Boundary behaviour of functions which possess universal Taylor series Stephen J. Gardiner Abstract It is known that, for any simply connected proper subdomain of the complex plane and any point in , there are holomorphic func- tions on that possess “universal” Taylor series expansions about ; that is, partial sums of the Taylor series approximate arbitrary polyno- mials on arbitrary compacta in C that have connected complement. This paper establishes a strong unboundednessn property for such func- tions near every boundary point. The result is new even in the case of the disc, where it strengthens work of several authors. 1 Introduction Throughout this paper denotes a simply connected proper subdomain of the complex plane C and is a point in . A holomorphic function f on is said to belong to the collection ( , ), of functions with universal Taylor series expansions about , if the partialU sums N (n) f () n SN (f, ) = (z ) n! n=0 X of the power series expansion of f about have the following property: for every compact set K C with connected complement and every func- n tion g which is continuous on K and holomorphic on K, there is a subse- quence (SNk (f, )) that converges to g uniformly on K. Nestoridis [16], [17] has shown that possession of such universal Taylor se- ries expansions is a generic property of holomorphic functions on simply connected domains; that is, ( , ) is a dense G subset of the space of all holomorphic functions on U endowed with the topology of local uniform 0 2010 Mathematics Subject Classification 30B30, 30E10. This research was supported by Science Foundation Ireland under Grant 09/RFP/MTH2149, and is also part of the programme of the ESF Network “Har- monic and Complex Analysis and Applications”(HCAA). 1 convergence. Further, Müller, Vlachou and Yavrian [14] (see also Theorem 9.1 in [12]) have shown that the collection ( , ) is independent of the choice of the centre of expansion . U A significant avenue of investigation concerns the boundary behaviour of functions in ( , ): see [17], [13], [6], [10], [12], [5], [14], [1], [3], [4]. For U example, in the case where is the unit disc D, Nestoridis [17] showed that (D, 0) does not intersect the Hardy space H1(D). This was subsequently strengthenedU by Melas, Nestoridis and Papadoperakis [13], as follows. Theorem A The collection (D, 0) does not intersect the Nevanlinna class; U that is, for every f (D, 0), we have 2 U 2 log+ f(rei) d (r 1 ). (1) j j ! 1 ! Z0 Bayart [3] considered the local boundary behaviour of functions in (D, 0) and answered a question raised by Armitage and Costakis [1] by establish-U ing the next result. We recall that a subset Z of the unit circle @D is called residual if @D Z is of first Baire category relative to @D. n Theorem B If f (D, 0), then there is a residual subset Z of @D such that f(rw) : 0 < r2 < U1 is unbounded for every w Z. f g 2 A complementary result of Armitage and Costakis [1] tells us that, for any set Y @D which is of first category relative to @D, there are functions f in (D, 0) such that f(rw) : 0 < r < 1 is bounded for every w Y . ThereU is no implicationf in either directiong between the conclusions2 of Theorems A and B. Much less is known about boundary behaviour in the case of general sim- ply connected domains . Müller, Vlachou and Yavrian [14] (see also [12]) have shown that no member of ( , ) is holomorphically extendable be- yond . However, it has remainedU an open question whether such functions are necessarily unbounded. (I am grateful to Vassili Nestoridis for drawing this problem to my attention.) The purpose of this paper is to show that functions with universal Taylor series expansions have a strong unbounded- ness property near every boundary point. We denote by D(w, r) the open disc of centre w and radius r. Theorem 1 Let f ( , ). Then, for any w @ , any r > 0 and any component U of D(w,2 r U) , the function log+ f 2does not have a harmonic \ j j majorant on U, and thus C f(U) is polar. n Corollary 2 If f ( , ), then f is unbounded near every point of @ . 2 U 2 Theorem 1 is new even in the case of the disc. It contains Theorem A because condition (1) is equivalent to saying that the subharmonic function + log f does not have a harmonic majorant on the whole of D. The special j j case of the corollary where = D is equivalent to Theorem B because, by the Collingwood Maximality Theorem (Proposition 2.1 in [18]), the radial cluster set of f agrees with the full cluster set at a residual subset of @D. Theorem 1 fails if we drop our overall assumption that is simply connected, as is clear from Corollary 1 in Melas [11]. The proof of Theorem 1 presents challenges because no special assump- tions are made concerning the boundary of . In particular, since we have not assumed that @ is locally connected, we cannot rely on having a con- formal mapping g : D with continuous extension to D. We will make extensive use of potential! theoretic tools to overcome such diffi culties, and refer to the book [2] for the background theory. 2 A preparatory lemma Let GU denote the Green function of a non-empty open set U C when it exists, that is, when the complement of U is non-polar. Under[ f1g the U same assumption we denote by H the (Perron-Wiener-Brelot) solution to the Dirichlet problem on U with resolutive boundary function . We also write A for the characteristic function valued 1 on a set A and 0 elsewhere. Lemma 3 Let U be a simply connected proper subdomain of C and 0 U, 2 and let ! be a non-empty open subset of U. Suppose that (vk) is a decreasing sequence of harmonic functions on U such that v1/GU (0, ) is bounded above on ! and lim v < 0 on U. If H! 1, then there exists· k such that k U 0 N 6 !1 2 the open set !1 = ! vk < 0 is non-empty and H 1. \ f 0 g U 6 If A U and u is a positive superharmonic function on U, we define the reduced function URA = inf w : w is positive and superharmonic on U and w u on A . u f ≥ g U U ! ! U U ! In the context of the above lemma, since R n = H on ! and R n = 1 1 U 1 ! U U ! on U !, the hypothesis that H 1 is equivalent to saying that R n 1. n U 6 1 6 Now let H denote the upper halfplane and g : H U be a conformal map- ! ping such that g(i) = 0. Since superharmonicity is preserved by conformal mappings, 1 U U ! H g (!) R n g = HR n . 1 1 1 U U ! H g (!) Thus R n 1 if and only if HR n 1. Also, GU ( , g(z)) = G (i, z). 1 6 1 6 0 H It is therefore enough to prove the lemma in the case where U = H and 3 H H ! 0 = i. Starting with the assumption that R1 n 1, we will show that H !1 6 HR n 1 for a suitable choice of k . 1 6 0 Let Z = R 0 3. For any function : H [ , + ] we define f g ! 1 1 (x1, y(x2, x3, x4)) 4 (x1, ..., x4) = ((x1, ..., x4) R Z), y(x2, x3, x4) 2 n where 2 2 2 y(x2, x3, x4) = x2 + x3 + x4. q If C2(H), then a straightforward calculation shows that 2 ( )(x1, y(x2, x3, x4)) ( )(x1, ..., x4) = . (2) y(x2, x3, x4) Thus the transformation preserves harmonicity, and even (by a smoothing argument) superharmonicity.7! In fact, it establishes a one-to- one correspondence between superharmonic functions on H and Z-axially symmetric superharmonic functions on R4 Z, the inverse transformation being given by , where n 7! y y(x1, x2) = x2 (x1, x2, 0, 0) (cf. [9]). The function u defined by 1 u(x1, ..., x4) = y(x2, x3, x4) is superharmonic on R4 and harmonic on R4 Z. Further, by the above H ! n transformation, our hypothesis that HR n 1 implies that 1 6 R4 Z R4 (W Z) n R n [ u, u 6 where W = (x1, ..., x4):(x1, y(x2, x3, x4)) ! . f 2 g Indeed, since the polar set Z is removable for positive superharmonic func- tions on R4 Z, we have n R4 R4 W R n u. (3) u 6 The function u can be expressed as a Newtonian potential, 1 u = u dt, (4) t ZR where 1 ut(x1, ..., x4) = 2 2 2 2 (t R). (x1 t) + x + x + x 2 2 3 4 4 Hence 4 4 1 4 4 R RR W = R RR W dt. (5) u n ut n ZR (The interchange of the integration and the reduction is justified by Tonelli’s theorem because, as we have previously noted, in the set W the reduction can be expressed as a Dirichlet solution and so can be viewed as an integral against harmonic measure for W .) Let denote Lebesgue measure on R and let A be the Borel set defined by R4 R4 W A = t R : Ru n ut . 2 t 6 n o Then (3), (4) and (5) together show that (A) > 0.