omega)$. Using the density result by Koskela and Zhang, we define a surjective operator \mbox{$Tr: W^1_1(\Omega_K)\rightarrow X(\Omega_K)$}, where $\Omega_K$ is von Koch's snowflake and $X(\Omega_K)$ is a trace space with the quotient norm. Since $\Omega_K$ 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 $Tr$, i.e. a linear operator $S: X(\Omega_K) \rightarrow W^1_1(\Omega_K)$ such that $Tr \circ S= Id_{X(\Omega_K)}$. 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 $\ell_1$. As an additional consequence of our approach we obtain a simple proof of the Peetre's theorem about non-existence of the right inverse for domain $\Omega$ with regular boundary, which explains Banach space geometry cause for this phenomenon.

It was shown by Gagliardo ([8]) that the trace operator maps the space W 1 1 (Ω) onto L 1 (∂Ω) 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 S : L 1 (∂Ω) → W 1 1 (Ω) such that T r • S = Id.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 Ω 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.
Actually the proof of Theorem 1 presented here works with weaker assumptions.Whenever 1. one can define reasonable trace operator i.e. functions which are continuous up to the boundary are dense in W 1 1 (Ω) and the Trace is just the restriction operator on such functions; when ∂Ω is a Jordan curve this is provided by Koskela, Zhang theorem ( [13]), 2. trace space X contains isomorphic copy of L 1 .
In [10] Hajlasz and Martio studied the existence of a right inverse to the trace operator in the case of Sobolev spaces W p 1 (Ω) for p > 1 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 p = 1 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 Ω K .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 ℓ 1 .Indeed, it follows from Theorem 1 and Theorem 3 of Chapter VI of [11] that the trace space is complemented subspace of B 1  1,1 (R 2 ).By Proposition 7 p.200 of [17] the space B 1  1,1 (R 2 ) is isomorphic to ℓ 1 .By Pełczyński's theorem complemented subspaces of ℓ 1 are isomorphic to ℓ 1 [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 ℓ 1 .The existence of a right inverse operator is just a property of ℓ 1 .In the proof we use the structure of a specific Whitney covering of Ω K described in the Appendix, which is quite interesting in itself.To summarize: Theorem 2. Let T r : W 1 1 (Ω K ) → X(Ω K ) be a trace operator, where X(Ω K ) is a trace space defined by (2).Then X(Ω K ) ≃ ℓ 1 and there exists a continuous, linear operator S : 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 BV (Ω) needed in the proof.

Properties of BV (Ω) and trace operator
From now on we assume that Ω ⊂ R 2 , ∂Ω 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 W 1 1 (Ω).Let us recall a notion of (slightly generalized) Whitney covering of Ω.
Definition 3. We call the family of polygons A a Whitney decomposition of an open set Ω ⊂ R 2 if it satisfies: 1.For A ∈ A the boundaries ∂A are uniformly bi-lipschitz images of S 1 .
2. Q∈A Q = Ω and elements of A have pairwise disjoint interiors.
4. If ∂A ∩ ∂B ha a positive one dimensional Hausdorff measure then where l(•) denotes length of a curve, and vol 2 denotes the area of the polygon.

For a given polygon
For the purpose of this article we will also assume that polygons of A are uniformly star shaped in the following sense 6.For every A ∈ A there exists a point x ∈ A and positive numbers λ, τ s.t.B(x, λ) ⊂ A ⊂ B(x, τ ), where λ τ is fixed and the polygon A is star shaped with respect to x.We call such point a center of A.
Let A be such covering then we can define a graph describing it's geometry.We denote by BV (Ω) a space of measures of bounded variation i.e.
We introduce some special subspaces of BV (Ω).

Definition 5.
Let A be a Whitney decomposition of Ω.We define the following subspaces of BV (Ω) It is a known fact that for a given Whitney decomposition the space BV A,0 is a complemented subspace of BV (Ω).A proof of this fact can be found in ( [21], [6]).Lemma 6.For any domain Ω: Let us observe that we can easily calculate the norm of function f Let Ω be a simply connected planar domain with Poincare inequality for p = 1 and A be its Whitney decomposition.We will call a spanning tree T = (V T , E T ) of the graph G(A) a Whitney tree of A if it satisfies the following conditions: 2. for every point x on the boundary there is an infinite branch br(x) of T s.t.br(x) ∼ = Z + and dist(A n , x) → 0 as n → ∞, where A n ∈ br(x).For a sequence of real numbers {a An } we call a limit lim n→∞ a An a limit along the branch br(x).
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 G(A).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 BV
Using the above notation we define trace of f ∈ W 1 1 (Ω).Since Ω 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 Lip(R 2 ) are dense in W 1 1 (Ω).For f ∈ C(Ω) ∩ W 1 1 (Ω) we define the trace operator as a restriction of f to the boundary.We define a trace space X(Ω) as completion of a space T r(C(Ω) ∩ W 1 1 (Ω)) with respect to the norm ∥ • ∥ X , where Since Lipschitz functions on Ω are dense in W 1 1 (Ω) we can define the trace operator on the whole space W 1 1 (Ω).It is obvious that T r : W 1 1 (Ω) → X(Ω) is a continuous linear operator and it is surjective.We want to extend the trace operator to BV (Ω) (see also [14]).Lemma 8.There exists a continuous, linear operator Φ : Proof.We begin with a construction of an operator which smooths out the function from BV (Ω) inside of a fixed ball B ⊂ Ω.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 B ⊂ Ω ⊂ R 2 there exists a bounded linear operator with a constant independent of the radius of a ball.Moreover for f ∈ BV (R d ) ∩ L ∞ (B) we have Proof of Lemma 9. We will use operator T 1 from Proposition 4.2 in [20].It is a continuous operator from This operator is given by a convolution like formula: For every x ∈ ∂B 0 := B(0, 1) we have We can choose a finite cover of B 0 by {V x j } N j=1 and B(0, r 0 ) with r 0 < 1.There exists a decomposition of unity corresponding to this covering.We denote it by ϕ 0 , ϕ x 1 , . . ., ϕ x N .We define our operator by the formula where ψ is a mollifier with supp ψ ⊂ B(0, ϵ), r + 2ϵ < 1.
From properties of the operator T 1 we have Moreover from the properties of the operator T 1 and convolution for f Let J s be a homothety with a center at zero and scale s we define First we estimate the norm of the gradient Note that T 1 is a bounded linear operator on L 1 (R d + ).Thus H is bounded linear operator on L 1 (B(0, 1)).We estimate the norm of the function Note that from the properties of H B follows that T B 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 {B x,A } where A ∈ A and x ∈ ∂A.We put B x,a = B(x, r A ), where r A = (diam A) 4 .This collection is a Besicovitch covering of A∈A ∂A.There are at most c 2 families F j of disjoint balls, which cover A∈A ∂A.We define operators It is clear, that we smooth out the function on neighbourhoods of every point x ∈ A∈A ∂A.Hence for f ∈ BV G we get a function Φf ∈ W 1 1 (Ω).Moreover where C(A) depends on constants from Definition 3.3.Balls from F c 2 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 The estimates above remain comparable on the neighbouring cubes.Repeating this argument c 2 times we get The function is modified only on the set P = c 2 j=1 B∈ F j B. The radii of balls intersecting A ∈ A are comparable to (diam A) 4 .Thus for C depending on the bi-lipschitz constant (Def 3.1) we have Let P : BV (Ω) → BV G be a projection from BV (Ω) onto BV G .We define T r : BV (Ω) → X(Ω) by the formula Therefore its trace is a restriction of Φ (P f ) to the boundary.However the value of the restriction at point x ∈ ∂Ω for the function from C(Ω) is equal to the limit of − A Φ(P f (y))dy along the branch br(x).From (3) and the definition of the space and T rf = T rf the operator T r is an extension of the trace operator to BV (Ω).We will abuse the notation and from now on we will denote T r by T r.From the definition of the trace operator it follows that 2 Proof of Peetre's theorem In this section we will give a proof of Theorem 1.

Proof.
Since Ω has Lipschitz boundary by theorem of Gagliardo X(Ω) ∼ = L 1 (∂Ω) -the space of functions integrable with respect to the 1-dimensional Hausdorff measure.Let us denote by P : BV (Ω) → BV G the projection onto BV G .Assume there exists S : From ( 6) and the theorem of Gagliardo we conclude that T r| BV G is onto L 1 (∂Ω).On the other hand, T r Since the measure on the boundary is non atomic, L 1 (∂Ω) ∼ = L 1 (T).However, it is well known that L 1 could not be embedded in ℓ 1 .(To see this, note that by Khintchine inequality, Radamacher functions span ℓ 2 in L 1 space.The space ℓ 2 could not be embedded in ℓ 1 because every subspace of ℓ 1 contains a copy ℓ 1 ([15], Proposition 1.a.11).

Trace operator on von Koch's snowflake
Let Ω K be a domain bounded by von Koch's curve.Since Ω K 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 S : , where Φ is an operator from Lemma 8.It is a well known fact that Ω K 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).
where |µ| Ω K is a total variation of a measure µ on Ω K .This inequality implies

where
ḂV T = BV T /P 0 , where P 0 is the space of constant functions on Ω K .In this case the norm is equal to Similarly X(Ω K ) = R ⊕ Ẋ(Ω K ) for the quotient space Ẋ(Ω K ) = X/P 0 .We have already established that for all g ∈ Ẋ(Ω K ) we have We reduce the problem to finding a right inverse operator to the trace T r : ḂV T → Ẋ(Ω K ).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 We take a covering A K as shown on the Figure 1.This covering of von Koch's snowflake is easy to describe if we look at its Whitney tree T K .The root of T K is a six pointed star with six "pants" shaped descendants.We denote it by R. 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 D n+1 are similar to polygons from D n with a scale 1  3 .The tree T K is the tree from Definition 7. Indeed let G be a Whitney graph of this decomposition.We denote by However for {A, B} ∈ H n we have from the triangle inequality We get by induction The geometric sequence with quotient 1  3 is convergent.Therefore for such Whitney covering the norm of Further we will use the above formula as a norm on ḂV T K .We want to study the norm on Ẋ(Ω K ).
To be precise, we want to define and calculate the norm of ∥ j a j 1 [x j ,y j ] ∥ Ẋ(Ω K ) .
Definition 11.Let us denote by D ∞ (A) a cylinder of A, i.e.D ∞ (A) = {x ∈ ∂Ω K : A ∈ br(x)}.We call an arc [x, y] rational if there exists a finite sequence ) and we say that points x,y are rational points.

For a given arc [x, y] there exists a sequence of vertices
and sets D ∞ (A k ) are pairwise disjoint.Moreover this sequence can be taken maximal in the sense that if a vertex A is in the sequence then there exists z ∈ D ∞ (A ↓), which is not in [x, y].Such a sequence is unique for [x, y].Let n(k) be a natural number such that We introduce an auxiliary metric on the boundary ∂Ω K It is easy to check that d K (x, y) 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 [x, y] 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 ḂV T K .We call an arc [x,y] a short arc iff 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 x, y ∈ ∂Ω K and [x, y] be a short arc.Let function F : ∂Ω K → R be a monotone and continuous function on the arc [x, y] and supp(F ) ⊂ [x, y].There exists h ∈ ḂV T K such that Proof.First we prove the existence of s h the good extension for characteristics functions on arcs [s, y] ⊂ [x, y].Since arc [x, y] is a short arc then an arc [s, y] is a short arc and can be written as a countable sum M k=1 D ∞ (A k ) in a unique way mentioned in the definition of d K .From this assumption it is clear that #{A k : A k ∈ D n } ⩽ 10.Let us put Clearly along every infinite branch br(z) the limit of lim A∈br(z) A→z s h A exists and it is equal to 1 [s,y] (z).
We need to estimate the total variation of s h.From the definition of A k it follows that Let us assume that F is an increasing function.For F let µ be its Lebesgue-Stieltjes measure µ i.e. µ((a, b]) = F (b) − F (a). From the definition of the measure µ and the assumptions on F we get Proof.In order to construct such function we proceed inductively. .This procedure allows us to define f on a dense subset.We extend f to the whole arc.The function f has desired properties.

It follows from the definition of h that
In the lemma below we prove the existence of a class of functions in ḂV , 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 x, y ∈ ∂Ω K .There are sequences of functions ḂV (Ω K ), and h n ∈ ḂV (Ω K ) such that: Proof.We use Lemma 13.For every ε and every rational arc [x, y], the characteristic function of [x, y] can be written as sum of a Lipschitz function g and a two monotone Lipschitz functions p 1 , p 2 , with supports in arcs [t 1 , x], [y, t 2 ] respectively.Moreover t 1 , t 2 are rational, |t 1 − x| + |t 2 − y| ⩽ ε and the monotone functions p i are bounded uniformly by one.Hence from the Lemma 12 for every function p i there exists a function f i such that ∥f i ∥ ḂV (Ω K ) ⩽ Cε and for every z ∈ ∂Ω K lim A∈br(z) A→z Any Lipschitz extension of g to Ω K is in W 1 1 (Ω K ).Hence g is in the trace space.From the definition of the trace space there exists a g ε + h ε has desired properties.The limits along br(z) of − A g ε (y)dy exist and are equal to where the term Cε is the estimate on the norms of the functions f i .For every n we choose suitable ε and we get the desired properties.The sequence T rg n is Cauchy sequence.Indeed for a given function g n and m > n there exists a continuous piecewise monotone function q with support on a small set on the boundary such that From Lemma 12 there exists a function q ∈ ḂV (Ω K ) with a small norm such that T r(g n + q) = T rg m .
The size of the support of q depends only on g n .Therefore for sufficiently large n, m.
The Cauchy sequence {T r g n } defines an element u ∈ Ẋ(Ω K ).From the analogous argument as in the above Lemma if f ∈ ḂV (Ω K ) satisfies 1 Since the projection from ḂV onto ḂV T K preserves the trace, we may assume that functions f n are from ḂV T K .Therefore the function g = j a j 1[x j , y k ], whose arcs [x j , y j ] are rational, satisfies where L ⊂ ḂV T K consists of such f that the limit lim

A∈br(x) A→x
f A exists for every x ∈ ∂Ω K and it is equal to T rf (x).
Remark 15.In the above lemmas we abuse the notation a bit.For rational points x there are two branches br(x).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 A. 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 Ẋ(Ω K ).We introduce, a metric on von Koch's curve by the formula d(x, y) where 1 [x,y] is a characteristic function of an arc on the von Koch's curve which connects x and y.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.
is a norm, d is a metric on the boundary.For a given metric space (Y, d Y ) we define the Arens-Eells space ( [22]).
Definition 16.Let (Y, d Y ) be a metric space.We call a function f : Y → R a molecule if it has finite support and y∈Y f (y) = 0. Let x, y ∈ Y .We define special type of a molecule -an atom : , where 1 a is a characteristic of a set {a}.Let m be a molecule, i.e. m = M j=1 a j m x j y j , then the Arens-Eells norm of m is where the infimum is taken over all possible representations of m as a sum of m pq .The Aerens-Eells space is the completion of molecules with respect to the norm ∥ • ∥ AE .
We want to show that Ẋ(Ω K ) is isomorphic to the Arens-Eells space with the metric d.We will denote by M ( d) the linear space of molecules.Clearly it is a non-complete norm space.By the definition it is dense in AE( d).We define the candidate for the isomorphism on the a linearly dense subsets of both spaces.We set Ψ : AE( d) → Ẋ(Ω K ) by the formula Lemma 17. Ψ : AE( d) → Ẋ(Ω K ) is an isomorphism between Banach spaces.
Proof.By the triangle inequality and the definitions of d(x, y) and Arens-Eells space, it follows that Proving the estimate from below is more involved.In the trace space we have following density result.
Proof.From [13] we know that the restrictions of Lipschitz functions on R 2 are dense in W 1 1 (Ω K ).Therefore Lipschitz functions are dense in Ẋ(Ω K ).Hence for any f ∈ Ẋ(Ω K ) there exists a sequence of Lipschitz functions f n such that So it is enough to approximate Lipschitz functions with piecewise constant functions.Let f be a Lipschitz function.We define a piecewise constant function , where x j are rational points of order k i.e. ∃A ∈ D k such that [x j , x j+1 ] = D ∞ (A).We define a function h by the formula

The function h satisfies
T r h = f − g k .
We will estimate the ḂV T K norm of the function h.The function f is also Lipschitz with respect to the metric d K .Let K be the Lipschitz constant of f with respect to d K .Observe that due to the Lipschitz continuity of the function f there are positive numbers {b i } 5 i=1 , {c i } 3 i=1 such that for every pants shaped polygon A ∈ D n and n ⩾ k we have Similarly for palace shaped polygon B Let ρ := max{b 1 , . . ., b 5 , c 1 , c 2 , c 3 }.We can prove inductively that #D j (A) ≲ 4 j .Let A ∈ D k we have the following estimate on the variation on the sub-tree D↑ (A), starting with A ↓ We sum the above inequalities over all A ∈ D k and we get The left hand side tends to zero with k → ∞.Hence Ψ(M ( d)) is dense in Ẋ(Ω K ).
To show that Ψ is an isomorphism we need to prove the estimate from below on the norm of Ψ(m).The next auxiliary lemma reduces our problem to a finite tree.We can assume that for B ∈ D ↑ (A 0 ) the value f B does not exceed one.Indeed if B is such that f B↓ ⩽ 1 and f B > 1 then we define an auxiliary function h The function h has the same trace as f and differs from f only on D ↑ (B).Since and h is constant on D ↑ (B) it follows that We can assume that f is monotone (non-decreasing) on D↑(A 0 ) with respect to the descendancy relation i.e. if B ∈ D ↑ (A 0 ) and C is a descendant of B then f B ⩽ f C .Indeed suppose that f C < f B < 1 for some C ∈ D 1 (B).Since for functions in L the value of trace T rf (x) is defined as the limit along br(x), but for x ∈ D ∞ (A) the limit is one.Therefore on every branch br(x) such that x ∈ D ∞ (C) there exists a vertex Q such that f Q ⩾ f B and f Q↓ < f B .We denote by ω(C) the set of all such vertices.Let T (C) be a tree with a root C and the set of leafs is equal to {Q ↓: We define an auxiliary function p by the formula On the tree T (C) the variation of p is equal to the weighted sum of differences on leafs.However for every We have reduced our problem to the set of functions Y (f ) ⊂ L such that h ∈ Y (f ) iff it is a non-decreasing function on D↑(A 0 ) with respect to the descendancy relation, h B = f B for every B ∈ V T K \D↑(A 0 ) and T r h(x) = 1 for x ∈ D ∞ (A 0 ).We introduce a partial order on Y (f ).For h, z ∈ Y (f ) Hence for every n we can choose a sequence Taking the limit with n → ∞ we get Since every chain in Y (f ) has an upper bound in Y (f ) by the Kuratowski-Zorn Lemma, there exists an element of Y (f ) maximal with respect to ⪯.Let w ∈ Y (f ) be the maximal element.By the monotonicity of w, it follows that w Q↓ ⩽ w Q for every Q ∈ D↑(A 0 ).Since for every Q ∈ V T K the set of direct descendants D 1 (Q) has at least three elements, The function w is maximal with respect to ⪯, hence w Therefore there is an infinite branch br(x) such that x ∈ D ∞ (A 0 ) and w is constant on br(x) ∩ D↑ (A 0 ).However for x ∈ D ∞ (A 0 ) the limit over any branch br(x) is equal to one.Hence h B = 1 for every B ∈ D↑(A 0 ).We have proven that changing the values of f to one on the descendants of A 0 does not increase the total variation.It remains to consider the value at the point A 0 .By the triangle inequality and the fact that for every vertex Therefore changing the value of f on A 0 and its descendants to one, will not increase the total variation.Since the only assumption on A 0 was that D ∞ (A 0 ) ⊂ [x, y] we have desired estimate Proof.For any From the Lemma 19 it follows that The right hand side of the inequality is the total variation of a function p given by the formula Let us observe that the set of functions j a j 1 [x j ,y j ] , where x j , y j are rational, is dense in Ẋ(Ω K ).Indeed for every irrational arc [x, y] there exists a sequence of points t n , z n such that Similarly we observe that molecules j a j m x j y j , where x j , y j are rational, are dense in Arens-Eells space.
We fix g = j a j 1 [x j ,y j ] , where arcs [x j , y j ] are rational and pairwise disjoint.Let f ∈ L be any function such that T rf = g.There exists 0 = n 0 (g) such that for A ∈ D n 0 either there exists an arc [x j , y j ] such that D ∞ (A) ⊂ [x j , y j ] or D ∞ (A) and [x k , y k ] are disjoint.We define the function W f ∈ L by It is easy to observe that T rf = T rW f .Moreover from Lemma 19 it follows that ∥W f ∥ ḂV T K ⩽ ∥f ∥ ḂV T K .
Since we minimize the total variation over the set {T rf = g and f = W f }, the values f A are fixed for A ∈ D k , k > n 0 .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 ψ.We define by γ A ∈ ḂV T K in other cases.
Therefore by Abel's summation formula A simple calculation gives us The function ∥ψ∥ ḂV T K minimize the variation for a given trace, hence ∥T rf ∥ Ẋ(Ω K ) = ∥ψ∥ ḂV T K .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 K n be n-th approximation of von Koch's snowflake.By G n we denote the set covered by polygons from n-th generation of the construction, where G 0 is just the six pointed star in middle of von Koch's snowflake.The polygons on the n-th step of the construction almost cover the set K n+2 (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 1  3 n−2 for n ⩾ 2. Therefore there exists a constant C > 0 such that Since for k ∈ N we have K n ⊂ K n+k and for any x ∈ K n the sequence dist(x, ∂K n+j ) is nonincreasing.We get Therefore where Ω K is von Koch's snowflake.Obviously

Definition 4 .
Let A be a Whitney decomposition.We call a graphG := G(A) = (V (A), E(A)) =: (V, E) a graph of A if V := Aand {A, B} ∈ E only if boundaries of A and B have intersection of positive one dimensional Hausdorff measure.

Figure 1 :
Figure 1: Self similar Whitney decomposition of von Koch's snowflake

Figure 2 :
Figure 2: On the left "pants" shaped polygon and its descendants, on the right "palace" shaped polygon and its descendants

Since s h A ⩽ 1 Lemma 13 .
it follows from Lebesgue's dominated convergence theorem that lim dµ(s) = F (x) + y x 1 [s,y] (z)dµ(s) = F (z).For every x, y ∈ ∂Ω K and a, b ∈ R there is a monotone Lipschitz function f , with respect to the euclidean metric, on the arc [x, y] such that f (x) = a and f (y) = b.
[x,y] (z) = lim A∈br(z) A→z − A f (y)dy for every z on the boundary then T rf = u.To simplify the notation we denote u = 1 [x,y] .From the point 6. of the Lemma 14 it follows

Lemma 19 .
Let f ∈ L and T rf (z) = c for every z ∈ [x, y].Function f ∈ L given by the formula y].Without loss of generality we assume that f A 0 = 0 and c = 1.If B is a descendant of A 0 it follows from the definition that D ∞ (B) ⊂ [x, y].
is a chain with respect to the relation ⪯ then it has an upper bound in Y (f ).Indeed the function z ∈ Y (f ) defined by the formulaz A = sup u∈C u Ais an upper bound.Function z is a supremum of non-decreasing functions hence it is non-decreasing.If every non-decreasing sequence b k α is convergent to one as k → ∞ then sup α b k α converges to one.Therefore z has the same trace as functions in Y (f ).In particular T r h :

Figure 6 :
Figure 6: Construction step for the yellow region.

Figure 7 :
Figure 7: Second generation of the polygons.

Figure 8 :
Figure 8: Blue and yellow regions for the third step of construction.