RATIONAL APPROXIMATION NEAR ZERO SETS OF FUNCTIONS

RATIONAL APPROXIMATION NEAR ZERO SETS OF FUNCTIONS

The main result of this paper (Theorem 2 in §2) was obtained in Moskow and was discussed in some seminars and conferences in USSR.It was announced in the proceedings of the 1983 Kiev conference on complex analysis, but no proof was provided .The other results were found while the author was visiting the CRM in Barcelona in the fall of 1987.
1 .Let X be a compact in C, and R(X) the closure in C(X) (with sup-norm) of the space of functions which are holomorphic on X .
For f E C(X) we will write f E R(X,x) for some x E X, if these exists neighbourhood U of x such that f ¡y,--u -E R(X n U) .a We let N(f) stand for {x E X I f(x) = 0} -the zero set of f.We will consider the following two closely related problems .Power problem (P-Pr) .Is it true that for any function f E R(X) the condition f a E C(X) implies f a E R(X) (same fixed branch)?.Zero set problem (Z-Pr) .Let f E C(X) and f E R(X,x) for X \N(f) .
Is it true that f E R(X) ? .It is clear that these problems are trivial for N(f) = 0, and that a positive answer to the Z-Pr implies the same answer to the P-Pr.Also it is not difficult to see, that P-Pr and Z-Pr are equivalent for f in the class L(X) = U,>o Lip(P, X) .

andq>0 all x E
The author is grateful to the CRM for making possible his visit to Barcelona, and to the faculty of the Department of Mathematics of the UAB for many interesting discussions and remarks in seminars and private conversations.
For the P-Pr the simpler case q > 1 was considered in [51 .It was proved there that this problem has a positive answer for all X and all f E L(X), but really the proof presented in [5] works only under the additional assumption f q-' E L(X).One can find a simple example (X C_ R), for which f E L(X ), q > 1, f v and f v -1 E C(X ), but f v and f q-1 are not in L(X) (the branches are fixed and corresponding) .
Given two compact sets X and Y, Y C X and h(z) E C(X) .We will write if there exist p > 0 and a constant c > 0, such that for every z E X and w E Y we have Jh(z) -h(w)j < ciz -wiP .
The following Theorem has a proof absolutely like Theorem 1 in [5], except for a small change, wich will be described below.
The corollaxy follows applying Theorem 1 to h = fq-1 .To prove Theorem 1 we proceed as in [5] observing that we just need to worry about the squares (coming from the Vitushkin localization procedure) which intersect N(f).To deal with these squares we use Lemma 3 in [5, p. 416] which turns out to be true under our weaker hypothesis .Concretely what we need is the following Lemma.

Lemma. Leí h be as in Theorem 1 . There exisi p = p(h) > 0 and A = A(h),
such that for every square T6 with side length 6, T6 fl N(f) and every function g(z) E Có(T6), we have where a(-) is C-analytic capacüy and M1+p( .) is the Hausdorf content of order 1 -fp (see 16], p. 145) .
We close this section by stating an open problem dealing with RP(X), the closure in Lip(p, X) of the space of holomorphic functions on X .Problem.Let f E RP(X ), and assume that f 9 E lip(p, X) for some q > 1.Is it true that f q E RP(X)? .
2. In [3,4] we proved, that the Z-Pr has a positive answer for all X and all functions f E Lip(p, X) , p > z .Now we are going to prove one theorem concerning this problem for all p > 0, but for some special compact sets X.This result points in the direction that the Z-Pr has a positive answer also for all X and all f E L(X) .We need some notation .
Let X be compact and x E X.We say that x is a point of stability (of the capacity of C \ X) if one of the following two conditions holds.
i) There exists lim log6(a(T(x, 6) \ X)), where T(x, 6) (here and below) is 6-.0 the square with center x and side length 6 .
Let now X* = {x E X 1 x is unstable} .
Corollary 1. Le¡ f E L(X) and assume f E R(X, x) for all x E X \ N(f) .
Then f E R(X, x) also for all x E X \ (N(f) n Y*) .
The proof of Theorem 2 and the main idea in the proof of Lemma 2 [5] allow to get also the following corollary: Corollary 2. Let f E Lip(p, X) , p > 0, and assume f E R(X, x) for all x E X\N(f ).If We remark that for p > the last hypothesis is automatically satisfied and z so we obtain the main result of [3,4] .Some preliminary results and remarks.For T6 = T(z, 6) we will write r -T6 = T(z, r -6), for each r > 1 .As in [4] we will need the following Theorem 0. Let T be a square and 0 < T < 1.If f E C(4T), llfIl4T < 1 and f is analytic on 4T°\ ME (f ), where ME(f) = {z E 4TI lf(z)l < e}, then where ME(f) = {z E G1 lf(z)j < e}, f E C(G) and f is analytic on G\ME(f ) .Problem.Is estimate (2) true if we take G = D? .
In applications to rational approximation we really don't need the condition G = D, but it seems useful to have (2) with T = 0.
As it tums out estimates ( 2) and ( 1) are not true for r = 0. We present an example here .
Let S be a compact with m(S) > 0 (where m(.) is plane Lebesgue measure), with empty interior and with connected complement .Set W(z) = f dm(C) s z-(, so that cp E C(C), ep(oo) = 0, and cp is analytic on C \ S. Let now T be a square, containing S. Then cp(z)dz = 27ri lim zcp(z) = 27rim(S) ~0 .

aT zoo
By the Mergelian's Theorem [2] for any e > 0 there exists polynomial P(z) such that 11w(z) -P(z)lis < e.

Theorem V1 . Le¡ f E C(C) .
Then f E R(X) provided there exist r >_ 1 and a( 6) ) 0 as 6 -> 0, such that for any square T6 of side length 6 we have lJ f(z)dzl < a(6)[a(rT6 \ X) + 62] .aT6 Conversely if f E R(X) we obtain (3) with a(6) = cw(f, 6), r = 1 and without 62 in the right hand-side.Theorem V2.Leí E be a bounded set with a(E) = a > 0, and {Ej}w1 a finite number of sets Ej C E such that any square Ta with side a intersecis at most p (p > 1 is a fixed integer) Ej '.s .Then for come absolute constant c Proof of Theorem 2: Now we fix a compact X with X* = 0, a function f E Lip(p, X), p > 0, and let us suppose, that f 1 R(X ).After several lemmas we will have a contradiction.
We will denote by C an absolute constant and by A a constant depending only on f (on p).
Both of them may vary from an inequality to another.Let T6 be any square with side 6 and put a(5T6 \ X) -5 -613.Then fl > 1 .Take al = 6,6 and consider a non-overlapping family {Tj } of squares of side 61 and centers {zj }, covering the plane.Let (pj E Có(2Tj ), IV~pj j < C611 and Ej cpj = 1 on C. We denote by j' the indexes j for which f l3Tj E R(X n 3Tj) and 3Tj C 5T6. N a(Ej) < c -pa(E) .j-1 Lemma 1 .Let t be a point in 4T6 such that a((T(t,e)n5T6)\X) <C .e 1 for some, v, 0 < v < 1, and all e, 61 < e < 106.where M is defined by am < a < a,y+1 .
A computation now gives the sublema.
For n = 1 we let S1 be any square such that f J R(X1), where X1 = S1 n X.
Then there exists a square Q2 C_ Q1 such that f ¡y, q R(Y 2 ), where Y2 = Y1 n Q2 = X n Q2, and for every square TE C_ Q2, satisfying the condi- tion we have f E R(Tf (1 Y2) .

Remark 1 .
Assume D is Jordan domain with rectifiable boundary 8D, and G any open neighbourhood of D. If the estimate of the integral due to Melnikov and Vitushkin ([6], p. 158) is satisfied in D, then it is easily seen that we also have r) el-r lif lJca(ME(f) n D) (4) will follow from Theorem V2 and the following elementary result .Sublemma.Let 0 < al < a2, . . ., and al < a .If a sequence {an, a >_ 0} satisfes ¡he conditions Er 1 ap < am for all m > 1 and rn 1 % < a