Hartogs companions and holomorphic extensions in arbitrary dimension
aa r X i v : . [ m a t h . C V ] S e p HARTOGS COMPANIONS AND HOLOMORPHIC EXTENSIONSIN ARBITRARY DIMENSION
VLAD TIMOFTE
Abstract.
We show that every holomorphic map f ∈ H (Ω \ K ) ( K ⊂ Ω ⊂ C n ,with K compact, Ω open, and n ≥ Hartogs companion ”˜ f ∈ H (Ω) matching f on an open subset C K, Ω ⊂ Ω \ K . Furthermore, ˜ f extends f , if and only if C n \ K is a connected set; this equivalence provesthe converse implication from the Hartogs Kugelsatz. The existence of vector-valued Hartogs companions in any dimension yields a Hartogs-type extensiontheorem for Gˆateaux holomorphic maps f ∈ H G (Ω \ K, Y ) on finitely opensets in arbitrary complex vector spaces. The equivalence is very similar to thatfor K ⊂ Ω ⊂ C n and leads to a corresponding Hartogs Kugelsatz in arbitrarydimension and to extension theorems for five types of holomorphy (Gˆateaux,Mackey/Silva, hypoanalytic, Fr´echet, locally bounded). We also show thatthe range ˜ f (Ω) of a vector-valued Hartogs companion cannot leave a domainof holomorphy containing f (Ω \ K ). We establish a boundary principle formaps f ∈ H G (Ω , Y ) ∩ C (Ω , Y ) on finitely bounded open sets. For Y = C , theprinciple states that f (cid:0) Ω (cid:1) = f ( ∂ Ω) (hence sup x ∈ Ω | f ( x ) | = sup x ∈ ∂ Ω | f ( x ) | ).Several results require a new identity theorem, which yields a maximum normprinciple and a “max-min” seminorm principle. Introduction
The famous Hartogs extension theorem is a striking and deep result emphasizingthe difference between the theory of holomorphic functions of one and of severalcomplex variables:
Theorem 1 (Hartogs extension) . Let n ≥ , an open set Ω ⊂ C n , and a compactsubset K ⊂ Ω . If Ω \ K is connected, then every map f ∈ H (Ω \ K ) has a uniqueextension ˜ f ∈ H (Ω) . A different version of this theorem is the following:
Theorem 2 (Hartogs Kugelsatz) . Let n ≥ , an open set Ω ⊂ C n , and a compactsubset K ⊂ Ω . If C n \ K is connected, then the restriction ρ : H (Ω) → H (Ω \ K ) , ρ ( f ) = f | Ω \ K , is an isomorphism of C -algebras. The surjectivity of the restriction operator ρ is equivalent to the existence of anextension ˜ f ∈ H (Ω) for every map f ∈ H (Ω \ K ). According to the above results,such extensions exist if one of the sets Ω \ K and C n \ K is connected. Date : September 7, 2020.2010
Mathematics Subject Classification.
Primary 32D15; Secondary 46G20; 46E50.
Key words and phrases.
Several complex variables; Holomorphic extension; Kugelsatz; Hartogscompanion; Gˆateaux holomorphy; finite open topology.
In this paper we improve and unify Theorems 1 and 2 as an equivalence of fourstatements and we generalize them to spaces of infinite dimension. To accomplishthis we prove several facts, not necessarily in the order listed below. We show that every map f ∈ H (Ω \ K ) has an extension ˜ f ∈ H (Ω), if and onlyif C n \ K is connected (Theorem 8). Therefore, the “ right condition ” in Theorem 1is the connectedness of C n \ K , which is also equivalent to the surjectivity of therestriction map ρ from Theorem 2. It turns out that extending every f ∈ H (Ω \ K )only depends on K , and not on the surrounding open set Ω ⊃ K . Furthermore, inTheorem 1 we may replace compactness by significantly weaker assumptions (seeCorollary 47) allowing K to be unbounded and not closed, and K ∪ ( C n \ Ω) to bepath-connected (that is, K “breaks the boundary” of Ω), as in Example 48. Without the connectedness assumption from Theorem 1, every f ∈ H (Ω \ K )still has a unique “Hartogs companion” ˜ f ∈ H (Ω) matching f on a coincidence set(Theorem 3). The resulting association f ˜ f is a left inverse for the restrictionoperator ρ defined as in Theorem 2. The range inclusion ˜ f (Ω) ⊂ f (Ω \ K ) holds, andif f has an extension from H (Ω), that must be ˜ f . This leads to a striking compactexcision property of C -valued holomorphic maps of several variables: removinga compact from the domain does not change the range (Corollary 26). Droppingconnectedness is essential for the construction of vector-valued Hartogs companionsin arbitrary dimension (Theorem 33, obtained by a slicing technique with linearvarieties of finite dimension ). This construction leads to the Hartogs Kugelsatzequivalence in infinite dimension (Theorem 40 and Corollary 41). In the process, we also prove a needed identity theorem for Gˆateaux holomorphicmaps on polygonally connected 2-open sets (Theorem 16). As byproducts we geta maximum norm principle (Theorem 17) and a surprising “max-min” seminormprinciple (Theorem 19). For f ∈ H G (Ω , Y ), the latter states that if p ◦ f has a localmaximum value M > p on Y , then p ◦ f ≥ M everywhere (the “min” conclusion) and even though p ◦ f may not be constant, the map f vanishes nowhere and its range has empty interior. Hartogs companions in arbitrary dimension lead to several extension theoremsfor K ⊂ Ω ⊂ X and Gˆateaux holomorphic maps f ∈ H G (Ω \ K, Y ). We assumeat most that Ω is finitely open, K is finitely compact, and Ω \ K is polygonallyconnected (Theorems 40, 42, 44, and the four corollaries from the last section). Evenfor X = C n and Y = C some of these results are more general than Theorem 1,since K may not be closed or bounded and Ω may not be open (Example 14(c)),while Gˆateaux holomorphy still can be considered on 1-open sets Ω ⊂ C n . Several regularity properties (local boundedness, continuity, hypocontinuity,holomorphy, hypoanalyticity, Mackey holomorphy) are inherited from a map by itsHartogs companion (Theorem 38), hence also by Gˆateaux holomorphic extensionswhenever these exist. We thus get Hartogs-type extension results for five differenttypes of holomorphy (Corollary 41). Every domain of holomorphy containing the range of a map f ∈ H G (Ω \ K, Y ),also contains the range of the Hartogs companion ˜ f ; we call this range inertia . Inparticular, Hartogs-type extensions have this property. Intersections of a connected set with linear varieties may be disconnected. In all three theorems listed here, at least one of the three assumptions is weakened.
ARTOGS COMPANIONS AND HOLOMORPHIC EXTENSIONS 3 Viewing every map f ∈ H G (Ω , Y ) as the Hartogs companion of its restrictions ofthe form f | Ω \ K leads to general boundary principles (Theorem 28 and Corollary 29);according to the latter, f (Ω) = f ( ∂ Ω) for every f ∈ H G (Ω) ∩ C (Ω), where Ω is a2-bounded open set in a Hausdorff topological vector space.Due to its very interesting properties and consequences, we may conclude thatthe Hartogs companion is a new flexible tool which deserves further investigation.2. Hartogs companions in finite dimension
Hartogs companions in dimension at least two.
For arbitrary complexvector space X , sets A, B ⊂ X and S ⊂ C , and elements u ∈ X , λ ∈ C , it isconvenient to write A + B := { a + b | a ∈ A, b ∈ B } , S · A := { sa | s ∈ S, a ∈ A } ,A + u = u + A := A + { u } , S · u := S · { u } , λ · A := { λ } · A. Setting 1.
Throughout Section 2.1, for arbitrary integer n ≥ we consider anopen set Ω ⊂ C n and a compact subset K ⊂ Ω . For shortness, the connected components of any subset of C n will be simplycalled components. We denote by Υ Ω the set of all components of Ω.By removing connectedness assumptions, the following result accomplishes theconstruction of Hartogs companions in finite dimension. Furthermore, its last partyields both Theorems 1 and 2. As Theorem 8 will show, the Hartogs phenomenonfor holomorphic functions is characterized by the connectedness of C n \ K . Theorem 3 (Hartogs companions in finite dimension) . Let us define the coinci-dence (open) set of the inclusion K ⊂ Ω as C K, Ω := [ ω ∈ Υ Ω ( ω ∩ K u ω ) ⊂ Ω \ K, where K u ω denotes the unbounded component of C n \ ( K ∩ ω ) . Let an arbitrary map f ∈ H (Ω \ K ) . Then (a): There exists a unique map ˜ f ∈ H (Ω) (which will be called the Hartogscompanion of f ), such that ˜ f | C K, Ω = f | C K, Ω . (1) Furthermore, we have the range inclusion ˜ f (Ω) ⊂ f (Ω \ K ) . (2) If f has an extension ¯ f ∈ H (Ω) , then ¯ f = ˜ f . (b): If K ⊂ K ⊂ Ω ⊂ Ω , with K compact and Ω open, then ˜ f | Ω is theHartogs companion of f | Ω \ K . (c): For arbitrarily fixed a ∈ Ω and u ∈ C n \ { } , let a bounded open set G ⊂ C with the boundary ∂G consisting of finitely many piecewise C Jordancurves, and satisfying the following condition denoted by C G,u ( a ):0 ∈ G, a + G · u ⊂ Ω , K ∩ ( a + C · u ) ⊂ a + G · u (any set G with the above properties will be called ( a, u ) -admissible ). Then Ω G,u := { x ∈ Ω | C G,u ( x ) holds } is an open neighborhood of a . We have the Or more specific, whenever needed: ( a, u )-admissible for the inclusion K ⊂ Ω. VLAD TIMOFTE inclusion Ω G,u + ∂G · u ⊂ Ω \ K and the representation formula (where theboundary ∂G is oriented such that G lies to the left of ∂G ) ˜ f ( x ) = 12 π i Z ∂G f ( x + ζu ) ζ d ζ, for every x ∈ Ω G,u . (3) Hence for every linear variety L ⊂ C n of dimension at least one, such that Ω L := Ω ∩ L = ∅ , the restriction ˜ f | Ω L is uniquely determined by f | Ω L \ K . (d): If C n \ K is connected, then C K, Ω = Ω \ K , and hence ˜ f | Ω \ K = f .Proof. To shorten notation, we write C K, Ω and Υ Ω simply as C and Υ, respectively.There is no loss of generality in assuming K = ∅ .(a). The uniqueness of ˜ f follows easily by (1) and the identity theorem appliedto ˜ f | ω for each component ω ∈ Υ, since ω ∩ C = ω ∩ K u ω = ∅ is an open subsetof ω . Therefore, we only need to prove the existence part for an arbitrarily fixed ω ∈ Υ, its compact subset K ω := K ∩ ω , and the restriction f | ω \ K ω ∈ H ( ω \ K ω ).Consequently, there is no loss of generality in assuming that Ω is connected. Thus C = Ω ∩ K u , where K u denotes the unbounded component of C n \ K . Let us fix χ ∈ C ∞ (Ω), with χ ≡ K (throughout the proof, anysuch χ will be called a K -map). Set K χ := supp χ ⊂ Ω. The construction below of the map ˜ f χ ∈ H (Ω) follows the idea from the proof ofTheorem 2.3.2 from H¨ormander [5] (p.30) . There is a (unique) map f χ ∈ C ∞ (Ω),such that f χ = (1 − χ ) f on Ω \ K and f χ | K ≡
0. Hence f χ = f on Ω \ K χ . Since f ∈ H (Ω \ K ), for the smooth complex differential (0 , ∂f χ = n X j =1 ∂f χ ∂ ¯ z j d¯ z j , we have supp ¯ ∂f χ ⊂ K χ . Therefore, ¯ ∂f χ extends (by 0) to a differential (0 , g = P nj =1 g j d¯ z j , with g , . . . , g n ∈ C ∞ ( C n ) and supp g ⊂ K χ . It follows that¯ ∂g | Ω = ¯ ∂ f χ ≡ ∂g | C n \ K χ ≡
0, that is, ¯ ∂g ≡
0. For the map h ∈ C ∞ ( C n )defined by (see Th. 2.3.1 from H¨ormander [5] and its proof) h ( z , . . . , z n ) = 12 π i Z C g ( ζ, z , . . . , z n ) ζ − z d ζ ∧ d¯ ζ, (4)we have ¯ ∂h = g . Since K ⊂ K χ ⊂ Ω, for the unbounded component K u χ of C n \ K χ it is easily seen that K u χ ⊂ K u , Ω ∩ K u χ = ∅ . On K u χ ⊂ C n \ supp g we have ¯ ∂h = g ≡
0, and so h | K u χ ∈ H ( K u χ ). As h ∈ C ∞ ( C n )vanishes on the open set K u χ \ supp h = ∅ , by the identity theorem we get h | K u χ ≡ f χ := f χ − h | Ω ∈ C ∞ (Ω), we have ¯ ∂ ˜ f χ = ¯ ∂f χ − g | Ω ≡
0, that is, ˜ f χ ∈ H (Ω).Clearly, ˜ f χ | Ω ∩ K u χ = f χ | Ω ∩ K u χ − h | Ω ∩ K u χ = f | Ω ∩ K u χ . Let us observe that the map˜ f := ˜ f χ ∈ H (Ω) does not depend on the choice of χ . Indeed, for every K -map η with K η := supp η ⊂ K χ , we have K u χ ⊂ K u η . This yields ˜ f χ | Ω ∩ K u χ = f | Ω ∩ K u χ = ˜ f η | Ω ∩ K u χ ,which forces ˜ f χ = ˜ f η , by the identity theorem. Hence ˜ f has the property that˜ f | Ω ∩ K u χ = f | Ω ∩ K u χ , for every K -map χ. (5) The cited proof uses the ¯ ∂ technique initiated by Ehrenpreis [3]. ARTOGS COMPANIONS AND HOLOMORPHIC EXTENSIONS 5
In order to show that ˜ f | C = f | C , let us fix a ∈ C = Ω ∩ K u . Choose c ∈ C n , with k c k > max x ∈ K k x k . Thus δ := [1 , ∞ [ · c ⊂ K u . As K u is a connected open set, thereis a polygonal chain Λ ⊂ K u joining a to c . Hence ∆ := δ ∪ Λ ⊂ K u . There existsa K -map χ , such that K χ ∩ ∆ = ∅ . Since ∆ is an unbounded connected set, itfollows that ∆ ⊂ K u χ , and so a ∈ Ω ∩ K u χ . By (5) we deduce that ˜ f ( a ) = f ( a ). As a was arbitrary, we conclude that ˜ f | C = f | C . We thus have proved the existenceand uniqueness of the map ˜ f ∈ H (Ω) satisfying (1). Let us observe that whenever f has an extension ¯ f ∈ H (Ω), then ¯ f | C = f | C yields ¯ f = ˜ f , by the uniqueness of ˜ f . The range inclusion (2). Let us fix z ∈ C \ f (Ω \ K ). For f z , h, g ∈ H (Ω \ K ) and p ∈ H (Ω), defined by f z := f − z , h := f z , g := f z h , p ≡
1, we have ( ˜ f z ˜ h ) | C = g | C and ( ˜ f − z ) | C = f z | C and p | C = g | C . By the uniqueness of the Hartogs companionswe get ˜ g = ˜ f z ˜ h and ˜ f z = ˜ f − z and ˜ g = p . It follows that ( ˜ f − z )˜ h = ˜ f z ˜ h = ˜ g = p ≡ z ∈ C \ ˜ f (Ω). We thus conclude that ˜ f (Ω) ⊂ f (Ω \ K ). It f has anextension from H (Ω), that must be ˜ f . In this case, f (Ω \ K ) = ˜ f (Ω \ K ) ⊂ ˜ f (Ω), andso ˜ f (Ω) = f (Ω \ K ). For another proof of the range inclusion (2), see Proposition 21(independent of any preceeding results) and the comment at the end of its proof.(c). Let us consider a ∈ Ω and u ∈ C n \ { } , and an ( a, u )-admissible set G ⊂ C .Clearly, such sets G exist (in C · u ≃ C , the compact ( K − a ) ∩ ( C · u ) may becovered by a finite union of open balls with the closures contained in the open set(Ω − a ) ∩ ( C · u )). For every x ∈ C n , the last two conditions from C G,u ( x ) may bewritten as (where G c := C \ G and A c := C n \ A for every A ⊂ C n ) x + G · u ⊂ Ω ⇐⇒ (cid:0) x + G · u (cid:1) ∩ Ω c = ∅ ⇐⇒ x / ∈ Ω c − G · u,K ∩ ( x + C · u ) ⊂ x + G · u ⇐⇒ K ∩ ( x + G c · u ) = ∅ ⇐⇒ x / ∈ K − G c · u, and these obviously yield x ∈ Ω and x + ∂G · u ⊂ Ω \ K . We thus getΩ G,u = (cid:0) Ω c − G · u (cid:1) c ∩ (cid:0) K − G c · u (cid:1) c ⊂ Ω , Ω G,u + ∂G · u ⊂ Ω \ K. All four sets G · u , K , Ω c , G c · u , are closed in C n , and the first two are compact.Hence both Ω c − G · u and K − G c · u are closed, and so Ω G,u is open. Clearly, a ∈ Ω G,u . In order to prove (3), let us fix x ∈ Ω G,u . By using a linear change ofcoordinates in C n , we may assume u = (1 , , . . . , x := ( x , . . . , x n ) ∈ C n − and L := x + C · u = C × { x } . Thus Ω L = { z ∈ C | ( z , x ) ∈ Ω } . Let us chooseanother ( x, u )-admissible set G ⊂ C , such that G ⊂ G . Hence both conditions C G,u ( x ) and C G ,u ( x ) hold, and so0 ∈ G , x + G · u ⊂ Ω L , K ∩ L ⊂ x + G · u. Since K := K ∪ ( x + G · u ) is compact, Ω := Ω \ ( x + G c · u ) is open, and K ⊂ Ω , there exists χ ∈ C ∞ (Ω), such that χ ≡ ⊂ Ω of K ⊃ K and K χ = supp χ ⊂ Ω . Let us observe that K ∩ L = x + G · u, K χ ∩ L ⊂ x + G · u. By the construction from (a) it follows that ˜ f = f χ − h | Ω , where h is defined by (4)and g ∈ C ∞ ( C n ) is the extension by 0 of ∂f χ ∂ ¯ z ∈ C ∞ (Ω). Hence supp g ⊂ K χ \ Ω .For S := G \ G ⊂ C , we have S = G \ G and ∂S = ∂G ∪ ∂G , and so x + S · u = ( x + G · u ) \ ( x + G · u ) ⊂ ( x + G · u ) \ ( K ∩ L ) ⊂ Ω \ K, supp ( g | Ω L ) ⊂ supp g ∩ L ⊂ ( K χ \ Ω ) ∩ L ⊂ ( K χ ∩ L ) \ ( K ∩ L ) ⊂ ( x + G · u ) \ ( x + G · u ) = x + S · u = ( x + S ) × { x } . VLAD TIMOFTE
Hence supp [ g ( · , x )] ⊂ x + S . Since f χ ≡ ⊃ K ⊃ x + ∂G · u and f χ = f on Ω \ K χ ⊃ Ω L \ ( K χ ∩ L ) ⊃ x + ∂G · u , by the definitions of ˜ f , f χ , g , h , togetherwith (4) and Green’s formula (for the C ∞ differential 1-form ζ f χ ( ζ + x ,x ) ζ d ζ on C \ { } ⊃ S and the compact set S with piecewise C boundary), it follows that˜ f ( x ) = − h ( x ) = 12 π i Z C g ( ζ, x ) ζ − x d¯ ζ ∧ d ζ = 12 π i Z x + S ∂f χ ∂ ¯ z ( ζ, x ) ζ − x d¯ ζ ∧ d ζ = 12 π i Z S ∂f χ ∂ ¯ z ( ζ + x , x ) ζ d¯ ζ ∧ d ζ = 12 π i Z S ∂∂ ¯ ζ (cid:18) f χ ( ζ + x , x ) ζ (cid:19) d¯ ζ ∧ d ζ = 12 π i Z ∂S f χ ( ζ + x , x ) ζ d ζ = 12 π i Z ∂S f χ ( x + ζu ) ζ d ζ = 12 π i Z ∂G f ( x + ζu ) ζ d ζ. We thus have proved the representation formula (3). Now let us fix a linear variety L ⊂ C n as in (c). For a ∈ Ω L and u ∈ ( L − a ) \ { } , we may write ˜ f ( a ) as in (3).Since a + C · u ⊂ L , we conclude that ˜ f ( a ) only depends on the restriction f | Ω L \ K .(b). According to (a), the map f := f | Ω \ K has a unique Hartogs companion˜ f ∈ H (Ω ). For fixed a ∈ Ω , let us choose u ∈ C n \ { } and a set G ⊂ C , whichis ( a, u )-admissible for the inclusion K ⊂ Ω , and hence also for K ⊂ Ω. By usingthe representation formula (3) for both ˜ f and ˜ f it follows that ˜ f ( a ) = ˜ f ( a ). As a was arbitrary, we conclude that ˜ f | Ω = ˜ f .(d). Assume K c = C n \ K is connected. We claim that for every ω ∈ Υ, the set C n \ K ω is connected. To show this, let us fix ω ∈ Υ. There is no restriction inassuming Υ = { ω } (otherwise, ω = Ω and C n \ K ω = K c is connected). We have C n \ K ω = K c ∪ ω c = K c ∪ Ω c ∪ (Ω \ ω ) = K c ∪ (Ω \ ω ) = [ ω ′ ∈ Υ \{ ω } ( K c ∪ ω ′ ) . For every ω ′ ∈ Υ, the set K c ∪ ω ′ is connected, since so are both K c and ω ′ , and K c ∩ ω ′ = ω ′ \ K = ∅ . Since T ω ′ ∈ Υ \{ ω } ( K c ∪ ω ′ ) ⊃ K c = ∅ , the set C n \ K ω isconnected. Our claim is proved. Hence ω ∩ K u ω = ω \ K ω = ω \ K . As ω wasarbitrary, we get C = S ω ∈ Υ ( ω \ K ) = Ω \ K , which yields ˜ f | Ω \ K = f , by (1). (cid:3) Remark . Since Ω ∩ K u ⊂ C K, Ω , the map ˜ f from Theorem 3(a)also satisfies the simpler, but weaker condition ˜ f | Ω ∩ K u = f | Ω ∩ K u , (6)where K u denotes the unbounded component of C n \ K . The above condition yieldsthe uniqueness of ˜ f , if and only if ω ∩ K u = ∅ for every component ω ∈ Υ Ω . Forinstance, with the notation B r := B C n (0 , r ) for r >
0, the open sets ω := B and ω := B \ B are the components of Ω := ω ∪ ω , which contains the compact set K := ∂B . We have ω ∩ K u ω = B and ω ∩ K u ω = Ω ∩ K u = B \ B . Hence forarbitrarily given f ∈ H (Ω \ K ), the condition (6) determines ˜ f | ω , but not ˜ f | ω . Remark . By Theorem 3(a), the
Hartogs companion operator H C : H (Ω \ K ) → H (Ω) , H C ( f ) = ˜ f , The conditions (1) and (6) coincide if Ω is connected.
ARTOGS COMPANIONS AND HOLOMORPHIC EXTENSIONS 7 is a morphism of C -algebras and a left inverse for the restriction operator ρ definedas in Theorem 2, that is, g ρ ( g ) = g, for every g ∈ H (Ω) . Proposition 6 (composition property of Hartogs companions) . Let an open set D ⊂ C and a map g ∈ H ( D ). Then ^ ( g ◦ f ) = g ◦ ˜ f , for every f ∈ H (Ω \ K ) with f (Ω \ K ) ⊂ D. Proof.
Let f ∈ H (Ω \ K ) as in the claimed property. Hence ˜ f (Ω) ⊂ f (Ω \ K ) ⊂ D by (2), and so g ◦ f ∈ H (Ω \ K ) and g ◦ ˜ f ∈ H (Ω). Since (cid:0) g ◦ ˜ f (cid:1) | C dK, Ω = ( g ◦ f ) | C dK, Ω ,by the uniqueness of the Hartogs companion we conclude that ^ ( g ◦ f ) = g ◦ ˜ f . (cid:3) We next establish some relevant equivalences for the connectedness assumptionsused by Theorems 1, 2 and 3(d).
Proposition 7 (connectedness conditions) . We have the equivalencesΩ \ K is connected ⇐⇒ C n \ K and Ω are connected . (7) C n \ K is connected ⇐⇒ C n \ K ω is connected for every ω ∈ Υ Ω (8) ⇐⇒ ω \ K is connected for every ω ∈ Υ Ω . (9) Proof.
There is no restriction in assuming K = ∅ . Throughout this proof the threeequivalences from the proposition will be referred to as “ ( i ) ⇔ ”, with i ∈ { , , } .“ ( ) ⇒ ”. Assume Ω \ K is connected. It is easily seen that for all a ∈ Ω and b ∈ C n \ K ,there exist a ′ , b ′ ∈ Ω \ K , such that [ a, a ′ ] ⊂ Ω and [ b, b ′ ] ⊂ C n \ K . This yieldsthe connectedness of both sets Ω and C n \ K , since in Ω \ K any to points can bejoined by a polygonal chain.“ ( ) ⇐ ”. On the contrary, assume both C n \ K and Ω are connected, but Ω \ K isnot. Thus Ω \ K = Ω ∪ Ω , for some disjoint nonempty open sets Ω , Ω ⊂ Ω. ByTheorem 3(d), the map f ∈ H (Ω \ K ) defined by f | Ω j ≡ j ( j ∈ { , } ) has a uniqueextension ˜ f ∈ H (Ω). For j ∈ { , } , since Ω is connected and ˜ f | Ω j = f | Ω j ≡ j , bythe identity theorem it follows that ˜ f ≡ j . We thus get 1 ≡ ˜ f ≡
2, a contradiction.We conclude that Ω \ K is connected. The first equivalence of the proposition isproved. This also yields “ ( ) ⇔ ”, since every ω ∈ Υ Ω is connected and ω \ K = ω \ K ω .“ ( ) ⇒ ”. If C n \ K is connected, then so is C n \ K ω for every component ω ∈ Υ Ω , asalready shown by the first part (the claim) of the proof of Theorem 3(d).“ ( ) ⇐ ”. Assume C n \ K ω is connected for every ω ∈ Υ Ω . By the third equivalence,all ω \ K are connected. Let us fix a, b ∈ C n \ K . Let a polygonal chain Λ ⊂ C n joining a to b . The set F Λ := { ω ∈ Υ Ω | Λ ∩ K ω = ∅} is finite, since the compactΛ ∩ K ⊂ Ω = S ω ∈ Υ Ω ω may be covered by finitely many ω ∈ Υ Ω , which are disjoint.Now let us choose Λ for which the cardinality of F Λ is the smallest possible. Weclaim that F Λ = ∅ . On the contrary, suppose there exists ω ∈ F Λ = ∅ . There is apath γ : [0 , → C n , such that γ (0) = a , γ (1) = b , and γ ([0 , ∩ K ω iscompact, ω is open, and a, b ∈ C n \ K ω , there exist t, s ∈ ]0 , t < s and γ ( t ) , γ ( s ) ∈ ω \ K, γ ([0 , t ]) ∪ γ ([ s, ⊂ C n \ K ω . As ω \ K is connected, there is a polygonal chain Λ ω ⊂ ω \ K joining γ ( t ) to γ ( s ).The polygonal chain Ψ = γ ([0 , t ]) ∪ Λ ω ∪ γ ([ s, a to b and F Ψ ⊂ F Λ \ { ω } , VLAD TIMOFTE which contradicts the choice of Λ. Hence F Λ = ∅ , that is, Λ ⊂ C n \ K . We thusconclude that C n \ K is connected. (cid:3) Summarizing, we can now unify Theorems 1, 2 and 3(a,d) as follows:
Theorem 8 (Hartogs extension/Kugelsatz in C n ) . Let n ≥ , an open set Ω ⊂ C n ,and a compact subset K ⊂ Ω . The following four statements are equivalent. (i): Every map f ∈ H (Ω \ K ) has a (unique) extension ˜ f ∈ H (Ω) . (i’): Every locally constant map g : Ω \ K → C has an extension ˜ g ∈ H (Ω) . (ii): The restriction ρ : H (Ω) → H (Ω \ K ) is an isomorphism of C -algebras. (iii): C n \ K is connected.In this case the isomorphisms ρ and H C are range preserving, that is, f (Ω) = f (Ω \ K ) , for every f ∈ H (Ω) . Proof.
Since the implications (iii) ⇒ (ii) ⇒ (i) ⇒ (i’) are clear, we only need to provethat (i’) ⇒ (iii). Assume (i’) holds, but C n \ K is disconnected. By Proposition 7, ω \ K = ω ∪ ω for some ω ∈ Υ Ω and disjoint nonempty open sets ω , ω ⊂ ω \ K . Leta locally constant map g : Ω \ K → C , such that g | ω j ≡ j for j ∈ { , } . Accordingto (i’), g has an extension ˜ g ∈ H (Ω). Since ω is connected, by the identity theoremwe get 1 ≡ ˜ g | ω ≡
2, a contradiction. Hence C n \ K is connected. We thus haveproved the equivalence of the four statements. By Theorem 3(a) we see that every f ∈ H (Ω) is the Hartogs companion of its restriction h := f | Ω \ K , which leads by(2) to f (Ω) = ˜ h (Ω) ⊂ h (Ω \ K ) = f (Ω \ K ) ⊂ f (Ω). Hence f (Ω) = f (Ω \ K ). (cid:3) Remark . The existence of a holomorphic extension for every map from H (Ω \ K )depends solely on the connectedness of C n \ K , and hence on the compact set K (the larger open set Ω ⊃ K is irrelevant!). Meanwhile, Hartogs companions maybe considered regardless of the existence of holomorphic extensions. Therefore, theHartogs phenomenon for holomorphic maps is a special case of existence of Hartogscompanions.2.2. Hartogs companions in dimension one.
The 1-companions defined belowwill be used for proving a very general Hartogs-type extension result (Theorem 44).
Definition 10 (Hartogs 1-companion) . Let an open set Ω ⊂ C and a compactsubset K ⊂ Ω. Any map f ∈ H (Ω \ K ) has a Hartogs -companion ˜ f ∈ H (Ω)defined as follows: for every bounded open set D ⊂ Ω with the boundary ∂D ⊂ Ωconsisting of finitely many piecewise C Jordan curves, ˜ f | D is given by˜ f ( z ) := 12 π i Z ∂D f ( ζ ) ζ − z d ζ, for every z ∈ D, (10)where ∂D is oriented such that D lies to the left of ∂D .The definition is consistent, since for every fixed a ∈ Ω, the integral defining˜ f ( a ) does not depend of the choice of the set D ⊃ K ∪ { a } . Indeed, let us consideranother bounded open set D ⊂ Ω with the boundary ∂D ⊂ Ω as in Definition 10,and such that D ⊂ D . For the cycles ∂D , ∂D ⊂ Ω a := Ω \ ( K ∪ { a } ) (we may The inverse of ρ is the Hartogs companion operator H C from Remark 5. ARTOGS COMPANIONS AND HOLOMORPHIC EXTENSIONS 9 view these as cycles by identifying any closed path γ : [0 , → Ω a whose restrictionto [0 ,
1[ is injective, with the Jordan curve γ ([0 , ⊂ Ω a ), it is easily seen thatInd ∂D ( z ) − Ind ∂D ( z ) = (cid:26) , z ∈ D \ D, , z ∈ C \ ( D \ D ) ⊃ C \ Ω a . Applying Cauchy’s theorem (the version from Rudin [6], Th. 10.35, p.218) for themap g ∈ H (Ω a ) defined by g ( ζ ) = f ( ζ ) ζ − a yields R ∂D g ( ζ )d ζ = R ∂D g ( ζ )d ζ . For anybounded open set D ⊂ Ω with the boundary ∂D ⊂ Ω as in Definition 10, choosing D ⊃ D ∪ D now forces R ∂D g ( ζ )d ζ = R ∂D g ( ζ )d ζ = R ∂D g ( ζ )d ζ . This proves theconsistency of the definition of the Hartogs 1-companion ˜ f .For arbitrarily fixed a ∈ Ω and D ⊃ K ∪{ a } as in Definition 10, the set G := D − a is ( a, f | Ω G, may be represented as in (3), with u = 1 ∈ C (such an integral representation is a common feature of the Hartogscompanions in any dimension).Let us note that whenever f has a holomorphic extension to Ω, that is ˜ f .The Hartogs 1-companion of a holomorphic map with finitely many singularitiesis the regular part of the map (obtained from it by subtracting the sum of theprincipal parts of all Laurent series about its singularities): Example 11.
Let an open set Ω ⊂ C , a finite subset S ⊂ Ω, and f ∈ H (Ω \ S ).For every singularity s ∈ S , let f s ∈ H ( C \ { s } ) denote the map defined by theprincipal part of the Laurent series of f about s . Then˜ f = f − X s ∈ S f s on Ω \ S. Indeed, for g := f − P s ∈ S f s ∈ H (Ω \ S ), all singularities from S are removable.Hence for arbitrarily fixed z ∈ Ω \ S and open set D ⊃ S ∪ { z } as in Definition 10,by Cauchy’s integral formula it follows that˜ f ( z ) = g ( z ) + 12 π i X s ∈ S Z ∂D f s ( ζ ) ζ − z d ζ = g ( z ) , because every map ζ f s ( ζ ) ζ − z belongs to H ( C \ { z, s } ) and has residue 0 at ∞ .For 1-companions, coincidence sets or range inclusions as in Theorem 3 do notexist. For instance, for f ∈ H ( C \ { } ) defined by f ( z ) = z , according to the aboveexample we have ˜ f ≡
0, but ˜ f ( C ) ∩ f ( C \ { } ) = ∅ .3. Hartogs companions in arbitrary dimension
For complex spaces of infinite dimension, various holomorphy types have beenconsidered, however, all imply Gˆateaux holomorphy (see Definition 37). Hence anyHartogs-type extension theorem provides at least a Gˆateaux holomorphic extension,which may have some regularity, depending on that of the extended map. Therefore,we first prove extension results for the weakest holomorphy type (Gˆateaux), andthen we show the regularity of such extensions. The extension results will followfrom the existence of vector-valued Hartogs companions in arbitrary dimension.
Setting 2.
From now on, X denotes a complex vector space with dim C ( X ) ≥ ,and Y = { } a sequentially complete Hausdorff complex locally convex space. In order to preserve generality, we avoid considering additional structures on X . Nonetheless, the main results will be illustrated by corollaries stated in aparticular, but more familiar setting: for sets in (and maps between) topologicalvector spaces. We avoid requiring topological compactness (as in Theorems 1,2,8) ininfinite dimension, since in such Hausdorff spaces compact sets have empty interior.3.1. Linear cuts, related topologies, and the identity theorem.
We nextdefine some appropriate notions; these derive from the vector space structure of X and will allow us to state our results in full generality.On linear varieties of finite dimension we always consider the (unique) Euclideantopology. For every d ∈ N ∗ , let Γ d ( X ) denote the set of all complex linear varieties L ⊂ X of dimension dim C ( L ) = d . For arbitrary integers n ≥ d ≥
1, setΓ d,n ( X ) := n [ k = d Γ k ( X ) , Γ d, ∞ ( X ) := [ k ≥ d Γ k ( X ) . A subset A ⊂ X is said to be parallel to u ∈ X (we write this as A k u ), if and onlyif A + C · u = A . If L ⊂ X is a linear variety, L k u is equivalent to L + u = L . Definition 12 (linear cuts and related notions) . Let d ∈ N ∗ and A ⊂ X . (a): For every linear variety L ⊂ X , the intersection A L := A ∩ L is calledthe L -cut of A . If L ∈ Γ d ( X ), we also call A L a d -cut of A , and if a ∈ A L ,then A L is said to be a d -cut of A through a . The set A is called (i): d -open , if and only if every cut A L with L ∈ Γ ,d ( X ) is open in L .In the same way we define d -closed/bounded/compact sets. (i’): finitely open , if and only if A is d -open for every d ∈ N ∗ . We define finitely closed/bounded/compact sets in the same way. (ii): polygonally connected , if and only if any to points from A can bejoined by a polygonal chain Λ ⊂ A . (b): A map f : A → Y is called d -continuous , if and only if the restriction f | A L is continuous, for every L ∈ Γ ,d ( X ). Set C ( d ) ( A, Y ) := { f : A → Y | f is d -continuous } , C ( d ) ( A ) := C ( d ) ( A, C ) . The above notions do not require any topological structure on X . Nonetheless,a subset A ⊂ X is d -open/closed, if and only if A is open/closed in the translationinvariant topology τ d defined by all d -open subsets of X . Thus X becomes atopological space, which will be denoted by X ( d ) . We have X ( d ) = lim −→ L ∈ Γ ,d ( X ) L ,where the inductive limit is considered in the category of topological spaces. Anysubset A ⊂ X may be considered as a topological subspace A ( d ) of X ( d ) . We have C ( A ( d ) , Y ) ⊂ C ( d ) ( A, Y ) , with equality if A is d -open or d -closed.On X we may also consider the finite open topology τ f of the inductive limit X f := lim −→ L ∈ Γ , ∞ ( X ) L , whose open sets are the finitely open sets. The finite opentopology is translation invariant and the multiplication C × X f → X f is continuous(see Herv´e [4], Prop. 2.3.4, p.37). To cover the case when d > dim C ( X ), we need to consider cuts of dimension d and lower. ARTOGS COMPANIONS AND HOLOMORPHIC EXTENSIONS 11
Remark . In X f the origin has a neighborhood base consisting of balanced (henceconnected) finitely open sets.For finitely open sets τ f -connectedness is equivalent to polygonal connectedness,and components of finitely open sets are finitely open.In Hausdorff topological vector spaces, open/closed/bounded/compact sets arefinitely open/closed/bounded/compact, but the converse is false (Example 14).Even in normed spaces, finitely open sets may not be open, and finitely compactsets may not be closed or bounded: Example 14. (a):
For arbitrary infinite set T , let us consider the direct sumnormed space (cid:0) C ( T ) , k k ∞ (cid:1) , a map ρ : T → ]0 , ∞ [, and the setsΩ ρ = (cid:8) u ∈ C ( T ) (cid:12)(cid:12) | u | < ρ pointwise (cid:9) ⊂ C ( T ) ,K ρ = (cid:8) u ∈ C ( T ) (cid:12)(cid:12) | u | ≤ ρ pointwise (cid:9) ⊂ Ω ρ . Then Ω ρ is finitely open and finitely bounded. Its subset K ρ is closed, andhence finitely compact. If inf ρ ( T ) = 0, then ˚Ω ρ = ∅ , and so Ω ρ is not open.If sup ρ ( T ) = ∞ , then K ρ is unbounded. (b): For arbitrary p ∈ ]0 , ∞ ] and A ⊂ B := B C (0 , ℓ p C (which is a Banach space, if p ≥ K A = { ( z n ) n ∈ N | z ∈ A } ⊂ ℓ p C is finitely compact (all d -cuts of K A are finite sets, for every d ∈ N ∗ ), and ℓ p C \ K A is finitely open. Furthermore, K A is closed (resp. bounded), if andonly if A ∩ B ⊂ A (resp. A ⊂ B ). For A = B ∩ Q , the set K A is finitelycompact, but neither closed, nor bounded, and hence ℓ p C \ K A is not open. (c): For arbitrary integers n > d ≥
1, polynomials p , . . . , p d +1 ∈ C [ Z ] \ C with mutually distinct degrees and deg p = 1, and set A ⊂ C , the subset K A = { ( p ( z ) , . . . , p d +1 ( z ) , , . . . , ∈ C n | z ∈ A } ⊂ C n is d -compact (all d -cuts of K A are finite sets), and C n \ K A is d -open.Furthermore, K A is ( d + 1)-closed (resp. ( d + 1)-bounded), if and only if A is closed (resp. bounded). For A = Q , the set K Q is d -compact, but neither( d + 1)-closed, nor ( d + 1)-bounded, and hence C n \ K Q is not ( d + 1)-open.The above (with d ≥ K ⊂ Ω as in Theorem 33.
Definition 15 (Gˆateaux holomorphy, holomorphy, the topologies τ k( d ) and τ k ) .(a): A map f : Ω → Y defined on a 1-open set Ω ⊂ X is called Gˆateauxholomorphic , if and only if for all a ∈ Ω, v ∈ X , and ϕ ∈ Y ∗ (the continuousdual of Y ), there exists r >
0, such that the map B C (0 , r ) ∋ λ ( ϕ ◦ f )( a + λv ) ∈ C is holomorphic. Let H G (Ω , Y ) denote the complex vector space consistingof all such Y -valued maps on Ω. Set H G (Ω) := H G (Ω , C ). Obviously, f ∈ H G (Ω , Y ) ⇐⇒ ϕ ◦ f ∈ H G (Ω) for every ϕ ∈ Y ∗ . We take by convention z = 1 for every z ∈ C . The usual definition of Gˆateaux holomorphy requires the set Ω to be finitely open. (b):
If Ω is an open subset of a Hausdorff complex locally convex space, wemay consider the vector spaces of all holomorphic functions H (Ω , Y ) := { f ∈ H G (Ω , Y ) | f is continuous } , H (Ω) := H (Ω , C ) . (c): For linear variety L ⊂ X and 1-open subset D ⊂ L , we define in thenatural way the vector spaces H G ( D, Y ) and H G ( D ). For L ∈ Γ , ∞ ( X )and open subset D ⊂ L , we may write these spaces as H ( D, Y ) and H ( D ),respectively. (d): Let a fixed integer d ≥
2. If Ω ⊂ X is d -open, then H G (Ω , Y ) k( d ) isa sequentially complete Hausdorff locally convex space, with the topology τ k( d ) defined by the seminorms p M : H G (Ω , Y ) → R + , p M ( f ) = sup x ∈ M p ( f ( x )) = max p ( f ( M )) , considered for all continuous seminorms p : Y → R + and compact subsets M ⊂ Ω L , with L ∈ Γ ,d ( X ). If Y is complete, then so is H G (Ω , Y ) k( d ) . (e): If Ω ⊂ X is finitely open, then H G (Ω , Y ) k is a sequentially completeHausdorff locally convex space, with the topology τ k of uniform convergenceon all finite dimensional compact subsets M ⊂ Ω. This topology is definedby all seminorms as in (d), with L ∈ Γ , ∞ ( X ). If Y is complete, then so is H G (Ω , Y ) k .If Ω ⊂ X is d -open, then H G (Ω , Y ) ⊂ C ( d ) (Ω , Y ) (Dineen [1], Lemma 2.3, p.54),and so the seminorms p M from the above definition are well-defined. Every map f ∈ H G (Ω , Y ) is Gˆateaux differentiable, that is, the limit lim C ∋ λ → f ( a + λv ) − f ( a ) λ exists in Y for all a ∈ Ω and v ∈ X (Dineen [2], Lemma 3.3, p.149). The vectorspace H G (Ω , Y ) is also a H G (Ω)-module.The construction of Hartogs companions in arbitrary dimension (Theorem 33)and the proofs of the generalized Kugelsatz and of several Hartogs-type extensionresults require a more special identity theorem. Theorem 16 (identity) . Let a polygonally connected -open set Ω ⊂ X and asubset C ⊂ Ω , such that C − c is real-absorbing for some c ∈ C . Then f (Ω) − f ( c ) ⊂ Sp( f ( C ) − f ( c )) , for every f ∈ H G (Ω , Y ) (in particular, f ≡ if and only if f | C ≡ ).Proof. Let us first prove the equivalence. Assume f | C ≡
0. For c ∈ C as in thetheorem, let us consider a linear segment [ c, a ] ⊂ Ω. We claim that f | A ≡ A ⊂ Ω, such that a ∈ A (then A − a is real-absorbing). Set A := { x ∈ X | [ c, x ] ⊂ Ω } . Hence a ∈ A ⊂ Ω. Suppose there exists a 1-cut A L , which is not open in L ∈ Γ ( X ).Consequently, there exists a sequence ( x n ) n ∈ N ⊂ L \ A , which converges in L tosome a ∈ A L ⊂ A . Therefore, [ c, a ] ⊂ Ω. For every n ∈ N we have x n / ∈ A , that is, ξ n := (1 − t n ) c + t n x n / ∈ Ω for some t n ∈ [0 , n →∞ t n = s ∈ [0 , L ∪ { c } ⊂ L ′ for some L ′ ∈ Γ ( X ). As Ω L ′ is open in L ′ and ( ξ n ) n ∈ N ⊂ L ′ \ Ω, a passage to the limit in L ′ yields (1 − s ) c + sa ∈ L ′ \ Ω, which contradicts [ c, a ] ⊂ Ω. We conclude that Every linear variety is a translated vector subspace. For every x ∈ X , there exists ε >
0, such that [0 , ε ] · x ⊂ C − c (this holds if C is 1-open). ARTOGS COMPANIONS AND HOLOMORPHIC EXTENSIONS 13 A is 1-open, and hence that A − a is a real-absorbing set. In order to show that f | A ≡
0, let us fix x ∈ A and θ ∈ Y ∗ R . Thus [ c, x ] ⊂ Ω and the map g : [0 , → R , g ( t ) = ( θ ◦ f )((1 − t ) c + tx ) , is real-analytic. As C − c is a real-absorbing set, we have [0 , ε ] · ( x − c ) ⊂ C − c for some ε ∈ ]0 , g | [0 ,ε ] ≡
0. By the identity theorem for real-analytic maps we get g | [0 , ≡
0. It follows that θ ( f ( x )) = 0 for every θ ∈ Y ∗ R , which yields f ( x ) = 0. Ourclaim is proved. Since Ω is polygonally connected, an easy induction (on the numberof linear segments from a polygonal chain in Ω joining c to other points x ∈ Ω) basedon the above claim shows that f ≡
0. We thus have proved the equivalence. In orderto show the inclusion for f (Ω), let the map g := f − f ( c ) ∈ H G (Ω , Y ), the closedvector subspace Y := Sp( g ( C )) ⊂ Y , the quotient Hausdorff locally convex space Y /Y = { ˆ y | y ∈ Y } , and the standard continuous linear surjection s : Y → Y /Y .Since s ◦ g ∈ H G (Ω , Y /Y ) and ( s ◦ g ) | C ≡ ˆ0, by the already proved equivalence weget s ◦ g ≡ ˆ0, that is, g (Ω) ⊂ Y . This yields f (Ω) − f ( c ) ⊂ Sp( f ( C ) − f ( c )). (cid:3) The identity theorem now allows us to prove a much more general vector-valuedversion of the maximum modulus principle.
Theorem 17 (maximum norm principle) . Assume Y is a strictly convex normedspace. Let a polygonally connected -open set Ω ⊂ X and a map f ∈ H G (Ω , Y ) .If k f ( · ) k has a τ (1) -local maximum, then f is constant.Proof. We have divided the proof into two steps.
Step 1 . Let us first show the theorem for Y = C . For f ∈ H G (Ω), assume that | f | has a τ (1) -local maximum at c ∈ Ω. Set Γ c = { L ∈ Γ ( X ) | c ∈ L } . For every L ∈ Γ c , we have | f ( c ) | = max x ∈ C L | f ( x ) | for some connected open neighborhood C L ⊂ Ω L of c in L . As f | C L ∈ H ( C L ), by the classical maximum modulus principlewe deduce that f | C L ≡ f ( c ). Set C := S L ∈ Γ c C L ⊂ Ω. Since C − c is an absorbingset and f | C ≡ f ( c ), Theorem 16 now yields f ≡ f ( c ). Step 2 . In the general case, assume k f ( · ) k has a τ (1) -local maximum at c ∈ Ω andset y := f ( c ). There is a 1-open neighborhood C ⊂ Ω of c in X (1) , such that k f ( x ) k ≤ k y k , for every x ∈ C. (11)Hence C − c is an absorbing set. According to the Hahn-Banach theorem, thereexists ϕ ∈ Y ∗ \ { } , such that ϕ ( y ) = k y k and | ϕ ( y ) | ≤ k y k for every y ∈ Y . Set g := ϕ ◦ f ∈ H G (Ω). For every x ∈ C , we have | g ( x ) | = | ϕ ( f ( x )) | ≤ k f ( x ) k ≤ k y k = | ϕ ( y ) | = | g ( c ) | , and so | g | has a τ (1) -local maximum at c . By the already proved theorem for Y = C it follows that g ≡ g ( c ), that is, f (Ω) ⊂ y + ker ϕ . This together with (11) and thestrict convexity of Y lead to f ( C ) ⊂ ( y + ker ϕ ) ∩ B Y (0 , k y k ) = { y } , and henceto f | C ≡ y . By Theorem 16 we conclude that f ≡ y . (cid:3) In the above theorem the strict convexity of Y cannot be dropped: Example 18.
Let us consider a complex vector space Y of dimension at least 2,a norm ν : Y → R + , a vector y ∈ Y and ϕ ∈ Y ∗ such that ϕ ( y ) = 1, andthe equivalent norm on Y defined by k y k = max { ν ( y − ϕ ( y ) y ) , | ϕ ( y ) |} . For fixed n ∈ N ∗ , let T ∈ L ( C n , ker ϕ ) \ { } and f ∈ H ( C n , Y ) defined by f ( x ) = T ( x ) + y . Here k f ( · ) k denotes the map Ω ∋ x
7→ k f ( x ) k ∈ R + . Then k f ( · ) k = max { ν ◦ T, } has a local maximum and a global minimum at c = 0.Nonetheless, both maps f and k f ( · ) k are nonconstant.Without the strict convexity of Y from Theorem 17, we still can show that if k f ( · ) k has a τ (1) -local maximum c ∈ Ω, then c is also a global minimum and f (Ω)has empty interior (and therefore is not a domain), and if f ( c ) = 0, then f vanishesnowhere on Ω. This still holds with the norm replaced by a continuous seminorm: Theorem 19 (max-min seminorm principle) . Let a polygonally connected -openset Ω ⊂ X , a map f ∈ H G (Ω , Y ) , and a continuous seminorm p on Y . Assumethat c ∈ Ω is a τ (1) -local maximum for p ◦ f . Then c is also a global minimum and ˚ f (Ω) = ∅ . In particular, if p ( f ( c )) > , then / ∈ f (Ω) .Proof. Let us consider the quotient normed space ˆ Y := Y /p − ( { } ) = { ˆ y | y ∈ Y } with the usual norm k ˆ y k = p ( y ), the standard linear surjection s : Y → ˆ Y , the mapˆ f := s ◦ f ∈ H G (Ω , ˆ Y ), and y := f ( c ). Thus k ˆ f ( · ) k has a τ (1) -local maximum at c .As in the proof of Theorem 17 (Step 2) we choose ϕ ∈ ˆ Y ∗ \ { } , for which we showthat ˆ f (Ω) ⊂ ˆ y + ker ϕ . Consequently, for every x ∈ Ω we have p ( f ( x )) = k ˆ f ( x ) k ≥ | ϕ ( ˆ f ( x )) | = | ϕ (ˆ y ) | = k ˆ y k = p ( y ) . Hence p ◦ f has a global minimum at c . The above inclusion for ˆ f (Ω) is equivalentto f (Ω) ⊂ y + ker( ϕ ◦ s ). As ϕ ◦ s ∈ Y ∗ \ { } , we have ˚ f (Ω) = y + ˚ker( ϕ ◦ s ) = ∅ .If p ( f ( c )) >
0, then 0 / ∈ p ( f (Ω)), and so 0 / ∈ f (Ω). (cid:3) Corollary 20 (minimum modulus principle) . Let a polygonally connected 2-openset Ω ⊂ X and a map f ∈ H G (Ω). If | f | has a τ (1) -local minimum at c ∈ Ω and f ( c ) = 0, then f is constant. Proof.
Let us consider Γ c as in the proof of Theorem 17. For every L ∈ Γ c , wehave | f ( c ) | = min x ∈ C L | f ( x ) | for some connected open neighborhood C L ⊂ Ω L of c in L . As g := f | CL ∈ H ( C L ) and | g | has a maximum at c , by the classicalmaximum modulus principle we get g ≡ g ( c ), that is, f | C L ≡ f ( c ). As in the proofof Theorem 17 (Step 1) it follows that f ≡ f ( c ). (cid:3) Range inclusions and inertia.
The next result on automatic extensions willbe used for proving the range inclusions for Hartogs companions and the continuityof the Hartogs linear operator, as well as for obtaining boundary principles.
Proposition 21 (range inclusions for Gˆateaux holomorphic extensions) . Let two1-open sets Ω , Ω ⊂ X and C ⊂ Ω ∩ Ω. Assume that for every f ∈ H G (Ω , Y ),the restriction f | C has a unique extension ¯ f ∈ H G (Ω , Y ). Then (a): The same extension property holds for C -valued maps and( h · f ) = ¯ h · ¯ f for all h ∈ H G (Ω ) , f ∈ H G (Ω , Y ) , ¯ h (Ω) ⊂ h (Ω ) for every h ∈ H G (Ω ) , ¯ f (Ω) ⊂ co( f (Ω )) for every f ∈ H G (Ω , Y ) . (b): If Y is a unital Banach algebra (not necessarily commutative), then H : H G (Ω , Y ) → H G (Ω , Y ) , H ( f ) = ¯ f , This condition is similar to that from Definition 23(b).
ARTOGS COMPANIONS AND HOLOMORPHIC EXTENSIONS 15 is a unital C -algebra morphism. For every f ∈ H G (Ω , Y ), we have f (Ω ) ⊂ U( Y ) = ⇒ ¯ f (Ω) ⊂ U( Y ) , where U( Y ) denotes the (open) set of all invertible elements of Y . Proof. (a). The proof is divided into three steps.
Step 1 . We first prove the existence and uniqueness property for C -valued maps.Let us fix h ∈ H G (Ω ). Choose y ∈ Y \ { } and ϕ ∈ Y ∗ , such that ϕ ( y ) = 1. For f := h · y ∈ H G (Ω , Y ) and the unique corresponding map ¯ f ∈ H G (Ω , Y ) as in thehypothesis, set ¯ h := ϕ ◦ ¯ f ∈ H G (Ω). Since ϕ ◦ f = h , it follows that¯ h | C = ϕ ◦ ¯ f | C = ϕ ◦ f | C = ( ϕ ◦ f ) | C = h | C . The existence of ¯ h ∈ H G (Ω) is proved. For the uniqueness part, let g ∈ H G (Ω),such that g | C = h | C . For g y := g · y ∈ H G (Ω , Y ), we see that g y | C = g | C · y = h | C · y = f | C = ¯ f | C . By the uniqueness of ¯ f we get g y = ¯ f , which yields g = ϕ ◦ g y = ϕ ◦ ¯ f = ¯ h . Hence¯ h is unique. We thus have proved the claimed existence and uniqueness property.For arbitrary h ∈ H G (Ω ) and f ∈ H G (Ω , Y ), and the unique corresponding maps¯ h ∈ H G (Ω) and ¯ f ∈ H G (Ω , Y ), we have ( h · f ) | C = ( h · f ) | C = (¯ h · ¯ f ) | C , which leadsby the uniqueness property to ( h · f ) = ¯ h · ¯ f . Step 2 . We next show the inclusion for ¯ h (Ω) ( h ∈ H G (Ω )). Let us fix λ ∈ C \ h (Ω ).Thus g := h − λ ∈ H G (Ω ) and f := ( h − λ ) g ≡
1. For the unique correspondingmaps ¯ h, ¯ g, ¯ f ∈ H G (Ω) and for F := (¯ h − λ )¯ g ∈ H G (Ω), we have F | C = f | C = ¯ f | C .Since f ≡
1, the uniqueness of ¯ f yields F = ¯ f ≡
1. Hence λ ∈ C \ ¯ h (Ω). As λ wasarbitrary, we conclude that ¯ h (Ω) ⊂ h (Ω ). Step 3 . We next show the inclusion for ¯ f (Ω) ( f ∈ H G (Ω , Y )). According to Step 1,for every ϕ ∈ Y ∗ , the unique map ¯ h ∈ H G (Ω) with the property that ¯ h | C = ( ϕ ◦ f ) | C is ¯ h := ϕ ◦ ¯ f . By the inclusion from Step 2 we see that ϕ ( ¯ f (Ω)) = ¯ h (Ω) ⊂ ϕ ( f (Ω )).Hence θ ( ¯ f (Ω)) ⊂ θ ( f (Ω )) for every θ ∈ Y ∗ R (the continuous dual of Y consideredas a real locally convex space Y R ), because Y ∗ R = { Re( ϕ ) | ϕ ∈ Y ∗ } . Now by theHahn-Banach separation theorem it follows that ¯ f ( a ) ∈ co( f (Ω )) for every a ∈ Ω,that is, ¯ f (Ω) ⊂ co( f (Ω )).(b). Let e denote the identity element of Y . Since all Gˆateaux holomorphic mapsare also Gˆateaux differentiable, both H G (Ω , Y ) and H G (Ω , Y ) are unital algebras.That H is a unital algebra morphism is immediate, by the uniqueness property.Let f ∈ H G (Ω , Y ), such that f (Ω ) ⊂ U( Y ). For g ∈ H G (Ω , Y ) defined by g ( x ) = f ( x ) − , we have f · g = g · f ≡ e . By applying the morphism H we deducethat ¯ f · ¯ g = ¯ g · ¯ f ≡ e , which yields ¯ f (Ω) ⊂ U( Y ). Alternative proof of the range inclusion (2) from Theorem 3(a) . We have alreadyproved that for every f ∈ H G (Ω \ K ), there is a unique ˜ f ∈ H G (Ω) (the Hartogscompanion of f ), such that ˜ f | C K, Ω = f | C K, Ω . Since C := C K, Ω ⊂ Ω := Ω \ K ⊂ Ω,by the first inclusion from Proposition 21(a) we get ˜ f (Ω) ⊂ f (Ω \ K ). (cid:3) For vector-valued Hartogs companions the range inclusion corresponding to (2)will have three versions, depending on the dimension of Y . In order to state itwithout mentioning cases, we next make an appropriate convention. Notation 1.
For arbitrary subsets
A, B ⊂ Y , we write A ⊏ B , if and only if oneof the following three conditions holds: (a): A ⊂ co( B ) and Y has infinite dimension. (b): A ⊂ co( B ) and Y has finite dimension at least 2. (c): A ⊂ B and Y has dimension 1.The preorder “ ⊏ ” on P ( Y ) is weaker than the inclusion. If A i ⊏ B i for every i ∈ I ,then S i ∈ I A i ⊏ S i ∈ I B i .The following vector-valued version of Theorem 3 is a needed ingredient for theconstruction of Hartogs companions in arbitrary dimension. Lemma 22 (vector-valued Hartogs companion) . Let an open set Ω ⊂ C n and acompact subset K ⊂ Ω . Theorem 3 holds for vector-valued maps f ∈ H (Ω \ K, Y ) ,with all spaces of the form H ( · ) replaced by the corresponding H ( · , Y ) and with therange inclusion (2) for ˜ f ∈ H (Ω , Y ) replaced by ˜ f (Ω) ⊏ f (Ω \ K ) . (12) Proof.
The statements of the lemma corresponding to those of Theorem 3 will bereferred to as Lemma 22(a,b,c,d).(a). According to Theorem 3(a), for every ϕ ∈ Y ∗ the map f ϕ := ϕ ◦ f ∈ H (Ω \ K )has a unique Hartogs companion ˜ f ϕ ∈ H (Ω). Let u ∈ C n \ { } be fixed. For every a ∈ Ω, choose a particular ( a, u )-admissible set G a ⊂ C and define˜ f ( a ) = 12 π i Z ∂G a f ( a + ζu ) ζ d ζ ∈ Y (the integral exists, since f is continuous, ∂G a is piecewise C , and Y is sequentiallycomplete). We thus have defined a map ˜ f : Ω → Y . By Theorem 3(c) we see that˜ f ϕ ( a ) = 12 π i Z ∂G a f ϕ ( a + ζu ) ζ d ζ = ϕ (cid:0) ˜ f ( a ) (cid:1) , for all ϕ ∈ Y ∗ , a ∈ Ω . Hence ϕ ◦ ˜ f = ˜ f ϕ ∈ H (Ω) for every ϕ ∈ Y ∗ , and so ˜ f ∈ H (Ω , Y ). Since every ϕ ◦ ˜ f is the Hartogs companion of ϕ ◦ f , it follows that (1) holds. The uniqueness of˜ f ∈ H (Ω , Y ) satisfying (1) is obvious. The inclusion (12) will be proved after (d).(b). For every ϕ ∈ Y ∗ , by (a) and Theorem 3(b) we see that ϕ ◦ ˜ f | Ω is the Hartogscompanion of ϕ ◦ f | Ω \ K . Therefore, ˜ f | Ω is the Hartogs companion of f | Ω \ K .(c). By Theorem 3(c), (3) holds for all ϕ ◦ ˜ f = ˜ f ϕ . Therefore, (3) also holds for ˜ f .(d). By Theorem 3(d) we get C K, Ω = Ω \ K , which yields ˜ f | Ω \ K = f , by (1). Proof of (12). Applying Proposition 21(a) for C := C K, Ω ⊂ Ω := Ω \ K ⊂ Ωyields ˜ f (Ω) ⊂ co( f (Ω \ K )), for arbitrary dimension of Y . If dim C ( Y ) = 1, then(2) holds. For dim C ( Y ) ∈ N ∗ \ { } , in order to show that ˜ f (Ω) ⊂ co( f (Ω \ K )), letus fix a ∈ Ω. Choose the sets K ⊂ Ω as in Theorem 3(b), such that a ∈ Ω and K ⊂ ˚ K , and Ω ⊂ Ω is compact. Since ˜ f | Ω is the Hartogs companion of f | Ω \ K and Ω \ ˚ K is compact, by the already proved part of (12) we get˜ f ( a ) ∈ ˜ f (Ω ) ⊂ co( f (Ω \ K )) ⊂ co (cid:0) f (cid:0) Ω \ ˚ K (cid:1)(cid:1) = co (cid:0) f (cid:0) Ω \ ˚ K (cid:1)(cid:1) ⊂ co( f (Ω \ K )) . As a was arbitrary, we conclude that ˜ f (Ω) ⊂ co( f (Ω \ K )). (cid:3) We consider domains of holomorphy as in Herv´e [4] (Def. 5.2.1(a), p.135), butwithout the connectedness assumption on the set:
Definition 23 (domain of holomorphy) . An open subset D ⊂ Y is called a domainof holomorphy , if and only if there are no open sets C, D ⊂ Y , such that ARTOGS COMPANIONS AND HOLOMORPHIC EXTENSIONS 17 (a): ∅ 6 = C ⊂ D ∩ D and D is connected, with D D . (b): For every g ∈ H ( D ), the restriction g | C has an extension ¯ g ∈ H ( D ). Theorem 24 (range inertia) . Let an open set Ω ⊂ C n , a compact subset K ⊂ Ω ,and f ∈ H (Ω \ K, Y ) . Then for every domain of holomorphy D ⊂ Y , we have f (Ω \ K ) ⊂ D = ⇒ ˜ f (Ω) ⊂ D. (13) Proof.
Let a domain of holomorphy D ⊂ Y , such that f (Ω \ K ) ⊂ D . Assume˜ f ( b ) ∈ Y \ D for some b ∈ Ω. Let us choose sets K and Ω as in Lemma 22(b),such that b ∈ Ω and K ⊂ ˚ K , and Ω ⊂ Ω is compact. The Hartogs companionof the restriction f := f | Ω \ K is ˜ f = ˜ f | Ω , by Lemma 22(b). Since the set f (cid:0) Ω \ ˚ K (cid:1) ⊂ f (Ω \ K ) ⊂ D is compact and D is open, there is an absolutelyconvex open neighborhood V ⊂ Y of 0, such that f (Ω \ K ) + V ⊂ D . Thus f ∈ H (Ω \ K , Y ) , f ∈ H (Ω \ K ) + V ⊂ D, ˜ f ( b ) / ∈ D. By replacing Ω , K, f, ˜ f , by Ω , K , f , ˜ f , respectively, we can assume that f (Ω \ K ) + V ⊂ D. There exist a component ω ∈ Υ Ω containing b and a path γ : [0 , → ω , such that c := γ (0) ∈ ω ∩ K u ω ⊂ C K, Ω and b = γ (1). Since ˜ f ( c ) = f ( c ) ∈ D and ˜ f ( b ) ∈ Y \ D ,there exists s ∈ ]0 , f ( γ ([0 , s ])) ⊂ D and ˜ f ( γ ( s )) + V D . SetΓ := γ ([0 , s ]) ⊂ ω, a := γ ( s ) ∈ Γ , y := ˜ f ( a ) ∈ D, D := y + V D. Thus D is a connected open set and ˜ f (Γ) ⊂ D . As Γ is compact and D, ω areopen, we can choose successively an absolutely convex open neighborhood V ⊂ V of 0 and an open ball B ⊂ C n centered at the origin, such that˜ f (Γ) + 2 V ⊂ D, ω := Γ + B ⊂ ω ∩ ˜ f − ( ˜ f (Γ) + V ) . Hence ω ⊂ ω is a connected open set with ˜ f ( ω ) + V ⊂ D and C := y + V ⊂ D ∩ D is a nonempty open set. We next show that the condition from Definition 23(b)holds (which will lead to a contradiction). To this end, let us fix g ∈ H ( D ). Forevery fixed z ∈ V , we have f z := f + z ∈ H (Ω \ K, Y ) and f z (Ω \ K ) ⊂ D . As g ∈ H ( D ), we also have h z := g ◦ f z ∈ H (Ω \ K ). According to Theorem 3(a), h z hasa unique Hartogs companion ˜ h z ∈ H (Ω). Hence ˜ h z | C K, Ω = h z | C K, Ω = g ◦ ˜ f z | C K, Ω .For every z ∈ V , by the identity theorem we get ˜ h z | ω = g ◦ ˜ f z | ω (the latter iswell-defined and the two maps coincide on the open set ω ∩ C K, Ω ∋ c ). Thus˜ h z ( x ) = g (cid:0) ˜ f ( x ) + z (cid:1) , for all z ∈ V , x ∈ ω . (14)Now let us define the map¯ g : D → C , ¯ g ( y ) = ˜ h y − y ( a ) . For every y ∈ C , since z y := y − y ∈ V , by (14) and the definition of ¯ g we seethat ¯ g ( y ) = ˜ h z y ( a ) = g ( y + z y ) = g ( y ). Hence ¯ g | C = g | C . In order to show that If f has an extension from H (Ω , Y ) (which must be ˜ f ), then equivalence holds in (13). ¯ g ∈ H ( D ), let us fix u ∈ C n \ { } and an ( a, u )-admissible set G ⊂ C . By usingthe representation formula (3) (at x = a ) we deduce that¯ g ( y ) = 12 π i Z ∂G h y − y ( a + ζu ) ζ d ζ = 12 π i Z ∂G g ( f ( a + ζu ) + y − y ) ζ d ζ, (15)for every y ∈ D . Since the map h : (Ω \ K ) × V → C , h ( x, y ) = g ( f ( x ) + y − y ) , is holomorphic, it follows that ¯ g is continuous and Gˆateaux y -differentiation underthe integral sign holds in (15). Hence ¯ g ∈ H ( D ). As the existence of the sets D and C as above leads to a contradiction, we conclude that ˜ f (Ω) ⊂ D . (cid:3) Boundary principle and 2-cuts properties.
In this section we explore thefirst consequences of Lemma 22 and Theorem 24. We will be mainly using 2-cuts,since topological assumptions on these are less restrictive and the 2-boundary ofany set (defined below) is smaller.
Proposition 25 (2-compact excision) . Let a 2-open set Ω ⊂ X and f ∈ H G (Ω , Y ).Let a subset K ⊂ Ω which has particular compact 2-cuts through all points of K .Then f (Ω) ⊏ f (Ω \ K ). For every domain of holomorphy D ⊂ Y , f (Ω) ⊂ D ⇐⇒ f (Ω \ K ) ⊂ D. If Y = C , then f (Ω) = f (Ω \ K ). Proof.
Let us fix a ∈ K , together with a compact 2-cut K L through a . Since f | Ω L ∈ H (Ω L , Y ) is the Hartogs companion of its restriction f | Ω L \ K L , by (12) itfollows that f ( a ) ∈ f (Ω L ) ⊏ f (Ω L \ K L ) ⊂ f (Ω \ K ). As a was arbitrary, we concludethat f ( K ) ⊏ f (Ω \ K ), and hence that f (Ω) ⊏ f (Ω \ K ). For the equivalence, weonly need to prove the implication “ ⇐ ”. Therefore, assuming f (Ω \ K ) ⊂ D , wenext show that f ( K ) ⊂ D . Let us fix again a ∈ K , together with a compact 2-cut K L through a . Since f | Ω L is the Hartogs companion of f | Ω L \ K , by Theorem 24 wesee that f ( a ) ∈ f (Ω L ) ⊂ D . We thus conclude that f ( K ) ⊂ D . (cid:3) Corollary 26 (excision) . Assume X is a Hausdorff topological vector space. Letan open set Ω ⊂ X and a 2-bounded closed subset K ⊂ Ω. Then f (Ω) = f (Ω \ K ) , for every f ∈ H G (Ω) . Proof.
Since Ω is 2-open and K is 2-compact, we may apply Proposition 25. (cid:3) Corollary 27 (level sets) . Let a 2-open set Ω ⊂ X , a map f ∈ H G (Ω), and S ⊂ C .Then for every L ∈ Γ ( X ), the set f − ( S ) ∩ L is not nonempty and relativelycompact in Ω L . In particular, f − ( { } ) has no nonempty compact 2-cuts. Proof.
Assume ∅ 6 = f − ( S ) ∩ L ⊂ K ⊂ Ω L , for some L ∈ Γ ( X ) and compactset K ⊂ L . Since f | Ω L ∈ H (Ω L ) extends its restriction f | Ω L \ K , by Corollary 26we deduce that f ( f − ( S ) ∩ L ) ⊂ f (Ω L ) = f (Ω L \ K ) ⊂ Y \ S . It follows that f ( f − ( S ) ∩ L ) ⊂ S ∩ ( Y \ S ) = ∅ , which yields f − ( S ) ∩ L = ∅ . (cid:3) This condition on K is fulfilled in particular by 2-compact sets. ARTOGS COMPANIONS AND HOLOMORPHIC EXTENSIONS 19
For the next theorem, we define the 2 -boundary of any subset A ⊂ X by ∂ A := [ L ∈ Γ ( X ) ∂A L , where each boundary ∂A L := ∂ ( A ∩ L ) is considered in L . If X is also a topologicalvector space, then ∂ A ⊂ ∂A, A ∪ ∂ A ⊂ A. Theorem 28 (2-boundary principle) . Let a -open set Ω ⊂ X having particularbounded open -cuts through all points of Ω and f ∈ H G (Ω , Y ) ∩ C (2) (Ω ∪ ∂ Ω , Y ) .Then f (Ω) ⊏ f ( ∂ Ω) . For every continuous seminorm p : Y → R + , sup x ∈ Ω p ( f ( x )) = sup x ∈ ∂ Ω p ( f ( x )) . For every domain of holomorphy D ⊂ Y , f ( ∂ Ω) ⊂ D = ⇒ f (Ω) ⊂ D. Proof.
We first show the inclusion f (Ω) ⊏ f ( ∂ Ω). To this end, let us fix y = f ( a ),with a ∈ Ω, together with a bounded 2-cut Ω L through a . Let us choose a sequence( K n ) n ∈ N of compact subsets of Ω L , such that K n ⊂ ˚ K n +1 (the interior is taken in L ) for every n ∈ N and S n ∈ N K n = Ω L . According to Proposition 25, for every n ∈ N we have y ∈ f (Ω L ) ⊏ f (Ω L \ K n ). We need to analyze three cases. Case 1 . If dim C ( Y ) = 1, then y ∈ f (Ω L ) = f (Ω L \ K n ), and so y = f ( b n ) for some b n ∈ Ω L \ K n . In L , the bounded sequence ( b n ) n ∈ N ⊂ Ω L has a subsequence whichconverges to some b ∈ ∂ Ω L ⊂ ∂ Ω ∩ L . Since f is 2-continuous, a passage to thelimit yields y = f ( b ) ∈ f ( ∂ Ω). Hence f (Ω) ⊂ f ( ∂ Ω).
Case 2 . For Y of arbitrary dimension, by the already proved inclusion we deducethat ϕ ( f (Ω)) ⊂ ϕ ( f ( ∂ Ω)) for every ϕ ∈ Y ∗ . As in the proof of Proposition 21(a)(Step 3) we conclude that f (Ω) ⊂ co( f ( ∂ Ω)).
Case 3 . If m := dim C ( Y ) ∈ N ∗ \ { } , then y ∈ f (Ω L ) ⊂ co( f (Ω L \ K n )), and so y = P mi =0 λ i,n f ( b i,n ) for some b i,n ∈ Ω L \ K n and λ i,n ∈ [0 ,
1] (0 ≤ i ≤ m ), suchthat P mi =0 λ i,n = 1. By taking a convergent subsequence in L m +1 × [0 , m +1 , wemay assume that lim n →∞ b i,n = b i ∈ ∂ Ω ∩ L in L and lim n →∞ λ i,n = λ i ∈ [0 ,
1] for0 ≤ i ≤ m . Since f is 2-continuous, it follows that y = P mi =0 λ i f ( b i ) ∈ co( f ( ∂ Ω)).Hence f (Ω) ⊂ co( f ( ∂ Ω)). By the above cases we conclude that f (Ω) ⊏ f ( ∂ Ω).To show the equality, let a continuous seminorm p : Y → R + . By the range inclusionwe see that sup p ( f (Ω)) ≤ sup p ( f ( ∂ Ω)). Now let us fix b ∈ ∂ Ω. Hence b is thelimit in some L ∈ Γ ( X ) of a sequence ( b n ) n ∈ N ⊂ Ω L . Since f | Ω L ∪ ∂ Ω L is continuous,it follows that p ( f ( b )) = lim n →∞ p ( f ( b n )) ≤ sup p ( f (Ω L )) ≤ sup p ( f (Ω)).In order to prove the last implication, assume f ( ∂ Ω) ⊂ D , that is, ∂ Ω ⊂ f − ( D ).Set K := Ω \ f − ( D ). For every L ∈ Γ ( X ) such that Ω L is bounded, the 2-cut K L = Ω L \ f − ( D ) = Ω L \ f − ( D ) is compact in L . It follows that K satisfies thecondition from Proposition 25. Since Ω \ K ⊂ f − ( D ), by the equivalence fromProposition 25 we conclude that f (Ω) ⊂ D . (cid:3) The following version of Theorem 28 is closer to a maximum modulus principle.
Corollary 29 (boundary principle) . Assume X is a Hausdorff topological vectorspace. Let a 2-bounded open set Ω ⊂ X and a map f ∈ H G (Ω , Y ) ∩ C (Ω , Y ). Then These conditions on Ω hold in particular for 2-open sets which are 2-bounded. f (Ω) ⊏ f ( ∂ Ω). In particular, for every continuous seminorm p : Y → R + , we havesup x ∈ Ω p ( f ( x )) = sup x ∈ ∂ Ω p ( f ( x )). For every domain of holomorphy D ⊂ Y , f (Ω) ⊂ D ⇐⇒ f ( ∂ Ω) ⊂ D. If Y = C , then f (Ω) = f ( ∂ Ω).
Proof.
Since Ω is 2-open and 2-bounded, ∂ Ω ⊂ ∂ Ω, and f is 2-continuous, theconclusion follows by Theorem 28. (cid:3) Remark . If Y a unital complex Banach algebra, in the above results we mayconsider the set D = U( Y ) of all invertible elements of Y . Example 31 (2-boundary principle) . Let us consider α >
1, a normed space X with dim C ( X ) ≥
3, a nonzero linear functional ϕ : X → C , and the setsΩ := { x ∈ X | k x k < | ϕ ( x ) | α } , F := { x ∈ X | k x k = | ϕ ( x ) | α } . Then Ω is finitely open and unbounded, ∂ Ω ⊂ F , and f (Ω) ⊂ f ( F ) , for every f ∈ H G (Ω) ∩ C (Ω ∪ F ) . Vector-valued Hartogs companions in arbitrary dimension.
Remark
32 (Hartogs 1-companion) . For nonempty open set Ω ⊂ C , compact subset K ⊂ Ω, and map f ∈ H (Ω \ K, Y ), we may consider the
Hartogs 1-companion ˜ f ∈ H (Ω , Y ) defined by (10) or by (3) ( Y -valued integrals). A similar constructionis possible if we replace C by any complex line L ∈ Γ ( X ).For the construction of the Hartogs companion in arbitrary dimension we use aslicing technique with linear varieties of finite dimension. More precisely, for fixedsets K ⊂ Ω ⊂ X and map f : Ω \ K → Y , and arbitrary L ∈ Γ , ∞ ( X ), we have K L ⊂ Ω L ⊂ L . If Ω L is open, K L is compact, and f L := f | Ω L \ K L ∈ H (Ω L \ K L , Y ),then Lemma 22(a) provides a unique Hartogs companion ˜ f L ∈ H (Ω L , Y ). Then wewill show that any two Hartogs companions ˜ f L and ˜ f L agree on Ω L ∩ Ω L , andso all ˜ f L may be patched together to define a map ˜ f ∈ H G (Ω , Y ). The coincidenceset will be the union of all coincidence sets of the inclusions K L ⊂ Ω L . Notation 2.
For every inclusion M ⊂ Ω ⊂ L ∈ Γ , ∞ ( X ), with Ω open and M compact in L , as in Theorem 3(a) we may consider the coincidence set C M, Ω := [ ω ∈ Υ Ω0 ( ω ∩ M u ω ) ⊂ Ω \ M, where Υ Ω stands for the set of all components of Ω in L (endowed with itsnatural topology) and M u ω denotes the unbounded component of L \ ( M ∩ ω ). ByLemma 22(a) we deduce that every map h ∈ H (Ω \ M, Y ) has a unique Hartogscompanion ˜ h ∈ H (Ω , Y ), and that ˜ h | C M, Ω0 = h | C M, Ω0 . Theorem 33 (Hartogs companion in arbitrary dimension) . Let a fixed integer d ≥ , a d -open set Ω ⊂ X , and a d -compact subset K ⊂ Ω . Let us define the d -coincidence set of the inclusion K ⊂ Ω as C dK, Ω := [ L ∈ Γ ,d ( X ) C K L , Ω L ⊂ Ω \ K. (16) Let a map f ∈ H G (Ω \ K, Y ) . Then ARTOGS COMPANIONS AND HOLOMORPHIC EXTENSIONS 21 (a):
There exists a unique map ˜ f ∈ H G (Ω , Y ) (which will be called the Hartogscompanion of f ), such that ˜ f | C dK, Ω = f | C dK, Ω . (17) For every L ∈ Γ ,d ( X ) , the restriction ˜ f | Ω L is the Hartogs companion of f | Ω L \ K L . Furthermore, ˜ f (Ω) ⊏ f (Ω \ K ) . For every domain of holomorphy D ⊂ Y , the implication (13) holds. (b): Let a linear variety E ⊂ X of dimension (possibly infinite) at least . If K E ⊂ K ⊂ Ω ⊂ Ω E , with K -compact and Ω -open in E , then ˜ f | Ω is the Hartogs companion of f | Ω \ K . (c): For arbitrarily fixed a ∈ Ω and u ∈ X \ { } , and ( a, u ) -admissible G ⊂ C ,the set Ω G,u := { x ∈ Ω | C G,u ( x ) holds } is ( d − -open and a ∈ Ω G,u . Wehave Ω G,u + ∂G · u ⊂ Ω \ K , and the representation formula (3) holds. (d): If Ω \ K is polygonally connected, then ˜ f | Ω \ K = f .If Ω is finitely open and K is finitely compact, the theorem holds with a few changes: Γ ,d ( X ) is replaced by Γ , ∞ ( X ) , the coincidence set from (16) is defined by C K, Ω := [ L ∈ Γ , ∞ ( X ) C K L , Ω L = [ d ≥ C dK, Ω , the set Ω G,u from (c) is finitely open, and the statement (d) is replaced by (d’): If X \ K is polygonally connected, then C K, Ω = Ω \ K and ˜ f | Ω \ K = f .Proof. For every L ∈ Γ ,d ( X ), the open subset Ω L of L , contains the compact K L ,and so the map f L := f | Ω L \ K L ∈ H (Ω L \ K L , Y ) has a unique Hartogs companion˜ f L ∈ H (Ω L , Y ) as in Lemma 22(a). If Ω L = ∅ , both f and ˜ f are empty maps. Toshorten notation, we write C K L , Ω L simply as C L .(a). The uniqueness of ˜ f . If (17) holds for some map ˜ f ∈ H G (Ω , Y ), then for every L ∈ Γ ,d ( X ) we have C L ⊂ C dK, Ω ∩ Ω L , and so ˜ f | C L = f | C L = f L | C L = ˜ f L | C L . Bythe uniqueness of the Hartogs companions ˜ f L , it follows that˜ f | Ω L = ˜ f L , for every L ∈ Γ ,d ( X ) . (18)Since Ω = S L ∈ Γ ( X ) Ω L , the above condition yields the uniqueness of ˜ f . The existence of ˜ f . Let us show that a map ˜ f ∈ H G (Ω , Y ) can be defined by (18).We claim that for all varieties L , L ∈ Γ ,d ( X ) such that L := L ∩ L = ∅ , wehave ˜ f L | Ω ∩ L = ˜ f L | Ω ∩ L (the Hartogs companions “agree”). In order to prove this,let us fix such L , L , L , together with a ∈ Ω ∩ L . We next analyze two cases. Case 1 . If L = { a } , then L ∈ Γ ,d ( X ) and a ∈ Ω L . By the representation (3) fromLemma 22(c) we get ˜ f L ( a ) = ˜ f L ( a ) (both are uniquely determined by f | Ω L \ K ). Case 2 . If L = { a } , then L ∩ L = { a } 6 = L ∩ L for some L ∈ Γ ( X ), such that a ∈ L . By the conclusion of the first case we deduce that ˜ f L ( a ) = ˜ f L ( a ) = ˜ f L ( a ).Our claim is proved. Consequently, there exists a unique map ˜ f ∈ H G (Ω , Y )defined by (18). This map satisfies (17), since for every L ∈ Γ ,d ( X ) we see that˜ f | C L = ˜ f L | C L = f L | C L = f | C L . We thus have proved the existence and uniqueness Applying the theorem for another integer s ≥ s < d leads to the same map ˜ f . of ˜ f satisfying (17). By (12) and (18) it follows that˜ f (Ω) = [ L ∈ Γ ,d ( X ) ˜ f L (Ω L ) ⊏ [ L ∈ Γ ,d ( X ) f L (Ω L \ K L ) = f (Ω \ K ) . If D ⊂ Y is a domain of holomorphy and f (Ω \ K ) ⊂ D , then by Lemma 22(a) wededuce that ˜ f L (Ω L ) ⊂ D for every L ∈ Γ ,d ( X ), which yields ˜ f (Ω) ⊂ D .(c). Let us fix a ∈ Ω and u ∈ X \ { } , and an ( a, u )-admissible set G ⊂ C . As inthe proof of Theorem 3(c) we see that Ω G,u = A c ∩ B c , with A := Ω c − G · u and B := K − G c · u (where G c = C \ G and the other complements are considered in X ). We next show that A and B are ( d − L ∈ Γ ,d − ( X ).Clearly, L + C · u ⊂ L ′ for some L ′ ∈ Γ ,d ( X ). It is easy to check that A ∩ L = (Ω c − G · u ) ∩ L = (Ω c ∩ L ′ − G · u ) ∩ L,B ∩ L = ( K − G c · u ) ∩ L = ( K ∩ L ′ − G c · u ) ∩ L. In the vector space span( L ′ ), all four sets G · u , K ∩ L ′ , Ω c ∩ L ′ , G c · u are closed,and the first two are compact. Therefore, Ω c ∩ L ′ − G · u and K ∩ L ′ − G c · u areclosed in both span( L ′ ) and L ′ . Hence A ∩ L and B ∩ L are closed in L ⊂ L ′ . As L was arbitrary, we conclude that A and B are ( d − G,u is ( d − d = 2), the map f := f | Ω \ K has a uniqueHartogs companion ˜ f ∈ H G (Ω , Y ). As in the proof of Theorem 3(b), by usingthe representation formula from (c) for both ˜ f and ˜ f we deduce that ˜ f | Ω = ˜ f .(d). Assume Ω \ K is polygonally connected. Let us choose c ∈ Ω \ { } , such that δ := [1 , ∞ [ · c ⊂ X \ K . Set C := { x ∈ X | [ c, x ] ⊂ Ω \ K } . Thus c ∈ C ⊂ Ω \ K andthe set C − c is real-absorbing, since Ω is also 1-open. We claim that ˜ f | C = f | C . Inorder to prove this, let us fix x ∈ C . Clearly, ∆ := [ c, x ] ∪ δ ⊂ L for some L ∈ Γ ( X ).The subset ∆ ⊂ L \ K is unbounded and connected in L , and so ∆ ⊂ K u L . Hence x ∈ [ c, x ] ⊂ Ω L ∩ K u L ⊂ C L . Since ˜ f | Ω L = ˜ f L , it follows that ˜ f ( x ) = ˜ f L ( x ) = f ( x ).Our claim is proved. As Ω \ K is 2-open and polygonally connected, by Theorem 16we conclude that ˜ f | Ω \ K = f .From now on we assume that Ω is finitely open and K is finitely compact. Forevery integer d ≥
2, by applying the already proved result we get the coincidenceset C d := C dK, Ω and a unique map ˜ f d ∈ H G (Ω , Y ), such that ˜ f d | C d = f | C d . Forarbitrary integers s ≥ d ≥ C d ⊂ C s , and so ˜ f s | C d = f | C d , which leads bythe uniqueness of the Hartogs companion ˜ f d to ˜ f s = ˜ f d . For ˜ f := ˜ f ∈ H G (Ω \ K, Y ),we deduce that ˜ f = ˜ f d and ˜ f | C d = f | C d for every d ≥
2, and so ˜ f | C K, Ω = f | C K, Ω . Itis easily seen that all statements from (a,b,c) hold for every d ≥
2. Therefore, theset Ω
G,u from (c) is finitely open.(d’). Let us fix x ∈ Ω \ K . There exists c ∈ Ω \ { } , such that δ := [1 , ∞ [ · c ⊂ X \ K .As X \ K is polygonally connected, there is a polygonal chain Λ ⊂ X \ K joining x to c . Clearly, ∆ := Λ ∪ δ ⊂ L for some L ∈ Γ , ∞ ( X ). The subset ∆ ⊂ L \ K isunbounded and connected in L , and so ∆ ⊂ K u L . Hence x ∈ Ω L ∩ K u L ⊂ C L . Wethus have proved the claimed set equality. It follows that ˜ f | Ω \ K = f . (cid:3) We next show a weaker version of the Hartogs Kugelsatz in arbitrary dimension.For X = C n and Y = C , this version is equivalent to Theorem 2. ARTOGS COMPANIONS AND HOLOMORPHIC EXTENSIONS 23
Corollary 34 (Kugelsatz) . Let a finitely open set Ω ⊂ X and a finitely compactsubset K ⊂ Ω. The
Hartogs companion operator H Y : H G (Ω \ K, Y ) k → H G (Ω , Y ) k , H Y ( f ) = ˜ f , is linear, continuous, and surjective. A right inverse of H Y is the restriction operator ρ : H G (Ω , Y ) k → H G (Ω \ K, Y ) k , ρ ( g ) = g | Ω \ K . If X \ K is polygonally connected, then H Y and ρ are isomorphisms of locallyconvex spaces. Proof.
The linearity of H := H Y is immediate. By Theorem 33(a) we deduce that H ( ρ ( g )) = H ( g | Ω \ K ) = g for every g ∈ H G (Ω , Y ). Hence ρ is a (continuous) rightinverse of H , which is surjective. In order to show that H is continuous, let us fixa seminorm p M as in Definition 15(e), for L ∈ Γ , ∞ ( X ), compact subset M ⊂ Ω L ,and continuous seminorm p : Y → R + . In L let us choose the sets K ⊂ Ω , with K compact and Ω open, such that K L ⊂ ˚ K and M ⊂ Ω , and Ω ⊂ Ω L iscompact. According to Theorem 33(b), for every f ∈ H G (Ω \ K, Y ), the map ˜ f | Ω is the Hartogs companion of f | Ω \ K . By (12) we get successively˜ f ( M ) ⊂ ˜ f (Ω ) ⊏ f (Ω \ K ) ⊂ f (Ω \ ˚ K ) , ˜ f ( M ) ⊂ co( f (Ω \ ˚ K )) . Consequently, p M ( ˜ f ) ≤ p Ω \ ˚ K ( f ) for every f ∈ H G (Ω \ K, Y ). Therefore, H iscontinuous. If X \ K is polygonally connected, by Theorem 33(d’) it follows that ρ ( H ( f )) = ˜ f | Ω \ K = f for every H G (Ω \ K, Y ). We thus conclude that both H and ρ are continuous isomorphisms of locally convex spaces. (cid:3) Remark
35 (Kugelsatz) . The above corollary still holds for d -open Ω and d -compact K (with d ≥ H Y : H G (Ω \ K, Y ) k( d ) → H G (Ω , Y ) k( d ) , and with thepolygonal connectedness of Ω \ K instead of that of X \ K .Indeed, the proof remains valid if we consider L ∈ Γ ,d ( X ) (instead of Γ , ∞ ( X ))and we use Theorem 33(d) (instead of (d’)) for the connectedness of Ω \ K . Proposition 36 (multiplication and composition property of Hartogs companions) . Let d ≥
2, a d -open set Ω ⊂ X , and a d -compact subset K ⊂ Ω. Then (a):
We have ] ( αf ) = ˜ α ˜ f for all α ∈ H G (Ω \ K ) and f ∈ H G (Ω \ K, Y ). (b): Let a domain of holomorphy D ⊂ Y , a sequentially complete complexHausdorff locally convex space Z , and a map g ∈ H ( D, Z ). Then ^ ( g ◦ f ) = g ◦ ˜ f , for every f ∈ H G (Ω \ K, Y ) with f (Ω \ K ) ⊂ D. Proof. (a). For α, f as in (a), αf ∈ H G (Ω \ K, Y ). Since (˜ α ˜ f ) | C dK, Ω = ( αf ) | C dK, Ω ,by the uniqueness of the Hartogs companion we conclude that ] ( αf ) = ˜ α ˜ f .(b). Let f ∈ H G (Ω \ K, Y ) as in (b). Hence ˜ f (Ω) ⊂ D , by Theorem 33(a). Wehave g ◦ f ∈ H G (Ω \ K, Z ) and g ◦ ˜ f ∈ H G (Ω , Z ) (the composition property fromHerv´e [4], Th. 3.1.10, p.57). Since ( g ◦ ˜ f ) | C dK, Ω = ( g ◦ f ) | C dK, Ω , by the uniqueness ofthe Hartogs companion we conclude that ^ ( g ◦ f ) = g ◦ ˜ f . (cid:3) Regularity of Hartogs companions
Regularity results for Hartogs companions may be obtained by showing that therepresentation (3) from Theorem 33(c) holds locally, on neighborhoods of points.Let us note that under the hypothesis of Theorem 38 below, the sets Ω
G,u fromTheorem 33(c) are not necessarily open, unless K is bounded.Let us recall four holomorphy types stronger than Gˆateaux holomorphy: Definition 37 (holomorphy/analyticity types) . Assume X is a Hausdorff locallyconvex space. A map f ∈ H G (Ω , Y ) on an open subset Ω ⊂ X is called (LB): locally bounded holomorphic , if and only if f is locally bounded. (FR): holomorphic (or Fr´echet analytic ), if and only if f is continuous. (HY): hypoanalytic , if and only if f is hypocontinuous (that is, all restrictions f | M to compact subsets M ⊂ Ω are continuous). (MS):
Mackey/Silva holomorphic , if and only if for all a ∈ Ω and boundedsubset B ⊂ X , there exists ε >
0, such that f (Ω ∩ ( a + εB )) is bounded.The following inclusions hold (with the standard notations for the four vector spacesconsisting of maps as in (LB)–(MS); see Dineen [1], p.62): H LB (Ω , Y ) ⊂ H (Ω , Y ) ⊂ H HY (Ω , Y ) ⊂ H M (Ω , Y ) ⊂ H G (Ω , Y ) . Without assuming that X is locally convex, we next show that the conditionsfrom the above definition are inherited from a map by its Hartogs companion. Theorem 38 (regularity) . Assume X is a Hausdorff topological vector space. Letan open set Ω ⊂ X , a -bounded closed set K ⊂ Ω , a map f ∈ H G (Ω \ K, Y ) , andits Hartogs companion ˜ f ∈ H G (Ω , Y ) as in Theorem 33(a). (a): For arbitrary a ∈ Ω and u ∈ X \ { } , and ( a, u ) -admissible set G ⊂ C , let D G,u denote the component of a in the open set (Ω c − G · u ) c ∩ ( K − ∂G · u ) c (hence D G,u is an open neighborhood of a ). Then D G,u + ∂G · u ⊂ Ω \ K and the representation formula (3) holds for every x ∈ D G,u . (lb): If f is locally bounded, then so is ˜ f . If p : Y → R + is a continuousseminorm and p ◦ f is locally bounded, then so is p ◦ ˜ f . (fr): If f is continuous, then so is ˜ f . (hy): If f is hypocontinuous, then so is ˜ f . (ms): If f satisfies the condition from Definition 37(MS), then so does ˜ f .If X is locally convex, any holomorphy from Definition 37 is inherited by ˜ f from f .Proof. Let S Y denote the set of all continuous seminorms on Y .(a). For fixed a , u , and G as in (a), set A := Ω c − G · u and B := K − ∂G · u , and D := A c ∩ B c1 . Thus a ∈ A c ⊂ Ω. For every x ∈ X , we have the equivalences x ∈ D ⇐⇒ (cid:26) x / ∈ Ω c − G · ux / ∈ K − ∂G · u ⇐⇒ (cid:26) x + G · u ⊂ Ω x + ∂G · u ⊂ Ω \ K. Hence D + ∂G · u ⊂ Ω \ K . Since G is ( a, u )-admissible, it follows that a ∈ D . Allfour sets G · u , ∂G · u , Ω c , K are closed, and the first two are compact. Hence both For several other conditions equivalent to holomorphy, see Herv´e [4], Def. 3.1.1, p.52. This is not the definition, but an equivalent condition (Dineen [1], Prop. 2.18(a,b), p.61).
ARTOGS COMPANIONS AND HOLOMORPHIC EXTENSIONS 25 A and B are closed, and so D is open. Therefore, D G,u is an open neighborhoodof a . Let us define the map¯ f : D G,u → Y, ¯ f ( x ) = 12 π i Z ∂G f ( x + ζu ) ζ d ζ. By the representation formula (3) we get ¯ f | C = ˜ f | C , where C := Ω G,u ∩ D G,u ∋ a .As in the proof of Theorem 44 (Step 4) we deduce that ¯ f ∈ H G ( D G,u , Y ). Accordingto Theorem 33(c), the set Ω
G,u is 1-open, and hence so is C . Since the set D G,u is open and connected, and hence polygonally connected, by Theorem 16 it followsthat ¯ f = ˜ f | D G,u . We conclude that the representation (3) holds for every x ∈ D G,u .(lb). Assume f is locally bounded. For fixed a ∈ Ω, let us choose u and G as in (a).Since a + ∂G · u is compact and f is locally bounded, by a standard compactnessargument we find a neighborhood U ⊂ D G,u of a , such that P := f ( U + ∂G · u ) isbounded. We claim that ˜ f ( U ) is bounded. In order to prove this, let us fix p ∈ S Y .As 0 ∈ G , we have B C (0 , r ) ⊂ G for some r >
0. By the representation formula (3)on D G,u ⊃ U it follows that p ( ˜ f ( x )) ≤ ℓ ( ∂G )2 πr sup ζ ∈ ∂G p ( f ( x + ζu )) ≤ ℓ ( ∂G )2 πr sup p ( P ) , for every x ∈ U, where ℓ ( ∂G ) denotes the length of the boundary ∂G (which consists of finitely manypiecewise C Jordan curves). We thus conclude that ˜ f ( U ) is bounded, and hencethat ˜ f is locally bounded. The proof of the second statement from (lb) is similar.(fr). In order to show that ˜ f is continuous, let us fix a ∈ Ω and p ∈ S Y , and ε > a , let us choose u and G as in (a), together with r >
0, such that B C (0 , r ) ⊂ G .By the representation formula (3) from (a) it follows that˜ f ( x ) − ˜ f ( a ) = 12 π i Z ∂G f ( x + ζu ) − f ( a + ζu ) ζ d ζ, for every x ∈ D G,u . Since a + ∂G · u is compact and f is continuous, by a standard compactness argumentwe find a neighborhood U ε ⊂ D G,u of a , such that p ( f ( x + ζu ) − f ( a + ζu )) < rεℓ ( ∂G ) , for all x ∈ U ε , ζ ∈ ∂G. Consequently, for every x ∈ U ε we have p ( ˜ f ( x ) − ˜ f ( a )) ≤ ℓ ( ∂G )2 πr sup ζ ∈ ∂G p ( f ( x + ζu ) − f ( a + ζu )) ≤ ε π < ε. Hence ˜ f is continuous at a . It follows that ˜ f is continuous on Ω.(hy). Let us fix a compact subset M ⊂ Ω and a ∈ M , together with p ∈ S Y and ε >
0. For a , let us choose u and G as in (a), and a closed neighborhood U ⊂ D G,u of a . The set M := U ∩ M + ∂G · u ⊂ Ω \ K is compact, and so f | M is uniformlycontinuous. Consequently, there is a balanced neighborhood V ⊂ X of 0, such that W ε := a + V ⊂ U and p ( f ( y ) − f ( z )) < rεℓ ( ∂G ) , for all y, z ∈ M , with y − z ∈ V. For every x ∈ W ε ∩ M we see that x + ∂G · u ⊂ M and x − a ∈ V , which lead asin the proof of (fr) to p ( ˜ f ( x ) − ˜ f ( a )) ≤ ε π < ε . Hence ˜ f | M is continuous at a . Itfollows that ˜ f | M is continuous, and hence that ˜ f is hypocontinuous.(ms). Let us fix a ∈ Ω and a nonempty balanced bounded subset B ⊂ X . For a , choose u and G as in (a). The balanced set B u := B + B C (0 , · u is bounded and B ⊂ B u . For each ζ ∈ ∂G and for a ζ := a + ζu ∈ Ω \ K , there is ε ζ >
0, such that a + ε ζ B u ⊂ D G,u , f ( a ζ + ε ζ B u ) is bounded . As ∂G is compact, there is a finite subset F ⊂ ∂G , such that ∂G ⊂ S ξ ∈ F B C ( ξ, ε ξ ).For ε := min ξ ∈ F ε ξ >
0, we have a + εB ⊂ D G,u and P := S ξ ∈ F f ( a ξ + ε ξ B u ) isbounded. Let us observe that a + εB + ∂G · u ⊂ [ ξ ∈ F [ a + εB + B C ( ξ, ε ξ ) · u ] = [ ξ ∈ F [ a ξ + εB + B C (0 , ε ξ ) · u ] ⊂ [ ξ ∈ F [ a ξ + ε ξ B + ε ξ B C (0 , · u ] = [ ξ ∈ F ( a ξ + ε ξ B u ) , and hence that f ( a + εB + ∂G · u ) ⊂ P . Since P is bounded, as in the proof of (lb)(with a + εB instead of U ) it follows that ˜ f ( a + εB ) is bounded. (cid:3) Hartogs-type extension theorems.
To avoid repetition, let us first note that unless Theorem 44, all results from thissection hold together with the following:
Additional conclusions.
For f and its unique extension ˜ f as above, we have therange inclusion ˜ f (Ω) ⊏ f (Ω \ K ) . Furthermore, for every domain of holomorphy D ⊂ Y , we have the equivalence f (Ω \ K ) ⊂ D ⇐⇒ ˜ f (Ω) ⊂ D. If Y = C , then ˜ f (Ω) = f (Ω \ K ) .Remark
39 (connectedness conditions) . All three equivalences from Proposition 7also hold for a finitely open set Ω ⊂ X and a finitely compact subset K ⊂ Ω.Indeed, the proof of Proposition 7 remains valid if we replace C n by X f and we useTheorems 16, 33(d’) instead of the classical identity theorem and Theorem 3(d).The next result is the correspondent of Theorem 8 in arbitrary dimension. Theorem 40 (extension/Kugelsatz) . Let a finitely open set Ω ⊂ X and a finitelycompact subset K ⊂ Ω . The following four statements are equivalent (where in (i’) we consider X equipped with the finite open topology τ f ). (i): Every map f ∈ H G (Ω \ K, Y ) has a (unique) extension ˜ f ∈ H G (Ω , Y ) . (i’): Every locally constant map g : Ω \ K → C has an extension ˜ g ∈ H G (Ω) . (ii): The restriction ρ : H G (Ω , Y ) k → H G (Ω \ K, Y ) k is an isomorphism ofcomplex vector spaces. (iii): X \ K is polygonally connected.In this case, ρ is an isomorphism of locally convex spaces whose inverse is theHartogs companion operator H Y , and the additional conclusions hold.Proof. (i) ⇒ (ii). The extension of every f ∈ H G (Ω \ K, Y ) is the Hartogs companion H Y ( f ), and so ρ ( H Y ( f )) = H Y ( f ) | Ω \ K = f . Consequently, for H Y the restriction ρ is a left inverse, but also a right inverse, by Corollary 34. Since both ρ and H Y are τ k -continuous, ρ is an isomorphism of locally convex spaces.(ii) ⇒ (i). Since the restriction operator ρ is surjective, (i) holds.(iii) ⇒ (i) obviously holds by Theorem 33(a,d’).(i) ⇒ (i’). Let a locally constant map g : Ω \ K → C . Let us choose y ∈ Y \ { } and ARTOGS COMPANIONS AND HOLOMORPHIC EXTENSIONS 27 ϕ ∈ Y ∗ , such that ϕ ( y ) = 1. For f := g · y ∈ H G (Ω \ K, Y ) and its extension ˜ f asin (i), we have ˜ g := ϕ ◦ ˜ f ∈ H G (Ω) and ˜ g | Ω \ K = ϕ ◦ ( ˜ f | Ω \ K ) = ϕ ◦ f = g .(i’) ⇒ (iii). The proof is the same as that of the corresponding implication fromTheorem 8 (with C n replaced by X f ) and uses Remark 39 and Theorem 16 insteadof Proposition 7 and the classical identity theorem.The additional conclusions follow by Theorem 40(a). (cid:3) The topological version of the above theorem allows K to have nonempty interior(for instance, when X is normable and K is closed and bounded). Corollary 41 (extension/Kugelsatz) . Assume X is a Hausdorff locally convexspace. Let an open set Ω ⊂ X and a finitely bounded closed subset K ⊂ Ω. Thefollowing six statements are equivalent. (i):
Every map f ∈ H G (Ω \ K, Y ) has an extension ˜ f ∈ H G (Ω , Y ). (ii): Every map f ∈ H M (Ω \ K, Y ) has an extension ˜ f ∈ H M (Ω , Y ). (iii): Every map f ∈ H HY (Ω \ K, Y ) has an extension ˜ f ∈ H HY (Ω , Y ). (iv): Every map f ∈ H (Ω \ K, Y ) has an extension ˜ f ∈ H (Ω , Y ). (v): Every map f ∈ H LB (Ω \ K, Y ) has an extension ˜ f ∈ H LB (Ω , Y ). (vi): X \ K is connected.In this case, all extensions are unique and the additional conclusions hold. Proof.
As Ω and X \ K are finitely open, K is finitely compact, and (vi) is equivalentto the polygonal connectedness of X \ K , by Theorem 40 we see that (i) ⇔ (vi).That (i) yields all (ii)–(v) follows by Theorem 38. Finally, any of (ii)–(v) impliesthe condition from Theorem 40(i’) (locally constant maps g : Ω \ K → C are of allfour holomorphy types), which is equivalent to (i), by Theorem 40. We thus haveproved the equivalence of all six statements. In (i)–(v) the extension of f is theHartogs companion, which if unique. (cid:3) The following extension theorem only involves 2-cuts of both sets K and Ω. Thisresult is more general than Theorem 1 even for X = C n and Y = C , since 2-compactsets may not be closed or bounded and 2-open sets may not be open. Theorem 42 (2-cuts extension) . Let a -open set Ω ⊂ X and a -compact subset K ⊂ Ω . If Ω \ K is polygonally connected, then every map f ∈ H G (Ω \ K, Y ) hasa unique extension ˜ f ∈ H G (Ω , Y ) and the additional conclusions hold.Proof. The conclusion is immediate, by Theorem 33(a,d) for d = 2. (cid:3) Corollary 43 (holomorphic extensions) . Assume X is a Hausdorff locally convexspace. Let an open set Ω ⊂ X and a 2-bounded closed subset K ⊂ Ω, such thatΩ \ K is connected. Then the statements (i)–(v) from Corollary 41 hold togetherwith the additional conclusions. Proof.
The conclusion follows by applying successively Theorems 42 and 38. (cid:3)
The above four results deal with inner Gˆateaux holomorphic extensions; we callthese “inner”, since for every variety L ∈ Γ ( X ) the set K L ⊂ Ω L is compact in L (we may say that K L is a “compact hole” in Ω L ). We next establish a theorem Here and in Corollary 43 topological boundedness of K is unnecessary and more restrictive. suitable for outer (that is, not inner) extensions. For some fixed u ∈ X \ { } , thisresult only involves particular cuts with linear varieties of the form L a ( u ) := a + C · u, L a ( u, v ) := a + C · u + C · v, and with closed linear -strips parallel to u , defined by L a ( u, v [ ε ]) := L a ( u ) + B C (0 , ε ) · v. Theorem 44 (outer extension) . Let the sets K ⊂ Ω ⊂ X , such that Ω and Ω \ K are -open and Ω \ K is polygonally connected. Assume there exists u ∈ X \ { } ,such that: (i): For all a ∈ Ω and v ∈ X , there exists ε > , such that K ∩ L a ( u, v [ ε ]) iscompact in L a ( u, v ) . (ii): K ∩ L c ( u ) = ∅ for some c ∈ Ω .Then every map f ∈ H G (Ω \ K, Y ) has a unique extension ˜ f ∈ H G (Ω , Y ) . We have ˜ f (Ω) ⊂ co( f (Ω \ K )) . If Y = C , then ˜ f (Ω) = f (Ω \ K ) .Proof. There is no loss of generality in assuming K = ∅ . The uniqueness of ˜ f willfollow by Theorem 16, if we show that Ω \ K = ∅ and Ω is polygonally connected. Weclaim that every a ∈ Ω can be joined in Ω by a linear segment to some a ′ ∈ Ω \ K = ∅ .Clearly, it suffices to prove this for a ∈ K . For such a , in L = L a ( u ) ∈ Γ ( X ) the setΩ L is open and K L is compact (which follows from (i) for v = 0), with a ∈ K L ⊂ Ω L .Therefore, there exists t >
0, such that a ′ := a + tu / ∈ K L and [ a, a ′ ] ⊂ Ω L . Ourclaim is proved. Hence for arbitrarily fixed a, b ∈ Ω, we have [ a, a ′ ] ∪ [ b, b ′ ] ⊂ Ω forsome a ′ , b ′ ∈ Ω \ K . Since Ω \ K is polygonally connected, a ′ and b ′ can be joinedby a polygonal chain Λ ⊂ Ω, and so [ a, a ′ ] ∪ Λ ∪ [ b ′ , b ] ⊂ Ω. We thus conclude thatΩ is polygonally connected. Now the uniqueness of ˜ f follows by Theorem 16. Theproof of the existence part is divided into five steps, among which the first four areonly based on the condition (i). Step 1 . We first show the following needed technical property:
For every nonempty closed subset H ⊂ C the set S := Ω \ ( K − H · u ) is -open. On the contrary, suppose there exists a 1-cut S L , which is not open in L ∈ Γ ( X ).Hence there is a sequence ( s ′ n ) n ∈ N ⊂ L \ S , convergent in L to some s ∈ S L ⊂ Ω L .Thus L = L s ( v ) for some v ∈ X \ { } . Since Ω L is open in L and s ∈ Ω L , by theabove convergence we may assume ( s ′ n ) n ∈ N ⊂ Ω L \ S ⊂ K − H · u . Consequently,there exist two sequences ( k n ) n ∈ N ⊂ K and ( h n ) n ∈ N ⊂ H , such that s ′ n = k n − h n u, for every n ∈ N . (19)According to (i), there exists ε >
0, such that K ε := K ∩ L s ( u, v [ ε )] is compactin L s ( u, v ). As lim n →∞ s ′ n = s in L , by taking subsequences we may also assume( s ′ n ) n ∈ N ⊂ s + B C (0 , ε ) · v . By (19) we see that k n = s ′ n + h n u ∈ K ε for every n ∈ N . Since K ε is compact in L s ( u, v ), by taking subsequences we may assumethe existence of the limit k := lim n →∞ k n ∈ K ε ⊂ K in L s ( u, v ). By (19) we deducethat ( h n ) n ∈ N ⊂ H converges and h := lim n →∞ h n ∈ H . Now a passage to the limitin L s ( u, v ) in the equality (19) forces s = k − hu ∈ K − H · u , which contradicts s ∈ S . We thus have proved the claimed property. Step 2 . We next show that all sets Ω
G,u defined as in Theorem 33(c) are 1-open.Let us fix a ∈ Ω and an ( a, u )-admissible set G ⊂ C . Such sets G indeed exist, sincein L = L a ( u ) ∈ Γ ( X ) the set K L ⊂ Ω L is compact and Ω L = ∅ is open. As in the Here and in Corollaries 45, 47, the set K may not be 2-closed or 2-bounded. ARTOGS COMPANIONS AND HOLOMORPHIC EXTENSIONS 29 proof of Theorem 33(c) we see that Ω
G,u = A c ∩ (Ω \ B ), where A = Ω c − G · u and B = K − G c · u , and A c is 1-open. As G c is closed in C , according to the propertyfrom Step 1, Ω \ B is also 1-open. We thus conclude that Ω G,u is 1-open.
Step 3 . We next define the map ˜ f . As the set Γ (Ω , u ) := { L a ( u ) | a ∈ Ω } ⊂ Γ ( X )consists of mutually disjoint complex lines, (Ω L ) L ∈ Γ (Ω ,u ) is a partition of Ω. Forevery L ∈ Γ (Ω , u ), the open set Ω L = ∅ contains the compact K L . Consequently,the restriction f L := f | Ω L \ K L ∈ H (Ω L \ K L , Y ) has a unique Hartogs 1-companion˜ f L ∈ H (Ω L , Y ) as in Remark 32. Let us define the map˜ f : Ω → Y, ˜ f | Ω L = ˜ f L for every L ∈ Γ (Ω , u ) . Hence for arbitrarily fixed a ∈ Ω and ( a, u )-admissible set G ⊂ C , the restriction˜ f | Ω G,u may be represented by (3).
Step 4 . We next show that ˜ f ∈ H G (Ω , Y ). To this aim, let us fix a ∈ Ω, togetherwith v ∈ X \ { } and ϕ ∈ Y ∗ . Choose an ( a, u )-admissible set G ⊂ C . Since Ω G,u is 1-open and a ∈ Ω G,u , there exists r >
0, such that a + B C (0 , r ) · v ⊂ Ω G,u . Hence a + B C (0 , r ) · v + ∂G · u ⊂ Ω \ K . As Ω \ K is 2-open, the set D := { ( λ, ζ ) ∈ C | a + λv + ζu ∈ Ω \ K } is open in C and B C (0 , r ) × ∂G ⊂ D . Let us define the map F ∈ H ( D ) , F ( λ, ζ ) = ( ϕ ◦ f )( a + λv + ζu ) . By the integral representation (3) of ˜ f | Ω G,u it follows that( ϕ ◦ ˜ f )( a + λv ) = 12 π i Z ∂G ( ϕ ◦ f )( a + λv + ζu ) ζ d ζ = 12 π i Z ∂G F ( λ, ζ ) ζ d ζ, for every λ ∈ B C (0 , r ). Since differentiation under the integral sign with respect to λ holds (and a, v, ϕ were arbitrarily fixed), we conclude that ˜ f ∈ H G (Ω , Y ). Step 5 . We finally show that ˜ f | Ω \ K = f . Choose c ∈ Ω as in (ii) and set C := { x ∈ Ω | K ∩ L x ( u ) = ∅} . It is easy to check that c ∈ C and C = Ω \ ( K − C · u ) ⊂ Ω \ K . By the propertyfrom Step 1 we deduce that C is 1-open, and so C − c is a real-absorbing set. Inorder to prove that ˜ f | C = f | C , let us fix x ∈ C . For L = L x ( u ) ∈ Γ (Ω , u ) we have K L = ∅ , and so ˜ f L = f L . By the definition of ˜ f it follows that ˜ f | Ω L = f L = f | Ω L ,which yields ˜ f ( x ) = f ( x ). Hence ˜ f | C = f | C . Since Ω \ K is 2-open and polygonallyconnected, by Theorem 16 we conclude that ˜ f | Ω \ K = f . The proof of the existenceand uniqueness of the extension ˜ f is now complete. Since every f ∈ H G (Ω \ K, Y )has a unique extension ˜ f , the statements on ˜ f (Ω) (the general range inclusion andthe equality for Y = C ) follow by Proposition 21(a). (cid:3) The following corollary strengthens Theorem 42 by weakening its compactnessassumption on K (only 2-cuts parallel to a given vector need to be compact). Corollary 45 (outer extension) . Let the sets K ⊂ Ω ⊂ X , such that Ω and Ω \ K are 2-open and Ω \ K is polygonally connected. Assume there exists u ∈ X \ { } ,such that the cut K ∩ L a ( u, v ) is compact for all a ∈ Ω and v ∈ X . Then everymap f ∈ H G (Ω \ K, Y ) has a unique extension ˜ f ∈ H G (Ω , Y ) and the additionalconclusions hold. Proof.
Since conditions (i,ii) from Theorem 44 are fulfilled, every f ∈ H G (Ω \ K, Y )has a unique extension ˜ f ∈ H G (Ω , Y ). We next prove the additional conclusions. The range inclusion . For every fixed a ∈ Ω, there exists L ∈ Γ ( X ), such that a ∈ L and L k u . Since K L is compact and Ω L is open in L , the restriction ˜ f | Ω L isthe Hartogs companion of f | Ω L \ K L . By (12) it follows that˜ f ( a ) ∈ ˜ f (Ω L ) ⊏ f (Ω L \ K L ) ⊂ f (Ω \ K ) . As a was arbitrary, we conclude that ˜ f (Ω) ⊏ f (Ω \ K ). The equivalence for the domain of holomorphy D ⊂ Y . Assume f (Ω \ K ) ⊂ D . Forevery fixed a ∈ Ω and for L ∈ Γ ( X ) as above, we have f (Ω L \ K L ) ⊂ D , and so˜ f ( a ) ∈ ˜ f (Ω L ) ⊂ D , by Theorem 24. We thus conclude that ˜ f (Ω) ⊂ D . (cid:3) For some inclusions K ⊂ Ω ⊂ X Theorem 44 applies, while Corollary 45 fails:
Example 46 (outer extension) . Assume dim C ( X ) ≥ (a): For e ∈ X \ { } and 3-compact subset B ⊂ X , let K := B + R + · e . Thenevery map f ∈ H G ( X \ K, Y ) has a unique extension ˜ f ∈ H G ( X, Y ). (b): Assume X is a normed space. Let a linear functional ϕ : X → C , anupper semicontinuous map r : C → R and K := { x ∈ X | k x k ≤ r ( ϕ ( x )) } .Then every map f ∈ H G ( X \ K, Y ) has a unique extension ˜ f ∈ H G ( X, Y ). (c): Assume X is an inner product space. Let two open balls B , B ⊂ X ,such that B B , B B , and B ∩ B = ∅ . Set Ω := co( B \ B ) ⊂ X .Then every map f ∈ H G ( B \ B , Y ) has a unique extension ˜ f ∈ H G (Ω , Y ). (d): Assume X is an inner product space. Let ϕ ∈ X ∗ \{ } and two open balls B ( Ω ⊂ X centered at 0. Then every map f ∈ H G (cid:0) Ω \ (ker ϕ \ B ) , Y (cid:1) hasa unique extension ˜ f ∈ H G (Ω , Y ). On the other hand, for fixed y ∈ Y \ { } ,the map ϕ · y ∈ H G (Ω \ ker ϕ, Y ) has no extension from H G (Ω , Y ); thisshows that the condition (ii) from Theorem 44 cannot be dropped.In each case we indicate how to use Theorem 44 (Corollary 45 does not apply).(a). For u ∈ X \ ( C · e ) and c = − te , with t > u ∈ ker ϕ \ { } and c = te , with e ∈ ker ϕ \ ( C · u ) and t > a, b ∈ X denote the centers of the two balls. We may apply the theoremfor K := B ∩ Ω and u ∈ { b − a } ⊥ \ { } , and c ∈ ( B \ B ) ∩ L a ( b − a ).(d). For K := Ω ∩ ker ϕ \ B and u ∈ (ker ϕ ) ⊥ \ { } , and c = 0.As pointed out in the introduction, in Theorem 1 we may replace the compactnessrequirement on K by significantly weaker assumptions. Corollary 47 (Hartogs extension) . Let n ≥
2, and the sets K ⊂ Ω ⊂ C n , such thatΩ and Ω \ K are open and Ω \ K is connected. Assume there exists u ∈ C n \ { } ,with the property that K ∩ L is compact, for every L ∈ Γ ( C n ) , L k u. Then every map f ∈ H (Ω \ K ) has a unique extension ˜ f ∈ H (Ω). Furthermore,˜ f (Ω) = f (Ω \ K ). Proof.
The conclusion is immediate, by Corollary 45. (cid:3)
Example 48 (Hartogs extension) . For every open set Ω ⊂ C , the subset K := Ω ∩ { ( z, z , | z ∈ C } ARTOGS COMPANIONS AND HOLOMORPHIC EXTENSIONS 31 satisfies the condition from Corollary 47 with u = (0 , , K may not bebounded or closed. (a): K is bounded and not closed, and K ∪ ( C n \ Ω) is path-connected forΩ = B C (0 , (b): K is unbounded and closed for Ω = C . (c): K is unbounded and not closed, and K ∪ ( C n \ Ω) is path-connected forΩ = { ( z , z , z ) ∈ C | Re( z ) > } . References
1. S. Dineen, Complex analysis in locally convex spaces. North-Holland Mathematics Studies,57, North-Holland Publishing Co., Amsterdam-New York, 1981.2. S. Dineen, Complex analysis on infinite dimensional spaces. Monographs in Mathematics,Springer-Verlag, 1999.3. L. Ehrenpreis,
A new proof and an extension of Hartogs’ theorem , Bull. Amer. Math. Soc. (1961), 507–509.4. M. Herv´e, Analyticity in infinite-dimensional spaces. De Gruyter Studies in Mathematics, 10,Walter de Gruyter Co., Berlin, 1989.5. L. H¨ormander, An introduction to complex analysis in several variables (Third edition). NorthHolland, Amsterdam-New York-Oxford-Tokyo, 1990.6. W. Rudin, Real and complex analysis (Third edition). McGraw-Hill Inc., 1987. Institute of Mathematics “Simion Stoilow” of the Romanian Academy, P.O. Box 1-764, RO-014700 Bucharest, Romania
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