Topological genericity of nowhere differentiable functions in the disc and polydisc algebras
aa r X i v : . [ m a t h . C V ] N ov TOPOLOGICAL GENERICITY OF NOWHEREDIFFERENTIABLE FUNCTIONS IN THE DISC ANDPOLYDISC ALGEBRAS
Alexandros Eskenazis Konstantinos Makridis
Abstract
In this paper we examine functions in the disc algebra A ( D ) and thepolydisc algebra A ( D I ), where I is a finite or countably infinite set.We prove that, generically, for every f ∈ A ( D ) the continuous peri-odic functions u = Ref | T and ˜ u = Imf | T are nowhere differentiableon the unit circle T . Afterwards, we generalize this result by prov-ing that, generically, for every f ∈ A ( D I ), where I is as above, thecontinuous periodic functions u = Ref | T I and ˜ u = Imf | T I have nodirectional derivatives at any point of T I and every direction v ∈ R I with k v k ∞ = 1. Strange functions have attracted the interest of mathematicians since Weier-strass, who first gave an explicit example of a function u : R → R which iscontinuous, periodic but not differentiable at any real number. Since thenthere have been proven many results indicating the existence of functionswith strange behaviour, such as non differentiability and universality in var-ious senses.In fact, many times these strange properties come to be generic, even ifno explicit example of such a function is known. This is usually proven withsome arguments involving Baire’s Category Theorem in a suitable completemetric space or Fr´echet space. For the role of Baire’s Category Theoremin Analysis we refer to [7] and [9]. An old example of such a techniqueis the classical result of Banach and Mazurkiewicz (see [3]) which statesthat, generically, every continuous function on a compact interval J of R isnowhere differentiable.In section 2 of this paper we will prove an analogue of this theorem forfunctions in the disc algebra A ( D ), i.e. continuous functions defined onthe closed unit disc D which are holomorphic on D . For every function f ∈ A ( D ) we can naturally construct a 2 π − periodic continuous function h : R → C defined by h ( θ ) = f ( e iθ ); often, by abuse of notation, we write1 f ( θ ) instead of h ( θ ). Taking into consideration the above result of Banachand Mazurkiewicz it is natural to ask whether there are functions f ∈ A ( D )such that the corresponding function h is nowhere differentiable. Of course,since h takes complex values, this means that for every point θ ∈ R , either u = Reh will not be differentiable on θ or ˜ u = Imh will not be differentiableon θ . In Theorem 2.1, we will prove the stronger fact that, generically, forevery function f ∈ A ( D ) both functions u = Ref | T and ˜ u = Imf | T arenowhere differentiable.In section 3 we extend the above result in the context of several com-plex variables. A function f : D I → C belongs to the algebra A ( D I ) if itis continuous on D I , endowed with the cartesian topology, and separatelyholomorphic in D I . Equivalently, it is well known that f ∈ A ( D I ) if andonly if, f is a uniform limit of polynomials on D I , where every polynomialis meant to depend only on finitely many variables. We consider functions f ∈ A ( D I ), where I is a finite or countably infinite set, such that neithertheir real part u = Ref | T I , nor their imaginary part ˜ u = Imf | T I have direc-tional derivatives at any point of T I , where T is the unit circle, for any di-rection v ∈ R I with k v k ∞ = 1. We prove that, generically, every f ∈ A ( D I )has the above properties. Finally, we prove that a direct generalization ofthis result cannot hold if I is uncountable.We mention that the previous result, valid for every v ∈ R I with k v k ∞ =1, in the case where I is a finite set implies the same result for every direction v ∈ R I \ { } . Thus, in this case our result is the strongest possible. Inthe case where I is infinite countable we have not been able to prove sucha strong result. The set of directions with respect to which there is nodirectional derivative contains ℓ ∞ ( I ) \ { } which is dense in R I , with respectto the cartesian topology.The above results establish the topological genericity of nowhere differ-entiable functions in the disc and polydisc algebras. It remains an openquestion to determine other forms of genericity for this class of functions.For example, the dense lineability of this class, i.e. the existensc of a denselinear subspace V of A ( D ) (or in general A ( D I )), every non-zero elementof which is nowhere differentiable. Also, it would be interesting to examinethe spaceability of this class of functions, i.e. the existence of a closed, in-finite dimensional linear subspace W of A ( D ) (or in general A ( D I )), everynon-zero element of which is nowhere differentiable. Finally, since A ( D ) isan algebra, it would be interesting to study the algebrability of this class,i.e. to examine whether the above questions hold in the case where linearsubspaces were replaced by subalgebras. These questions will hopefully beexamined in future papers. For related results we refer to [2], [4] and [6]. We start with the well known example of Weierstrass: the function u : R → R defined by u ( x ) = ∞ X n =0 a n cos( b n x ) , (1)for 0 < a < b , is a 2 π − periodic continuous functionthat is not differentiable at any real number x if the condition ab > π holds. In fact, for the differentiability of u something even stronger holds(see [13], pp. 351-354): for every x ∈ R (cid:12)(cid:12)(cid:12)(cid:12) lim sup t → x + u ( t ) − u ( x ) t − x (cid:12)(cid:12)(cid:12)(cid:12) = + ∞ . (2)The continuity of u follows easily from Weierstrass’ M-Test, since ∞ X n =0 | a n cos( b n x ) | ≤ ∞ X n =0 a n < + ∞ , because 0 < a <
1. Since u is 2 π − periodic, we can see it as a function u : T → R , where T is the unit circle, and thus, it is well known that (see[1], pp. 168-171) there is a continuous extension of u on the closed unit disc D which in harmonic on D . Of course, from the form of u we can easilysee that u ( z ) = Re ∞ X n =0 a n z b n ! , | z | ≤ . (3)Hence, using once again the M-Test, the harmonic conjugate ˜ u of u can beextended continuously on D and its restriction on T is given by the formula˜ u ( x ) = ∞ X n =0 a n sin( b n x ) . (4)Using these elementary results we will provide a simple proof of the following: Theorem 2.1.
There is a π − periodic continuous function u : R → R suchthat:(i) u is nowhere differentiable on R .(ii) It’s harmonic conjugate ˜ u extends continuously on D and(iii) The π − periodic continuous functions ˜ u | T is nowhere differentiable on R . In fact, the class of functions f ∈ A ( D ) , such that the real part of f has theabove properties is residual, i.e. it contains a G δ dense subset of A ( D ) . Our proof makes use of the Weierstrass function defined above and itsproperties. A more technical proof without use of Weierstrass’s function u can be found in [5]. We will need some lemmas. First, for n ∈ N , considerthe sets D n = n u ∈ C R ( T ) : for every θ ∈ R there is a y ∈ (cid:18) θ, θ + 1 n (cid:19) such that | u ( y ) − u ( θ ) | > n | y − θ | o , (5)where, as usual, we interpret functions defined on T as 2 π − periodic functionsdefined on R . Afterwards, we will also need the sets E n = { f ∈ A ( D ) : Ref | T ∈ D n } , n ∈ N . (6) Lemma 2.2.
For every n ∈ N , E n is an open set in A ( D ) , endowed withthe supremum norm.Proof. Let n ∈ N be fixed. We will prove that A ( D ) \ E n is closed in A ( D ).Let { f m } be a sequence in A ( D ) \ E n and f ∈ A ( D ) such that f m → f uniformly on D . Since f m / ∈ E n , for each m, there is a θ m ∈ R such that (cid:12)(cid:12)(cid:12)(cid:12) u m ( y ) − u m ( θ m ) y − θ m (cid:12)(cid:12)(cid:12)(cid:12) ≤ n, (7)for every y ∈ (cid:0) θ m , θ m + n (cid:1) , where u m = Ref m | T . Since each u m is 2 π − periodic,we can assume that θ m ∈ [0 , π ] for every m and hence there is a subsequence { θ k m } of { θ m } and a θ ∈ [0 , π ] such that θ k m → θ .If y ∈ (cid:0) θ, θ + n (cid:1) , then for large enough m it holds y ∈ (cid:0) θ k m , θ k m + n (cid:1) .Thus, applying (7) for these indices { k m } and then letting m → ∞ we havethat (cid:12)(cid:12)(cid:12)(cid:12) u ( y ) − u ( θ ) y − θ (cid:12)(cid:12)(cid:12)(cid:12) ≤ n, (8)since the convergence of { f m } to f is uniform (again u = Ref | T ). Hence (8)holds for every y ∈ (cid:0) θ, θ + n (cid:1) and thus f / ∈ E n . So, A ( D ) \ E n is closed orequivalently E n is open.Therefore, the intersection T ∞ n =1 E n is a G δ set in A ( D ). Lemma 2.3.
The set S = ∞ \ n =1 E n (9) is dense in A ( D ) . Proof.
Let u : T → R be the Weierstrass function mentioned above, ˜ u itsharmonic conjugate and f ∈ A ( D ) the function f = u + i ˜ u defined on D . From (2), we can deduce that for every θ ∈ R and n ∈ N we can find y n = y n ( θ ) ∈ (cid:0) θ, θ + n (cid:1) such that (cid:12)(cid:12)(cid:12)(cid:12) u ( y n ) − u ( θ ) y n − θ (cid:12)(cid:12)(cid:12)(cid:12) > n, (10)i.e. u ∈ D n and thus f ∈ E n for every n ∈ N . Hence, S 6 = ∅ .In order to prove that S is dense in A ( D ) consider a polynomial p ( z )restricted on D and we will prove that f + p is also in S . Let n ∈ N befixed. The function u = Rep | T is a C ∞ real valued 2 π − periodic functiondefined on R and thus the same holds for its derivative u ′ . In particular, u ′ is bounded by a constant M >
0. Also, for every y, θ ∈ R with y = θ thereis a ξ between y and θ such that u ( y ) − u ( θ ) y − θ = u ′ ( ξ ) . (11)Hence, for every y, θ ∈ R , y = θ we have (cid:12)(cid:12)(cid:12)(cid:12) u ( y ) − u ( θ ) y − θ (cid:12)(cid:12)(cid:12)(cid:12) ≤ M. (12)Consider now a large N >
N > M + n and some θ ∈ R . Since u ∈ D N there is a y ∈ (cid:0) θ, θ + N (cid:1) such that (cid:12)(cid:12)(cid:12)(cid:12) u ( y ) − u ( θ ) y − θ (cid:12)(cid:12)(cid:12)(cid:12) > N. (13)Of course, it is true that θ < y < θ + n and hence we have (cid:12)(cid:12)(cid:12)(cid:12) Re ( f + p )( y ) − Re ( f + p )( θ ) y − θ (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) u ( y ) − u ( θ ) y − θ + u ( y ) − u ( θ ) y − θ (cid:12)(cid:12)(cid:12)(cid:12) ≥ (cid:12)(cid:12)(cid:12)(cid:12) u ( y ) − u ( θ ) y − θ (cid:12)(cid:12)(cid:12)(cid:12) − (cid:12)(cid:12)(cid:12)(cid:12) u ( y ) − u ( θ ) y − θ (cid:12)(cid:12)(cid:12)(cid:12) > N − M > n. (14)Hence we conclude that f + p ∈ E n for the arbitrary n and thus f + p ∈ S .It is well known that the set of polynomials is dense in A ( D ); it follows thatthe set of translations { f + p : p polynomial } ⊆ S (15)is also dense. From this we derive the density of S in A ( D ). Lemma 2.4.
The set T = S ∩ i S = ∞ \ n =1 (cid:0) E n ∩ iE n (cid:1) (16) is also G δ and dense in A ( D ) .Proof. Since multiplication by i is a homeomorphism and S is G δ and densein A ( D ), it follows that i S is also G δ and dense in A ( D ). Hence, from Baire’sCategory Theorem their intersection T is also G δ and dense in A ( D ).We can now proceed to the proof of our main result: Proof of Theorem 2.1.
Let E the class of functions f ∈ A ( D ) such that boththe real part u and the imaginary part ˜ u of f are nowhere differentiable. Wewill prove that T = ∞ \ n =1 (cid:0) E n ∩ iE n (cid:1) ⊆ E (17)and from this our result will follow. Let f ∈ T . Since f ∈ S we concludethat Ref = u ∈ D n for every n ∈ N . Hence, for θ ∈ T and for every n wecan find a y n ∈ (cid:0) θ, θ + n (cid:1) such that (cid:12)(cid:12)(cid:12)(cid:12) u ( y n ) − u ( θ ) y n − θ (cid:12)(cid:12)(cid:12)(cid:12) > n. (18)So y n → θ and thus, from (18), u is not differentiable at θ . For the imaginarypart ˜ u of f , since f ∈ i S we have that − if ∈ S and thus Re ( − if ) = Imf ∈ D n for every n ∈ N . It follows that ˜ u is also nowhere differentiable on T .So, the inclusion (17) has been proven and Theorem 2.1 now follows fromLemma 2.4. ✷ Remark 2.5.
Lemma 2.4 could be ommited if we had used some moredifficult results. Hardy has proven in [8] that if 0 < a < b ∈ N suchthat ab ≥
1, then both functions u ( x ) = ∞ X n =0 a n cos( b n x ) and ˜ u ( x ) = ∞ X n =0 a n sin( b n x ) (19)are continuous and also belong in all the sets D n defined above with a muchmore technical and difficult proof from the one of Weierstrass. Hence, thefunction f : D → C defined by f ( z ) = ∞ X n =1 a n z b n , | z | ≤ A ( D ) such that Ref = u has the properties listed inTheorem 2.1, i.e. f ∈ E . Thus, using essentially the same density argumentof Lemma 2.3 we could prove Theorem 2.1. It is of course obvious that theproof which we presented above is much more elementary.We close this section with a result concerning the size of the class of2 π − periodic real valued functions u that satisfy the properties listed inTheorem 2.1: we will prove that, in contrast with the result of Theorem 2.1,this subset of the space of 2 π − periodic functions is actually topologicallysmall: Proposition 2.6.
Let L be the subspace of C R ( T ) consisting of all continuous π − periodic functions u : R → R which satisfy the properties (i), (ii) and(iii) of Theorem 2.1. Then L is a dense set of first category in C R ( T ) .Proof. For the density of L , consider a function u ∈ C R ( T ) and an ε > u its harmonic extension on D which is continuous on D ;due to the uniform continuity of u on the compact D , there is an r ∈ (0 , | u ( z ) − u ( rz ) | < ε , (21)for every z ∈ D . If ˜ u is a harmonic conjugate of u , then the function ϕ : D → R defined by ϕ ( z ) = u ( rz ) has a harmonic conjugate defined bythe formula ˜ ϕ ( z ) = ˜ u ( rz ) , (22)which extends continuously on D and thus there is a function f ∈ A ( D )such that ϕ = Ref . Using the density of T proven above, we conclude thatthere is a function g ∈ T such that k f − g k ∞ < ε/
2. If ψ = Reg , it holdsthat ψ ∈ L and k u − ψ k ∞ ≤ k u − ϕ k ∞ + k ϕ − ψ k ∞ < ε k f − g k ∞ < ε. (23)Hence, L is indeed dense in C R ( T ).To prove that L is of first category it suffices to prove that L c ≡ C R ( T ) \ L contains a G δ dense subset. Of course, for every function u ∈ L , theharmonic conjugate ˜ u is bounded on D . Hence, for the set Z = { u ∈ C R ( T ) : sup | z | < | ˜ u ( z ) | = ∞} = ∞ \ n =1 { u ∈ C R ( T ) : sup | z | < | ˜ u ( z ) | > n } (24)we conclude that Z ⊆ L c . We will prove that Z is G δ and dense in C R ( T ).In order to prove that Z is G δ we will prove that for every n , the set { u ∈ C R ( T ) : sup | z |≤ | ˜ u ( z ) | > n } is open in C R ( T ). This follows easily fromthe fact that for every z ∈ D , the linear operator C R ( T ) ∋ u ˜ u ( z ) ∈ R (25)is continuous. Indeed, it is well known (see [1] pp. 169) that | ˜ u ( z ) | ≤ | z | − | z | k u k ∞ , (26)from which the result follows.For the density of Z , we will proceed in a similar way as in the proof ofTheorem 2.1. Using essentially the same argument, we can see that if Z 6 = ∅ then Z is dense in C R ( T ). Indeed, if u ∈ Z , then for every real trigonometricpolynomial p : T → R it will also be true thatsup | z | < | ˜ u ( z ) + ˜ p ( z ) | = ∞ (27)because ˜ p is bounded on D . So the problem is to find an element u ∈ Z .Consider the domain Ω in C bounded by the line segments x = − x = 1the x − axis and the graph of the function ϕ ( x ) = x , i.e.Ω = (cid:26) ( x, y ) ∈ R : 0 < | x | < < y < x (cid:27) ∪ { (0 , y ) : y > } . (28)Of course Ω is simply connected and not equal to the whole complex planeand thus, from the Riemann Mapping Theorem, there exists a conformalmapping f : D → Ω. Let u = Ref and ˜ u = Imf . We will prove that u ∈ Z .Consider the map ψ : Ω ∪ {∞} → C defined by ψ ( z ) = z + i , which is aconformal map from Ω to ψ (Ω). Then, the image of ∂ Ω ∪ {∞} under ψ is aJordan curve in C . Thus, the function ψ ◦ f : D → ψ (Ω) is also a coformalmap from the unit disc onto a Jordan domain. Hence, from Caratheodory’sTheorem (see [10] pp. 36-40), ψ ◦ f has a continuous extension from D to ψ (Ω) which is a homeomorphism and maps T on ∂ψ (Ω) = ψ ( ∂ Ω ∪ {∞} ).Thus, f maps continuously the unit circle T except one point to ∂ Ω and thelimit of u at this point is 0.Hence, u can be continuously extended on D and thus u | T will be con-tinuous and take values inside the interval [ − , u of u , restricted on the open disc D , will take every value of theinterval (0 , + ∞ ) and thus it is unbounded. In other words u ∈ Z and theproof is complete. We will now extend the results of the preceding section in the context offinitely many as well as infinite countably many complex variables. Let I be a finite or countably infinite set. We will consider functions defined onthe closed polydisc D I = { ( z i ) i ∈ I : | z i | ≤ , for every i ∈ I } (29)which belong in the algebra of this polydisc A ( D I ), i.e. they are holomorphicin D I and continuous on D I , where D I is endowed with the cartesian topol-ogy. The distinguished boundary of this polydisc is the set T I and a function F : T I → C can equivalently be considered as a function F : R I → C whichis 2 π − periodic in every principal direction. We are interested in functions f ∈ A ( D I ) such that their restriction f | T I has no directional derivatives inany point of T I for a large set of directions, the largest we can prove. Themain theorem of the preceding section takes the following form: Theorem 3.1.
Let I be a finite or countably infinite set. There is function f ∈ A ( D I ) such that for every point θ ∈ T I and every direction v ∈ R I with k v k ∞ = 1 both the functions u = Ref | T I and ˜ u = Imf | T I are notdifferentiable at θ in the direction v . In fact, the class of such functions f is residual in A ( D I ) . Following the steps of Section 2, for n ∈ N we consider the sets D n = n u ∈ C R ( T I ) : for every θ ∈ R I and every direction v ∈ R I with k v k ∞ = 1 there is a y ∈ (cid:18) θ − n · v, θ + 1 n · v (cid:19) such that | u ( y ) − u ( θ ) | > n k y − θ k ∞ o , (30)and E n = { f ∈ A ( D I ) : Ref | T I ∈ D n } . (31)Since in the definition of D n we are interested in directions v ∈ R I with k v k ∞ = 1 one can easily see that D n = \ k ∈ I D ( k ) n , (32)where each D ( k ) n is defined as follows: D ( k ) n = n u ∈ C R ( T I ) : for every θ ∈ R I and every direction v = ( v i ) i ∈ I ∈ R I with k v k ∞ = 1 and | v k | ≥
12 there is a y ∈ (cid:18) θ − n · v, θ + 1 n · v (cid:19) such that | u ( y ) − u ( θ ) | > n k y − θ k ∞ o . (33)Afterwards, we define naturally E ( k ) n = { f ∈ A ( D I ) : Ref ∈ D ( k ) n } . (34)0 Proceeding as in the proof of Theorem 2.1 our objective is to show thatthe set ∞ \ n =1 (cid:0) E n ∩ iE n (cid:1) = ∞ \ n =1 \ k ∈ I (cid:0) E ( k ) n ∩ iE ( k ) n (cid:1) (35)is G δ dense and its elements satisfy the desired properties of Theorem 3.1.We first prove that: Lemma 3.2.
For every n ∈ N and k ∈ I , E ( k ) n is an open set in A ( D I ) ,endowed with the supremum norm.Proof. As in section 2, in order to prove that A ( D I ) \ E ( k ) n is closed weconsider a sequence { f m } such that f m / ∈ E ( k ) n for every m and a function f ∈ A ( D I ) such that f m → f uniformly on D I . Since f m / ∈ E ( k ) n , for each m ,there is a θ ( m ) ∈ T I and a direction v ( m ) = (cid:0) v ( m ) i (cid:1) i ∈ I ∈ R I with k v ( m ) k ∞ = 1and | v ( m ) k | ≥ such that | u m ( y ) − u m ( θ ( m ) ) | ≤ n k y − θ ( m ) k ∞ , (36)for every y ∈ (cid:0) θ ( m ) − n · v ( m ) , θ ( m ) + n · v ( m ) (cid:1) , where u m = Ref m | T I . Butboth T I and { v = ( v i ) i ∈ I ∈ R I : k v k ∞ = 1 and | v k | ≥ } (equipped withthe product topologies) are metrizable (since I is at most countable) andcompact from Tychonoff’s Theorem. Hence there is a strictly increasingsequence of positive integers { k n } , a point θ ∈ T I and a direction v =( v i ) i ∈ I ∈ R I with k v k ∞ = 1 and | v k | ≥ such that θ ( k m ) → θ and v ( k m ) → v .If y ∈ (cid:0) θ − n · v, θ + n · v (cid:1) there is some − n < s < n such that y = θ + s · v = lim m →∞ (cid:0) θ ( k m ) + s · v ( k m ) (cid:1) . (37)Thus, applying (36) for these indices { k m } , y ( k m ) = θ ( k m ) + sv ( k m ) and letting m → ∞ we conclude that | u ( y ) − u ( θ ) | ≤ n k y − θ k ∞ , (38)since the convergence of { f m } to f is uniform (again u = Ref | T I ). Hence(38) holds for every y ∈ (cid:0) θ − n · v, θ + n · v (cid:1) and thus f / ∈ E ( k ) n . So A ( D I ) \ E ( k ) n is closed or equivalently E ( k ) n is open.Hence, the intersection T n,k E ( k ) n is a G δ set in A ( D I ). Lemma 3.3.
For every k ∈ I , the set S ( k ) = ∞ \ n =1 E ( k ) n (39) is dense in A ( D I ) . Proof.
Fix some k ∈ I . We consider the function f k : D I → C defined by f k (cid:0) ( z i ) i ∈ I (cid:1) = f ( z k ) , (40)where f ∈ A ( D ) is the function defined in the proof of Lemma 2.3. It isobvious that f k ∈ A ( D I ); we will prove that f k ∈ S ( k ) . Indeed, for some θ = ( θ i ) i ∈ I ∈ T I and v = ( v i ) i ∈ I with k v k ∞ = 1 and | v k | ≥ the quotient (cid:12)(cid:12)(cid:12)(cid:12) Ref k ( θ + tv ) − Ref k ( θ ) t (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) u ( θ k + tv k ) − u ( θ k ) t (cid:12)(cid:12)(cid:12)(cid:12) (41)does not remain bounded as t → v k = 0 and u has the propertiesmentioned in the previous section. Hence, S ( k ) = ∅ .Next, as in Lemma 2.3, we will prove that for every complex polynomial p (cid:0) ( z i ) i ∈ I (cid:1) the function f k + p is also in S ( k ) . We remind the reader that p is a polynomial in a set of variables ( z i ) i ∈ I if it is a polynomial in the set( z j ) j ∈ J for some finite subset J ⊆ I . Let θ ∈ T I and v = ( v i ) i ∈ I ∈ R I with k v k ∞ = 1 and | v k | ≥ . The function h : [ − , → R defined by h ( t ) = u ( θ + tv ) , (42)where u = Rep | T I , is of course of class C ∞ and thus its derivative h ′ isbounded on [ − ,
1] by a constant
M >
0. Thus, for every y ∈ [ θ − v, θ + v ], y = θ there is a ξ ∈ ( − ,
1) such that | u ( y ) − u ( θ ) |k y − θ k ∞ = | h ′ ( ξ ) | ≤ M. (43)Thus, proceeding now as in the proof of Lemma 2.3 we can easily prove that f k + p ∈ S ( k ) . Then, using the fact that the space of polynomials is dense in A ( D I ) (see [11] for example) we conclude that also S ( k ) is dense A ( D I ).Combining now these two lemmas, we conclude that each S ( k ) is a G δ dense subset of A ( D I ) and thus, since I is at most countable, from Baire’sCategory Theorem (in the complete metric space A ( D I )) the same holds fortheir intersection S = \ k ∈ I S ( k ) = ∞ \ n =1 \ k ∈ I E ( k ) n . (44)From this, the analogue of Lemma 2.4 follows immediately: the set T = S ∩ i S = ∞ \ n =1 \ k ∈ I (cid:0) E ( k ) n ∩ iE ( k ) n (cid:1) (45)is also a G δ dense subset of A ( D I ). With this result at hand we proceednow to the proof of the main result of this section:2 Proof of Theorem 3.1.
Let E be the class of functions f ∈ A ( D I ) such thatboth the real part u and the imaginary part ˜ u of f are nowhere differentiablein the sense of Theorem 3.1. We will prove that T ⊆ E and from this ourresult will follow.Let f ∈ T , a point θ ∈ T I and a direction v ∈ R I with k v k ∞ = 1. Since f ∈ S , for every n there is a y n ∈ (cid:0) θ − n · v, θ + n · v (cid:1) , y n = θ such that | u ( y n ) − u ( θ ) |k y n − θ k ∞ > n, (46)where u = Ref | T I . Hence y n → θ and (46) yields that u is not differentiableat θ in the direction v . Using the fact that f ∈ i S we have the sameconclusion for the function ˜ u = Imf | T I and the proof is complete. ✷ Remark 3.4. If I is an uncountable set the result of Theorem 3.1 cannothold: it is true that if f : D I → C is a continuous function, then f dependsonly on a countable number of coordinates (see [11] for example). Thus,if z j is a coordinate which does not belong in this countable set then thepartial derivative of f with respect to z j will be of course zero at everypoint of T I .Nevertheless, one can prove that generically for every f ∈ A ( D I ) thefollowing holds: For every n ∈ N , there is a countably infinite subset J n ⊆ I such that forevery point θ ∈ T I and every direction v ∈ R I with k v k ∞ = 1 and v | J n there are points y , y ∈ (cid:0) θ − n · v, θ + n · v (cid:1) with | u ( y ) − u ( θ ) | > n k y − θ k ∞ and | ˜ u ( y ) − ˜ u ( θ ) | > n k y − θ k ∞ . (47) where u = Ref | T I and ˜ u = Imf | T I . One can also prove that, the set of functions such that the above sets J n coincide is dense in A ( D I ). However, this set may not be G δ , and thus thiswould not be a generic property of A ( D I ). Aknowledgements:
Both authors would like to express their gratidute toProfessor Vassili Nestoridis for presenting the problem as well as for his helpand guidance throughout the creation of this paper.
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Department of Mathematics, University of Athens,Panepistimioupolis, 157 84, Athens, Greece.E-mail: [email protected]
Konstantinos Makridis: