Nowhere differentiable functions of analytic type on products of finitely connected planar domains

Using Laurent decomposition and Mergelyan's theorem combined with Baire's category theorem, we prove generic nowhere differentiability on the distinguished boundary of functions of analytic type on products of planar domains bounded by finitely many disjoint Jordan curves. The parametrization of the boundaries are those induced by the natural Riemann maps.


Introduction
In the nineteenth century, it was widely believed that every continuous function defined on R has derivative at a significant set of points. Weierstrass was the first one to disprove this claim by providing an explicit example of a function u : R → R which is continuous, 2π-periodic, but not differentiable at any real number. Since then the existence of such functions attracted the interest of many mathematicians.
In fact, as Banach and Mazurkiewicz independently proved, in [1] and [6] respectively, generically every continuous function on a compact interval J of R is nowhere differentiable. In [2], the above result has been generalized for functions in the disc algebra A(D), i.e. continuous functions defined on the closed unit disc D, which are also holomorphic on D. In particular, it is proved that generically for every function f ∈ A(D) both functions Ref ↾ T and Imf ↾ T are nowhere differentiable. The result is even extended for functions f : D I → C that belong to the algebra A(D I ), namely f is continuous on D I , which is endowed with the cartesian topology, and separately holomorphic on D I .
In section 4, we prove a similar result in the case of a domain Ω bounded by a finite number of disjoint Jordan curves. If Ω 0 , Ω 1 , ...Ω n are the complements of the connected components of (C ∪ {∞}) \ Ω, then the Riemann mappings φ i : D → Ω i , i = 1, ..., n yield a parametrization of ∂Ω i , i = 0, 1, ..., n, with parameter θ ∈ R. We will prove that generically for every f ∈ A(Ω) the functions Re(f • φ i ) ↾ T and Im(f • φ i ) ↾ T are nowhere differentiable with respect to θ ∈ R. In order to do so, we combine the Laurent decomposition of every function f ∈ A(Ω) with the results of the previous sections which deal with the case of the disc or more generally with the case of Jordan domains.
In section 5, we generalize the above result in the context of several complex variables. Specifically, if we consider sets Ω which are the finite or countably infinite cartesian products of domains in C as in section 4, then a parametrization of the distinguish boundary of Ω arises from the Riemann mappings of the factors Ω i of the product Ω. We prove that, generically, for every f ∈ A(Ω) neither the real part Ref , nor the imaginary part Imf have directional derivatives at any point of the distinguished boundary of Ω = i∈I Ω i with respect to the above parametrization, for any direction v ∈ l ∞ \ {0}. We first treat the case of finite products that is when the set I is finite. In the case of infinitely many variables, we use the result for the case of finitely many variables and the fact that every f ∈ A(Ω) can be uniformly approximated by functions depending on a finite set of variables ( [3], [5]).

The open unit disk
We denote the open unit disk in C by D and T = ∂D. Let Ω be a domain in C.
Although the only domains we will deal with in this sections are D andĈ \ D we will give a general definition which will be useful more generally.
it is continuous on Ω and holomorphic on Ω. We endow A(Ω) with the topology of uniform convergence on compact subsets of Ω, where the closure of Ω is taken in C.
We note that a function defined on T can equivalently be thought as a 2π-periodic function defined on R. Thus, by abuse of notation we write u(y) instead of u(e iy ) for y ∈ R.
In [2], it has been proven that the set of functions f ∈ A(D) such that f ↾ T belongs to Z, which we denote by Z(D), is a dense and G δ subset of the disk algebra A(D).
It will be useful to restrict ourselves to functions f ∈ A(D), such that f (0) = 0. We denote the space of such functions by A 0 (D) and we endow it with the subspace topology. Using the above result we are going to prove that the functions f ∈ A 0 (D) such that f ↾ T ∈ Z is dense and G δ in A 0 (D). Denote this set by Z 0 (D). Proposition 2.3. The set of functions f ∈ A 0 (D) such that f ↾ T belongs to Z is a dense and G δ subset of A 0 (D).
Proof. Observe that Z 0 (D) = Z(D) ∩ A 0 (D). Consider the mapping L is obviously continuous and it is easy to see that It is now obvious thatg(z) = g(z) − g(0) ∈ A 0 (D) andg ∈ Z 0 (D). Using the triangular inequality we deduce that which completes the proof.
We will now prove similar results for a subspace of A(Ĉ\D) denoted by A 0 (Ĉ\D). In general, if the domain Ω is the complement of a compact set, then a function f belongs to A 0 (Ω) if f ∈ A(Ω) and Proof. Consider the mappings We can easily see that l is continuous, onto A 0 (D) and Hence, l is an isometric isomorphism, which implies that the set We will now prove that Z 0 (Ĉ\D) coincides with the set of functions f ∈ A 0 (Ĉ\D) such that f ↾ T ∈ Z. This derives from the following equalities: where u, v are the real and imaginary part of a function f ∈ A 0 (D) respectively and θ, y ∈ R are such that y = θ.

Jordan Domains
In this section we replace the unit circle T by a Jordan curve J and we prove similar results for functions defined on the domain Ω, which is the bounded or unbounded connected component of the complement inĈ of J. For such domains Ω, we consider a Riemann mapping φ : D → Ω and we claim that generically for every function In what follows, we prove the aforementioned assertions.
It is easy to verify that L is an isometric isomorphism, which implies that the set L −1 (Z(D)) is a dense and G δ subset of A(Ω).
Proof. Consider the mapping Similarly to Proposition 2.4 the desired set is a dense and G δ subset of A(Ω ∪ {∞}).

Domains bounded by a finite number of disjoint Jordan curves
Now we consider a more general setting, where Ω is a bounded domain whose boundary consists of a finite number of disjoint Jordan curves. If V 0 , ..., V n−1 are the connected components ofĈ \Ω, ∞ ∈ V 0 and Ω 0 =Ĉ \V 0 , ..., Ω n−1 =Ĉ \V n−1 , then there exist Riemann mappings φ i : For every k ∈ N and i ∈ {0, 1, ..., n − 1} consider the sets k is an open subset of A(Ω).
Proof. Let k ∈ N and i ∈ {0, 1, ..., n − 1}. We will prove equivalently that where u m = Ref m . Since every u m is 2π-periodic we may assume that every θ m ∈ [0, 2π] and thus the sequence {θ m } has a convergent subsequence. Without loss of generality we assume that {θ m } converges to a θ ∈ [0, 2π].
Let y ∈ θ, θ + 1 k . We can easily deduce that there exists m 0 ∈ N, such that for every m ≥ m 0 it is true that y ∈ θ m , θ m + 1 k . Thus, applying (4.3) and using the convergence of {θ m } to θ and the uniform convergence of {f m } to f we have that where u = Ref . Hence (4.4) holds for every y ∈ θ, θ + 1 k and thus f ∈ A(Ω)\E k is a closed subset of A(Ω).
is a dense and G δ subset of A(Ω).

Proof. Lemma 4.1 indicates that the sets
are G δ . We will prove that S 0 is dense in A(Ω).
Similarly, the sets S 1 , S 2 , ..., S n−1 are dense and Baire's theorem completes the proof. Proof. It is easy to see that S ∩ iS is a subset of all functions f ∈ A(Ω) such that for every i ∈ {0, 1, ..., n − 1} the functions (Ref • φ i )↾ T and (Imf • φ i )↾ T have no derivatives at any point. Since S is G δ and dense in A(Ω) and multiplication by i is an automorphism of A(Ω), we conclude that iS is G δ and dense in A(Ω). Applying Baire's category theorem we have that S ∩ iS is G δ dense in A(Ω).

The case of several variables
Let I be a non empty finite or countably infinite set and {Ω i } i∈I be a family of domains in C, all of them bounded by a finite number of disjoint Jordan curves as in section 4. We denote by k i the degree of connectivity of Ω i . In accordance to the notation of the previous sections, we denote by V i,j , i ∈ I, j ∈ {0, ..., k i − 1} the connected components ofĈ \ Ω i and Ω i,j =Ĉ \ V i,j . Let also ∞ ∈ Ω i,0 and φ i,j : D → Ω i,j be some fixed Riemann maps.
We consider the spaces Ω = i∈I Ω i , Ω = i∈I Ω i and b 0 Ω = i∈I ∂Ω i , the distinguished boundary of Ω, all of them endowed with the product topology. According to [3], [4] and [5] we give the following definition.
Definition 5.1. A function f : Ω → C belongs to the space A(Ω) if it is continuous on Ω, endowed with the product topology, and separately holomorphic on Ω. By separately holomorphic we mean that if all coordinates but one are fixed, say the coordinate i 0 , then the restriction of f is a holomorphic function of the variable z i 0 in Ω i 0 . We endow the space A(Ω) with the supremum norm.
We also consider the set and the functions We are going to prove that, generically, for every f ∈ A(Ω) the functions Re(f • φ r ) ↾ T I and Im(f • φ r ) ↾ T I have no directional derivatives for any r ∈ R and any direction v ∈ ∞ (I) \ {0}. In order to do so, we will first treat the case where I is a finite set and separately the case where I is a countably infinite set.

Finitely many variables
Let I be a non empty finite set.
We consider the sets D n,k,s (I) = u ∈ C R (T I ) : for every θ ∈ R I and every direction (5.3) v ∈ R I with 1 s ≤ |v k | ≤ s and |v j | ≤ s for every j ∈ I there exists for some n, s ∈ N and k ∈ I. Also consider: for some r ∈ R.
Lemma 5.2. For every n ∈ I and r ∈ R the set E Proof. We will prove that the set A(Ω) \ E k | ≤ s, such that: But T I and the set {v ∈ R I : |v j | ≤ s, ∀j ∈ I and 1 s ≤ |v k | ≤ s} are metrizable and compact by Tychonoff's theorem. Thus, the sequences {θ (m) } m∈N and {v (m) } m∈N have converging subsequences. Without loss of generality we assume that θ (m) → θ ∈ R I and v (m) → v ∈ R I with |v j | ≤ s for all j ∈ I and 1 s ≤ |v k | ≤ s.
Since f m converges uniformly to f , it follows that u m converge uniformly to u, where u = Ref and φ r is continuous which implies that φ r (θ (m) +tv (m) ) → φ r (θ+tv), as m → ∞. The above combined with (5.5) imply that and therefore f ∈ A(Ω) \ E Proof. Let us show that S (r) k,s (I) is non empty. We have proved in a previous section that there is a function g ∈ A(Ω k ) such that lim sup for all θ ∈ R. The above implies that for all θ ∈ R: for any α ∈ R with 1 s ≤ |α| ≤ s.
Now consider the function f : Ω −→ C which maps (z i ) i∈I to g(z k ). We will prove that f ∈ S (r) k,s (I). Let θ ∈ R I and v ∈ R I such that |v j | ≤ s for all j ∈ I and 1 s ≤ |v k | ≤ s. We observe that is a dense and G δ subset of A(Ω).
Consequently, S ∩ iS is dense and G δ in A(Ω) and it is a subset of the set of all functions f ∈ A(Ω) such that both Re(f • φ r ) ↾ T I and Im(f • φ r ) ↾ T I are nowhere differentiable for any r ∈ R and any direction v ∈ l ∞ (I) \ {0}.

Infinitely many variables
Let I be an infinitely countable set.
We consider again the sets D n,k,s (I) = {u ∈ C R (T I ) : for every θ ∈ R I and every direction (5.18) v ∈ R I with |v j | ≤ s for any j ∈ I and 1 s ≤ |v k | ≤ s there exists for n, s ∈ N and k ∈ I. Proof. We will prove that the set A(Ω) \ E n,k,s (I) is closed. We consider a sequence {f m } such that every f m ∈ A(Ω) \ E n,k,s (I), which converges to a function f ∈ A(Ω). For each m there exist r (m) ∈ R, θ (m) ∈ T N and a direction v (m) ∈ R N with |v (m) j | ≤ s for all j ∈ I and 1 s ≤ |v the set {v ∈ R N : |v j | ≤ s, ∀j ∈ I and 1 s ≤ |v k | ≤ s} are metrizable and compact by Tychonoff's theorem. Thus, they have converging subsequences. Without loss of generality we assume that r (m) → r ∈ R, θ (m) → θ ∈ R N and v (m) → v, where v ∈ R N and |v j | ≤ s for all j ∈ I and 1 s ≤ |v k | ≤ s.
If y ∈ θ − 1 n v, θ + 1 n v , then there exists a t ∈ R with − 1 n < t < 1 n such that: Let i ∈ I. Since R = i∈I {0, ..., k i − 1} is a product of finite sets and r (m) → r ∈ R, there exists m 0 ∈ N, such that Since φ i,r i is continuous, we have that in the product topology. Therefore, applying (3.5), using (3.9) and letting m → ∞ we conclude that is a dense and G δ subset of A(Ω).
Proof. Lemma 5.5 implies that S is a G δ . Thus, we only have to prove that S is dense. Let F ∈ A(Ω). From [3] there exist a sequence of finite sets {I n }, I n ⊂ I, and a sequence {F n } in A(Ω), such that each F n depends only on the variables z i , i ∈ I n and F n → F . Let f n be the restriction of F n to i∈In Ω i . From Theorem 5.4 there exists a sequence g n ∈ S(I n ) = r∈R(In) k∈In ∞ s=1 S (r) k,s (I n ), where R(I n ) = i∈In {0, ..., k i − 1}, such that f n − g n ∞ → 0. It is easy to see that the functions G n : Ω → C which map (z i ) i∈I to g n ((z i ) i∈In ) respectively belong to S(I) = k∈I ∞ s=1 Remark 5.7. In [2] it is claimed that the set of functions f ∈ A(D N ) for which for every θ ∈ R N and every direction v ∈ R N with v ∞ = 1 and 1 2 ≤ |v k | ≤ 1 for a fixed k ∈ N there exists y ∈ θ − 1 n v, θ + 1 n v such that |u(θ) − u(y)| > n y − θ ∞ (5.

28)
is open in A(D N ). The authors' proof uses the wrong statement that the set {v ∈ R N with v ∞ = 1 and 1 2 ≤ |v k | ≤ 1} for a fixed k ∈ N is compact in R N endowed with the product topology. However, their proof works just fine for every compact set of directions with non zero k-coordinate. According to our Theorem 5.6 their statement is correct.