AN INDESTRUCTIBLE BLASCHKE PRODUCT IN THE LITTLE BLOCH SPACE

CHRISTOPHER J . BISHOP The little Bloch space, 130 , is the space of all holomorphic functions f on the unit disk such that lim1 z 1l (f'(z)j(1 Iz12) = 0. Finite Blaschke products are clearly in 130, but examples of infinite products in 80 are more difficult to obtain (there are now several constructions due to Sarason, Stephenson and the author, among others) . Stephenson has asked whether 130 contains an infinite, indestructible Blaschke product, Le., a Blaschke product B so that (B(z) a)/(1 QB(z)), is also a Blaschke product for every a E D. In this paper we give an afirmative answer to his question by constructing such a Blaschke product. We also answer a question of Carmona and Cufí by constructing a VMO function, f, so that Ilf J¡ . = 1 and whose range set, R(f, a) = {w : there exists zn, ~ a, f(z~) = w}, equals the open unit disk for every a E T .

where E(1 -Izn1) G oo. Finite Blaschke products are clearly in BO , but examples of infinite products in BO are not so obvious .Such functions do exist, as is shown in [1], [8], [10] .A more explicit example, as well as a characterization of such products in terms of the zero sequence {zn}, has been given in [3] .That result answers several questions about 130, but does not resolve the following question from [10] : does Bo contain an infinite, indestructible Blaschke product, Le., a Blaschke product B so that _ B(z) -a Ba(z) 1 -úB(z)' is also a Blaschke product for every a E D?
The question arises because of Frostman's theorem.An inner function F is a holomorphic function on D with boundary values of absolute value 1 a.e. on T .Any such function can be written as a product _ I I ¡o F(z) = B(z)S(z) = (fl 1 n zn z~)(exp(-f e ie ± zdp(e)), of a Blaschke product and a singular inner function (M is a finite, positive measure singular to dB) .Frostman's theorem states that for any inner function F on D, Fa F(z) -a (z) = 1 -úF(z)' is a Blaschke product for every a E D\E where E is an exceptional set of zero logarithmic capacity [5,Theorem 11 .6 .4] .The constructions of Blaschke products in [8], [10] first build an inner function in Bo and then apply Frostman's theorem.Stephenson asked if this was unavoidable, e.g., does 13o contain any indestructible Blaschke products?The example in [3] is built by constructiog the zero set, so does not use Frostman's theorem.Furthermore, a variant of Stephenson's construction gives a Blaschke product without using Frostman's theorem (see next section) .In this note we expand on this observation to give a "cut and paste" construction of an indestructible Blaschke product in 130 .
One could also try to produce such an example by finding a sufficient condition on the zeros for the product to be indestructible and which includes some sequences satisfying the 13o condition from [3] .One such sufficient condition for indestructibility is that the sequence be thin, Le., zk -zn kan 1 -znzk However, this condition is incompatible with the BO condition.An even more ambitious problem is to characterize indestructibility in terms of the zero-set .In [6], Morse has constructed a destructible Blaschke product which becomes indestructible when a single point is deleted from its zero-set .This indicates any characterization of indestructibility in terms of the zero set would be very delicate (and probably very dificult).A related problem has been solved, however.In some sense, finite products of interpolating Blaschke products are the "conformally invariant" class of Blaschke products.In [7] Nicolau has given a zeroset characterization of those Blaschke products B so that Ba is a finite product of interpolating Blaschke products for every a in the disk.Thus he has solved the conformally invariant version of the problem of characterizing indestructibility.
Our construction gives a Blaschke product whose singular set (the accumulation set of the zeros) has measure zero.If we could construct an example whose singular set was the entire circle, this function would also have the property that its range set, R(f, a) = {w : there exists zna, f (z n) = w}, equals the whose disk for every a E T. Carmona and   Cufí had asked in [4] if there was a function in H°°n BO with this property.I believe the construction can be modified to give such an example, but rather than do this, I will sketch the construction of a function f E H°°n VMO with 11 f jj,, = 1 and R(f, a) = D for every a E T. Since VMO C BO, this is an even stronger result (again answering a question of Carmona and Cufí).I thank Arturo Nicolau for bringing this question to my attention and our discussions on it.I also thank the referee for his helpful comments .His suggestions have clarified the exposition in several places .

. The BO construction
The idea is quite simple; we will build a simply connected Riemann surface by taking copies of the unit disk with slits and "gluing" different copies along the slits.A simple example of this idea is to take infinitely many copies of D\ [2,1), and identifying the "top" edge of one copy with the "bottom" edge of the next .See  et So be the initial sheet, which we also refer to as the "zeroth sheet" .This sheet contains a point corresponding to zero in the unit disk and we refer to this point as "0" on the surface .Let S, be nth stage of this construction (the union of copies -n to n) and S = UnSn the limiting surface .For each of there surfaces the point "0" refers to the point 0 on So.In the rest of this paper we shall assume that any Riemann mapping of the unit disk to a constructed surface like Sn or S maps 0 in the disk to the point 0 on the surface .Whenever we talk about harmonic on the surface it is the push forward of normalized Lebesgue measure on the circle under such a Riemann mapping, Le., harmonic measure will always be with respect to the point 0 on the zeroth sheet .
S is simply connected so there is a Riemann mapping ob : D --> S and there is an obvious holomorphic projection P : S -3 D. We claim that F = P o D must be an inner function because all the harmonic measure for S lives on the part of the boundary above the the unit circle .To prove this we consider Sn and show that the harmonic measure of the two radial slits in its boundary are O(ñ).To do this we map S, to a half infinite strip W = {(x, y) : -oo < x < 0, -(2n -f-1)7r < y < (2n + 1)7r} by the mapping z --> log( z 2 ) 1-2z which has a well defined branch on S,.The point 0 on the surface is mapped to -1/2 and the radial edges are mapped to the horizontal edges of the strip.Standard estimates (e.g., map the strip to a halfplane via sin(z/i) and use the Poisson integral) show that the harmonic measure of the horizontal edges of the strip with respect to the point -1/2 are approximately 1/n.By conformal invariance of harmonic measure the claim about Sn is proved .Thus the circular part of OSn has measure >_ 1 -C/n.Taking n --> oo we see that the circular part of OS has full measure, Le, ¡Fl = 1 a.e. on the unit circle .
In fact, F must be Blaschke product .To see this, recall that a function f in the unit ball of HI(D) is a Blaschke product iff the least harmonic majorant of log lf 1 is 0 (e .g., [5,Theorem II.2.4]) .This says that F is a Blaschke product iff 0 is the least harmonic majorant of log ¡P(z) 1 on S. Let u be the least harmonic majorant of logjP(z)1 on S. Then u restricted to S,,, is harmonic and has boundary values 0 on P-1(T) and >_ log 2 on the two radial slits in OS, .Thus 0 >_ u(0) >_ 2 log 2 .Since this holds for any n, u(0) = 0 (recall that the "0" in u(0) refers to the designated point on the zeroth sheet) .Since u is nonpositive this implies u -0 and so F is a Blaschke product .
Let a E D and Ta(z) = (z -a)/(1 -áz) .An argument like the one above shows shows that if a :,¿ 2 then Fa = Ta o F is a Blaschke product .However, since no point of S covers the point { 2 }, F2 is never zero, so must be a singular inner function (in fact, since F is continuous except for one boundary point, up to rotations it must be exp(A i+z ), for some A > 0, Le., the singular inner function corresponding to a positive point mass) .
To build an indestructible Blaschke product we will have to vary the construction, adding sheets which cover the omitted points of earlier generations, and in particular, so that the least harmonic majorant of log 1To,(P(z))1 on S is 0 for any choice of a E D .This says that not only is each point covered infinitely often, but there is some sense in which it is "frequently" covered .To get F into the little Bloch space imposes another constraint: given any e > 0 only finitely many of the sheets we attach may contain a disk of radius e.This arises because of a geometric characterization of ,Cio due to Stegenga and Stephenson [9] .For f analytic on D and a E D, r > 0 they define SZa(r) to be the component of f -1 (D (f (a), r)) containing a, Fa (r) = Oga(r) n T and rf(a) = sup{r : Fa (r) = 0} .Then f E Bo iff rf(a) = o(1) as ¡al --> 1.In particular, if the Riemann surface of f is obtained by identifying copies of D along slits, then the endpoints of any such slit are in the ideal boundary of the surface .Therefore f will be in the little Bloch space if for every E > 0, these endpoints of pasted edges are E-dense in D for all but finitely many sheets .
We will inductively construct a sequence of positive numbers {En,} tending to zero, a sequence of finite point sets {E,}, a collection of radial line segments T,, and two sequences of integers {gn}, {h } tending to infinity.The sets {Ej, {T} will satisfy (1) Tn C Tra+l, En C En+1, En C T .
(2) The endpoint of each segment in T,, is in E,,,.
(3) E,/E ,+1 is an even integer .(4) Adjacent points of & en a segment of T,, are at distante E, from each other .(5) SupzED dist(z, En,) G lOE .See Figure 2 .An "edge" I of T denotes a subinterval en T, connecting two adjacent points of E,,, Le., a component of T,\En .We let en, denote the number of edges in T,.In the construction below each such edge will be treated as two separate pieces of the boundary of the domain Rn = D\T,, corresponding to its two sides.One side will be pasted te a sheet of previous generation, the other pasted te one or more sheets in the next higher generation.

Figure 2. E , T,, Rn
Given an edge I in the boundary of R, we can either attach another copy of Rn,, or divide the edge into m = En/En+1 edges in T ..+1 ( since En C En+1) and attach m copies of R,+1 .Given a sequence of integers {gi} we could build a Riemann surface as follows .Start with one copy of R1 and attach 2e1 copies of of R1 along (both sides of) each edge of T1 .Call this 51 .Then attach more copies of R1 along each edge in OS, to obtain S 2 and continuing for g1 generations, obtaining a nested sequence of surfaces S1 C S2 C . . .C Sgl .The term "generations" refers to the fact that to connect the point 0 in the zeroth sheet So to any of the unpasted edges of Sk a path must pass though at least k + 1 different sheets (i.e., copies of R1) belonging to So, S1\So, . . ., Sk\Sk_1 .
We have obtained Sgl by pasting together identical sheets, Le., copies of Rl.To get the next surface, Sgl+1, we attach to each unpasted edge of Sgl El/E2 copies of the sheet R2 .We obtain Sgl+2 by pasting a copy of R2 to each unpasted edge of Ssl+1 .We continue in this way for 92 generations, obtaining a surface S91+92 .
The next surface Sg1+g2+1, is constructed by attaching copies of R3 to the unpasted edges of Sg1+g2 .Thus given the sequence of integers {gk} (which tells us for how many generations to attach copies of Rk) and continuing in the obvious manner, we obtain an increasing, nested sequence of simply connected surfaces, {Sn } .Then S = UnSn, is a simply connected connected Riemann surface .If -P : D -S is the Riemann map (mapping 0 to 0 on So), and P : S --> D the projection then F = P o <P is a holomorphic function on the unit disk which we claim is an infinite Blaschke product in 130, if the parameters are chosen correctly .This is essential Stephenson's construction in [10] .The fact that F E BO follows from the characterization of Stegenga and Stephenson mentioned earlier .If the sequence {gi} grows quickly enough, Stephenson shows the mapping F is an inner function.If dist(o,Tn ) >_ En then F is actually a Blaschke product (again if gn / oo fast enough) .To prove this, consider the least harmonic majorant u of log IP(z)j restricted to SN = Sgl+ ...+g" The boundary breaks into two pieces aSN = 01SN U a2SN corresponding respectively to P-1 (T) and the radial edges .Then u has boundary values 0 on 81 SN and u >_ 109 En on á2SN .The set a2SN can be made to have as small harmonic measure as we wish by taking gn large enough, so we may take if gn is large enough (recall that as before, harmonic measure refers to the harmonic measure with respect to the point 0 on the zeroth sheet C .J. BISHOP So) .Thus F is a Blaschke product, but it cannot be indestructible since it only takes values in each E n finitely often.As Stephenson points out, this example shows the exceptional set in Frostman's theorem may be dense in the unit disk.
To make F indestructible, we modify the construction slightly.Associated to each E,, define another set Fn, by replacing each z E En by a point w E En,+1 with Izw i = á Era and such that w is on same radius as z .The sets F, satisfy approximately the same density conditions as the En (with En replaced by 2c,,) .Our idea is to modify the construction by alternating the use of the sets E, and F in the construction .Since E, f1 Fn, = 0 this means our surface will cover the whole disk and since max(dist(z, En), dist(z, Fn)) >_ c,/4 for every point z in the disk, we should be able to prove our function is indestructible by estimating harmonic measure either on the "E,-sheets" or 'T,,-sheets" (depending on whether z is far from E, or far from Fn) .However since E", f1 F,, = 0, we need some further modifications to to able to attach an "F,-sheet" to an "En -sheet" .This is how we attach a F,-sheet to an En,-sheet .Consider a component interval I of Tn with endpoints in E, .Let Tn, be the analogue of T, for the set F, and let R,, = D\Tn, .Assume (without loss of generality) that F, has been chosen so T, C T,. Let {ao, al, . . . .an} = I n E ,+1, listed in order (e .g., ao, ara are the endpoints of I) .Let F,j = Fn U {aj, a j+ 1} .Along each interval (aj, aj+1) attach a copy of Rn .To this sheet we attach copies of Ñ,, along all component intervals of t,,\Fj .We continue in this way, attaching copies of Rn along intervals of Tn\Fn, except for those sheets reached by either looping around aj or around aj+1, in which case we are forced to attach copies of Rn along intervals of the form T,,\FnU{aj } (or Tn\F,U{aj+1}).Some of these identifications are illustrated in Figure 3.More precisely, Figure 3 shows regions on four sheets, labeled I, II, III, IV.Sheet I is pasted to sheet II along the edge [aj, aj +1] .Sheet II is pasted to sheet III along edge [q, aj ] and to sheet IV along the edge [p, q], where p, q are points of F,,, adjacent to aj .The solid and doted curves illustrate paths from sheet I to sheets III and IV respectively which (must) pass through sheet III .Notice that the point A E En in the ideal boundary of sheet I is covered when sheets II and IV are pasted along [p, q] .Similarly ao E En is covered when II is pasted to III along [q, aj] (assuming j 0 0; otherwise it would be covered by some sheet attached to sheet IV) .Suppose we have already constructed a surface S, whose boundary consists of arcs covering T or edges of T,, .To each component interval I of T, ,n\En we attach copies of R, as described above.Do this for g,+1 generations .The resulting sheets cover En, (the only sheets which do not cover every point of En, are those attached along subintervals of the form (ao, al) or (an_I, an) in the construction above) .We call the resulting surface Sn,.To the boundary of S,z attach copies of R n+ 1 = D\Tn+1 along component intervals of Tn+1\En+1 for hn+1 generations (this poses no difficulties since En , Fn and all points of the form aj in the previous stage of construction were in En+1 ; thus every radial interval in the boundary of Sn has endpoints in En+l) .The resulting surface is called Sn+I and satisfies the induction hypothesis .The union over n of those (nested) surfaces is denoted S and we obtain the desired function by mapping the disk to S and then projecting back to the disk.All that remains is to choose the sequences {En}, {gn} and {hn} so that the harmonic measure estimates hold.We will first choose gn, then En+1 and then hn+1 Let u be the least harmonic majorant of log ITa o P(z) 1 .We want to show u -0 .If dist(a, E,) >_ 4c,,then we estimate u on S, by an argument similar to the one used before .More precisely, let W be S,,, minus the componente of P-1(D\D(a, 8En)) which hit BS,, (recall that D(a, 8En) n E,, = Ql so they can only hit OS in radial edges if at all) .
Divide the boundary up into two pieces áW = 01 W U 02W according to whether the points lie over T or T, U OD(a, 8En .The function u is zero on ál W and u >_ log $ Era on á2W.Moreover we can make the harmonic measure of a2 W as small as we wish by taking g,, large enough .We choose g, so that Then 0 > u(0) > -1 -n , in this case.
If dist(a, En ) <_ 11 1 En then we do the computation on Sn .In this case let W be 5,,, minus the componente of P-1(D(a, 8En+1)) which hit 0Sn.Divide the boundary into three pieces áW = 81W U á2 W U á3W, according to whether the points lie over T, Tn or OD(a, -8 1 6J .As before u = 0 on á1W.On the rest of the boundary 0 >_ u >_ log(En+l/8) .
The harmonic measure of á3 W n S n can be made as small as desired by taking En+l small enough (in fact the measure decays like C(En+1) a for some cti depending on S, (it depende on the number of sheets in Sn)) .The boundary piece 83W\Sn can only be reached by leaving Sn through at most two of the intervals in Tn\E n+ 1 (Le ., intervals of the form (aj , aj+1) described above) .These intervals also have harmonic measure in S, dominated by C(En+1) a .Therefore we can choose En+1 so that 0 > log(1E,,+1)W(a3W) > -1/n .
The harmonic measure of 82W can be made as small as we wish by taking hn+1 large enough, Le., choose hn+1 so that Thus u(0) = 0 and so u -0 on S. Hence F is an indestructible Blaschke product, as desired .

. The VMO Construction
In this section we will construct an f E VMO with Ilf 11,, = 1 so that R(f, a) = D for every a E T .The construction is a "cut and paste" argument like the one in the previous section .As before we obtain the surface by identifying copies of the unit disk along radial slits, although the details are somewhat different.
First we will describe how to construct a function f E H°°n VMO whose range sets R(f, a) are dense in D for every a E T. Then we will then indicate how to modify the construction to obtain a function whose range sets equal D for every a.Recall that a dyadic subinterval of [0,1] is an interval of the form [j2-% (j+1)2 -n] for some n >_ 0 and 0 <_ j < 2n .A dyadic subinterval of a line segment [a, b] denotes an image of such an interval under the afñne mapping from [0,1] to [a, b] .
For each n = 1, 2, 3 . . . .we construct a collection of radial line segments Tn with one endpoint on the unit circle .Each line segment in T n is subdivided into a countable number of subintervals Inj .The collection of all endpoints of some such In,7 is denoted by En .It is a simple exercise to construct such sets so that (1) TnCTn+l, E n CEn+l " (2) The radial segments in Tn have arguments which are rational multiples of 27r .(3) Every dyadic subinterval of Inj is also of the form I n,k for some m > n. ( 4) diam(Inj) < dist(I,, j , T) .
To each interval In,j C T, associate the Cantor set Cn, j C In , j which is the affine image of C under the obvious map from [-1,1] to Inj .The component intervals of In,j\Cn,j will be denoted Jn,j,k (which is as many subscripts as I dare use) .Each of there is a dyadic subinterval of In,j and hence is equal to some I,n,a for some m and s.We build our surface by identifying copies of the disk along such segments .
More precisely, let Rr, = D\Tn, Le., R is the disk with countable many radial slits removed .To start the construction let SI = Ri .For each interval of the form Jn,j,k in the boundary of 51 find the m for which Jn,j,k = Im,s and attach a copy of R,, to S2 along this common segment (as in the previous section we think of different sides of a boundary slit as being different boundary points ; thus each Jn,j,k corresponds to two boundary intervals).Do this for every Jn,j,k gives S2 .We obtain S3 in the same way, by attaching the appropriate Rn to each of the boundary intervals Jn,j,k for each copy of Rn in S2\Sl .In an obvious manner we obtain an increasing sequence Sl C S2 C . . . of simply connected Riemann surfaces.The union, S, is the desired surface .Let f be the function obtained by composing the covering map D --> S (say one that maps 0 E D to 0 E Sl) with the projection map from S to D. Clearly f a holomorphic function with Ilf li c , = 1.
For a point w E S let r(w) denote the radius of the largest disk centered at w contained in S. Properties (7) and (8) of the Cantor set C imply that there is an absolute constant M so that for any e > 0 w(w, áS n D(w, Mr(w)), S) > 90 (Since R, is simply connected, a Brownian path starting at w hits BR,,, nD(w, Mr(w)) with high probability (Beurling's theorem says with probability > 1 -CM -1 1 2 ) .If it never leaves Rn then it certainly hits áS n D(w, Mr(w)) .If it does leave R n then estimates ( 7) and ( 8) say that it hits a third sheet with probability less than 10-4.Thus it hits as n D(w, Mr(w)) with high probability.) In terms of f this means w(z, {x E T : I f (x) -f (z) 1 > Mr(f (z))}, D) :100 Since the set of points where r(f (z)) is larger than some fixed number is a compact subset of the disk we see that 1 -IzI < b = 8(e) implies {x E Iz : 1f(x) -.f (z) I > E} <-10141 where Iz denotes the interval centered at z/Iz1 of length 1 -IzI .This condition is well known to imply f E VMO (e.g.[5,Exercise VI.4]) .
Thus we have constructed a f E VMO with II f II,, = 1 .We claim that the range set R(f, a) is dense in D for every boundary point a .To see this, note that each edge we paste along corresponds to a cross cut in the unit disk and that the diameter of the cross cut is comparable to its harmonic measure with respect to zero, which equals the harmonic measure of the edge in the surface with respect to the point 0 in the zeroth sheet.Thus a path which crosses infinitely many sheets corresponds to a path in the disk crossing infinitely many such cross cuts whose harmonic measures (and hence diameters) tend to zero, and thus defines a boundary point of the disk.The range set at that point will be dense in the disk since each sheet is open and dense and is covered by the part of the disk separated from the origin by one of the cross cuts and hence covered by a neighborhood of the boundary point.Since a countable intersection of dense open sets is dense, the range set at that boundary point is dense (in fact, a dense Ga) .The set of boundary points corresponding to paths which cross infinitely many sheets is dense (if a path approaches the boundary through one sheet we can always change it arbitrarily close to the boundary so that it crosses infinitely many sheets)) .Therefore the range set is dense for every boundary point (since a countable intersection of dense Gó's is dense) .However, we do not have R(f, a) = D because the copies of the Cantor set are omitted .To take tare of this problem we can modify the construction slightly.Instead of attaching a copy of R,, to R, as above, attach a "perturbed" copy R,, of R,. .We define R,. = D\T,a, where T,,, E.,, have the same properties as before except that T, and T, only intersect along one radial segment; the one where we want to attach R, to R n .For example, T, could be obtained from T,, by simply changing the argument of each radial segment by some homeomorphism of the circle which only fixes the one desired angle and maps the other rationals C .J .BISHOP (e .g., the arguments of the other radial segments of T,,) onto irrational values.Such a map is easy to construct, e.g., 9 ---> 0 + a6(0 -27r) with a small irrational .\will fix 0 and map other rational multiples of 27r to irrational ones.
When we attach such a modified sheet R,, to R, the union R, U R-,, covers the whole disk except for a subset of the radial segment Tn, n T, .This exceptional set can be covered when we attach a normal" sheet Rp to R,, along any radial segment, except one with the same argument as the radial segment where we attached R-,, to R,.Thus by alternating "normal" and "perturbed" versions of R, in the construction we obtain a VMO function with 11f1j,,, = 1 and R(f, a) = D for every a E 8D .(One sheet does not cover the disk; however a given sheet and all sheets separated by it from the zeroth sheet do cover the whole disk.Thus the cross cut argument above shows the range set is the whole disk at every boundary point corresponding to a path crossing infinitely many cross cuts .Since such points are dense on the boundary, it is easy to see R(f, a) = D for every a.)

C
fiz1 < 1} denote the unit disk.The little Bloch space, BO, is the space of holomorphic functions f on D such that lim 1f'(z)1(1 -Iz12) = 0. Iz1-1 Basic facts about BO can be found in [21 .A Blaschke product is a holomorphic function of the form This work was supported by NSF Grant DMS 91-00671 (D (D (D . . .

Figure 3 .
Figure 3 .Modifications to cover En