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www.elsevier.com/locate/jmaa
Mityagin’s
extension
problem.
Progress
report
✩Alexander Goncharov∗, Zeliha Ural
a r t i cl e i n f o a b s t r a c t
Articlehistory:
Received19September2016 Availableonline16November2016 Submittedby J.Bonet Keywords: Whitneyfunctions Extensionproblem Hausdorffmeasures Markov’sfactors
GivenacompactsetK⊂ Rd,letE(K) denotethespaceofWhitneyjetson K.The
compactsetK issaidtohavetheextensionpropertyifthere existsa continuous linearextension operatorW :E(K) −→ C∞(Rd).In1961B.S. Mityaginposeda
problem to givea characterizationof theextension propertyingeometric terms. We show that there is no such complete description in terms of densities of Hausdorffcontents or related characteristics.Also theextension propertycannot becharacterizedintermsofgrowthofMarkov’sfactorsfortheset.
©2016ElsevierInc.Allrightsreserved.
1. Introduction
By the celebrated Whitney theorem [24], for each compact set K ⊂ Rd, by means of a continuous
linear operator one can extend jets of finite order from Ep(K) to functions defined on the whole space,
preserving the order of differentiability. In the case p = ∞, the possibility of such extension crucially depends on thegeometry of the set.Following [21], letus say thatK has theextension property (EP) if there exists a linear continuous extension operator W : E(K) −→ C∞(Rd). Clearly, there always exists
a linear extension operator (one can individually extend the elements of a vector basis in E(K)) and a continuousextension operator,byWhitney’sconstruction.Numerousexamplesshowthataset K hasEP provided “local thickness” of K.For example, any set K with anisolated point doesnot have EP ([14], Prop. 21).
B.S.Mityagin posedin1961([14], p. 124)thefollowing problem(inourterms): What isageometriccharacterizationof theextensionproperty?
WeshowthatthereisnocompletecharacterizationofthatkindintermsofdensitiesofHausdorffcontents ofsetsor analogousfunctionsrelatedtoHausdorffmeasures.
Thisissimilar tothestateinPotential TheorywhereR.Nevanlinna [15]andH. Ursell [22]provedthat thereisnocompletecharacterizationofpolarityofcompactsets intermsofHausdorffmeasures.Thescale ofgrowthrateoffunctionsh,whichdefinetheHausdorffmeasureΛh,canbedecomposedintothreezones.
✩ TheauthorsarepartiallysupportedbyagrantfromTÜBITAK:115F199. * Correspondingauthor.
E-mailaddresses:goncha@fen.bilkent.edu.tr(A. Goncharov),zeliha.ural@bilkent.edu.tr(Z. Ural).
http://dx.doi.org/10.1016/j.jmaa.2016.11.001
Forh fromthefirstzoneofsmallgrowth,if0< Λh(K) thentheset K isnotpolar.Forh fromthezoneof
fastgrowth,ifΛh(K)<∞ thentheset K ispolar. However,there isazone ofuncertaintybetweenthem.
Itispossibletotaketwofunctionswithh2≺ h1fromthiszoneandthecorrespondingCantor-typesetsKj
with 0< Λhj(Kj)<∞ for j ∈ {1,2},suchthatthelarge(with respectto theHausdorffmeasure) set K2 is polar,whereasthesmallerK1isnotpolar.
Here wepresent asimilar example of two Cantor-type sets: thesmaller set hasEP whereas thelarger set doesnot.
Of course, such global characteristics as Hausdorff measures or Hausdorff contents cannot be used, in general,todistinguishEP ,whichisdefinedbyalocal structureoftheset.Onecansuggestforthisreason tocharacterizeEP intermsoflowerdensitiesofHausdorffcontentsofsets,because(seeSection9)densities of Hausdorff measures cannot be used for this aim. We analyze a wide class of dimension functions and show thatlowerdensitiesof HausdorffcontentsdonotdistinguishEP .
Neither EP canbe characterized intermsof growth rate ofMarkov’s factors(Mn(·))∞n=1 forsets. Two
setsarepresented,K1withEP andK2withoutit,suchthatMn(K1) growsessentiallyfasterthanMn(K2)
asn→ ∞.Itshouldbenotedthat,byW.Pleśniak [17],anyMarkovcompactset(withapolynomialgrowth rateofMn(·))hasEP .AllexamplesaregivenintermsofthesetsK(γ) introducedin [10].Thepapersums
uptheresearchrelatedto theproblem bythefirstauthor inthelast twodecades.
Theorganizationofthepaperisthefollowing.Section2isashortreviewofmainmethodsofextension; initwealsoconsider theTidten–VogtlineartopologicalcharacterizationofEP .InSection3wegivesome auxiliary results about the weakly equilibrium Cantor-type set K(γ). In Section4 we use local Newton interpolations to construct an extension operatorW . Section5 contains the main result, namely a char-acterization of EP forE(K(γ)) in termsof asequence related to γ.Insection 6we compareW with the extension operatorfrom [12],whichis givenbyindividual extensionsofelements ofSchauderbasisforthe space E(K(γ)). InSection 7we consider two examplesthat correspond respectively to regular and irreg-ular behaviourof the sequence γ. In Section8 we calculate Λh(K(γ)) for the dimension function h that
corresponds tothesetandshowthatΛh|K(γ) coincideswiththeequilibriummeasure ofK(γ). Alsointhis
section we presentUrsell’s typeexamplefor EP .InSection9we consider Hausdorffcontentsand related characteristics. InSection10wecomparethegrowth ofMarkov’sfactorsandEP forK(γ).
For the basic facts about the spaces of Whitney functions defined on closed subsets of Rd see e.g. [3], the concepts of the theory of logarithmic potential can be found in[18]. Throughout the paper, log denotes the natural logarithm. Given compact set K, Cap(K) stands for the logarithmic capacity of K, Rob(K)= log(1/Cap(K))≤ ∞ is theRobin constantfor K. IfK is not polarthen μK is itsequilibrium
measure. For each A⊂ R, let #(A) be thecardinality of A, |A| bethe diameterof A. Given afinite set A= (am) andx∈ R,by(dk(x,A)) wedenotedistancesfromx tothepointsofA arrangedinnondecreasing
order, sodk(x,A)=|x− amk|.Also,[a] isthegreatestintegerina,nk=m(· · · )= 0 and
n
k=m(· · · )= 1
ifm> n.Thesymbol∼ denotesthestrongequivalence:an∼ bn meansthatan = bn(1+ o(1)) forn→ ∞.
2. Threemethodsofextension
Let K ⊂ Rd be a compactset, α = (αj)j=1d ∈ Nd0 be amulti-index. Let I bea closed cube containing
K and F(K,I) = {F ∈ C∞(I) : F(α)|
K = 0, ∀α} be the ideal of flat on K functions. The Whitney
space E(K) ofextendable jetsconsistsof tracesonK of C∞-functionsdefinedonI,so itisafactorspace of C∞(I) and the restriction operator R : C∞(I) −→ E(K) is surjective. This means that the sequence 0−→ F(K,I)−→ CJ ∞(I)−→ E(K)R −→ 0 isexact. If itsplits, then the right inverseto R is the desired linearcontinuousextension operatorW and K hasEP .
In [21] M. Tidten applied D.Vogt’s theory of splitting of short exactsequences of Fréchet spaces (see e.g. [13], Chapter 30) and presented the following important linear topological characterization of EP :
a compact setK has theextensionproperty ifandonly ifthespaceE(K) has adominatingnorm(satisfies thecondition (DN)).
RecallthataFréchetspaceX withanincreasingsystemofseminorms(||·||k)∞k=0hasadominatingnorm
||· ||pifforeachq∈ N there existr∈ N andC≥ 1 suchthat|| · ||2q ≤ C ||· ||p||· ||r.
Concerningthequestion“HowtoconstructanoperatorW ifitexists?”,wecanselectthreemainmethods thatcanbeappliedforwidefamiliesofcompactsets.
The first method goes back to B.S. Mityagin [14]: to extend individually the elements (en)∞n=1 of a
topologicalbasisofE(K).Thenforf =∞n=1ξn·en takeW (f )=
∞
n=1ξn·W (en).SeeTheorem2.4in [23]
aboutpossibilityofsuitablesimultaneousextensionsofen inthecasewhenK hasanonemptyinterior.The
main problem withthis methodisthatwe donotknowwhethereach spaceE(K) hasatopologicalbasis, eventhoughE(K) iscomplementedinC∞(I).ThisisaparticularcaseofthesignificantMityagin–Pełczyński problem:SupposeX isanuclearFréchetspacewithbasisandE isacomplementedsubspaceofX.DoesE possessabasis? ThespaceX = s ofrapidly decreasingsequences,whichisisomorphic toC∞(I),presents themostimportantunsolvedcase.
Thesecondmethodwassuggestedin [16],whereW.PawłuckiandW.Pleśniakconstructedanextension operatorW inthe form ofatelescopingseries containingLagrange interpolationpolynomials with Fekete nodes.Theauthorsconsideredthefamilyofcompactsetswithpolynomialcusps,butlater,in [17],theresult wasgeneralizedtoanyMarkovset.Infact(seeT.3.3in [17]),foreachC∞determiningcompactsetK,the operatorW iscontinuousintheso-calledJackson topologyτJ ifandonlyifτJ coincideswith thenatural
topologyτ of thespace E(K) and this happensif and onlyif theset K is Markov.We remark thatτJ is
notstrongerthan τ andthatτJ alwayshasthedominatingnormproperty,seee.g. [2].Thus,inthecaseof
non-MarkovcompactsetwithEP ([5,2]),thePawłucki–PleśniakextensionoperatorisnotcontinuousinτJ,
yetthis doesnotexcludethepossibilityfor itto bebounded inτ . Atleastfor somenon-Markovcompact sets, thelocal versionofthisoperatorisboundedinτ ([2]).
In [4] L. Frerick, E. Jordá, and J. Wengenroth showed that, provided some conditions, the classical WhitneyextensionoperatorforthespaceofjetsoffiniteordercanbegeneralizedtothecaseE(K).Instead ofTaylor’spolynomials intheWhitneyconstruction,theauthors usedakindofinterpolation bymeansof certain local measures.A linear tame extension operatorwas presented for E(K), provided K satisfies a localform ofMarkov’sinequality.
TherearesomeothermethodstoconstructW forclosedsets, forexampleSeeley’sextension [19]froma halfspaceorStein’sextension [20]fromsetswiththeLipschitzboundary.Howeverthesemethods,inorder to define W (f,x) at somepoint x, essentiallyrequire existenceof aline through x with aray where f is defined,sothese methodscannotbe appliedforcompactsets.
Hereweconsider rathersmallCantor-typesetsthatareneitherMarkovnorlocalMarkov.Wefollow [2]
inourconstruction,soW isalocalversionofthePawłucki–Pleśniakoperator.Itisinterestingthat,atleast for small sets, W canbe consideredalso as an operator extendingbasis elements of the space. Thus, for suchsets,thefirstmethodandalocal versionofthesecondmethodcoincide.
3. Notationsandauxiliaryresults
Inwhatfollows wewill consider onlyperfect compactsets K⊂ I = [0,1], sothe Fréchettopologyτ in thespaceE(K) canbe givenbythenorms
f q=|f|q,K+ sup |(Rq yf )(k)(x)| |x − y|q−k : x, y∈ K, x = y, k = 0, 1, . . . , q
for q ∈ N0, where |f|q,K = sup{|f(k)(x)| : x ∈ K,k ≤ q} and Rqyf (x) = f (x)− Tyqf (x) is the Taylor
Givenf ∈ E(K),let|||f|||q= inf |F|q,I,wheretheinfimumistakenoverallpossibleextensionsoff to
F ∈ C∞(I).BytheLagrangeformoftheTaylorremainder,wehave||f||q≤ 3|F|q,I foranyextensionF .
ThequotienttopologyτQ,givenbythenorms(|||· |||∞q=0),iscompleteand, bytheopenmappingtheorem,
is equivalenttoτ . Hence,forany q thereexist r∈ N,C > 0 suchthat
||| f |||q≤ C || f ||r (1)
for any f ∈ E(K).In general,extensions F thatrealize |||f|||q for agiven functionf , essentiallydepend
on q. Of course, the extension property of K means the existence of a simultaneous extension which is suitableforallnorms.
Our main subject is the set K(γ) introduced in [10]. For the convenience of the reader we repeat the relevant material. Given sequence γ = (γs)∞s=1 with 0< γs < 1/4, letr0 = 1 andrs = γsr2s−1 for s∈ N.
Define P2(x) = x(x− 1),P2s+1 = P2s(P2s + rs) and Es = {x ∈ R : P2s+1(x) ≤ 0} for s ∈ N. Then Es = ∪2
s
j=1Ij,s, where the s-th level basic intervals Ij,s are disjoint and max1≤j≤2s|Ij,s| → 0 as s → ∞. Here, (P2s + rs/2)(Es) = [−rs/2,rs/2], so the sets Es are polynomial inverse images of intervals. Since Es+1⊂ Es,wehaveaCantor-typeset K(γ):=∩∞s=0Es.
Inwhatfollowswewillconsider onlyγ satisfyingtheassumptions
γk ≤ 1/32 for k ∈ N and ∞
k=1
γk <∞. (2)
Thelengthslj,softheintervalsIj,s ofthes-thlevelarenotthesame,but,provided (2),wecanestimate
them intermsoftheparameterδs= γ1γ2· · · γs ([10],L.6):
δs< lj,s< C0δs for 1≤ j ≤ 2s, (3)
where C0= exp(16∞k=1γk).Each Ij,s containstwo adjacent basicsubintervalsI2j−1,s+1 andI2j,s+1.Let
hj,s= lj,s− l2j−1,s+1− l2j,s+1 be thedistance betweenthem.ByLemma4in [10],
hj,s> (1− 4γs+1)lj,s≥ 7/8 · lj,s> 7/8· δs for all j≤ 2s. (4)
Inaddition,byT.1in [10],theleveldomainsDs={z ∈ C: |P2s(z)+ rs/2|< rs/2} formanestedfamily and K(γ) = ∩∞s=0Ds. The value Rs = 2−slog 2+sk=12−klogγ1k represents the Robin constant of Ds.
Therefore,thesetK(γ) isnon-polarifandonlyifRob(K(γ))=∞n=12−nlogγ1 n =
∞
n=12−n−1log
1
δn <∞. We decompose all zeros of P2s into s groups. Let X0 = {x1,x2} = {0,1},X1 = {x3,x4} = {l1,1,1− l2,1},· · · ,Xk ={l1,k,l1,k−1− l2,k,· · · ,1− l2k,k} fork≤ s− 1.Thus,Xk ={x: P2k(x)+ rk = 0} contains all zerosof P2k+1 thatare notzerosof P2k. Set Ys =∪sk=0Xk. ThenP2s(x) =
xk∈Ys−1(x− xk). Clearly, #(Xs)= 2sfors∈ N and#(Ys)= 2s+1 fors∈ N0.Werefers-thtype pointsto theelementsofXs.
ThepointsfromYscanbeorderedusing, asin [8],theruleofincrease oftype.Firstwetakepointsfrom
X0 and X1 inthe ordering given above.To put inorder the set X2, for 1≤ j ≤ 4, we take xj+4 as the
pointofthesecondtypewhichistheclosesttoxj.Thus,x5= x1+ l1,2,x6= x2− l4,2,· · · andtheordered
set X2 is{x5,x7,x8,x6}.Inother words, theordered set X2 canbe obtainedfrom X0∪ X1 ifwe arrange
this set inincreasing way andenlarge everyindex ofx by4. Similarly,eachXk ={x2k+1,· · · ,x2k+1} can be ordered.See [12]formoredetails.
In thesameway, any N points canbe chosen oneachbasic interval.Suppose 2n ≤ N < 2n+1 and the
points Z = (xk,j,s)Nk=1 arechosenonIj,s bythisrule.Then Z includesall2n zerosofP2s+n onIj,s (points ofthetype≤ s+ n− 1)andsomeN− 2n pointsofthetypes+ n.Inwhatfollows,wewriteZ = (z
k,j,s)Nk=1
or Z = (zk)Nk=1,whennoconfusioncanarise,forthesameset intheorder ofincreasing.
Let2n≤ N < 2n+1andabasicintervalI
j,s begiven.SupposeZN = (xk,j,s)Nk=1andZN +1= (xk,j,s)N +1k=1
arechosenonIj,sbytheruleofincreaseoftype.WriteC1= 8/7· (C0+ 1).
LemmaA.(Lemma2.2 from[12]) Foreach x∈ R with dist(x,K(γ)∩ Ij,s))≤ δs+n andz∈ ZN +1 wehave
δs+n Nk=2dk(x,ZN)≤ C1N
N +1
k=2 dk(z,ZN +1).
Let (zk)N +1k=1 be the sameset ZN +1 butarranged inascending order. Forq = 2m− 1 with m< n and
1≤ j ≤ N + 1− q,letJ ={zj,· · · ,zj+q} be2m consecutive pointsfrom ZN +1. Given j, we consider all
possible chainsof strictembeddings ofsegments ofnaturalnumbers:[j,j + q]= [a0,b0]⊂ [a1,b1]⊂ · · · ⊂
[aN−q,bN−q]= [1,N + 1], whereak = ak−1, bk = bk−1+ 1 orak = ak−1− 1,bk = bk−1 for1≤ k ≤ N − q.
Every chain generates theproduct N−qk=1(zbk − zak). For fixed J , letΠ(J ) denote the minimum of these productsforallpossiblechains.
Lemma B.(Lemma2.3 from[12]) Foreach J ⊂ ZN +1 there existsz˜∈ J suchthat
N +1
k=q+2dk(˜z,ZN +1)≤
Π(J ).
Wewillcharacterize EP ofK(γ) intermsofthevaluesBk= 2−k−1· logδ1k thathavePotentialTheory
meaning:Rob(K(γ))=∞k=1Bk.Themainconditionis(comparewith(3)in [9]):
Bn+s
n+s k=sBk
→ 0 as n → ∞ uniformly with respect to s. (5) Weseethatthisconditionallowspolarsets.
Example1. Letγ1= exp(−4B) andγk = exp(−2kB) fork≥ 2, where B ≥ 14log 32,so (2)is valid. Here,
Bk = B forallk.Hence (5)issatisfiedandthesetK(γ) ispolar.
Thecondition (5)meansthat
∀ε ∃s0,∃n0: Bs+n< ε(Bs+· · · + Bs+n) for n≥ n0, s≥ s0. (6)
Clearly,insteadof∃s0 onecantakeabove∀s0.Letusshow that (6)isequivalent to
∀ε1 ∀m ∈ N0∃N : Bs+n−m+· · · + Bs+n< ε1(Bs+· · · + Bs+n−m−1), n≥ N, s ≥ 1. (7)
Indeed,thevaluem= 0 in (7)gives (6)atonce. Fortheconverse,remark thatin (7)wecantakeε1(Bs+
· · ·+Bs+n) ontherightside,sohereweconsider (7)inthisnewform.Suppose (6)isvalid.Givenε1and m,
take ε = ε1/(m+ 1) and the corresponding value n0 from (6). Take N = n0+ m. Then for n ≥ N and
0≤ k ≤ m wehaven− k ≥ n0,so Bs+n−k < ε(Bs+· · · + Bs+n−k)< ε(Bs+· · · + Bs+n).Summingthese
inequalities,weobtainanewform of (7).
Itfollowsthatthenegationofthemainconditioncanbe writtenas
∃ε ∃m : ∀N ∃n > N : s+n s+n−m Bk > ε s+n−m−1 s Bk for s = sj↑ ∞. (8) Also, (6)isequivalentto ∀ε ∃m, n0, s0: Bs+n< ε(Bs+n−m+· · · + Bs+n−1) for n≥ n0, s≥ s0. (9)
Indeed,comparisonofrightsidesofinequalitiesshowsthat (9)implies (6).Conversely,givenε,taken0such
that (6)isvalidwithε/(1+ε) insteadofε.Takem= n0.Thenforn≥ n0,s≥ s0wehaves = s˜ +n−m≥ s0
and, by (6),Bs+n = B˜s+m< 1+εε (B˜s+· · · B˜s+m),whichis (9).
Wewill usea“geometric”versionof (9)intermsof(δk)
∀M ∃m, n0, s0: δs+n−1δs+n−22 · · · δ2
m−1
s+n−m< δMs+n for n≥ n0, s≥ s0. (10)
4. ExtensionoperatorforE(K(γ))
Here,asin [2],weusethemethodoflocal Newtoninterpolations. LetK be shorthandforK(γ).Wefix anondecreasingsequenceofnaturalnumbers(ns)∞s=0withns≥ 2 andns→ ∞.Givenfunctionf onK,we
interpolate f at2n0 points thatarechosenbythe ruleofincreaseof typeonthewholeset.Halfofpoints
are located onK∩ I1,1. Wecontinue interpolationon thisset upto the degree2n1. Separatelywe dothe
same onK∩ I2,1. Continuingin this fashion,we interpolatef withhigher and higherdegrees onsmaller
and smaller basicintervals. At eachstep the additionalpoints are chosen bythe rule of increaseof type. InterpolationonIj,s doesnotaffectotherintervals ofthesameleveldue tothefollowingfunction.
Lett> 0 andacompactsetE onthelinebegiven.Thenu(·,t,E) isaC∞-functionwiththeproperties: u(·,t,E)≡ 1 onE,u(x,t,E)= 0 fordist(x,E)> t andsupx∈R|u(p)xp(x,t,K)|≤ cp t−p,wheretheconstant cp dependsonlyonp.Letcp.
For any interval I and points (zk)N +1k=1 ⊂ I, let Ω(x) =
N +1
k=1(x− zk),ωk(x) = (x−zΩ(x)k)Ω(zk) and LN(f,x,I)=N +1k=1 f (zk)ωk(x) betheinterpolatingpolynomialwithnodesat thesepoints.
We define Ns= 2ns − 1 andMs= 2ns−1−1− 1 fors≥ 1, M0 = 1. Now, forfixed s,we takeMs+ 1≤
N ≤ Ns,so 2n≤ N < 2n+1with n∈ {ns−1− 1,· · · ,ns− 1}. ForsuchN ands lettN := δs+n.Fix j with
1≤ j ≤ 2s.Next, wechoosetheset Z
N +1 = (xk,j,s)N +1k=1 = (zk)k=1N +1 onIj,s bytherule ofincrease oftype
and consider,forgivenf ∈ E(K(γ)) andx∈ R,thevalue
AN,j,s(f, x) := [LN(f, x, Ij,s)− LN−1(f, x, Ij,s)] u(x, tN, Ij,s∩ K).
We call Aj,s(f,x) := NN =Ms s+1AN,j,s the accumulation sum. The last term here corresponds to the
interpolation onIj,s at2ns points.InordertocontinueinterpolationonsubintervalsofIj,s,letusconsider
thetransition sum
Tk,s(f, x) := [LMs+1(f, x, Ik,s+1)− LNs(f, x, Ij, s)] u(x, δs+ns−1, Ik, s+1∩ K), where wesuppose1≤ k ≤ 2s+1,j = [k+1
2 ].Of course,Ik,s+1⊂ Ij, s.
As above,werepresentthedifferenceinbracketsinthetelescopingform:
[LMs+1− LNs] =−
2ns−1
N =2ns−1
[LN(f, x, Ij, s)− LN−1(f, x, Ij, s)].
Here,theinterpolatingsetforLN consistsofMs+1+ 1 pointsofYs+ns−1∩ Ik,s+1andN− Ms+1pointson Ii,s+1. Thesecond parameterofu issmallerthanthemeshsize ofZ,so Tk,s(f,x) = 0 only nearIk,s+1.
Consideralinearoperator
W (f,·) = LM0(f,·, I1, 0) u(·, 1, K) + ∞ s=0 2s j=1 Aj,s(f,·) + 2s+1 k=1 Tk,s(f,·) . (11)
Weremarkattheoutsetthat,forfixedx∈ R ands,becauseofthechoiceofparametersforthefunction u, at mostonevalueAj,s doesnotvanish.Thesameisvalid forTk,s.
Let us show that W extends functions from E(K), provided asuitable choice of (ns)∞s=0. Define n0 =
n1= 2 andns= [log2logδ1s] fors≥ 2.Thenns≤ ns+1and 1 2log 1 δs < 2ns ≤ log 1 δs for s≥ 2. (12)
Lemma4.1.Let(ns)∞s=0begivenasabove.Thenforanyf ∈ E(K(γ)) andx∈ K(γ) wehaveW (f,x)= f (x).
Proof. Letusfixanaturalnumberq withq > 2+ log(8C0/7),whereC0isdefinedin (3).Bythetelescoping
effect,
W (f, x) = lim
s→∞LMs(f, x, Ij,s), (13)
wherej = j(s,x) ischosen insuchaway thatx∈ Ij,s.Asin [7],
| LMs(f, x, Ij,s)− f(x)| ≤ || f ||q
2n
k=1
| x − zk|q| ωk(x)|. (14)
Here n is shorthand for ns−1− 1 and s is suchthat Ms = 2n− 1 > q. The interpolating set (zk)2
n
k=1 for
LMs consists of allpoints of the type ≤ s+ n− 1 on Ij,s. Given point x, we consider thechain of basic intervals containing it: x∈ Ijn,s+n ⊂ · · · ⊂ Ij1,s+1 ⊂ Ij,s. Wesee that Ijn,s+n contains oneinterpolating point,Ijn−1,s+n−1\ Ijn,s+ndoesonemorezi,Ijn−2,s+n−2\ Ijn−1,s+n−1 containstwosuchpoints,etc.Thus,
forfixedk, weget
| x − zk|q 2n i=1,i=k |x − zi| ≤ lj,sq−1· ljn,s+n· ljn−1,s+n−1· l2jn−2,s+n−2· · · l2 n−1 j,s .
By (3),thisdoesnotexceedC02n+q−1δs+nδs+n−1δs+n−22 · · · δ2
n−2
s+1 δ2
n−1+q−1
s .
Ontheotherhand,byasimilarargument, forthedenominatorof|ωk(x)| wehave
| zk− z1| · · · | zk− zk−1| · | zk− zk+1| · · · | zk− z2n| ≥ lqn−1,s+n−1· h2q
n−2,s+n−2· · · h
2n−1
j,s
forsomeindicesqn−1,qn−2,· · · .Thelastproductexceeds(7/8)2n−2
δs+n−1δ2 s+n−2· · · δ2 n−1 s ,by (4).Itfollows that LHS of(14)≤ || f ||q2nC0q−1(8C0/7)2 n δs+nδsq−1.
The expression on the right side approaches zero as s → ∞. Indeed, 2n < log(1/δs−1), by (12), and
2n(8C
0/7)2
n
δq−1s < 1 duetothechoiceofq.Thusthelimitin (13)existsand equalsf (x). 2 5. ExtensionpropertyofweaklyequilibriumCantor-typesets
Weneedtwomorelemmas.
Lemma5.1. Letγ satisfy (2),q = 2m, r = 2n with m< n and Z = (z
k)rk=1 with z1<· · · < zr be allpoints
ofthetype≤ s+ n− 1 on I1,s forsomes∈ N0.Letf (x)=
r k=1(x− zk) forx∈ K(γ)∩ I1,s andf = 0 on K(γ)\I1,s.Then|f|0,K(γ)≤ C0r·δn+s·δn+s−1·δn+s2 −2· · · δ2 n−1 s , |f(q)(0)|≥ q!·(7/8)r−q·δ2 m n+s−m−1· · · δ2 n−1 s and||f||r≤ 2· r!.
Proof. Fix x that˜ realizes |f|0,K(γ) and a chain of basic intervals containing this point: x˜ ∈ Ij0,n+s ⊂ Ij1,n+s−1 ⊂ · · · ⊂ Ijn,s = I1,s. Arguing as in Lemma 4.1, we see that |f|0,K(γ) ≤ lj0,n+s · lj1,n+s−1 · l2
j2,n+s−2· · · l
2n−1
1,s ,which, by (3),givesthedesiredbound.
In order to estimate |f(q)(0)|, letus remark thatf(q)(x) is a sumof r q
products, each product hasa coefficient q! and consists of r− q terms (x− zk). One of these products is g(x) :=rk=q+1(x− zk). All
productsarenonnegativeatx= 0,sincer−q iseven.Fromhere,|f(q)(0)|≥ q!·g(0).Takingintoaccountthe
locationofpointsfromZ,wegetg(0)=rk=q+1zk> h2
m 1,n+s−m−1· · · h2 n−1 1,s > (7/8)r−q· δ2 m n+s−m−1· · · δ2 n−1 s ,
by (4). Thebound of||f||r isevident. 2
Inthe nextLemma,forgiven2n ≤ N < 2n+1,weconsider Ω
N(x)=Nk=1(x− zk) withZN = (zk)Nk=1,
where thepointsarechosenonIj,s bytheruleofincreaseoftype.Letu(x)= u(x,δs+n,Ij,s∩ K(γ)).
Lemma 5.2. The bound |(ΩN· u)(p)(x)|≤ 2p(C0+ 1)cpδ−p+1s+n Np
N
k=2dk(x,ZN) is validforeach p< N
and x∈ R.
Proof. By Leibnitz’s formula, |(ΩN · u)(p)(x)| ≤pi=0
p i
|Ω(i)N(x)|cp−iδs+n−p+i. Since dk increases, we have
|Ω(i)
N(x)|≤ (NN !−i)!
N
k=i+1dk(x,ZN).Thisgives
|(ΩN· u)(p)(x)| ≤ 2pcpδ−ps+n· max 0≤i≤p(N δs+n) i N k=i+1 dk(x, ZN). (15)
ThesetZN consistsof2nendpointsofsubintervalsofthelevels+ n−1 coveredbyIj,sandN−2n points
of thetypes+ n. Here,dist(x,Ij,s∩ K)=|x− x0|≤ δs+n forsomex0. Letx0 ∈ Ii,s+n⊂ Im,s+n−1. Then
Im,s+n−1 containsfrom 2to 4pointsofZN.Inallcases,d1(x,ZN)≤ li,s+n+ δs+n≤ (C0+ 1)δs+n,by(3).
Also, δs+n/2≤ d2 ≤ (C0+ 1)δs+n−1. Here the lower bound corresponds to the case#(Ii,s+n∩ ZN)= 2,
whereas the upper bound deals with #(Im,s+n−1∩ ZN) = 2. Similarly, d3 ≥ hm,s+n−1− δs+n. From (4)
and (2)itfollowsthatd3≥ 7/8 δs+n−1− δs+n≥ 27δs+n.Thisgivesδs+ni−1di+1· · · dN ≤ (C0+ 1)d2· · · dN for
0≤ i≤ p and,by (15),thelemmafollows. 2 Wecannowformulateourmain result.
Theorem 5.3.Suppose γ satisfies(2).ThenK(γ) hastheextension propertyif andonly if (5) isvalid. Proof. Recallthattheextensionpropertyofasetisequivalenttothecondition(DN ) ofthecorresponding Whitney space. Due to L.Frerick [3,Prop. 3.8], E(K) satisfies(DN ) if andonly iffor any ε> 0 and for any q∈ N thereexist r∈ N and C > 0 such that|· |1+ε
q ≤ C|· |0,K ||· ||εr. Hence,inorder to provethat (5) isnecessary forEP ofK(γ),wecanshowthat (8)impliesthelackof(DN ) forE(K(γ)), thatis there exist ε> 0 andq suchthatforany r∈ N onecanfindasequence (fj)⊂ E(K(γ)) with
| fj|1+εq | fj|−10,K(γ) || fj||− εr → ∞ as j → ∞.
Letusfixε andm fromthecondition (8)andtakeq = 2m.Foreachfixedlarger (clearly,wecantakeitin
theform r = 2n)and sj definedby (8),weconsiderthefunctionfj givenin Lemma 5.1fors= sj.Then
C| fj|1+εq | fj|−10,K(γ) || fj||− εr ≥ (δn+s· δn+s−1· δ2n+s−2· · · δ2 m−1 n+s−m)−1(δ2 m n+s−m−1· · · δ2 n−1 s )ε,
where C doesnotdependonj.Therightsideheregoestoinfinity.Indeed,itslogarithm is2n+s{2B n+s+
Therefore, the wholevalue exceeds 2n+sB
n+s = 12logδs+n1 , which goesto infinitywhen s = sj increases.
Thus,EP ofK(γ) implies (5).
Fortheconverse,weconsidertheextensionoperatorW fromSection4,where(ns) arechosenasin (12).
We proceed to show that W isbounded provided(10), which isequivalent to (5). Let us fix any natural numberp. This p and C1 from Lemma A define M = 2p+ 2+ log(2C1).We fix m∈ N that corresponds
to M in the sense of (10). Letq = 2m− 1 and r = r(q) bedefined by (1). Wewill show that thebound
|(W (f,x))(p)|≤ C ||f||
risvalidforsomeconstantC = C(p) andallf ∈ E(K), x∈ R.
Given f andx,letusconsider termsofaccumulationsums.Forfixed s∈ N we choosej ≤ 2s suchthat x∈ Ij,s. FixN with 2n ≤ N < 2n+1 forns−1− 1≤ n≤ ns− 1, so Ms+ 1≤ N ≤ Ns. For largeenough
s the value N exceeds p and q. We takeZN and ZN +1, as in Lemma A. ByNewton’s representation of
interpolatingoperators intermsofdivideddifferences,wehave
AN,j,s(f, x) = [z1,· · · , zN +1]f · ΩN(x) u(x),
whereΩN andu aretakenas in Lemma 5.2.Weaimtoshow that
Ns|A(p)N,j,s(f, x)| ≤ s−2||f||r (16)
forlarges.Thisgivesconvergenceoftheaccumulationsums. Forthedivideddifferenceweuse(4)from [2]:
| [z1,· · · , zN +1]f| ≤ 2N− q|||f|||q (Π(J0))−1, (17)
where Π(J0)= min1≤j≤N+1−qΠ(J ) forΠ(J ) definedinLemma B.Fixz˜∈ J0 thatcorresponds to thisset
inthesense of Lemma B.
Applying Lemma 5.2and Lemma Aforz = ˜z yields
|(ΩN· u)(p)(x)| ≤ C δs+n−p NpC1N
N +1
k=2
dk(˜z, ZN +1) with C = 2p(C0+ 1) cp.
Ontheotherhand, (17)and Lemma BforJ0 give
| [z1,· · · , zN +1]f| ≤ 2N− q|||f|||q N +1
k=q+2
d−1k (˜z, ZN +1).
Combiningthese weseethat
|A(p) N,j,s(f, x)| ≤ C |||f|||qδs+n−p Np(2C1)N q+1 k=2 dk(˜z, ZN +1). (18)
RecallthatthesetZN +1includesallpointsofthetype≤ s+n−1 onIj,sandN +1−2npointsofthetype
s+ n.Wecanonlyenlargetheproductq+1k=2dk(˜z,ZN +1) ifwewillconsideronlydistancesfromz to˜ points
fromYs+n−1∩ Ij,s.Arguingasin Lemma 4.1,wegetq+1k=2dk(˜z,ZN +1)≤ C0qδs+n−1δs+n−22 · · · δ2
m−1
s+n−m.We
observethatd1(˜z,ZN +1)= 0 isnotincludedintotheproductontheleftside.By (10),q+1k=2dk(˜z,ZN +1)≤
C0qδM s+n.
Inordertoget (16),itisenoughtoshowthat
Here, by (12), NsNp < 2ns(p+1) ≤ log(1/δs)p+1 < δ−p−1s . Also, (2C1)N < δs− log(2C1). Clearly, we can
replace δs+n in (19) by δs. Then, becauseof the choice of M , the product in (19)does notexceed s2δs,
whichapproaches 0as s→ ∞,since,by (2),δs≤ 32−s.
Similarargumentsareused fortermsofthetransition sums. 2 6. Extensionof basiselements
An extension operatorforthe spacesE(K(γ)) can also be constructedby meansof suitable extensions of basis elements of the space. It is interesting that for sufficiently small sets with EP both approaches coincide.
Lete0≡ 1 andeN(x)=
N
1(x− xk) forN∈ N,where thepoints(xk)∞1 arechosenonK(γ) bytherule
of increaseoftype.Then, byTheorem3.3in [12],(eN)∞N =0 isaSchauderbasisinE(K(γ)),provided
∀Q ∃m, k0: Q≤ Bk−m+· · · + Bk for k≥ k0. (20)
Thus, inthiscase,thespacepossessesastrictpolynomialbasis.If,inaddition, ∀M, Q ∃m, k0: Q + M· Bk ≤ Bk−m+· · · + Bk for k≥ k0,
thenonecantakesimultaneous(suitableforallnorms)extensionse˜N= eN·u(·,δn,K(γ)),where2n ≤ N <
2n+1.ThebiorthogonalfunctionalsaregivenbythedivideddifferencesξN(f )= [x1,x2,· · · ,xN +1]f .Here,
theMityaginmethodgivestheoperatorW (f )=∞n=0ξn(f )· ˜eN forf =∞n=0ξn(f )· en.Inthenotations
ofSection4,W (f )= L0(f,·,I1, 0)u(·,1,K(γ))+
∞
n=1AN,1,0(f,·),whichisexactlythePawłucki–Pleśniak
operator,if(xk)N1 aretheFeketepointsontheset.Weconjecturethat,atleastforN = 2n,thisisthecase.
In general, without (20), (eN)∞N =0 does not have the basis property. Here a basis can be constructed
by meansoflocal interpolations. Thecondition (5)provides existenceof extensionsof basiselementsthat correspond to the accumulation sums in (11). However, the terms of transition sums do not havesimple representations intermsofsuchextensions.
7. Twoexamples
First we consider regular sequences (Bk)∞k=1. Let βk = (log Bk)/k. We say that (Bk)∞k=1 is regular if,
for somek0,both sequences(Bk)∞k=k0 and(βk)
∞
k=k0 are monotone.Recall that(Bk)
∞
k=1hassubexponential
growth if βk → 0 ask→ ∞.
For example,givena> 1,letγk(1)= k−a,γk(2) = a−k,γk(3) = exp(−ak) forlargeenoughk. Thenγ(j) for
1≤ j ≤ 3 generateregularB(j)withBk(1)∼ 2−k−1ak log k,Bk(2)∼ 2−k−2k2log a,Bk(3)∼ (a/2)k+1/(a− 1). Here,βk(1),β(2)k −log 2 andβ(3)k → −log(a/2),soB(j) arenotofsubexponentialgrowth,exceptB(3) for
a= 2.Weseethat (5)isvalidinthefirsttwo casesand inthethirdcasewitha≤ 2.
Moregenerally, (5)isvalidforeachmonotoneconvergent(Bk)∞k=1.Indeed,ifBk B ≥ 0,thenLHSof (5)doesnotexceed(n+ 1)−1.IfBk B,thenwetakes0withBs> B/2 fors≥ s0.ThenBs+· · · Bs+n≥
(n+ 1)B/2 andLHS of (5)< 2(n+ 1)−1. This coversthecaseof regular sequences(Bk)∞k=1 whenβk are
negative. Let us show that(5) is valid as well for divergent regular sequences (Bk)∞k=1 of subexponential
growth.
Theorem7.1. Let(Bk)∞k=1beregularwithpositivevaluesofβk.Then (5) isvalidifandonlyif(Bk)∞k=1has
Proof. A regular sequence (Bk)∞k=1 is not of subexponential growth, provided βk > 0, in the following
three cases: βk β < ∞,βk ∞ and βk ε0 > 0. We aim to show that (5) is not valid under the
circumstances.
Inthefirstcase,givens andn,letb= exp βs+n.Thenb− 1≥ exp β1− 1> β1> 0 and b≤ exp β.Here,
s+n
k=sBk < bs+n+1/(b− 1) as Bk = exp(kβk)≤ bk for suchk. Therefore, Bs+n/s+nk=sBk > (b− 1)/b>
β1/exp β,whichcontradicts (5).
Ifβk ∞ then,bythesameargument,Bs+n/s+nk=sBk > (b− 1)/b> 1/2 fors≥ s0, wheres0 isfixed
withexp βs0 > 2.
Supposeβk ε0.Wefixindicess1< s2<· · · suchthattheintervalsIj connectingpoints(sj,βsj) and (sj+1,βsj+1) form aconvexenvelopeoftheset (k,βk) ontheplane. Westartfroms1= max{s: βs= β1}. Ifsj ischosen, thenwe takesj+1with heproperty:foreachk withsj ≤ k ≤ sj+1 thepoint (k,βk) is not
overIj.Atany stepwecantakethenextvaluesolargethattheslopesofIj increasestozero.Inaddition,
givensj,wetakesj+1 suchthat
(4− 2 sj/sj+1)βsj+1≥ (3 − sj/sj+1)βsj, (21) whichispossibleas βk decreasestoapositivelimit.
For fixed j, we take s = sj and s+ n = sj+1. Let β˜k = ak + b with a = −(βs− βs+n)/n and b =
βs+ (βs− βs+n)s/n for s≤ k ≤ s+ n,so thepoints (k,β˜k) are locatedjust onthe interval Ij. Also, let
g(x)= ax2+ bx andB˜
k = exp g(k)= exp(k ˜βk) on[s,s+ n].Ofcourse, B˜s= BsandB˜s+n= Bs+n.
It is easy to check that the function g increases on this interval. Hence, s+nk=sBk ≤
s+n k=sB˜k < s+n s g(x)dx+ Bs+n.Byintegrationbyparts, s+n s g(x)dx= g(n+ s)·[2a(n+ s)+ b]−1−g(s)·[2as+ b]−1+
2a ss+ng(x)(2ax+ b)−2dx.Weneglectthelastterm,asa< 0,andthesecondterm,as2as+ b= g(s)> 0. Also,2a(n+ s)+ b= (2+ s/n)βs+n− (1+ s/n)βs≥ βs/2≥ ε0/2,by (21).Hence
s+n
s g(x)dx< 2Bs+n/ε0
andBs+n/s+nk=sBk > ε0/(2+ ε0),so (5)isnotvalid.
Weproceedtoshowthat (5)isvalidforβk 0,thatisinthecaseofsubexponentialgrowthof(Bk)∞k=1.
Here, for fixed large s and n, we estimate s+nk=sBk from below. Let b = exp βs+n. Then Bk ≥ bk for
s≤ k ≤ s+ n.Therefore,Bs+n/
s+n k=sBk ≤b
n(b−1)
bn+1−1. Ifbn< 2 forthegivens andn then bn+1− 1> (n+ 1)β
s+n.On theotherhand,exp βs+n− 1< 2βs+n
forβs+n < 1.Thusthefraction abovedoesnotexceed4/(n+ 1).
Otherwise, bn ≥ 2 and bn < 2(bn+1− 1). Here the fraction does not exceed 4βs+n. It follows that
Bs+n/s+nk=sBk ≤ max{4/(n+ 1),4βn},whichisthedesiredconclusion. 2
Our next objective is to consider irregular sequences (Bk)∞k=1 (compare with Ex.6in [11]). Given two
sequences, (kj)∞j=1 ⊂ N with kj+1− kj ∞ and (εj)∞j=1 with εj 0, let γk = (k + 5)−2 for k = kj
and γkj = (kj + 5)−2εj. Then γ satisfies (2) with δk = (5!/(k + 5)!)
2ε
1ε2· · · εj for kj ≤ k < kj+1. Let
Aj := logε1ε21···εj.Wewillconsider onlysequenceswiththeproperty
kj+12 · A−1j → 0 as j → ∞. (22)
Providedthiscondition,Bk= 2−klog(k+5)!5! + 2−k−1Aj∼ 2−k−1Aj forkj ≤ k < kj+1.Inaddition,aneasy
computationshowsthatforlargej,
Bkj + Bkj+1+· · · + Bkj+1−1< 3 Bkj. (23) NowwecanconstructdifferentexamplesofcompactsetsK(γ) withoutextensionproperty.
Example 2.Let Aj = 2kj, so εj = exp(−2kj + 2kj−1) for j ≥ 2 and ε1 = exp(−2k1). In this case, (22)is
valid under mild restriction 2−kjk2
2−kj+1−1A
j+1= 1/2 and,by (23),Bs+· · ·+Bs+n−1 < 3Bs< 4·2−kj−1Aj= 2.Thisgives (8)withε= 1/4
and m= 0.
8. Extensionpropertyof K(γ) and Hausdorffmeasures
Fromnowon,h isadimensionfunction,whichmeansthath: (0,T )→ (0,∞) iscontinuous, nondecreas-ingand h(t)→ 0 ast→ 0.Theh-HausdorffcontentofE⊂ R isdefinedas
Mh(E) = inf{
h(|Gi|) : E ⊂ ∪Gi}
and theh-HausdorffmeasureofE is Λh(E) = lim inf
δ→0 {
h(|Gi|) : E ⊂ ∪Gi,|Gi| ≤ δ}.
Here weconsiderfinite orcountablecoveringsofE byintervals(openorclosed).
ItiseasilyseenthatMh(E)= 0 ifandonlyifΛh(E)= 0.Wewriteh1≺ h2ifh1(t)= o(h2(t)) ast→ 0.
Leth1≈ h2 ifC−1h1(t)≤ h2≤ Ch1(t) forsomeconstantC≥ 1 and0< t≤ t0< T .Wewilldenote byh0
thefunctionh0(t)= (log1t)−1 with0< t< 1,whichdefines thelogarithmicmeasureofsets.
AsetE iscalleddimensional ifthereisatleastonedimensionfunctionh thatmakesE anh-set,thatis 0< Λh(E)<∞.Inourcase,thesetK(γ) isdimensional.In [1],followingNevanlinna [15],thecorresponding
dimensionfunctionwaspresented.Letη(δk)= k fork∈ Z+withδ0:= 1 andη(t)= k + logδtk/logδk+1δk for
δk+1< t< δk.Thenh(t):= 2−η(t)for0< t≤ 1.Clearly,h(δk)= 2−k.
Lemma 8.1.Letγ satisfies (2)andh be definedas above.ThenΛh(K(γ))= 1.
Proof. Taket= C0δk,whereC0 isgivenin (3).Thenδk < t= C0γkδk−1< δk−1 forlargeenoughk.Here,
η(t)= k− log C0/log(1/γk) and h(t)= 2−kak withak := explog Clog(1/γ0·log 2k) .Since γk → 0, given ε> 0, there
is k0 such thatak < 1+ ε fork ≥ k0. From (3) it follows that 1= 2kh(δk) <2
k
j=1h(lj,k) < 2kh(t)<
1+ ε provided thatk ≥ k0. Of course, Λh(K(γ)) ≤
2k
j=1h(lj,k) for each k. Since ε is arbitrary, we get
Λh(K(γ))≤ 1.
Letus showthatΛh(K(γ))≥ 1.Fixε> 0 andchoosek0suchthat
ε log 1/γk>− log 2 · log(1 − 4γk) for k≥ k0. (24)
This canbedoneasγk→ 0.Takeanyopencovering∪Gi ofK(γ).Givenε,wecanconsider coveringsonly
with |Gi|< δk0 foreachi.Wechooseafinite subcover∪
N
i=1Gi of K(γ).
Fix i ≤ N andk with δk+1 <|Gi|≤ δk.By (3) and (4), thedistance betweenany two basicintervals
from Ek+1 exceeds (1− 4γk+1)δk. If|Gi|< (1− 4γk+1)δk thenGi canintersectat mostoneinterval from
Ek+1. In this case we canconsider only |Gi| ≤ max1≤j≤2k+1lj,k+1 ≤ C0δk+1, by (3). Thus there are two possibilities:δk+1<|Gi|≤ C0δk+1or (1− 4γk+1)δk<|Gi|≤ δk.
In the firstcase we haveh(|Gi|) > 2−k−1. Here, Gi intersects at mostoneinterval from Ek+1 and, by
construction,atmost2m−k−1 intervalfrom Emform> k.Inturn,inthelattercase,h(|Gi|)> 2−k(1− ε).
Indeed,here,η(|Gi|)< k−log(1−4γk+1)/log(1/γk+1) andh(|Gi|)> 2−ka,wherea= explog(1log(1/γ−4γk+1k+1)·log 2) >
(1− ε),by (24).NowGi intersectsat mosttwo intervalfromEk+1 andatmost2m−k intervalfrom Em.
Let us choose m so large that each basic interval from Em belongs to some Gi. We decompose all
intervals from Em into two groups corresponding to the cases consideredabove. Counting intervals gives
2m≤ i2m−k−1+ i 2m−k< 2m[ ih(|Gi|)+
i h(|Gi|)(1− ε)−1]. Fromthiswe seethat
ih(|Gi|)>
Thesamereasoningappliesto apartofK(γ) oneachbasicinterval.
Corollary8.2. Letγ andh beas inProposition above,k∈ N,1≤ j ≤ 2k.ThenΛ
h(K(γ)∩ Ij,k)= 2−k.
Theorem8.3. Supposeγ satisfies (2),h is definedasabove andK(γ) is notpolar.Then μK= Λh|K(γ).
Proof. Here,byCorollary 3.2in [1],μK(γ)(Ij,k)= 2−k,so thevaluesofμK(γ) andtherestrictionofΛh on
K(γ) coincideoneachbasicinterval.Fromhere,byLemma3.3in [1],thesemeasuresareequalonK(γ). 2 Thus,anon-polarsetK(γ) satisfying (2)isindeedequilibrium Cantor-typesetifweacceptfordefinition of this concept the condition μK = Λh|K(γ), which is more naturalthan the definition suggested in[10],
Section 6.
We recall thatthere is no complete characterization of polarity of compact sets interms of Hausdorff measures, see e.g. Chapter V in [15]. On the one hand, a set is polar if its logarithmic measure is fi-nite. This defines a zone Zpol in the scale of growth rate of dimension functions consisting of h with
lim inft→0h(t)/h0(t) > 0. If h ∈ Zpol and Λh(K) < ∞ then Cap(K) = 0. On the other hand, functions
with 0h(t)/tdt <∞ forma non-polarzone Znp: ifh∈ Znp and Λh(K) > 0 then Cap(K)> 0. But, by
Ursell [22],theremaindermakesupazoneZuofuncertainty.Onecantaketwo functionsinthiszonewith
h2 ≺ h1 and sets K1,K2, where Kj is ahj-set, such thatK2 is polar, K1 is not, though inthe sense of
Hausdorffmeasurethe setK2 islargerthanK1. Indeed,Λh2(K2)> 0,butΛh2(K1)= 0 orΛh1(K2)=∞,
butΛh1(K1)<∞.
Letusshowthatasimilar circumstanceisvalidwiththeextensionproperty.
Proposition 8.4. There are two dimensionfunctions h2 ≺ h1 andtwo sets K1,K2, where Kj is an hj-set
forj∈ {1,2},suchthat thesmallerset K1 has theextension property,whereasthelarger setK2 doesnot.
Proof. TakeK1from Example 1.Letusshowthatthecorrespondingfunctionh1= 2−η1 isequivalenttoh0.
ItisenoughtofindC > 0 suchthatη0(t)−C ≤ η1(t)≤ η0(t)+C forsmallt.Here,η0(t)= (log log 1/t)/log 2,
soh0(t)= 2−η0(t).ForthesetK1wehaveδk = exp(−2k+1B) andη0(δk)= k +log 2B/log 2.Ifδk+1< t≤ δk
forsomek,thenk≤ η1(t)< k + 1 andk + log 2B/log 2≤ η0(t)< k + 1+ log 2B/log 2,whichgivesh1≈ h0.
In turn, let K2 be as in Example 2 with Aj = 2kj2−j and εj = exp(−Aj + Aj+1) for j ≥ 2. Here
we suppose that (kj)∞j=1 satisfies 2−kj2jkj+12 → 0 as j → ∞. Then (22) and (23) are valid, which, as
in Example 2, gives the lack of the extension property for K2. Let us show that h2 ≺ h0. It is enough
to check thatη2(t)− η0(t) → ∞ as t → 0. Let δk < t ≤ δk−1 with kj ≤ k < kj+1 for large enough j.
Then log 1/δk = 2log((k + 5)!/5!)+ Aj < 2Aj and η0(t) < η0(δk) < kj + 1− j. On the other hand,
η2(t)≥ η2(δk−1)= k− 1≥ kj− 1.Therefore,η2(t)− η0(t)> j− 2,whichcompletestheproof. 2
Onecan suppose that,at leastfor the consideredfamily ofsets, thescale of growth rate of dimension functionscanbedecomposedas aboveinto threezones.IfK(γ) isanh-setforafunctionh withmoderate growth then the set has EP . If thecorresponding functionh is largeenough, then EP fails. Proposition aboveshowsthatthezoneofuncertaintyhereisnotempty.
Wesee thath= h0 is notthe largestfunctionwhichallowsEP forh-sets K(γ).Ifwe takeBk ∞ of
subexponential growth,asintheregularcase,then δk= exp(−2k+1Bk) andh0(δk)= 2−k−1Bk−1,whichis
essentiallysmallerthanh(δk)= 2−kforthecorresponding functionh.
Example 3. Let log(m)t denote the m-th iteration log· · · log t for large enough t. The sequence Bk =
exp(k/log(m)k) hassubexponential growth.Then the corresponding sequence (γk)∞k=1 satisfies (2),as for
largek we have γk = δk/δk−1 < exp(−2kBk)< exp(−2k) andfor theprevious k wecan takeγk = 1/32.
Wewill searchit intheform h(t)= hα(t)0 (t).Lett= δk. Thenlog 1/t= 2k+1Bk, sok∼ (log log 1/t)/log 2.
On the other hand,h(t)= 2−k = (2k+1Bk)−α(t),which givesα(t)∼ 1− (log 2· log(m)k)−1 ∼ 1− (log 2·
log(m+2)1/t)−1.Clearly,h h0.
The next Proposition generalizes Example 3. We restrict our attentionto strictly increasing functions h of the form h = hα
0, where α isa monotone function on [0,t0]. As we will be interested inconsidering
dimension functionsexceedingh0 inthe nextsections,letus supposethatα(t)≤ 1.Thenh tσ foreach
fixed σ > 0.
Inadditionweassumethatasymptotically
h(t)≤ 2 h(t2), (25)
whichisvalid fortypicaldimensionfunctionscorresponding tothecases a) α(t)= α0∈ (0,1],
b) α(t)= α0+ ε(t) with α0∈ [0,1),
c) α(t)= 1− ε(t). Here,
ε(t) 0 with ε(t) log log 1/t ∞ as t 0, (26) sinceforslowlyincreasingε wegethα0±ε≈ hα0
0 .
By (25), for the inverse function h−1, we have h−1(τ ) ≤ (h−1(2τ ))2 and h−1 ≺ τM for M given
beforehand. From this, γk = h−1(2−k)/h−1(2−k+1) defines a sequence satisfying (2). We denote the
corresponding set by Kα(γ). Our aim is to check EP for this set provided regularity of the sequence
Bk = 2−k−1log(1/h−1(2−k)). Wesee at once thatBk increases. Initsturn, βk 0 ifα0 = 1 in thecase
(a),βk 1/α0− 1 in(b) andin(a) withα0< 1.Concerning(c),themonotonicityofβkrequiresadditional
rather technicalrestrictionsonε.Atleastforε(t)= εm(t):= (log(m)1/t)−1 wehaveβk 0.Here,m≥ 3,
as h≈ h0 form∈ {1,2}.
Proposition 8.5. Let Kα(γ) be defined by a function h, as above, with a regular sequence (B
k)∞k=1. Then
Kα(γ) hastheextension propertyif andonly if
log 1 h−1(2−k)
1/k
→ 2 as k → ∞.
Proof. Let us find h−1 for the case α(t) = 1− ε(t). If h(t) = τ then [1− ε(t)]log log 1/t = log 1/τ . Let us define a function δ by the condition log log 1/t = [1+ δ(τ )]log 1/τ . Then [1− ε(t)][1+ δ(τ )] = 1, so δ(τ ) 0 as τ 0. Then t = h−1(τ ) = exp[−(1/τ)1+δ(τ )] and log(1/h−1(2−k)) = 2k(1+δ(2−k)). The
k-th root of this expression tends to 2. On the other hand, (Bk)∞k=1 here has subexponential growth as
βk = (δ(2−k)− 1/k)log 2→ 0.By Theorem 7.1,Kα(γ) hasEP .
Similarly, if α(t) = α0 + ε(t) with 0 < α0 < 1 then h−1(τ ) = exp[−(1/τ)1/α0−δ(τ)]. Here,
(log(1/h−1(2−k)))1/k = 2(1/α0−δ(2−k)) 2 and β
k 0, there is no EP . In the case (a), the function
δ vanishes.
Lastly, α0 = 0 in (b) gives h−1(τ ) = exp[−(1/τ)Δ(τ )] with Δ(τ ) ∞ as τ 0. Here,
9. ExtensionpropertyanddensitiesofHausdorffcontents
To decide whether a set K has EP , we have to consider a local structure of the most rarefied parts of K.Obviously, suchglobalcharacteristicsasHausdorffmeasuresorHausdorffcontentscannotbeapplied in general for this aim. Instead, one can suggest to describe EP in terms of lower densities of Mh or
related functions. Given a dimension function h, a compact set K,x ∈ K and r > 0, let ϕh,K(x,r) :=
Mh(K ∩ B(x,r)) and ϕh,K(r) := infx∈Kϕh,K(x,r), where B(x,r) = [x− r,x+ r]. One can suppose
that K has EP if and only if the corresponding function ϕh,K is not very small, in asense, as r → 0.
Essentially, this is similar to analysis of thelower density of theHausdorffcontent, whichcan be defined as φh(K) := lim infr→0infx∈K MMh(Kh(B(x,r))∩B(x,r)). Indeed, Mh(B(x,r)) = h(2r) for h with h(t) t and the
expressionaboveislim infr→0ϕh,K(r)
h(2r) .
In order to distinguish EP by means of φh, we have to consider large enough dimension functions h.
Indeed, if for some h1 with h1 h there exists h1-set K1 with EP , then h cannot be used for this
aim, because Λh(K1) = 0 implies Mh(K1) = 0 and the corresponding density vanishes contrary to our
expectations.Therefore,wecanconsideronlyfunctionsexceedingh0.
Weremark thatΛh-analogs ofϕh,K or φh cannotbe appliedingeneralfordistinguishing EP ,sincefor
fatsets(K = Int(K))we haveΛh(K∩ B(x,r))=∞ providedh(t) t.
Interestingly, it turns out thatthe lower density φh can be used to characterize EP forthe family of
compactsetsconsideredin [6].
Example 4. Given two sequences bk 0 (for brevity, we take bk = e−k) and Qk with Qk ≥ 2, let
K ={0}∪∞k=1Ik,whereIk = [ak,bk],|Ik|= bQkk.Inwhatfollowswewillconsidertwocases:Qk ≤ Q with
someQ andQk ∞ withQk < log k for large k. ByTheorem 4in[6],K has theextension property in
thefirstcaseanddoesnothaveitforunbounded(Qk).
Inthenextlemma weconsider concavedimensionfunctionsh= hα
0 forthecases(a),(b),as above,and
formoregeneral
c) α(t)= α0− ε(t) withα0∈ (0,1].
We suppose now thatε is amonotone differentiable functionon [0,t0] with 0< ε(t)< 1− α0 in(b) and
0< ε(t)< α0/2 in(c).As before,weassume (26). Adirectcomputationshowsthat
h(t) < h(t) h0(t) α(t)/t for the cases (a), (b) and h(t) < h(t) h0(t)/t for (c). (27)
Lemma 9.1. Suppose intervals Ik are given as in Example 4 and n is large enough. Then Mh(∪∞k=nIk) =
h(bn). This means that the covering of the set ∪∞k=nIk by the interval [0,bn] is optimal in the sense of
definitionof Mh.
Proof. Letus fixacoveringof K byopenintervals, chooseafinite subcovering∪M
i=1Gi andenumerate Gi
from left to right.We can suppose thatG1 covers∪k=N∞ Ik for someN ≥ n. Indeed, if G1 covers as well
somepartof IN−1, then other partofIN−1 is coveredby G2. Inthis case,association of G1 and G2 into
oneinterval will givebetter covering, sinceh(b) ≤ h(x)+ h(b− x) for 0≤ x≤ b, byconcavity of h. For thesamereason,wesuppose thateachGi coversentire numberofIk.After thiswereduceeachGi to the
minimal closed interval Fi containing thesame intervals Ik. Thus, F1 = [0,bN] and F2 = [aN−1,bq] with
somen≤ q ≤ N − 1.Ouraim istoshowthat
so replacingF1∪ F2with[0,bq] ispreferable.Weusethemeanvaluetheoremand thedecrease ofh. Note
thath(bk)= k−α(bk).
Considerfirstthevalueq = N− 1.Wewill showh(bN−1)− h(bN)< h(|IN−1|).
In thecases(a),(b),by (27), LHS < h(bN)e−N(e− 1)< N−1−α(bN)α(bN)(e− 1). Ontheother hand,
h(|IN−1|) = [QN−1(N − 1)]−α(|IN−1|). Here, α(|IN−1|) < α(b
N), so we reduce the desired inequality to
(QN−1/N )α(bN)α(b
N)(e− 1)< 1.Itisvalid,sinceforα0> 0 thefirsttermontheleftgoestozero,whereas
forα0= 0 in(b) wehaveα(bN)= ε(bN)→ 0 asN → ∞.
Similarly,inthe case(c) theinequality[QN−1(N− 1)]α0−ε(|IN−1|)(e− 1)< N1+α0−ε(bN) isvalid, as is easyto check.
Supposenow thatq≤ N − 2.Wewrite (28)ash(bq)− h(bq− aN−1)< h(bN).
Here, in all cases,by (27), LHS < h(bq− aN−1)aN−1 < h(bq)h0(bq)bqa−aN−1N−1, where the last fraction
does not exceed bbN−1
q−bN−1. On the other hand, h(bN) ≥ N
−1 as α(b
N) ≤ 1. Hence it is enough to show
thatN < (eN−q−1− 1)q1+α(bq).Weneglectα(b
q) andnoticethat(eN−q−1− 1)q≥ (e− 1)(N − 2),which
completes theproofof (28).
Continuinginthis manner,weseethath(bn)≤Mi=1h(|Fi|). 2
Corollary 9.2.Suppose bn+1≤ r ≤ bn− bn+1.Thenϕh,K(r)= h(|In|).
Proof. Clearly, ϕh,K(x,r)= h(|In|) foreach x∈ In. If x∈ K ∩ [0,bn+1] then B(x,r) coversall intervals
Ik withk≥ n+ 1.ByLemma,ϕh,K(x,r)= h(bn+1)> h(|In|).Of course,forx∈ Ik with k < n thevalue
ϕh,K(x,r) alsoexceedh(|In|). 2
Remark.Thecoveringoftwo(orsmallnumberof)intervalsIkbyoneintervalisnotoptimal,sinceMh(Ik∪
Ik+1)= h(|Ik|)+ h(|Ik+1|)< h(bk− ak+1).
Weproceed tocharacterizeEP forgivencompactsets intermsoflowerdensitiesφh forh= hα0, where
α(t) = α0∈ (0, 1] or α(t) = α0± εm(t) (29)
with 0< α0< 1 andεm(t)= (log(m)1/t)−1 form> 2,so (26)isvalid.
Proposition 9.3. LetK be from thefamily of compactsets givenin Example 4andh be asabove. ThenK has theextension propertyif andonly ifφh(K)> 0.
Proof. Suppose first that Qk ≤ Q with some Q, so K has EP . We aim to show lim infr→0ϕh(2r)h,K(r) > 0.
Let e−k−1 ≤ r < e−k for some k. Then, as ϕh,K increases, ϕh,K(r) ≥ ϕh,K(e−k−1), which is h(|Ik|) =
(k· Qk)−α(|Ik|),by Corollary 9.2.Ontheother hand,h(2r)< h(2e−k)= (k− log 2)−α(2e
−k)
.Therefore, ϕh,K(r)/h(2r) > Q−α(|Ik k|) kα(2e
−k)−α(|I
k|) (1− log 2/k)−α(2e−k).
The first termon theright converges to Q−α0 as k→ ∞. The second andthe thirdterms converge to1.
Hence,φh(K)≥ Q−α0.Besides,this valueisachievedinthecaseQk = Q bythesequencerk= bk− bk+1.
Thus, φh(K)= Q−α0 > 0.
Similar arguments apply to the case Qk ∞, when K does not have EP . Here, φh(K) ≤
limkϕh,K(rk)/h(2rk) forrk as above.By Corollary 9.2,ϕh,K(rk)= h(|Ik|). Also,h(2rk)> h(e−k).Hence,
ϕh,K(rk)/h(2rk)< Q−αk 0/2kα(e
−k)−α(|I
k|),whichconvergesto 0as k increases. 2
