Letdenote an open and bounded subset of Cand L be a compact subset of .
Consider the open setG=\L,p ∈ {0,1, . . .} ∪ {∞}and f0∈ Ap(G). Then there
are two cases:
(i) Either there is a functionF0∈ Ap(), such thatF0|G= f0
(ii) or there exists noF0∈ Ap(), such thatF0|G = f0.
Theorem 7.1 Let p∈ {0,1, . . .} ∪ {∞},be an open and bounded subset ofCand L be a compact subset of. Let also G = \L. Suppose that there is a function f0 ∈ Ap(G)for which there exists no F0 ∈ Ap(), such that F0|G = f0. Then the
set A(p,G)of functions f ∈ Ap(G)for which there exists no F ∈ Ap()such that F|G = f is an open and dense subset of Ap(G).
Proof First, we will prove that the setS = Ap(G)\ A(p,G)is closed. Let(hn)n≥1
be a sequence inSconverging in the topology ofAp(G)to a functionh∈ Ap(G). By the maximum modulus principle, the extensions Hnofhnform a uniformly Cauchy
sequence on. Thus, the limitHof Hnonis an extension ofh. Thereforeg ∈ S
andSis a closed subset ofAp(G).
Now we will prove that the set S has empty interior. If S does not have empty interior, then there is a function f in the interior ofS andl ∈ {0,1, . . .},l ≤ p and d >0, such that f ∈ Ap(G):sup z∈G f(j)( z)−g(j)(z)<d,0≤ j ≤l| ⊂S. Obviously, f0is not identically equal to zero which implies
m=max sup z∈G |f0(j)(z)|,j =0,1, . . . ,l >0.
From the definitions, the functionh(z)= f(z)+ d
2mf0(z),z∈Gbelongs toS. Since f,h belong to S, there are F,H ∈ Ap(), such that F|G = f, H|G = h. Then
2m
d (H(z)−F(z))belongs to A
p()and is equal to f
0inGwhich contradicts our
Remark 7.2 If the interior ofLinCis non-empty, then there always exists a function f0∈ Ap(G)for which there does not exist a functionF0∈ Ap(), such thatF0|G=
f0|G.
Remark 7.3 From the previous results, we have a dichotomy: Either every f ∈ Ap(G)
has an extension inAp()or generically all functions f ∈ Ap(G)do not admit any extension inAp(). The first case holds ifap(L)=0 and the second case ifap(L) >0
(Theorem 3.12).
Remark 7.4 In a similar way, we can prove that ifLis a compact set contained in the open setU, then either everyf ∈ ˜Ap(U\L)has an extension inA˜p(U)or generically every f ∈ ˜Ap(U\L)does not have an extension inA˜p(U),p∈ {0,1,2, . . .} ∪ {∞}. The first horn of this dichotomy holds if and only ifa˜p(L)=0 which is equivalent
with the fact that the interior ofL is void inC(Theorem3.16). Now we present some local versions of the above results.
Definition 7.5 LetLbe a compact subset ofCandUbe an open subset ofC, such that L ⊆U. Let alsoz0∈∂L. A function f ∈H(U\L)is extendable atz0if there exists
r>0 andF ∈ H(D(z0,r))such thatF|(U\L)∩D(z0,r)= f|(U\L)∩D(z0,r). Otherwise, f is not extendable atz0.
Below we will use the above definition of extendability.
Proposition 7.6 Let L be a compact subset ofCand U be an open subset ofC, such that L ⊆ U . Let also M and r be positive real numbers, p ∈ {0,1,2, . . .} ∪ {∞} and z0 ∈ ∂L. The set EM,p,U,L,z0,r of functions f ∈ A
p(U \L)such that there is
F ∈H(D(z0,r))with F|(U\L)∩D(z0,r) = f|(U\L)∩D(z0,r)andF∞≤M is a closed subset of Ap(U\L). Also, if there exists f0∈ Ap(U\L)which is not extendable at
z0, then the interior of EM,p,U,L,z0,r is void in A
p(U\L).
Proof We will first prove that the setEM,p,U,L,z0,r is a closed subset ofA
p(U\L).
Let(fn)n≥1be a sequence inEM,p,U,L,z0,r converging in the topology ofA
p(U\L)
to a function f ∈ Ap(U \ L). Without of loss of generality we assume thatU is bounded. This implies that fnconverges uniformly onU\Lto f and that there exists
a sequence (Fn)n≥1 in H(D(z0,r))such that Fn|(U\L)∩D(z0,r) = fn|(U\L)∩D(z0,r) and F ∞≤ M for everyn ≥1. By Montel’s Theorem there exists a subsequence of(Fn),(Fkn), which converges uniformly on the compact subsets ofD(z0,r)to a
functionFwhich is holomorphic and bounded byMonD(z0,r). SinceFknconverges
to f on(U\L)∩D(z0,r), the functions f andFare equal on(U\L)∩D(z0,r). Thus f belongs toEM,p,U,L,z0,r andEM,p,U,L,z0,r is a closed subset ofA
p(U\L).
If there exists f0 ∈ Ap(U \ L) which is not extendable at z0, the interior of
EM,p,U,L,z0,r is void in A
p(U \ L), the proof of which is similar to the proof of
Theorem7.1.
Here we have another dichotomy which is a local version of the first one.
Theorem 7.7 Let L be a compact subset ofCand U be an open subset ofC, such that L ⊆ U and let p ∈ {0,1,2, . . .} ∪ {∞}, z0 ∈ ∂L. The set Ep,U,L,z0 =
∞ M=1 ∞ n=1 EM,p,U,L,z
0,n1 is the set of extendable functions of A
p(U\L)at z
(i) either every function f ∈ Ap(U\L)is extendable at z0
(ii) or generically all functions f ∈ Ap(U\L)are not extendable at z0.
Proof If(i)is not true, then Proposition7.6shows thatEM,p,U,L,z
0,1n is closed with
empty interior for all natural numbersn ≥1,M ≥1. Then
Ap(U\L)\Ep,U,L,z0 = ∞ M=1 ∞ n=1 Ap(U\L)\EM,p,U,L,z 0,1n
is the intersection of a countable number of open and dense subsets ofAp(U\L)and Baire’s Theorem shows that Ap(U \L)\Ep,U,L,z0, which coincides with the set of non-extendable functions ofAp(U\L)atz0, is a dense andGδsubset ofAp(U\L).
In what follows, we compare two notions: local extendability and existence of a holomorphic extension. At first, we examine the case of a compact setLwith empty interior.
Proposition 7.8 Let L be a compact subset ofCand U be an open subset ofC, such that L ⊆U and L◦ = ∅. Let also f ∈ H(U \L). Then f is extendable at every z0∈∂L if and only if there exists a holomorphic extension F of f on U . If additionally
f ∈ Ap(U\L)for some p∈ {0,1, . . .} ∪ {∞}, then F ∈ Ap(U).
Proof If there exists a holomorphic extension F of f on U, then obviously f is extendable at everyz0∈∂L =L.
Conversely, if f is extendable at everyz0 ∈ L, then for everyz0 ∈ L there is a
holomorphic functionFz0 defined onD(z0,rz0)⊆Usuch thatFz0|(U\L)∩D(z0,rz0)= f|(U\L)∩D(z0,rz0). Let z1,z2 ∈ L with V = D(z1,rz1)∩ D(z2,rz2) = ∅. Since L◦ = ∅,V \L is a non-empty open set,Fz1,Fz2 are holomorphic on the domainV and coincide with f onV \L. By analytic continuation,Fz1 = Fz2 onV. So, the functionF defined onU such thatF(z)= Fz(z)for everyz∈ L andF(z)= f(z)
for every z ∈ U \ L is a holomorphic extension of f on U. Obviously, if f ∈
Ap(U\L), thenF∈ Ap(U).
Remark 7.9 WheneverL◦= ∅, the equivalence of Proposition7.8is not true. Indeed ifw∈ L◦= ∅, then the holomorphic function f(z)= z−1w forz∈U\Lcannot be extended to a holomorphic function onU, but it is extendable at everyz0∈∂L.
We consider again a compact setL ⊆Cand an open setU⊆C, such thatL ⊆U and a p ∈ {0,1,2, . . .} ∪ {∞}. Now we want to find a similar connection between ap(L)andap(L∩ D(z0,r)); that is, is the conditionap(L) = 0 equivalent to the
conditionap(L∩D(z0,r))=0 for allz0∈L?
If we suppose thatL◦= ∅, then there existz0andr >0 such thatD(z0,r)⊆L.
Thusap(L)andap(L∩D(z0,r))are strictly positive.
So, we need not assume that L◦ = ∅, since it follows from both the conditions ap(L)=0 andap(L∩D(z0,r))=0 for everyz0 ∈ L and for somer =rz0 >0. Also, the first condition obviously implies the second one.
Probably Theorem3.6holds even for p ≥1. Specifically, ifap(L)=0 andV is
an open set, then every functiong ∈ Ap(V \L)belongs to Ap(V). This leads us to believe that the above conditions are in fact equivalent. However, this will be examined in future papers.
Acknowledgements We would like to thank A. Borichev, P. Gauthier, J.P. Kahane, V. Mastrantonis, P. Papasoglu, and A. Siskakis for helpful communications.
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