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A Construction of Outer Functions along Curves on the Sphere inCn
Journal of Functional Analysis FU2938 journal of functional analysis 140, 151169 (1996) article no. 0103
A Construction of Outer Functions along Curves on the Sphere in C n Joaquim Bruna* Departament de Matematiques, Facultat de Ciencies, Universitat Autonoma de Barcelona, 08193 Bellaterra, Spain Received June 1995
For a simple smooth curve 1 on the unit sphere in C n transverse at all its points, and a non-negative continuous function . on 1 satisfying 1 log . ds>& together with a finite number of necessary compatibility conditions, an holomorphic function f on the unit ball is constructed such that | f | =. on 1. The case of a general, not necessarily transverse, curve is also investigated. 1996 Academic Press, Inc.
1. STATEMENT OF RESULTS The main result of this paper, proved in section 2, is the following: Theorem 1. Let B be the unit ball in C n, S its boundary and let 1 be a simple curve of class C 1 lying on S, with arc length parametrization #(s). Assume 1 is transverse at every point, i.e., #* (s) } #(s){0 \s. Then there is a non-negative number d=d(1 ) and d real continuous functions u 1 , ..., u d on 1 with zero mean such that: for every non-negative continuous function . on 1 satisfying
|
log .(s) ds>&
(1)
1
|
u i (s) log .(s) ds=0,
i=1, ..., d
(2)
1
there is an holomorphic function f in B, with no zeros in B, such that | f | # C(B ) and | f | =. on 1. Moreover: (i)
d=0 if 1 is polinomially convex.
(ii) If . satifies a Lipschitz condition of order ;>0, (or more generally is Dini-continuous) then f can be chosen in the ball algebra A(B). * Partially supported by DGICYT Grant PB92-0804-C02-02 and by Grant GRQ94-2014 from the Comissionat per Universitats i Recerca de la Generalitat de Catalunya.
151 0022-123696 18.00 Copyright 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
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The function f can be said to be an outer function with data . on 1. One of the main motivations of this theorem concerns interpolation by holomorphic functions from sets on the curve. In these problems, it is often very important to have holomorphic functions with prescribed size on the curve. Usually, . is there some function vanishing with a controled rate on a set E/1. Several comments are in order: (a) The techniques developped by A. B. Aleksandrov to construct non trivial inner functions in B, and more specifically his theory of regular triples (see [A]), yield the following result: if a . is a strictly positive continuous function on S and + is any positive Borel measure on S, there is f # H (B) such that | f *| =. a.e. d+ (here f * denotes the radial limit of f ). This is a general result, but two features must be pointed out. First, it is an a.e. result, and secondly, the assumption .>0 is essential. As said above, precisely one of our main motivations is the case when . vanishes on certain sets of zero measure. (b) Of course there is a close relation with the subject of pluriharmonic interpolation developed in [BO], [BB] and [AB]. Let PHC denote the space of real pluriharmonic functions in B continuous on B . In [BB] it is proved that for 1 of class C 3, (transverse or not) PHC has a trace in C(1 ) which is closed and of finite codimension; in [BO] this result was previously proved for C 2 transverse curves and in [AB] for certain C 1 curves including the transverse ones. The relation consists simply in noticing that if u # PHC interpolates # C(1 ) and u~ is a pluriharmonic conjugate of u, so that u+iu~ is holomorphic, then f =exp(u+iu~ ) is holomorphic, | f | =exp u # C(B ) and | f | =exp on 1. Thus the case .>0 is essentially settled by the results in [AB], [BO]. Moreover, the proof will show that d equals the codimension of the trace of PHC in C(1 ). In fact, the functions u 1 , ..., u d have the property that for # C(1 ), # PHC | 1
|
u i (s) (s) ds=0,
i=1, ..., d.
(3)
1
(c) By Stolzenberg's theorem and its improvements ([S]), either 1 is polinomially convex or 1 "1 is a one dimensional analytic subvariety of B. In the later case, it is proved in [AB] that 1 "1 is locally a C 1 graph near 1, so that singularities do not accumulate on 1 and are thus finite. It is then possible to relate the number d, which as said before is the codimension of PHC in C(1 ), with the number of these singularities and the topology of 1 "1 (see [SH] for the case 1 is a C curve). In case 1 is polinomially convex the space of holomorphic polynomials, and so PHC as
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153
well, is dense in C(1 ) so that d=0, and this explains part (i) in the statement. (d) Concerning part (ii) in the statement, in general one cannot say that f # A(B) when . is just continuous. The obstruction for this is similar to that in one variable, the fact that the conjugation operator does not preserve continuity. The proof will show that, as in one variable, it is enough to assume that . satisfies a local lipschitz condition in 1 ". &1(0). (e) The condition (1) is known to be necessary, at least in some cases, i.e. if f is holomorphic and zero-free in B and | f | # C(B ), then
|
log | f (#(s))| ds>& .
(4)
1
This is a consequence of the following result of [R] (see also [CC]): if u is a positive pluriharmonic function in B and 1 is a transverse sufficiently smooth curve, then u has a.e. [ds] a radial limit u*, and u* # L 1(ds). An inspection of the proof, based in the construction of an almost analytic disk with boundary 1 as introduced in [NR], shows that 1 # C 2+= is enough. In section 3 we will show (4) for a C 1+=-curve which is not polynomially convex (Theorem 2 below). We do not know how to prove it for a C 1+= polynomially convex curve. For C 1-curves the result is probably false. (f) Condition (2), when 1 is not polynomially convex, is also necessary in a certain sense. The precise result, as expected after (3), is the following one, proved in Section 3. Theorem 2. Assume 1 of class C 1+= and not polynomially convex, and let f # A(B), zero-free in B. Then log | f | # L 1(ds) and there is a singular [ds] finite measure d& on 1, concentrated in 1 & f &1(0), such that d+=(log | f | ) ds+d& satisfies
|
u i (s) d+(s)=0,
i=1, ..., d.
1
A natural question that arises is whether the obstruction (2) remains or not in case the function f is allowed to have zeros in B. (g) Without the transversality assumption, that is, if 1 is allowed to have some complex-tangential points #(s) at which #* (s) } #(s)=0, we can only prove a partial result. Note first that if 1 is C 2 and has one complex tangential point then it is polinomially convex ([F]) (this is no longer true if 1 is C 1, see [R] and Theorem 3.1 in [AB]), and so no obstructions like (2) are to be expected.
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Our result in this case is the following one, whose proof we carry out in section 4. In the statement, the function T(s) is defined by #* (s) } #(s)=iT(s) and we call it the index of transversality of 1. Theorem 3. Assume 1 of class C 2+= and let E=[T=0]{< the set of its complex-tangential points. Let . be a non-negative continuous function on 1. Assume either (a) (b)
1 |log .(s)| p ds<+ for some p>1, or 1 log .(s)| ds<+ and 1 has the property (P):
T4 (s)=0
whenever T(s)=0
Then there is an holomorphic function f in B such that | f | =. in 1"(E & . &1(0)). By technical reasons, we cannot control the boundary behavior of | f (z)| when z approaches a complex-tangential point where . vanishes. Besides this, Theorem 3 is not sharp because of the following remarks. If 1 is a complex-tangential curve (meaning that all its points are complex-tangential) then 1 is interpolating for the ball algebra [RU] and so there exists f # A(B) with f =. on 1. This means that the obstructions on . for the problem to have a solution should localize in 1 "E. For instance the result of Ramey quoted before localizes so that log . # L 1loc(1 "E) is a necessary condition. An open natural problem that we propose is to find necessary conditions of the type
|
w(T(s)) log .(s) ds>&
1
with w(x)=o(1) as x 0; it is not hard to see that in some cases (e.g. is 1 is real-analytic) such conditions do exist. The right theorem would be with assumptions (a), (b) replaced by such one. In particular it should hold with the single condition log . # L 1(ds), but our technique does not work with this sole condition. The technical reason is a typical one dealing with harmonic analysis in L 1, the non-boundedness of a Cauchy-type operator in L 1. As a final remark we note that all the results here hold for very general domains; existence of good support functions and appropriate estimates for the -equation to globalize is all what is needed. In particular the results hold for (polinomially convex) strictly pseudoconvex domains. All functions spaces over 1 that appear are assumed to be over the reals. By L p(1 ), 1 p we mean with respect arc-length in 1.
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CONSTRUCTION OF OUTER FUNCTIONS
2. PROOF OF THE MAIN RESULT 2.1. First we will simplify and extend to the C 1 range the result on pluriharmonic interpolation of [BO]. Recall that T is defined by #* (s) } #(t)=iT(t); without loss of generality we assume that T(t)>0. Let K 0(t, z) be the positive kernel #(t) #* (t) 1 1 1 = T(t) Re K 0(t, z)= Im ? 1&z#(t) ? 1&z#(t) and P 0 the operator defined by P 0[](z)=
|
# L 1(1 ),
(t) K 0(t, z) dt,
z # B.
1
P 0[] is clearly pluriharmonic. In order to study the boundary behavior of P 0[], let s=s(z) be such that def
|1&z#(s)| =d(z) = min |1&z#(t)| t
so that |1&z#(t)| & d(z)+ |1&#(s) #(t)|. Note that Re(1&#(t) #(s)= 12 |#(t)&#(s)| 2 & |t&s| 2, Im(1&#(t) #(s))=Im
|
t
#(t) #* (x) dx= s
|
t
(T(x)+O( |x&t| )) dx & |t&s|, s
and hence |1&#(t) #(s)| & |t&s|,
|1&z#(t)| & d(z)+ |t&s|.
Besides, 2Re(1&z#(t))= |#(t)&z| 2 +1& |z| 2 = |#(t)&#(s)| 2 +O(|#(s)&z| 2 )+1& |z| 2 & |s&t| 2 +O(d(z)) and consequently K 0(t, z)=O(1)+O
d(z)
\d(z) +(s&t) + . 2
2
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(5)
156
JOAQUIM BRUNA
This shows that K 0(t, z) is the sum of a Poisson-type kernel and a bounded perturbation. In particular,
|
K 0(t, z) dt=O(1),
z#B
1
and K 0(t, #(s)) is bounded for s{t. Next we claim that for a point #(x) # 1 where is continuous (and assuming (x)=0 if 1 is an arc and #(x) one of its endpoints), lim P 0[](z)=(x)+
z #(x) z#B
|
K 0(t, #(x)) (t) dt.
(6)
1
Indeed, we evaluate the difference, with $>0 small, by
|
K 0(t, z) |(t)&(x)| dt+ |(x)| |x&t| $
+ +
|
|x&t| $
|
|x&t| $
}|
K 0(t, z) dt&1 |x&t| $
}
|(t)| K 0(t, #(x)) dt |(t)| (K 0(t, z)&K 0(t, #(x)) dt.
(7)
It is clear by what has been said that all but the second term can be made arbitrarily small for small enough $ and z close enough to #(x). For the second one we note that z#* (t) O( |z&#(t)| ) 1 + K 0(t, z)= Im ? |1&z#(t)| |1&z#(t)| =
1 d Im log(1&z#(t))+O( |1&z#(t)| &12 ). ? dt
(8)
The later term is O( |s&t| &12 ), a uniformly integrable singularity. Since the variation of the argument of (1&z#(t)) along |x&t| $ is arbitrarily close to &? the claim follows. Let K 0 denote the integral operator with kernel K 0(t, #(x)) K 0 (x)=
|
(t) K 0(t, #(x)) dt, 1
(in case 1 is an arc we must first replace 1 by a larger arc 1 0 and extend to a continuous function vanishing at the end-points of 1 0 ).
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CONSTRUCTION OF OUTER FUNCTIONS
Since the kernel is bounded and continuous off the diagonal, K 0 is a compact operator in C(1 ). We have seen that the trace of PHC on 1 contains the range of I+K 0 and therefore, by Fredholm's theory, is closed and has finite codimension. This is the result proved in [BO] for C 2 curves. 2.2. Next we need to have analogous properties for a space of smoother functions. For 0<:<1, let 4 :(1 ), 4 :(B) respectively denote the space of real functions satisfying a Lipschitz condition of order : in 1, B. Lemma 1. (ii)
P 0 maps 4 :(1 ) into 4 :(B), 0<: 12.
(i)
K 0 maps L (1 ) into 4 :(1 ), 0<:<1.
Proof. Part (i) is equivalent to |(ddr) P 0[](r!)| =O((1&r) :&1 ), ! # S, and will follow from
|
T(t) (t)
1
#(t) z (1&#(t) z) 2
dt=O((1& |z| ) :&1 ).
(9)
If s=s(z) is as above, we decompose the left-hand side in the form (s)
|
T(t)(#(t) z)
1
(1&#(t) z) 2
dt+
|
O(|t&s| : )
1
|1&#(t) z| 2
dt.
Note that iT(t)
#(t) z (1&#(t) z)
2
=
d
1
dt (1&#(t) z
+O
1
\ |1&#(t) z| + . 32
Integrating by parts and using (5) we see that (i) is a consequence of the estimate dt O(d &; ) = 1& ; O( |log d | ) 1 (d+ |t&s| )
if ;>0 if ;=0.
{
|
It is immediate to check that d
1
} dx K (t, #(x)) } =O \ |t&x| + 0
and that this implies |K 0 (x 1 )&K 0 (x 2 )| C && |x 1 &x 2 | |log |x 1 &x 2 | | proving thus (ii). K
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JOAQUIM BRUNA
Fix now some :, 0<: 12 , and let PH4 : be the space of pluriharmonic functions in 4 :(B). Since (ii) implies that K 0 is compact in 4 :(1 ), we can repeat the same argument as before and conclude that both (I+K 0 )(4 :(1 )) and the trace of PH4 : on 1 are closed and of finite codimensions k, d respectively in 4 :(1 ). We proceed to define now the functions u 1 , ..., u d in the statement of Theorem 1. Let R 2(I+K 0 ) be the range of I+K 0 as an operator in L 2(1 ); K 0 is also compact in L 2(1 ) and so R 2(I+K 0 ) is the orthogonal of * N(I+K 0*), the null-space of I+K * 0 . Since K * 0 has kernel K 0(x, #(t)), K 0 maps L 2(1 ) to C(1 ). In conclusion N(I+K 0*) is a finite dimensional space V of real continuous functions and # L 2(1 ) belongs to R 2(I+K 0 ) if and only if
|
(s) u(s) ds=0,
u # V.
1
The operator K 0 maps L 2(1 ) to C(1 ), and C(1 ) to 4 :(1 ) (Lemma 1(ii)). These imply that for a given # 4 :(1 ), the statements # (I+K 0 )(4 :(1 )), # R 2(I+K 0 ) are equivalent. So we have that # (I+K 0 )(4 :(1 )) if and only if
|
(s) u(s) ds=0,
u # V.
1
In particular, dim R V=k. A certain subspace W of V of real dimension d corresponds to PH4 : | 1 , i.e. # PH4 : | 1 if and only if
|
(s) u(s) ds=0,
u # W.
1
Let u 1 , ..., u k be an orthogonal basis of V,
|
u j u i ds=$ ji
1
such that u 1 , ..., u d is a basis of W. Let F be a topological complement of (I+K 0 )(4 :(1 )) in 4 :(1 ), that is, 4 :(1 ) is the topological direct sum of (I+K 0 )(4 :(1 )) and F, and dim R F=k. We choose functions v 1 , ..., v k # F such that
|
v i u j ds=$ ij .
1
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CONSTRUCTION OF OUTER FUNCTIONS
159
Then, for # 4 :(1 ), if * i = 1 u i ds, i=1, ..., k k
1 :=& : * i v i # (I+K 0 )(4 :(1 )) i=1
because 1 1 u i ds=0, i=1, ..., k. Therefore k
= 1 + : * i v i i=1
is the decomposition of in (I+K 0 )(4 :(1 )) F, and PH4 : | 1 =(I+K 0 )(4 :(1 )) ( v d+1 , ..., v k ). Of course the same can be said in C(1 ), i.e. # C(1 ) belongs to PHC | 1 if and only if 1 (s) u i (s) ds=0 i=1, ..., d (so (3) is shown) or PHC | 1 =(I+K 0 )(C(1 )) ( v d+1 , ..., v k ). The corresponding result that we need for L 1(1 ) is part (ii) of next lemma. Lemma 2.
(i)
For # L 1(1 ), k
& : i=1
(ii)
\| u ds+ v # (I+K )(L (1 )) 1
i
i
0
1
# (I+K 0 )(L 1(1 )) ( v d+1 , ..., v k ) is equivalent to
|
(s) u i (s) ds=0,
i=1, ..., d.
1
Proof. We note first that K 0 is also compact in L 1(1 ) and therefore (I+K 0 )(L 1(1 )) is closed, and equals the closure of (I+K 0 )(4 :(1 ). Let 1 n # 4 :(1 ) be such that n ww n in L (1 ). Then k
n & : i=1
\| u ds+ v # (I+K )(4 (1 )). :
n
i
i
0
1
But
}|
1
}
n u i ds c & n & L1(1 ) c<+
and the result follows considering a convergent subsequence. Trivially (ii) follows from (i). K
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JOAQUIM BRUNA
2.3. We finish now the proof of Theorem 1. By lemma 2(ii), if (1) and (2) hold, then k
log .=(I+K 0 )()+ :
* i vi
(10)
i=d+1
for some # L 1(1 ), with v d+1 , ..., v k # PH4 : | 1 and * d+1 , ..., * k # R. Let v~ j denote the pluriharmonic conjugate of v j with v~ j (0)=0 and h j =v j +iv~ j . It is a well known fact that v~ j # 4 :(B) if v j # 4 :(B), and hence h j # 4 :(B). Taking into consideration the definition of K 0(t, z) and P 0 we define the holomorphic function f as follows f (z)=exp
1
{? |
1
T(t) (t) 1&z#(t)
=
k
dt exp
* j h j (z),
:
z # B.
j=d+1
Then it is clear that f # C(B "1 ) and log | f (z)| =P 0[](z)+: * j v j (z). j
Fix one point #(x) # 1. If .(x){0, then log . is continuous at x, and therefore so is , by (10), because K 0 maps L 1(1 ) to C(1 ). As z approaches .(x), P 0[](z) approaches (I+K 0 )()(x) (by (6)) and then (10) shows that | f (z)| approaches .(x). If .(x)=0, we must show that P 0[](z) tends to & as z approaches #(x), because v j # C(B ). Again, (10) shows that is bounded above and takes the value & continuously at x. Then, the positivity of K 0(t, z) implies P 0[](z) O(1)+
|
(t) K 0(t, z) dt. |x&t| $
In paragraph 2.1 it was already shown that the mass of K 0(t, z) dt in |#&t| $ is bounded below for z close to #(x), and so P 0[](z) & as z tends to #(x). We have thus proved that | f | # C(B ) and | f | =. on 1. It remains to show that f # A(B) if . satisfies a Lipschitz condition of order ;>0. Since hj # 4 :(B), it is enough for this to see that the conjugate of P 0[], Q 0[](z)=
1
?|
(t) T(t) Im 1
1 1&z#(t)
dt
has a continuous extension at each point #(x) where .(x){0. Now, (10), lemma 1(ii) and a localization argument imply that will satisfy as well a Lipschitz condition of order ;>0 in the neighbourhood of #(x). We are thus led to show that Q 0[] extends continuously at each point #(x)
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161
where satisfies a local lipschitz condition. Multiplying by a test function supported in a neighbourhood of #(x) we may assume that is globally Lipschitz in which case Q 0[] # 4 :(B). This follows like in the proof of Lemma 1(i), because (ddr) Q 0[](r!) is given by the imaginary part of the left hand side of (9). Part (i) in the statement of Theorem 1 was already explaned in the first section. This ends the proof of Theorem 1.
3. PROOF OF THEOREM 2 We assume that 1 is a transverse, closed simple curve of class C 1+= which is not polynomially convex. Let f # A(B), f {0 in B, without loss of generality we suppose | f | 1. Let h=&log | f |, a positive pluriharmonic function. We will show that sup r
|
h(r#(s)) ds<+ .
(11)
1
Fatou's lemma implies then that
|
log | f (#(s))| ds>& .
1
Also, if d+ is a weak limit of h(r n #(s)) ds, r nZ1, one has, since h n(z)=h(r n z) is in PHC and because of (3),
|
u i (s) d+(s)=lim n
1
|
u i (s) h n ds=0.
1
It is enough to prove (11) locally, i.e. for each Q # 1 there is an open arc {/1 containing Q such that
| h(r#(s)) /(s) ds C &/&
{
whenever / is supported in {. We may assume that Q=(1, 0, ..., 0). By [AB, Thm 2.1], 1 is near Q parameterized by its projection on z 2 = } } } , z n =0, more precisely, there is a neighbourhood N of Q, a C 1+= domain U in the z 1 -plane and f: U C n&1 holomorphic, of class C 1+= up to U such that U B * [ (*, f (*))
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parametrizes N & 1. Let { be the arc 1 & N. Let F : 2 U be the Riemann mapping function from the unit disk to U, F(1)=0. Then we have the parametrization 8
2 w B * [ (F(*), G(*)),
G= f b F.
Let : be the arc in b2 parametrizing {. In this parametrization, ds=(|F $| 2 + |G$1 | 2 + } } } + |G$n&1 | 2 ) 12 dt on {, *=e it. Then dsdt is a nonvanishing continuous function, because U is of class C 1+= ([P]). Then, since h b 8 is a positive harmonic function on 2
|
h(8(re it ) /(t) dt &/& h(8(0)).
:
Using that dsdt is bounded we can replace dt by ds. To finish it is enough to see that h(r8(e it )) Kh(8(re it )). This is a consequence of Harnack's inequality. Namely, since 8(re it ) enters transversally B, it is always possible to find in the complex line through r8(e it ) and 8(re it ) a disc D(w, R) of center w and radious R & 1&r, such that both r8(e it ) and 8(re it ) are at distance cR of w, with c<1. The restriction of h to this disk is a positive harmonic function, and then the above follows with K=K(c) from Harnack's inequality.
4. PROOF OF THEOREM 3 4.1. We can assume that 1 is a C 2+= simple closed curve, with E{<. The first step is to introduce a replacement of the kernel K 0 . Let us explain first why K 0 needs to be replaced. Recall that iT=##* ; then ## =iT4 &1 and Taylor formula gives 1&#(s) #(t)=&iT(t)(s&t)+ 12 (1&iT4 (t))(s&t) 2 +O((s&t) 2+= ).
(12)
In particular, its real part is &|s&t| 2 and |1&#(s) #(t)| & |s&t| 2 + |T(s)| |s&t| replaces (5). These imply |K 0(t, #(s))| &
|T(t)| . [ |T(s)| + |s&t| ] 2
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(13)
163
CONSTRUCTION OF OUTER FUNCTIONS
If E{<, this is not bounded; it has not even an uniformly (in s) integrable kernel, so that the operator K 0 would not be compact in general. Thus we replace K 0(t, z) by another kernel #(t) #* (t)+(z&#(t)) } /(t) 1 K(t, z)= Im ? 1&z#(t) with / to be determined, so that K(t, #(s)) is uniformly integrable in t. The form of the modification O(|z&#(t)| ) is justified because we still want a delta mass at #(s) as the principal part of the limit of K(t, z) dt as z approaches #(s), as expressed by (6). This was done in [BB, section 3] for C 3-curves, and we start by generalizing it to C 2+= curves. Since |1&#(s) } #(t)| dominates (s&t) 2 all quadratic terms in the numerator of K will give bounded terms; calling A(t)=#* (t) } (t) it is enough to estimate Im
iT(t)+(s&t) A(t) 1&#(s) #(t)
=
Im[iT(t)+(s&t) A(t)][1&#(s) #(t)] |1&#(s) #(t)| 2
.
The Taylor development (12) shows that the numerator above is 1 2
T(t)(s&t) 2 &T(t) Re A(t)(s&t) 2 + 12 (s&t) 3 Im[A(t)(1&iT4 (t)] +O((s&t) 3+= )+T(t) O((s&t) 2+= ).
Choosing A(t)= 12 (1+iT4 (t)), e.g. /(t)= 12 (1+iT4 (t)) #* (t), the first three terms vanish, and hence by (13) K(t, #(s))=O(1)+O
\
|s&t| = =O( |s&t| =&1 ) ( |T(s)| + |s&t| )
+
(14)
an uniformly integrable singularity. Let K be the integral operator with kernel K(t, #(s)). By (14), K is compact in C(1 ) and in L p(1 ). Also note that K, for # L 1(1 ), is always continuous on 1 "E, because (14) shows that K(t, #(s)) is bounded for #(s) far from E. The kernel K(t, z) corresponding to the choice of / above is K(t, z)=
(1+iT4 (t)) z#* (t)+(1&iT4 (t)) #(t) #* (t) 1 . Im 2? 1&z#(t)
Another property of K(t, z) that will be needed is
|
|K(t, z)| dt C,
\z # B.
1
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(15)
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JOAQUIM BRUNA
To see this, let s=s(z) be defined, as in section 2.1, by |1&z#(s)| =d(z)=min |1&z#(t)| t
so that |1&z#(t)| & d(z)+ |1&#(t) #(z)| & d(z)+ |T(s)| |s&t| + |s&t| 2.
(16)
Then it can be shown (the proof of [BB, Lemma 2] also applies under the regularity assumption 1 # C 2+= ) that
|
|K(t, z)&K(t, #(s))| dt C
1
and hence (15) follows from (14). We define a corresponding operator P by P[](z)=
|
=(t) (t) K(t, z) dt,
# L 1(1)
1
where =(t)=1 if T(t)>0, =(t)= &1 if T(t)<0, =(t)=0 if T(t)=0. Then =(t) K(t, z)=
1 1 +O |T(t)| Re ? 1&z#(t)
|z&#(t)|
\ |1&z#(t)| +
and hence it is clear that the boundary behaviour of P[](z) as z approaches a transverse point #(x) is analogous to P 0 []. Namely, if is continuous at one such #(x) then lim P[](z)=(z)+ z #(x)
|
=(t) (t) K(t, #(x)) dt
1
and if takes the value & continuously at x, then lim P[](z)= & . z #(x)
If #(x) is a complex-tangential point, the principal part K 0 of K has a wild behaviour as z approaches #(x). However, it is still true that if is continuous at x and (x)=0, then lim P[](z)= z #(x)
|
=(t) (t) K(t, #(x)) dt.
1
This follows like in (7), where now the second term does not appear, using (14) and (15).
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CONSTRUCTION OF OUTER FUNCTIONS
4.2. We cannot control the boundary behaviour of P[](z) as z approaches a complex-tangential point where does not vanish. To overcome this difficulty we are led to arrange first so that this is accomplished, without losing of course the good properties of P[] close to transverse points. With this purpose we introduce now two auxiliary operators S and T. Recall that E=[x: T(x)=0]. Since 1 # C 2+=, T is a C 1-function and def therefore there exists =>0 such that |T(t)| & |T(t)| whenever t # I(x) = 1 [ |x&t| = |T(x)|]. For # L (1) we define S by on
S=
E,
S(x)=
1 2= |T(x)|
|
|(t)| p
|
(t) dt
on 1 " E.
I(x)
Then
|
|S(x)| p dx
1"E
|
1"E
t # I(x)
dx . 2= |T(x)|
Since t # I(x) implies |T(t)| & |T(x)| , S is bounded in L p (1), 1 p . Also note that S # C(1 " E) and obviously S is also bounded in C(1). The second operator T is an operator of holomorphic interpolation already used in [BB]. With a fixed q, 12
1 h(z)
|
1
(t) (1&z#(t)) q
dt,
z # B,
# L 1(1)
where h is given by h(z)=
|
1
dt (1&z#(t)) q
.
Using (16) it follows that |h(z)| &
|
1
dt |1&z#(t)| q
& (d(z)+T 2(s)) 12&q
(17)
so that h(z) &1 vanishes precisely at E. It is then immediate that if is continuous at a point x # E then lim T(z)=(x). z #(x)
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JOAQUIM BRUNA
Since q<1, the singularity of |1&#(x) #(t)| &q is integrable at a transverse point and this implies that if is continuous at one such point then lim T(z)= z #(x)
1 h(#(x))
|
(t) 1
(1&#(x) #(t)) q
dt.
Correspondingly we will also denote by T, # L 1(1), the function defined by the above right-hand side term on 1 " E and equal to on E. Let us study the boundednes properties of T. Clearly, by (13) and (17) |T(x)| |T(x)| 2q&1
|
|T(x)| dx
1"E
=
| |
|
1
|(t)| dt (|T(x)| |x&t| + |x&t| 2 ) q
|T(x)| 2q&1 1"E
|(t)| 1
{|
1"E
{|
1
(18)
|(t)| dt dx ( |T(t)| |x&t| + |x&t| 2 ) q
=
|T(x)| 2q&1 dx dt. ( |T(t)| |x&t| + |x&t| 2 ) q
=
Note that if 0 both terms are of the same size. Therefore T is bounded in L 1(1) if and only if
|
1
|T(x)| 2q&1 dx C [ |T(t)| |x&t| + |x&t| 2 ] q
\t.
(19)
It is clear that this can hold only if the curve 1 has the property (P): T$(t)=0 whenever T(t)=0. On the other hand property (P) quantifies as |T$(t)| 2 =O(|T(t)| ). Then from the Taylor development T(x)=T(t)+T$(t)(x&t)+O((x&t) 1+= ) we get |T(x)| |T(t)| +O((x&t) 1+= ). This implies that (19) is bounded by |T(t)| 2q&1
|
1
dx + [ |T(t)| |x&t| + |x&t| 2 ] q
|
|x&t| (2q&1)(1+=)&2q dx C. 1
Thus we have seen that T is bounded in L 1(1) if and only if 1 has the property (P). On the other hand, we shall prove now that T is always bounded in L p(1) if p>1. It is enough for this to see that |T| *, where
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CONSTRUCTION OF OUTER FUNCTIONS
* is the HardyLittlewood maximal function of . We use (18): if J k = J k (x)=[t : |t&x| 2 k |T(x)| ] for k # Z then
|
|T(x)| |T(x)| q&1
J0
|(t)| dt+ |T(x)| 2q&1 |x&t| q
|
1 " J0
|(t)| dt |x&t| 2q
+
|T(x)| q&1 : (2 &k |T(x)| ) &q k=0
|
|(t)| dt Jk
+
+ |T(x)| 2q&1 : (2 k |T(x)| ) &2q k=0
*(x)
{
+
+
|
|(t)| dt Jk
=
: (2 k ) q&1 + : (2 k ) 1&2q C*(x) k=0
k=0
by the choice of q. 4.3. We finish now the proof of Theorem 3. For # L 1(1) we define M(z) by S(t) 1 dt, z#B h(z) 1 (1&z#(t) q M is a pluriharmonic function on B, with a continuous extension to B " E, because S # C(1" E), and with boundary value S(x) at z=#(x) if is continuous at #(x) # E. We also denote by M the operator defined by the same expression at points z=#(x) E and equal to on E. The operator S being bounded in L p(1) for all p 1, we obtain that M is bounded in L p(1) if p>1 and in C(1), and also bounded in L 1(1) if 1 satisfies property (P). Next we define P 1 [] by M(z)=Re T(S)(z)=Re
|
P 1 []=P[&M](z)+M(z) =
|
=(t)[(t)&M(t)] K(t, z) dt+M(z).
1"E
Clearly P 1 [] is a pluriharmonic function, continuous in B " E. At a point #(x) E where is continuous, P 1 [] has limit (x)+
|
=(t)[(t)&M(t)] K(t, #(x)) dt
(20)
1
whereas if takes the value & continuously at x, then P 1 [] has limit & . If #(x) # E and is continuous at x then &M is a vanishing continuous function on x, hence P[&M] behaves well near #(x) and again P 1 [] has limit (20).
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Consider now the operator U given by the right-hand term of (20). It is compact in C(1), in L p(1) if p>1 and in L 1(1) if 1 has property (P). From compactness in C(1) it follows that R(I+U) and hence PHC | 1 , is closed and has finite codimension in C(1) (thus generalizing the result in [BB]). The curve 1 being polynomially convex, this codimension must be zero, that is, every # C(1) can be interpolated by some function in PHC. By general theory of compact operators, R(I+U) has a topological complement in L p(1) included in the null space of some power of I+U. A power of I+U has the form I+KA with A bounded in L p(1), 1 p . Now, by (14) and Young's inequality, K maps L q(1) into L r(1), r1= in C(1). By iteration, it then follows from the equation 0=(I+KA) ,
= &K(A),
# L p(1)
(21)
that is in fact continuous in 1. In conclusion, R(I+U) has in L p(1) a topological complement F consisting of continuous functions. Let v 1 , ..., v k # C(1) be a basis of F. Let . be a non-negative continuous function on 1 satisfying the hypothesis of Theorem 3. We can write l
log .= : * i v i +(I+U)()
(22)
i=1
for some # L p(1) and * 1 , ..., * k # R. Let us denote u 1 , ..., u k functions in PHC interpolating v 1 , ..., v k respectively, and let u be the pluriharmonic function k
u=P 1 []+ : * i u i . i=1
The function U is, by the properties of K, always continuous on 1 " E. Assume #(x) # E and .(x){0. In this case is also continuous at x. This follows applying to the equation (22) the same argument as with (21), after localizing in a suitable neighbourhood of #(x). In conclusion, P 1 [](z) has a nice behavior close to #(x) except in the case #(x) # E, .(x)=0, and u has limit, by (22) k
(I+U)()+ : * i v i =log .(x). r=1
This ends the proof of Theorem 3.
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169
REFERENCES [A]
[AB] [BB] [BO] [CC] [F] [NR] [P] [R] [Ro] [RU] [S] [SH]
A. B. Aleksandrov, Inner functions on compact spaces, Funkts. Anal. Prilozhen 18, No. 2 (1984), 113 [in Russian]; Funct. Anal. Appl. 18, No. 2 (1984), 8798 [Engl. transl.] H. Alexander and J. Bruna, Pluriharmonic interpolation and hulls of C 1 curves in the unit sphere, Rev. Mat. Iberoamericana 11, No. 3 (1995), 547568. B. Berndtson and J. Bruna, Traces of pluriharmonic functions on curves, Ark. Mat. 28 No. 2 (1990), 221230. J. Bruna and J. Ortega, Interpolation by holomorphic functions smooth to the boundary in the unit ball of C n, Math. Ann. 274 (1986), 527575. J. Chaumat and A. M. Chollet, Ensembles de zeros et d'interpolation a la frontiere de domaines strictement pseudoconvexes, Ark. Mat. 24 (1986), 2757. F. Forstneric, Regularity of varieties in strictly pseudoconvex domains, Publ. Mat. 32 (1988), 145150. A. Nagel and W. Rudin, Local boundary behaviour of bounded holomorphic functions, Can. J. Math. 30 (1978), 583592. C. Pommerenke, ``Univalent Functions,'' Vandenhoeck and Ruprecht, Gottingen, 1975. W. Ramey, Local boundary behaviour of pluriharmonic functions along curves, Amer. J. Math. 108 (1986), 171191. P. Rosay, A remark on a theorem by F. Forstneric, Ark. Mat. 28 (1990), 311314. W. Rudin, ``Function Theory in the Unit Ball of C n,'' Grundlehren d. Matematische Wissenschaft, Vol. 241, Springer-Verlag, New York. E. L. Stout, ``The Theory of Uniform Algebras,'' Bogden and Quigleg, 1971. N. Shcherbina, Traces of pluriharmonic functions on the boundaries of analytic varieties, Math. Z. 213 (1993), 171177.
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A Construction of Outer Functions along Curves on the Sphere in C n Joaquim Bruna* Departament de Matematiques, Facultat de Ciencies, Universitat Autonoma de Barcelona, 08193 Bellaterra, Spain Received June 1995
For a simple smooth curve 1 on the unit sphere in C n transverse at all its points, and a non-negative continuous function . on 1 satisfying 1 log . ds>& together with a finite number of necessary compatibility conditions, an holomorphic function f on the unit ball is constructed such that | f | =. on 1. The case of a general, not necessarily transverse, curve is also investigated. 1996 Academic Press, Inc.
1. STATEMENT OF RESULTS The main result of this paper, proved in section 2, is the following: Theorem 1. Let B be the unit ball in C n, S its boundary and let 1 be a simple curve of class C 1 lying on S, with arc length parametrization #(s). Assume 1 is transverse at every point, i.e., #* (s) } #(s){0 \s. Then there is a non-negative number d=d(1 ) and d real continuous functions u 1 , ..., u d on 1 with zero mean such that: for every non-negative continuous function . on 1 satisfying
|
log .(s) ds>&
(1)
1
|
u i (s) log .(s) ds=0,
i=1, ..., d
(2)
1
there is an holomorphic function f in B, with no zeros in B, such that | f | # C(B ) and | f | =. on 1. Moreover: (i)
d=0 if 1 is polinomially convex.
(ii) If . satifies a Lipschitz condition of order ;>0, (or more generally is Dini-continuous) then f can be chosen in the ball algebra A(B). * Partially supported by DGICYT Grant PB92-0804-C02-02 and by Grant GRQ94-2014 from the Comissionat per Universitats i Recerca de la Generalitat de Catalunya.
151 0022-123696 18.00 Copyright 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
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JOAQUIM BRUNA
The function f can be said to be an outer function with data . on 1. One of the main motivations of this theorem concerns interpolation by holomorphic functions from sets on the curve. In these problems, it is often very important to have holomorphic functions with prescribed size on the curve. Usually, . is there some function vanishing with a controled rate on a set E/1. Several comments are in order: (a) The techniques developped by A. B. Aleksandrov to construct non trivial inner functions in B, and more specifically his theory of regular triples (see [A]), yield the following result: if a . is a strictly positive continuous function on S and + is any positive Borel measure on S, there is f # H (B) such that | f *| =. a.e. d+ (here f * denotes the radial limit of f ). This is a general result, but two features must be pointed out. First, it is an a.e. result, and secondly, the assumption .>0 is essential. As said above, precisely one of our main motivations is the case when . vanishes on certain sets of zero measure. (b) Of course there is a close relation with the subject of pluriharmonic interpolation developed in [BO], [BB] and [AB]. Let PHC denote the space of real pluriharmonic functions in B continuous on B . In [BB] it is proved that for 1 of class C 3, (transverse or not) PHC has a trace in C(1 ) which is closed and of finite codimension; in [BO] this result was previously proved for C 2 transverse curves and in [AB] for certain C 1 curves including the transverse ones. The relation consists simply in noticing that if u # PHC interpolates # C(1 ) and u~ is a pluriharmonic conjugate of u, so that u+iu~ is holomorphic, then f =exp(u+iu~ ) is holomorphic, | f | =exp u # C(B ) and | f | =exp on 1. Thus the case .>0 is essentially settled by the results in [AB], [BO]. Moreover, the proof will show that d equals the codimension of the trace of PHC in C(1 ). In fact, the functions u 1 , ..., u d have the property that for # C(1 ), # PHC | 1
|
u i (s) (s) ds=0,
i=1, ..., d.
(3)
1
(c) By Stolzenberg's theorem and its improvements ([S]), either 1 is polinomially convex or 1 "1 is a one dimensional analytic subvariety of B. In the later case, it is proved in [AB] that 1 "1 is locally a C 1 graph near 1, so that singularities do not accumulate on 1 and are thus finite. It is then possible to relate the number d, which as said before is the codimension of PHC in C(1 ), with the number of these singularities and the topology of 1 "1 (see [SH] for the case 1 is a C curve). In case 1 is polinomially convex the space of holomorphic polynomials, and so PHC as
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CONSTRUCTION OF OUTER FUNCTIONS
153
well, is dense in C(1 ) so that d=0, and this explains part (i) in the statement. (d) Concerning part (ii) in the statement, in general one cannot say that f # A(B) when . is just continuous. The obstruction for this is similar to that in one variable, the fact that the conjugation operator does not preserve continuity. The proof will show that, as in one variable, it is enough to assume that . satisfies a local lipschitz condition in 1 ". &1(0). (e) The condition (1) is known to be necessary, at least in some cases, i.e. if f is holomorphic and zero-free in B and | f | # C(B ), then
|
log | f (#(s))| ds>& .
(4)
1
This is a consequence of the following result of [R] (see also [CC]): if u is a positive pluriharmonic function in B and 1 is a transverse sufficiently smooth curve, then u has a.e. [ds] a radial limit u*, and u* # L 1(ds). An inspection of the proof, based in the construction of an almost analytic disk with boundary 1 as introduced in [NR], shows that 1 # C 2+= is enough. In section 3 we will show (4) for a C 1+=-curve which is not polynomially convex (Theorem 2 below). We do not know how to prove it for a C 1+= polynomially convex curve. For C 1-curves the result is probably false. (f) Condition (2), when 1 is not polynomially convex, is also necessary in a certain sense. The precise result, as expected after (3), is the following one, proved in Section 3. Theorem 2. Assume 1 of class C 1+= and not polynomially convex, and let f # A(B), zero-free in B. Then log | f | # L 1(ds) and there is a singular [ds] finite measure d& on 1, concentrated in 1 & f &1(0), such that d+=(log | f | ) ds+d& satisfies
|
u i (s) d+(s)=0,
i=1, ..., d.
1
A natural question that arises is whether the obstruction (2) remains or not in case the function f is allowed to have zeros in B. (g) Without the transversality assumption, that is, if 1 is allowed to have some complex-tangential points #(s) at which #* (s) } #(s)=0, we can only prove a partial result. Note first that if 1 is C 2 and has one complex tangential point then it is polinomially convex ([F]) (this is no longer true if 1 is C 1, see [R] and Theorem 3.1 in [AB]), and so no obstructions like (2) are to be expected.
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JOAQUIM BRUNA
Our result in this case is the following one, whose proof we carry out in section 4. In the statement, the function T(s) is defined by #* (s) } #(s)=iT(s) and we call it the index of transversality of 1. Theorem 3. Assume 1 of class C 2+= and let E=[T=0]{< the set of its complex-tangential points. Let . be a non-negative continuous function on 1. Assume either (a) (b)
1 |log .(s)| p ds<+ for some p>1, or 1 log .(s)| ds<+ and 1 has the property (P):
T4 (s)=0
whenever T(s)=0
Then there is an holomorphic function f in B such that | f | =. in 1"(E & . &1(0)). By technical reasons, we cannot control the boundary behavior of | f (z)| when z approaches a complex-tangential point where . vanishes. Besides this, Theorem 3 is not sharp because of the following remarks. If 1 is a complex-tangential curve (meaning that all its points are complex-tangential) then 1 is interpolating for the ball algebra [RU] and so there exists f # A(B) with f =. on 1. This means that the obstructions on . for the problem to have a solution should localize in 1 "E. For instance the result of Ramey quoted before localizes so that log . # L 1loc(1 "E) is a necessary condition. An open natural problem that we propose is to find necessary conditions of the type
|
w(T(s)) log .(s) ds>&
1
with w(x)=o(1) as x 0; it is not hard to see that in some cases (e.g. is 1 is real-analytic) such conditions do exist. The right theorem would be with assumptions (a), (b) replaced by such one. In particular it should hold with the single condition log . # L 1(ds), but our technique does not work with this sole condition. The technical reason is a typical one dealing with harmonic analysis in L 1, the non-boundedness of a Cauchy-type operator in L 1. As a final remark we note that all the results here hold for very general domains; existence of good support functions and appropriate estimates for the -equation to globalize is all what is needed. In particular the results hold for (polinomially convex) strictly pseudoconvex domains. All functions spaces over 1 that appear are assumed to be over the reals. By L p(1 ), 1 p we mean with respect arc-length in 1.
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CONSTRUCTION OF OUTER FUNCTIONS
2. PROOF OF THE MAIN RESULT 2.1. First we will simplify and extend to the C 1 range the result on pluriharmonic interpolation of [BO]. Recall that T is defined by #* (s) } #(t)=iT(t); without loss of generality we assume that T(t)>0. Let K 0(t, z) be the positive kernel #(t) #* (t) 1 1 1 = T(t) Re K 0(t, z)= Im ? 1&z#(t) ? 1&z#(t) and P 0 the operator defined by P 0[](z)=
|
# L 1(1 ),
(t) K 0(t, z) dt,
z # B.
1
P 0[] is clearly pluriharmonic. In order to study the boundary behavior of P 0[], let s=s(z) be such that def
|1&z#(s)| =d(z) = min |1&z#(t)| t
so that |1&z#(t)| & d(z)+ |1&#(s) #(t)|. Note that Re(1&#(t) #(s)= 12 |#(t)&#(s)| 2 & |t&s| 2, Im(1&#(t) #(s))=Im
|
t
#(t) #* (x) dx= s
|
t
(T(x)+O( |x&t| )) dx & |t&s|, s
and hence |1&#(t) #(s)| & |t&s|,
|1&z#(t)| & d(z)+ |t&s|.
Besides, 2Re(1&z#(t))= |#(t)&z| 2 +1& |z| 2 = |#(t)&#(s)| 2 +O(|#(s)&z| 2 )+1& |z| 2 & |s&t| 2 +O(d(z)) and consequently K 0(t, z)=O(1)+O
d(z)
\d(z) +(s&t) + . 2
2
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(5)
156
JOAQUIM BRUNA
This shows that K 0(t, z) is the sum of a Poisson-type kernel and a bounded perturbation. In particular,
|
K 0(t, z) dt=O(1),
z#B
1
and K 0(t, #(s)) is bounded for s{t. Next we claim that for a point #(x) # 1 where is continuous (and assuming (x)=0 if 1 is an arc and #(x) one of its endpoints), lim P 0[](z)=(x)+
z #(x) z#B
|
K 0(t, #(x)) (t) dt.
(6)
1
Indeed, we evaluate the difference, with $>0 small, by
|
K 0(t, z) |(t)&(x)| dt+ |(x)| |x&t| $
+ +
|
|x&t| $
|
|x&t| $
}|
K 0(t, z) dt&1 |x&t| $
}
|(t)| K 0(t, #(x)) dt |(t)| (K 0(t, z)&K 0(t, #(x)) dt.
(7)
It is clear by what has been said that all but the second term can be made arbitrarily small for small enough $ and z close enough to #(x). For the second one we note that z#* (t) O( |z&#(t)| ) 1 + K 0(t, z)= Im ? |1&z#(t)| |1&z#(t)| =
1 d Im log(1&z#(t))+O( |1&z#(t)| &12 ). ? dt
(8)
The later term is O( |s&t| &12 ), a uniformly integrable singularity. Since the variation of the argument of (1&z#(t)) along |x&t| $ is arbitrarily close to &? the claim follows. Let K 0 denote the integral operator with kernel K 0(t, #(x)) K 0 (x)=
|
(t) K 0(t, #(x)) dt, 1
(in case 1 is an arc we must first replace 1 by a larger arc 1 0 and extend to a continuous function vanishing at the end-points of 1 0 ).
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CONSTRUCTION OF OUTER FUNCTIONS
Since the kernel is bounded and continuous off the diagonal, K 0 is a compact operator in C(1 ). We have seen that the trace of PHC on 1 contains the range of I+K 0 and therefore, by Fredholm's theory, is closed and has finite codimension. This is the result proved in [BO] for C 2 curves. 2.2. Next we need to have analogous properties for a space of smoother functions. For 0<:<1, let 4 :(1 ), 4 :(B) respectively denote the space of real functions satisfying a Lipschitz condition of order : in 1, B. Lemma 1. (ii)
P 0 maps 4 :(1 ) into 4 :(B), 0<: 12.
(i)
K 0 maps L (1 ) into 4 :(1 ), 0<:<1.
Proof. Part (i) is equivalent to |(ddr) P 0[](r!)| =O((1&r) :&1 ), ! # S, and will follow from
|
T(t) (t)
1
#(t) z (1&#(t) z) 2
dt=O((1& |z| ) :&1 ).
(9)
If s=s(z) is as above, we decompose the left-hand side in the form (s)
|
T(t)(#(t) z)
1
(1&#(t) z) 2
dt+
|
O(|t&s| : )
1
|1&#(t) z| 2
dt.
Note that iT(t)
#(t) z (1&#(t) z)
2
=
d
1
dt (1&#(t) z
+O
1
\ |1&#(t) z| + . 32
Integrating by parts and using (5) we see that (i) is a consequence of the estimate dt O(d &; ) = 1& ; O( |log d | ) 1 (d+ |t&s| )
if ;>0 if ;=0.
{
|
It is immediate to check that d
1
} dx K (t, #(x)) } =O \ |t&x| + 0
and that this implies |K 0 (x 1 )&K 0 (x 2 )| C && |x 1 &x 2 | |log |x 1 &x 2 | | proving thus (ii). K
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JOAQUIM BRUNA
Fix now some :, 0<: 12 , and let PH4 : be the space of pluriharmonic functions in 4 :(B). Since (ii) implies that K 0 is compact in 4 :(1 ), we can repeat the same argument as before and conclude that both (I+K 0 )(4 :(1 )) and the trace of PH4 : on 1 are closed and of finite codimensions k, d respectively in 4 :(1 ). We proceed to define now the functions u 1 , ..., u d in the statement of Theorem 1. Let R 2(I+K 0 ) be the range of I+K 0 as an operator in L 2(1 ); K 0 is also compact in L 2(1 ) and so R 2(I+K 0 ) is the orthogonal of * N(I+K 0*), the null-space of I+K * 0 . Since K * 0 has kernel K 0(x, #(t)), K 0 maps L 2(1 ) to C(1 ). In conclusion N(I+K 0*) is a finite dimensional space V of real continuous functions and # L 2(1 ) belongs to R 2(I+K 0 ) if and only if
|
(s) u(s) ds=0,
u # V.
1
The operator K 0 maps L 2(1 ) to C(1 ), and C(1 ) to 4 :(1 ) (Lemma 1(ii)). These imply that for a given # 4 :(1 ), the statements # (I+K 0 )(4 :(1 )), # R 2(I+K 0 ) are equivalent. So we have that # (I+K 0 )(4 :(1 )) if and only if
|
(s) u(s) ds=0,
u # V.
1
In particular, dim R V=k. A certain subspace W of V of real dimension d corresponds to PH4 : | 1 , i.e. # PH4 : | 1 if and only if
|
(s) u(s) ds=0,
u # W.
1
Let u 1 , ..., u k be an orthogonal basis of V,
|
u j u i ds=$ ji
1
such that u 1 , ..., u d is a basis of W. Let F be a topological complement of (I+K 0 )(4 :(1 )) in 4 :(1 ), that is, 4 :(1 ) is the topological direct sum of (I+K 0 )(4 :(1 )) and F, and dim R F=k. We choose functions v 1 , ..., v k # F such that
|
v i u j ds=$ ij .
1
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CONSTRUCTION OF OUTER FUNCTIONS
159
Then, for # 4 :(1 ), if * i = 1 u i ds, i=1, ..., k k
1 :=& : * i v i # (I+K 0 )(4 :(1 )) i=1
because 1 1 u i ds=0, i=1, ..., k. Therefore k
= 1 + : * i v i i=1
is the decomposition of in (I+K 0 )(4 :(1 )) F, and PH4 : | 1 =(I+K 0 )(4 :(1 )) ( v d+1 , ..., v k ). Of course the same can be said in C(1 ), i.e. # C(1 ) belongs to PHC | 1 if and only if 1 (s) u i (s) ds=0 i=1, ..., d (so (3) is shown) or PHC | 1 =(I+K 0 )(C(1 )) ( v d+1 , ..., v k ). The corresponding result that we need for L 1(1 ) is part (ii) of next lemma. Lemma 2.
(i)
For # L 1(1 ), k
& : i=1
(ii)
\| u ds+ v # (I+K )(L (1 )) 1
i
i
0
1
# (I+K 0 )(L 1(1 )) ( v d+1 , ..., v k ) is equivalent to
|
(s) u i (s) ds=0,
i=1, ..., d.
1
Proof. We note first that K 0 is also compact in L 1(1 ) and therefore (I+K 0 )(L 1(1 )) is closed, and equals the closure of (I+K 0 )(4 :(1 ). Let 1 n # 4 :(1 ) be such that n ww n in L (1 ). Then k
n & : i=1
\| u ds+ v # (I+K )(4 (1 )). :
n
i
i
0
1
But
}|
1
}
n u i ds c & n & L1(1 ) c<+
and the result follows considering a convergent subsequence. Trivially (ii) follows from (i). K
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JOAQUIM BRUNA
2.3. We finish now the proof of Theorem 1. By lemma 2(ii), if (1) and (2) hold, then k
log .=(I+K 0 )()+ :
* i vi
(10)
i=d+1
for some # L 1(1 ), with v d+1 , ..., v k # PH4 : | 1 and * d+1 , ..., * k # R. Let v~ j denote the pluriharmonic conjugate of v j with v~ j (0)=0 and h j =v j +iv~ j . It is a well known fact that v~ j # 4 :(B) if v j # 4 :(B), and hence h j # 4 :(B). Taking into consideration the definition of K 0(t, z) and P 0 we define the holomorphic function f as follows f (z)=exp
1
{? |
1
T(t) (t) 1&z#(t)
=
k
dt exp
* j h j (z),
:
z # B.
j=d+1
Then it is clear that f # C(B "1 ) and log | f (z)| =P 0[](z)+: * j v j (z). j
Fix one point #(x) # 1. If .(x){0, then log . is continuous at x, and therefore so is , by (10), because K 0 maps L 1(1 ) to C(1 ). As z approaches .(x), P 0[](z) approaches (I+K 0 )()(x) (by (6)) and then (10) shows that | f (z)| approaches .(x). If .(x)=0, we must show that P 0[](z) tends to & as z approaches #(x), because v j # C(B ). Again, (10) shows that is bounded above and takes the value & continuously at x. Then, the positivity of K 0(t, z) implies P 0[](z) O(1)+
|
(t) K 0(t, z) dt. |x&t| $
In paragraph 2.1 it was already shown that the mass of K 0(t, z) dt in |#&t| $ is bounded below for z close to #(x), and so P 0[](z) & as z tends to #(x). We have thus proved that | f | # C(B ) and | f | =. on 1. It remains to show that f # A(B) if . satisfies a Lipschitz condition of order ;>0. Since hj # 4 :(B), it is enough for this to see that the conjugate of P 0[], Q 0[](z)=
1
?|
(t) T(t) Im 1
1 1&z#(t)
dt
has a continuous extension at each point #(x) where .(x){0. Now, (10), lemma 1(ii) and a localization argument imply that will satisfy as well a Lipschitz condition of order ;>0 in the neighbourhood of #(x). We are thus led to show that Q 0[] extends continuously at each point #(x)
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CONSTRUCTION OF OUTER FUNCTIONS
161
where satisfies a local lipschitz condition. Multiplying by a test function supported in a neighbourhood of #(x) we may assume that is globally Lipschitz in which case Q 0[] # 4 :(B). This follows like in the proof of Lemma 1(i), because (ddr) Q 0[](r!) is given by the imaginary part of the left hand side of (9). Part (i) in the statement of Theorem 1 was already explaned in the first section. This ends the proof of Theorem 1.
3. PROOF OF THEOREM 2 We assume that 1 is a transverse, closed simple curve of class C 1+= which is not polynomially convex. Let f # A(B), f {0 in B, without loss of generality we suppose | f | 1. Let h=&log | f |, a positive pluriharmonic function. We will show that sup r
|
h(r#(s)) ds<+ .
(11)
1
Fatou's lemma implies then that
|
log | f (#(s))| ds>& .
1
Also, if d+ is a weak limit of h(r n #(s)) ds, r nZ1, one has, since h n(z)=h(r n z) is in PHC and because of (3),
|
u i (s) d+(s)=lim n
1
|
u i (s) h n ds=0.
1
It is enough to prove (11) locally, i.e. for each Q # 1 there is an open arc {/1 containing Q such that
| h(r#(s)) /(s) ds C &/&
{
whenever / is supported in {. We may assume that Q=(1, 0, ..., 0). By [AB, Thm 2.1], 1 is near Q parameterized by its projection on z 2 = } } } , z n =0, more precisely, there is a neighbourhood N of Q, a C 1+= domain U in the z 1 -plane and f: U C n&1 holomorphic, of class C 1+= up to U such that U B * [ (*, f (*))
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JOAQUIM BRUNA
parametrizes N & 1. Let { be the arc 1 & N. Let F : 2 U be the Riemann mapping function from the unit disk to U, F(1)=0. Then we have the parametrization 8
2 w B * [ (F(*), G(*)),
G= f b F.
Let : be the arc in b2 parametrizing {. In this parametrization, ds=(|F $| 2 + |G$1 | 2 + } } } + |G$n&1 | 2 ) 12 dt on {, *=e it. Then dsdt is a nonvanishing continuous function, because U is of class C 1+= ([P]). Then, since h b 8 is a positive harmonic function on 2
|
h(8(re it ) /(t) dt &/& h(8(0)).
:
Using that dsdt is bounded we can replace dt by ds. To finish it is enough to see that h(r8(e it )) Kh(8(re it )). This is a consequence of Harnack's inequality. Namely, since 8(re it ) enters transversally B, it is always possible to find in the complex line through r8(e it ) and 8(re it ) a disc D(w, R) of center w and radious R & 1&r, such that both r8(e it ) and 8(re it ) are at distance cR of w, with c<1. The restriction of h to this disk is a positive harmonic function, and then the above follows with K=K(c) from Harnack's inequality.
4. PROOF OF THEOREM 3 4.1. We can assume that 1 is a C 2+= simple closed curve, with E{<. The first step is to introduce a replacement of the kernel K 0 . Let us explain first why K 0 needs to be replaced. Recall that iT=##* ; then ## =iT4 &1 and Taylor formula gives 1&#(s) #(t)=&iT(t)(s&t)+ 12 (1&iT4 (t))(s&t) 2 +O((s&t) 2+= ).
(12)
In particular, its real part is &|s&t| 2 and |1&#(s) #(t)| & |s&t| 2 + |T(s)| |s&t| replaces (5). These imply |K 0(t, #(s))| &
|T(t)| . [ |T(s)| + |s&t| ] 2
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(13)
163
CONSTRUCTION OF OUTER FUNCTIONS
If E{<, this is not bounded; it has not even an uniformly (in s) integrable kernel, so that the operator K 0 would not be compact in general. Thus we replace K 0(t, z) by another kernel #(t) #* (t)+(z&#(t)) } /(t) 1 K(t, z)= Im ? 1&z#(t) with / to be determined, so that K(t, #(s)) is uniformly integrable in t. The form of the modification O(|z&#(t)| ) is justified because we still want a delta mass at #(s) as the principal part of the limit of K(t, z) dt as z approaches #(s), as expressed by (6). This was done in [BB, section 3] for C 3-curves, and we start by generalizing it to C 2+= curves. Since |1&#(s) } #(t)| dominates (s&t) 2 all quadratic terms in the numerator of K will give bounded terms; calling A(t)=#* (t) } (t) it is enough to estimate Im
iT(t)+(s&t) A(t) 1&#(s) #(t)
=
Im[iT(t)+(s&t) A(t)][1&#(s) #(t)] |1&#(s) #(t)| 2
.
The Taylor development (12) shows that the numerator above is 1 2
T(t)(s&t) 2 &T(t) Re A(t)(s&t) 2 + 12 (s&t) 3 Im[A(t)(1&iT4 (t)] +O((s&t) 3+= )+T(t) O((s&t) 2+= ).
Choosing A(t)= 12 (1+iT4 (t)), e.g. /(t)= 12 (1+iT4 (t)) #* (t), the first three terms vanish, and hence by (13) K(t, #(s))=O(1)+O
\
|s&t| = =O( |s&t| =&1 ) ( |T(s)| + |s&t| )
+
(14)
an uniformly integrable singularity. Let K be the integral operator with kernel K(t, #(s)). By (14), K is compact in C(1 ) and in L p(1 ). Also note that K, for # L 1(1 ), is always continuous on 1 "E, because (14) shows that K(t, #(s)) is bounded for #(s) far from E. The kernel K(t, z) corresponding to the choice of / above is K(t, z)=
(1+iT4 (t)) z#* (t)+(1&iT4 (t)) #(t) #* (t) 1 . Im 2? 1&z#(t)
Another property of K(t, z) that will be needed is
|
|K(t, z)| dt C,
\z # B.
1
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(15)
164
JOAQUIM BRUNA
To see this, let s=s(z) be defined, as in section 2.1, by |1&z#(s)| =d(z)=min |1&z#(t)| t
so that |1&z#(t)| & d(z)+ |1&#(t) #(z)| & d(z)+ |T(s)| |s&t| + |s&t| 2.
(16)
Then it can be shown (the proof of [BB, Lemma 2] also applies under the regularity assumption 1 # C 2+= ) that
|
|K(t, z)&K(t, #(s))| dt C
1
and hence (15) follows from (14). We define a corresponding operator P by P[](z)=
|
=(t) (t) K(t, z) dt,
# L 1(1)
1
where =(t)=1 if T(t)>0, =(t)= &1 if T(t)<0, =(t)=0 if T(t)=0. Then =(t) K(t, z)=
1 1 +O |T(t)| Re ? 1&z#(t)
|z&#(t)|
\ |1&z#(t)| +
and hence it is clear that the boundary behaviour of P[](z) as z approaches a transverse point #(x) is analogous to P 0 []. Namely, if is continuous at one such #(x) then lim P[](z)=(z)+ z #(x)
|
=(t) (t) K(t, #(x)) dt
1
and if takes the value & continuously at x, then lim P[](z)= & . z #(x)
If #(x) is a complex-tangential point, the principal part K 0 of K has a wild behaviour as z approaches #(x). However, it is still true that if is continuous at x and (x)=0, then lim P[](z)= z #(x)
|
=(t) (t) K(t, #(x)) dt.
1
This follows like in (7), where now the second term does not appear, using (14) and (15).
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CONSTRUCTION OF OUTER FUNCTIONS
4.2. We cannot control the boundary behaviour of P[](z) as z approaches a complex-tangential point where does not vanish. To overcome this difficulty we are led to arrange first so that this is accomplished, without losing of course the good properties of P[] close to transverse points. With this purpose we introduce now two auxiliary operators S and T. Recall that E=[x: T(x)=0]. Since 1 # C 2+=, T is a C 1-function and def therefore there exists =>0 such that |T(t)| & |T(t)| whenever t # I(x) = 1 [ |x&t| = |T(x)|]. For # L (1) we define S by on
S=
E,
S(x)=
1 2= |T(x)|
|
|(t)| p
|
(t) dt
on 1 " E.
I(x)
Then
|
|S(x)| p dx
1"E
|
1"E
t # I(x)
dx . 2= |T(x)|
Since t # I(x) implies |T(t)| & |T(x)| , S is bounded in L p (1), 1 p . Also note that S # C(1 " E) and obviously S is also bounded in C(1). The second operator T is an operator of holomorphic interpolation already used in [BB]. With a fixed q, 12
1 h(z)
|
1
(t) (1&z#(t)) q
dt,
z # B,
# L 1(1)
where h is given by h(z)=
|
1
dt (1&z#(t)) q
.
Using (16) it follows that |h(z)| &
|
1
dt |1&z#(t)| q
& (d(z)+T 2(s)) 12&q
(17)
so that h(z) &1 vanishes precisely at E. It is then immediate that if is continuous at a point x # E then lim T(z)=(x). z #(x)
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JOAQUIM BRUNA
Since q<1, the singularity of |1&#(x) #(t)| &q is integrable at a transverse point and this implies that if is continuous at one such point then lim T(z)= z #(x)
1 h(#(x))
|
(t) 1
(1&#(x) #(t)) q
dt.
Correspondingly we will also denote by T, # L 1(1), the function defined by the above right-hand side term on 1 " E and equal to on E. Let us study the boundednes properties of T. Clearly, by (13) and (17) |T(x)| |T(x)| 2q&1
|
|T(x)| dx
1"E
=
| |
|
1
|(t)| dt (|T(x)| |x&t| + |x&t| 2 ) q
|T(x)| 2q&1 1"E
|(t)| 1
{|
1"E
{|
1
(18)
|(t)| dt dx ( |T(t)| |x&t| + |x&t| 2 ) q
=
|T(x)| 2q&1 dx dt. ( |T(t)| |x&t| + |x&t| 2 ) q
=
Note that if 0 both terms are of the same size. Therefore T is bounded in L 1(1) if and only if
|
1
|T(x)| 2q&1 dx C [ |T(t)| |x&t| + |x&t| 2 ] q
\t.
(19)
It is clear that this can hold only if the curve 1 has the property (P): T$(t)=0 whenever T(t)=0. On the other hand property (P) quantifies as |T$(t)| 2 =O(|T(t)| ). Then from the Taylor development T(x)=T(t)+T$(t)(x&t)+O((x&t) 1+= ) we get |T(x)| |T(t)| +O((x&t) 1+= ). This implies that (19) is bounded by |T(t)| 2q&1
|
1
dx + [ |T(t)| |x&t| + |x&t| 2 ] q
|
|x&t| (2q&1)(1+=)&2q dx C. 1
Thus we have seen that T is bounded in L 1(1) if and only if 1 has the property (P). On the other hand, we shall prove now that T is always bounded in L p(1) if p>1. It is enough for this to see that |T| *, where
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CONSTRUCTION OF OUTER FUNCTIONS
* is the HardyLittlewood maximal function of . We use (18): if J k = J k (x)=[t : |t&x| 2 k |T(x)| ] for k # Z then
|
|T(x)| |T(x)| q&1
J0
|(t)| dt+ |T(x)| 2q&1 |x&t| q
|
1 " J0
|(t)| dt |x&t| 2q
+
|T(x)| q&1 : (2 &k |T(x)| ) &q k=0
|
|(t)| dt Jk
+
+ |T(x)| 2q&1 : (2 k |T(x)| ) &2q k=0
*(x)
{
+
+
|
|(t)| dt Jk
=
: (2 k ) q&1 + : (2 k ) 1&2q C*(x) k=0
k=0
by the choice of q. 4.3. We finish now the proof of Theorem 3. For # L 1(1) we define M(z) by S(t) 1 dt, z#B h(z) 1 (1&z#(t) q M is a pluriharmonic function on B, with a continuous extension to B " E, because S # C(1" E), and with boundary value S(x) at z=#(x) if is continuous at #(x) # E. We also denote by M the operator defined by the same expression at points z=#(x) E and equal to on E. The operator S being bounded in L p(1) for all p 1, we obtain that M is bounded in L p(1) if p>1 and in C(1), and also bounded in L 1(1) if 1 satisfies property (P). Next we define P 1 [] by M(z)=Re T(S)(z)=Re
|
P 1 []=P[&M](z)+M(z) =
|
=(t)[(t)&M(t)] K(t, z) dt+M(z).
1"E
Clearly P 1 [] is a pluriharmonic function, continuous in B " E. At a point #(x) E where is continuous, P 1 [] has limit (x)+
|
=(t)[(t)&M(t)] K(t, #(x)) dt
(20)
1
whereas if takes the value & continuously at x, then P 1 [] has limit & . If #(x) # E and is continuous at x then &M is a vanishing continuous function on x, hence P[&M] behaves well near #(x) and again P 1 [] has limit (20).
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Consider now the operator U given by the right-hand term of (20). It is compact in C(1), in L p(1) if p>1 and in L 1(1) if 1 has property (P). From compactness in C(1) it follows that R(I+U) and hence PHC | 1 , is closed and has finite codimension in C(1) (thus generalizing the result in [BB]). The curve 1 being polynomially convex, this codimension must be zero, that is, every # C(1) can be interpolated by some function in PHC. By general theory of compact operators, R(I+U) has a topological complement in L p(1) included in the null space of some power of I+U. A power of I+U has the form I+KA with A bounded in L p(1), 1 p . Now, by (14) and Young's inequality, K maps L q(1) into L r(1), r
= &K(A),
# L p(1)
(21)
that is in fact continuous in 1. In conclusion, R(I+U) has in L p(1) a topological complement F consisting of continuous functions. Let v 1 , ..., v k # C(1) be a basis of F. Let . be a non-negative continuous function on 1 satisfying the hypothesis of Theorem 3. We can write l
log .= : * i v i +(I+U)()
(22)
i=1
for some # L p(1) and * 1 , ..., * k # R. Let us denote u 1 , ..., u k functions in PHC interpolating v 1 , ..., v k respectively, and let u be the pluriharmonic function k
u=P 1 []+ : * i u i . i=1
The function U is, by the properties of K, always continuous on 1 " E. Assume #(x) # E and .(x){0. In this case is also continuous at x. This follows applying to the equation (22) the same argument as with (21), after localizing in a suitable neighbourhood of #(x). In conclusion, P 1 [](z) has a nice behavior close to #(x) except in the case #(x) # E, .(x)=0, and u has limit, by (22) k
(I+U)()+ : * i v i =log .(x). r=1
This ends the proof of Theorem 3.
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CONSTRUCTION OF OUTER FUNCTIONS
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A. B. Aleksandrov, Inner functions on compact spaces, Funkts. Anal. Prilozhen 18, No. 2 (1984), 113 [in Russian]; Funct. Anal. Appl. 18, No. 2 (1984), 8798 [Engl. transl.] H. Alexander and J. Bruna, Pluriharmonic interpolation and hulls of C 1 curves in the unit sphere, Rev. Mat. Iberoamericana 11, No. 3 (1995), 547568. B. Berndtson and J. Bruna, Traces of pluriharmonic functions on curves, Ark. Mat. 28 No. 2 (1990), 221230. J. Bruna and J. Ortega, Interpolation by holomorphic functions smooth to the boundary in the unit ball of C n, Math. Ann. 274 (1986), 527575. J. Chaumat and A. M. Chollet, Ensembles de zeros et d'interpolation a la frontiere de domaines strictement pseudoconvexes, Ark. Mat. 24 (1986), 2757. F. Forstneric, Regularity of varieties in strictly pseudoconvex domains, Publ. Mat. 32 (1988), 145150. A. Nagel and W. Rudin, Local boundary behaviour of bounded holomorphic functions, Can. J. Math. 30 (1978), 583592. C. Pommerenke, ``Univalent Functions,'' Vandenhoeck and Ruprecht, Gottingen, 1975. W. Ramey, Local boundary behaviour of pluriharmonic functions along curves, Amer. J. Math. 108 (1986), 171191. P. Rosay, A remark on a theorem by F. Forstneric, Ark. Mat. 28 (1990), 311314. W. Rudin, ``Function Theory in the Unit Ball of C n,'' Grundlehren d. Matematische Wissenschaft, Vol. 241, Springer-Verlag, New York. E. L. Stout, ``The Theory of Uniform Algebras,'' Bogden and Quigleg, 1971. N. Shcherbina, Traces of pluriharmonic functions on the boundaries of analytic varieties, Math. Z. 213 (1993), 171177.
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