Convergence of Rational Interpolants to Analytic Functions with Restricted Growth

Computational Methods and Function TheoryVolume 6 (2006), No. 1, 203–222

Convergence of Rational Interpolants to AnalyticFunctions with Restricted Growth

Anders Gustafsson

(Communicated by Eli Levin)

Abstract. Let D be a regular domain in C with bounded boundary. Foreach n ≥ 1, take points An = {ani}n

i=0 in D and points Bn = {bni}ni=1 in C,

such that⋃

Bn has no limit points in D. Let αn and βn be the normalizedpoint counting measures of An and Bn respectively and α′

n and β′n their swept

measures onto ∂D. Denote by Uμ the logarithmic potential of the measure μ.It has been shown by Ambroladze and Wallin, that if for every weak-star limitpoint α of {αn}, we have α(D) > 0 and also

limn→∞

[sup

z∈∂D

(n + 1

nUα′

n(z) − Uβ′

n(z)

)]= 0,

then for every bounded analytic function f on D, the rational interpolantsrn of degree n with poles at Bn interpolating to f at An, converge locallyuniformly with geometric degree of convergence to f on D. We show thatunder a slightly stricter condition on {Bn}, the boundedness requirement on fcan be replaced with the weaker growth condition

|f(z)| ≤ Ad(z, ∂D)q,

where A > 0, q < 0 and that this growth condition is also sharp in the sensedescribed in this paper.

Keywords. Rational interpolation, logarithmic potentials.

2000 MSC. Primary 30E10; Secondary 31A15.

1. Introduction

Consider a domain D ⊂ C = C ∪ {∞} and an analytic function f on D. Letfor each n ≥ 1, An = {ani}n

i=0 ⊂ D and Bn = {bni}ni=1 ⊂ C be two collections of

points (some ani or some bni may coincide). Then there exists a unique rationalfunction rn, of order n (the degree of the numerator and the denominator areat most n), with poles at Bn, interpolating to f at An, counting multiplicities(see [12, Sec. 8.1]). If An and Bn have some common points, we cancel them in

Received August 22, 2005, in revised form February 20, 2006.

ISSN 1617-9447/$ 2.50 c© 2006 Heldermann Verlag

204 A. Gustafsson CMFT

the construction of rn. In this paper, we study the following problem that datesback to Walsh (cf. [12]): given D, is it possible to find interpolation points An

and poles Bn such that for all analytic f on D, the corresponding sequence {rn}of rational interpolants converges locally uniformly to f on D?

In a series of papers, A. Ambroladze and H. Wallin (see [1, 2, 3, 4] or Section 5of this paper) relate this problem to balayage properties of the normalized pointcounting measures of An and Bn and the balayage (sweeping out) of their weakstar limits. In [4] they observe that in some situations the set of interpolationpoints

⋃n≥1 An must accumulate on the boundary of D, but then we cannot

expect the rational functions rn interpolating to f at An to converge to f for allanalytic f on D. They prove the following theorem that gives a condition thatguarantees convergence for all bounded analytic functions on D.

Theorem A. Let D ⊂ C be a regular domain with bounded boundary and let foreach n ≥ 1, An = {ani}n

i=0 ⊂ D and Bn = {bni}ni=1, be points such that

⋃n≥1 Bn

has no limit point in D. Denote by αn and βn the normalized point countingmeasures of these sets and by α′

n and β′n their sweeping out measures on ∂D.

Assume that for any measure α which is a weak-star limit point of the set {αn},we have

(1) α(D) > 0.

Furthermore, assume that

(2) limn→∞

[supz∈∂D

(n + 1

nUα′

n(z) − Uβ′

n(z)

)]= 0.

Then, for any bounded analytic function f in D the corresponding rational in-terpolant rn with poles at {bni}n

i=1, interpolating f at {ani}ni=0, converges to f

in D:rn → f as n → ∞

uniformly, with geometric degree of convergence, on compact subsets of D.

In [4], the factor (n+1)/n is missing from condition (2), but the factor is neededin the proof, and under the other conditions of this theorem, the correspondingcondition without this factor is not equivalent to (2) (a counterexample can befound in [6]). This means that the theorem should be stated as above.

It is the purpose of this paper to establish a result for rational interpolantswith prescribed poles, similar to Theorem A but with a less restrictive growthcondition than boundedness. This is achieved in Theorem 1 where, instead ofboundedness, we require that the function f satisfies a “polynomial growth”condition

(3) |f(z)| ≤ Ad(z, ∂D)q for z ∈ D,

for some constants A and q, where q is real and A is positive (the point is ofcourse that q can be negative).

6 (2006), No. 1 Convergence of Rational Interpolants 205

It is also true that, at least on the unit disc, we can not find a better growthcondition than (3) for Theorem 1. That is, if f is an analytic function forwhich (3) does not hold, then we can find interpolation points {An} and poles{Bn}, n = 1, 2, . . . such that (2) holds but the interpolating rational functions{rn} do not converge to f on D. This is the statement of Proposition 1. LikeTheorem A, Theorem 1 also contains the conditions (1) and (2). We will prove(Theorem 2) that (2) is necessary in Theorem 1. In summary, Theorem 1, to-gether with Proposition 1 and Theorem 2, show that the growth condition (3)and the conditions (2) and (1) are well adapted to each other for the purpose ofgranting convergence for the rational interpolants rn.

2. Definitions and notation

C the extended complex plane, C = C ∪ {∞}.D a domain in C, regular with respect to the Dirichlet problem.

An a set of points (interpolation points), An = {ani}ni=0, An ⊂ D.

Bn a set of points (poles), Bn = {bni}ni=1, Bn has no limit points in D.

Hn the polynomial of degree at most n + 1 with zeroes in An \ {∞},counting multiplicity: Hn(z) =

∏ni=0(z − ani).

Hnm the polynomial Hnm(z) =∏

i=0,...,n, i�=m(z − ani).

Qn the polynomial of degree at most n with zeroes in Bn \{∞}, count-ing multiplicity: Qn(z) =

∏ni=1(z − bni).

μ a probability measure on C.

supp μ the support of the measure μ.

δz the Dirac measure at z.

αn, βn the normalized point counting measures of An and Bn,

αn = 1/(n + 1)∑n

i=0 δani, βn = 1/n

∑ni=1 δbni

.

αnm the measure αnm = 1/n∑

i=0,...,n, i�=m δani.

τ the unit equilibrium measure on ∂D.

μnw∗−→ μ weak star convergence of measures,

∫φ dμn → ∫

φ dμ for every

continuous function φ on C.

Uμ(z) the logarithmic potential of the measure μ with compact supportin C, Uμ(z) = − ∫

log |z − t| dμ(t).

gD(t, z) the Green function of the domain D, ∂D non-polar, with pole atz ∈ D. We define gD(t, z) = 0 for t ∈ C \ D.

�(Γ) the length of the rectifiable curve Γ.

d(F1, F2) the distance between the sets F1 and F2. This notation will also beabused to denote distance between points.


diam F the diameter of the set F .

‖ · ‖K the supremum norm over K.

C(K) the space of continuous functions φ : K → R.

2.1. Sweeping out of measures. We define the sweeping out of a measureusing the solution to Dirichlet’s problem and the Riesz Representation Theorem.

For a regular domain D in C, and a finite positive measure μ with supp μ ⊂ D,we can define a positive linear functional L on C(∂D) in the following way. Forevery φ ∈ C(∂D) there is a unique bounded harmonic function hφ on D suchthat limz→ζ hφ(z) = φ(ζ) for every ζ ∈ ∂D (see [9, Cor. 4.2.6]). We define L by

φ → Lφ =

∫hφ dμ

From the Maximum Principle [9, Thm. 3.6.9] we see that φ ≥ 0 implies Lφ ≥ 0.Also, φ ≡ 1 → Lφ = μ(D). By the Riesz Representation Theorem [9, Thm. A.3.2]there is a unique measure μ′ on ∂D for which

Lφ =

∫φ dμ′

for all φ ∈ C(∂D) and μ′ has the same mass as μ.

Definition. μ′ is the sweeping out or balayage of μ onto ∂D.

We will also speak of the sweeping out onto ∂D of a measure μ not supportedon D. Then we mean by μ′ the sum of the measures obtained by sweeping out μ,restricted to the different connected components of C\∂D, onto ∂D respectively.

When ∂D and supp μ are compact subsets of C the following holds (see [10,Thm. 4.7])

(i) Uμ′(z) ≤ Uμ(z) + c(μ) for all z ∈ C;(ii) Uμ′(z) = Uμ(z) + c(μ) for every z /∈ D.

The constant c(μ) is non-negative. If D is bounded c(μ) = 0 and if ∞ ∈ D thenc(μ) =

∫gD(t,∞)dμ(t).

The sweeping out process is linear ((μ + ν)′ = μ′ + ν ′) and continuous in the

sense that μnw∗−→ μ as n → ∞ implies μ′

nw∗−→ μ′ as n → ∞.

In the case of a point mass μ = δz, z ∈ D it is well known that δ′z is theharmonic measure on ∂D evaluated at z and in the special case where z = ∞we get δ′∞ = τ , where τ is the equilibrium measure on ∂D.


3. Results

Our first theorem deals with convergence of rational interpolants with prescribedpoles and it is in the same spirit as Theorem A but with the boundedness condi-tion on f replaced by the more general growth condition (3). Unfortunately, welose some generality concerning the poles of rn, a problem that will be discussedin Section 5.

Theorem 1. Let D be a regular domain in C with bounded boundary in C. Forevery n ∈ N, take points An = {ani}n

i=0 in D and points Bn = {bni}ni=1 in C.

Assume that⋃

n≥1 Bn has no limit point on D and assume that

(4) limn→∞

[supz∈∂D

(n + 1

nUα′

n(z) − Uβ′

n(z)

)]= 0,

where α′n and β′

n denote the sweeping out of αn and βn onto ∂D, respectively.Assume also that for every weak-star limit measure α of the set {αn}, we haveα(D) > 0. Then, if f is an analytic function on D, such that for some A > 0,q ∈ R,

(5) |f(z)| ≤ Ad(z, ∂D)q for z ∈ D,

we have for the rational functions rn, with poles at Bn and interpolating to f atAn, that the rn converge to f uniformly and with geometric rate of convergence,on compact subsets of D.

It should be pointed out that in the context of this theorem, when⋃

n≥1 Bn has

no limit points on D, the factor (n + 1)/n in (4) can be omitted.

Our second theorem tells us that for Theorem 1 to hold, the condition (4) isnecessary.

Theorem 2. Let D, An, Bn, α′n and β′

n be as in Theorem 1 with the exceptionthat

(6) limn→∞

[supz∈∂D

(n + 1

nUα′

n(z) − Uβ′

n(z)

)]�= 0,

or that the limit does not exist. Then there exists an analytic function f on D,satisfying equation (5), for some A > 0, q ∈ R, and a point z0 ∈ D, such that

(7) lim supn→∞

|f(z0) − rn(z0)| = ∞,

where rn is the rational function of degree n, with poles at Bn and interpolatingto f at An.

In the above theorem, as seen from its proof, we may also allow⋃

n≥1 Bn to have

limit points on D.

The proposition below shows that, at least for the unit disc, the growth conditionon f is sharp.


Proposition 1. Let D = {z : |z| < 1}. Assume that f is analytic on D andthat there is a sequence Z = {zj}j≥1, zj ∈ D, such that |zj| → 1, as j → ∞ andsuch that for every q < 0, there is a δ > 0 such that

|f(zj)| > (1 − |zj|)q = d(zj, ∂D)q

for all zj with d(zj, ∂D) < δ.

Then, for every R, with 0 < R < 1, there are points An = {ani}ni=0 ⊂ D,

Bn = {bni}ni=1 ⊂ C, such that (4) holds and α(D) > 0 for every weak star limit α

of {αn}, but for the rational function rn of degree n, interpolating to f at An wehave

lim supn→∞

|f(z) − rn(z)| = ∞for all z ∈ D with |z| > R.

The rational functions rn in the above proposition will in our construction bepolynomials, i.e. bni = ∞, i = 1, . . . , n.

Remark. Although Theorem 1 does not guarantee convergence for f growing asin Proposition 1, there is however, by [7, Thm. 1], for every analytic function fon D, interpolation points in D giving convergence for that specific f (and forall analytic g on D that satisfy |g| ≤ |f |). Unfortunately, in this case we do nothave an explicit condition like (4) to check when such interpolation points andpoles give convergence.

4. Proofs

4.1. Proof of Theorem 1. The following lemma is similar to Lemma 1 in [4]and can be proved in the same way. We refer to [4] for the proof.

Lemma 1. Under the conditions of Theorem 1, assume that αn → α and βn → βin the weak star sense. Then α′ = β′.

We proceed with a lemma that lists some implications and equivalent formula-tions of (4).

Lemma 2. Let D be a regular domain in C with bounded boundary and let forevery n ∈ N, An = {ani}n

i=0 be points in D and Bn = {bni}ni=1 be points in C

such that⋃

n≥1 Bn has no limit point in D. Then the following statements areequivalent:

(i) limn→∞

[supz∈∂D

(n + 1

nUα′

n(z) − Uβ′

n(z)

)]= 0;

(ii) limn→∞

(n + 1

nUα′

n(z) − Uβ′

n(z)

)= 0,

uniformly on ∂D.


If also ∞ /∈ ⋃n≥1(An ∪Bn) and we let Ω be the component of C \ ∂D containing

the point at infinity, then the statements above are also equivalent to:

(iii) limn→∞

∣∣∣∣∣Hn(z)

Qn(z)exp

(n∑

i=1

[gΩ(bni,∞) − gΩ(ani,∞)] − gΩ(an0,∞)

)∣∣∣∣∣1/n

= 1,

uniformly on ∂D.

The statements (i) and (ii) also imply that:

(iv) given ε > 0, there is an N ∈ N, such that n ≥ N implies∣∣Uα′nm

(z) − Uβ′n(z)

∣∣ < ε,

for all z ∈ ∂D, m ∈ {0, . . . , n}.Condition (iv) is, if ∞ /∈ ⋃

n≥1(An ∪ Bn), equivalent to:

(v) given ε > 0, there is an N ∈ N, such that n ≥ N implies∣∣∣∣∣∣∣∣∣∣∣Hnm(z)

Qn(z)exp

(n∑

i=1

gΩ(bni,∞) −∑

i=0,...,n, i�=m

gΩ(ani,∞)

)∣∣∣∣∣1/n

− 1

∣∣∣∣∣∣ < ε,

for all z ∈ ∂D, m ∈ {0, . . . , n}.

Proof. We will prove the equivalences (i)⇔(ii), (ii)⇔(iii), (iv)⇔(v) and theimplication (ii)⇒(iv).

Without loss of generality, we assume that there are measures α and β such thatαn → α and βn → β in the weak star sense.

The implication (ii)⇒(i) is trivial.

Now, assume that (i) holds. To prove (ii), it is enough to show that

(8) limn→∞

[inf

z∈∂D

(n + 1

nUα′

n(z) − Uβ′

n(z)

)]= 0.

The inequality,

(9) lim supn→∞

[inf

z∈∂D

(n + 1

nUα′

n(z) − Uβ′

n(z)

)]≤ 0,

follows directly from (i).

From the assumptions on Bn, there is an open set U with bounded boundary,such that U ⊂ C\D and Bn ⊂ U for n larger than some n0. From our definition

of balayage it follows that if we denote by βn the sweeping out of βn onto ∂U ,then β′

n = (βn)′. Furthermore, we see that since supp βn lies on ∂U which isbounded and separated from ∂D which is also bounded, the family {Uβn

}n>n0


is a uniformly bounded family of harmonic functions, on some open set contain-ing ∂D. Consequently, by [8, Thm, 2.18], the family {Uβn

}n>n0 is equicontinuouson ∂D. Furthermore, from properties of balayage, we have

Uβ′n(z) − Uβ′

n(w) = Uβn

(z) − Uβn(w), for z, w ∈ ∂D,

and we see that the family {Uβ′n}n>n0 is equicontinuous on ∂D as well. It fol-

lows then, from the Arzela-Ascoli Theorem and the fact that {Uβ′n} converges

pointwise to Uβ′ on ∂D, that

(10) limn→∞

Uβ′n

= Uβ′ uniformly on ∂D.

For any sequence {zn} ⊂ ∂D converging to some point z0 ∈ ∂D, we have nowby (10), the Principle of Descent (cf. [11, Sec. A.III]) and Lemma 1, that

lim infn→∞

(n + 1

nUα′

n(zn) − Uβ′

n(zn)

)≥ Uα′(z0) − Uβ′(z0) = 0.

However, as pointed out in [11, Sec. 1.1], by the continuity (constant) of the righthand side of this lim inf relation, it will hold uniformly on ∂D and thus

lim infn→∞

[inf

z∈∂D

(n + 1

nUα′

n(z) − Uβ′

n(z)

)]≥ 0.

Combining this with (9) we see that (8) and hence also (ii) must hold.

Next, we will prove that (ii)⇒(iv). Assume that (ii) holds and let ε > 0. Forz ∈ ∂D, we have

Uα′nm

(z) − Uβ′n(z) =

n + 1

nUα′

n(z) − Uβ′

n(z) − 1

nUδ′anm

(z)

≤ n + 1

nUα′

n(z) − Uβ′

n(z) +

log (diam(∂D))

n,

so by (ii) we get for large n that,

(11) supz∈∂D

(Uα′

nm(z) − Uβ′

n(z)

)< ε,

for all m ∈ {0, . . . , n}. Assume now, for the sake of contradiction that thereis a subsequence {nk} of the natural numbers, a sequence {mk}, where mk ∈{0, . . . , nk} and points zk ∈ ∂D, such that

(12) Uα′nkmk

(zk) − Uβ′nk

(zk) < −ε, k = 1, 2, . . . .

By taking a subsequence and relabeling, we can also assume that there is a

z0 ∈ ∂D such that zk → z0 as k → ∞. However, since αnkmk

w∗−→ α as k tends toinfinity and Uβ′

nk→ Uβ′ , uniformly on ∂D, we have by the Principle of Descent

that,

lim infk→∞

(Uα′

nkmk(znk

) − Uβ′nk

(znk))≥ Uα′(z0) − Uβ′(z0) = 0


and we have a contradiction to (12). Thus, for large n,

infz∈∂D

(Uα′

nm(z) − Uβ′

n(z)

)> −ε,

for all m ∈ {0, . . . , n} and together with (11) we get (iv).

The equivalence (ii)⇔(iii) follows directly from the fact that

n + 1

nUα′

n(z) − Uβ′

n(z) =

1

nlog

∣∣∣∣Qn(z)

Hn(z)

∣∣∣∣ +1

ngΩ(an0,∞)

+1

n

n∑i=1

(gΩ(ani,∞) − gΩ(bni,∞)) , z ∈ ∂D.

Finally, the equivalence (iv)⇔(v) follows directly from the equality

Uα′nm

(z) − Uβ′n(z) =

1

nlog

∣∣∣∣ Qn(z)

Hnm(z)

∣∣∣∣ +1

n

∑i=0,...,n, i�=m

gΩ(ani,∞)

−1

n

n∑i=1

gΩ(bni,∞), z ∈ ∂D.

Lemma 3. Let D be a regular domain in C, with ∞ ∈ D. For every n ∈ N, takeAn = {ani}n

i=0 ⊂ D and Bn = {bni}ni=1 ⊂ C such that

⋃n≥1 An is bounded and⋃

n≥1 Bn has no limit point on D. Assume that there is a measure α, such that

αn → α in the weak star sense, let c :=∫

gD(t,∞) dα(t) and let a be a constantless than c. Finally, assume that (4) holds.

Then, for Dn := {z ∈ C : Uβn(z) − (n + 1)n−1Uαn(z) < a} we have that:

• Dn is open;• Uβn − (n + 1)n−1Uαn = a on the boundary Γn of Dn;

• Uβn − (n + 1)n−1Uαn > a on C \ Dn;• Dn ⊂ D for n large.

Also, for every constant C, with 0 ≤ C < 1, we have for large enough n that

(13) d(Γn, ∂D) ≥ Cn.

Proof. We have for large n that Bn = supp βn does not intersect An and thatAn = supp αn is a finite set in Dn. Since Uαn and Uβn are continuous out-side their supports, Dn is open and Uβn − (n + 1)n−1Uαn = a on ∂Dn. Also,

Uβn − (n + 1)n−1Uαn is superharmonic on an open environment to C \Dn, so by

the Minimum Principle and since Uβn − (n+1)n−1Uαn = ∞ on Bn ⊂ C \Dn, we

have Uβn − (n + 1)n−1Uαn > a on C \ Dn.

For all z ∈ ∂D,

Uβn(z) − n + 1

nUαn(z) = Uβ′

n(z) − n + 1

nUα′

n(z) + cn,


where

cn =n + 1

n

∫gD(t,∞) dαn(t)

and (since supp α ∩ {∞} = ∅) cn → c as n → ∞. Thus we get from (4), that

limn→∞

(inf

z∈∂D(Uβn(z) − n + 1

nUαn(z))

)= c.

From this equation and the superharmonicity of Uβn − (n+1)n−1Uαn on an open

environment to C \ D, we get from the Minimum Principle that Dn ⊂ D forlarge n.

To prove the last assertion of the lemma, we first note that Γn is a level set ofthe rational function ρn = Hn/Qn. That is, Γn = {z : |ρn(z)| = ena}. Fix pointsζn ∈ Γn and zn ∈ ∂D such that d(Γn, ∂D) = d(ζn, zn) and let γn be the straightline segment from ζn to zn. Since, for large n, the function ρn is analytic on anopen environment to D and γn ⊂ D, we have that

(14) |ρn(zn) − ρn(ζn)| =

∣∣∣∣∫

γn

ρ′n(t) dt

∣∣∣∣ ≤ supt∈γn

|ρ′n(t)|d(ζn, zn).

For ρnm := Hnm/Qn, we have by straightforward calculation,

(15) |ρ′n(z)| ≤

n∑m=0

|ρnm(z)| + |ρn(z)|n∑

m=1

1

|z − bnm| .

Since cn → c as n → ∞ and a < c, we can take a constant b, such that 0 < b < 1and 0 < ea−cn/b < 1 for large n. Then we take an ε such that 0 < ε < 1−ea−cn/bfor large n. By Lemma 2, we have for large n that

(16)∣∣|ρn(z)|1/ne−cn − 1

∣∣ < ε, for all z ∈ ∂D

and also, by Lemma 2 and since | ∫ gD(t,∞) dαnm(t)−cn| tends to zero uniformlyin m as n approaches infinity, that

(17)∣∣|ρnm(z)|1/ne−cn − 1

∣∣ < ε, for all z ∈ ∂D.

Now, since γn ⊂ D and ρ′n is analytic on an environment to D we get from the

Maximum Principle and by using (15), (16), (17) and the assumptions on Bn,that

(18) |ρ′n(t)| ≤ ‖ρ′

n‖∂D ≤ (n + 1)(1 + ε)nencn

(1 +

1

δ

), for t ∈ γn,

where δ is chosen such that for large n, d(Bn, ∂D) > δ. From (16) we also getfor large n, that

|ρn(z)| > (1 − ε)nencn >ena

bn=

|ρn(ζ)|bn

for z ∈ ∂D, ζ ∈ Γn.


Thus, using this equation together with the equations (14) and (18), we get forany C with C < (1 − ε)/(1 + ε) and n large, that

d(Γn, ∂D) = d(ζn, zn) ≥ |ρn(zn) − ρn(ζn)|supt∈γn

|ρ′n(t)| ≥ |ρn(zn)| − |ρn(ζn)|

(n + 1)(1 + ε)nencn(1 + 1δ)

>

(1 − ε

1 + ε

)n1 − bn

(n + 1)(1 + 1δ)

> Cn.

The last inequality holds for large n because the right hand side tends to zerofaster than the left hand side. Now, since ε can be chosen arbitrarily small, theinequality holds for any C < 1, choosing n (depending on C) sufficiently largeand the lemma is proved.

We will also make use of the following lemma from [4].

Lemma 4. Let R(z) = A(z)/B(z) be a rational function of order n, withmax{deg A, deg B} = n. For an arbitrary number ρ > 0 let γρ = {z : |R(z)| = ρ}be a level curve. Denote by diam(γρ) and �(γρ) the diameter and the total lengthof γρ respectively. Then

�(γρ) ≤ 4n diam(γρ).

And here is the proof of our main theorem.

Proof of Theorem 1. It is straightforward to show that the condition (4) isinvariant under Mobius transformations, so we can assume that ∞ ∈ D. Wewill first also assume that the set

⋃n≥1 An is bounded in C and return to the

general situation later. We can also, without loss of generality, assume that thereare measures α and β, such that αn → α and βn → β in the weak-star sense.According to Lemma 1, their swept measures α′ and β′ satisfy α′ = β′. Takec :=

∫gD(t,∞) dα(t). From properties of balayage, we have that

Uβ(z) − Uα(z) = c for z ∈ ∂D.

Since Uβ is harmonic in D\{∞}, Uα is superharmonic in D\{∞} and Uβ−Uα hasa removable singularity at the point ∞, we can regard Uβ −Uα as a subharmonicfunction in D and then the Maximum Principle implies that

(19) Uβ − Uα < c for z ∈ D,

where the strictness in the inequality comes from the fact that α(D) > 0. Fix acompact set K ⊂ D and take another compact F , with K ⊂ int F and ∞ ∈ int F .Take also a domain V with V ⊂ D and F ⊂ V . From the upper semi-continuityof Uβ − Uα in D, there is an ε > 0 such that

Uβ(z) − Uα(z) < c − ε for z ∈ V.


For every z ∈ V and every sequence {zn} such that zn → z as n → ∞, we haveby the Principle of Descent (see [11, Sec. A.III]) that

lim supn→∞

(Uβn(zn) − n + 1

nUαn(zn)

)≤ Uβ(z) − Uα(z) < c − ε.

Since the right hand side of the above inequality is a continuous function of z (aconstant), the limit superior relation above implies (see [11, p. 3]) that the samerelation holds uniformly on compact subsets of V . This means that there is aconstant b such that for large n and z ∈ F ,

(20) Uβn(z) − n + 1

nUαn(z) < b < c − ε

2.

Let

Dn :=

{z : Uβn(z) − n + 1

nUαn(z) < c − ε

2

}and let Γn := ∂Dn. According to Lemma 3, we have for large n, that Dn ⊂ Dand Uβn(t)− (n + 1)n−1Uαn(t) = c− ε/2 if and only if t ∈ Γn. We see also, using(20), that for large n, F ⊂ Dn. For large n, An ∩ Bn = ∅ and

Uβn(ani) − n + 1

nUαn(ani) = −∞,

so An ⊂ Dn. The boundary Γn, which is a level set of Hn/Qn, consists of a finitenumber of rectifiable Jordan curves (cf. [12, Sec. 3.3]), and equipping these withpositive orientations with respect to Dn, we get from [12, p. 186] that

(21) Rn(z) := f(z) − rn(z) =1

2πi

∫Γn

Hn(z)Qn(t)

Hn(t)Qn(z)

f(t) dt

t − z, for z ∈ F.

If z = ∞ the integral of course makes no sense but Rn does and is continuousthere. For all z ∈ K,

|2πRn(z)|1/n ≤∣∣∣∣Hn(z)

Qn(z)

∣∣∣∣1/n

supt∈Γn

∣∣∣∣Qn(t)

Hn(t)

∣∣∣∣1/n (

1

d(K, Γn)

)1/n (∫Γn

|f | ds

)1/n

,

where ds denotes the arc length measure on Γn. Taking logarithms, we get, forz ∈ K, that

log |2πRn(z)|1/n ≤(

Uβn(z) − n + 1

nUαn(z)

)

− supt∈Γn

(Uβn(t) − n + 1

nUαn(t)

)(22)

+ log d(K, Γn)−1/n + supt∈Γn

log |f(t)|1/n + log [�(Γn)]1/n.

Since K ⊂ int F and F ⊂ Dn for large n, we get

(23) lim supn→∞

d(K, Γn)−1/n ≤ 1.


We will now show that

(24) lim supn→∞

(supt∈Γn

|f(t)|1/n

)≤ 1.

For this, we can assume that the constant q in (5) is negative because if not,then f is bounded and (24) holds trivially. Using the growth condition (5) andLemma 3, we get for any C < 1, that

lim supn→∞

(supt∈Γn

|f(t)|1/n

)≤ lim sup

n→∞

(supt∈Γn

[Ad(t, ∂D)q]1/n

)= lim sup

n→∞[Ad(Γn, ∂D)q]1/n

≤ lim supn→∞

[ACnq]1/n = Cq

and since C can be taken arbitrarily close to 1, equation (24) follows.

For the last term of (22) we use that since ∞ ∈ int F and F ⊂ Dn, the diameterof Γn is bounded. From Lemma 4 it follows then that

(25) lim supn→∞

[�(Γn)]1/n ≤ 1.

Thus, using (22), (20), (23), (24), (25) and the fact that

Uβn − n + 1

nUαn = c − ε

2

on Γn, we have

lim supn→∞

log |2πRn(z)|1/n ≤ b −(c − ε

2

)< 0, for z ∈ K.

Thus Rn tends to zero, uniformly and with geometric rate of convergence, on K.

To handle the situation when⋃

n≥1 An is not a bounded set in C, we can usethe following argument from [4, p. 10-11]. Fix as before a closed set F in Dcontaining the point at infinity in its interior. For DR = {|z| > R}, we haveDR ⊂ D for large R. We replace the points ani that lie inside DR with pointsa′

ni on ∂DR in such a manner that (4) still holds. This is done by choosing thepoints a′

ni asymptotically as the swept measure of the part of αn inside of DR,onto ∂DR. With this new choice of interpolation points we have proven that thecorresponding rational interpolants r′n converge geometrically to f on F . Also,the difference in the error formula between f − r′n and f − rn can be arbitrarilysmall on ∂F and thus also on F , by taking R large. For more details, see thelast part of the proof of Theorem 1 in [4].

4.2. Proof of Theorem 2. Before we prove Theorem 2, we will prove thefollowing simple lemma.


Lemma 5. Let K be a compact subset of C and let {μn} and {νn} be sequencesof Borel probability measures on K. Assume also that there are probability mea-sures μ and ν on K such that μn → μ and νn → ν, in the weak star sense, asn → ∞. Then

lim infn→∞

[supz∈K

(n + 1

nUμn(z) − Uνn(z)

)]≥ 0.

Proof. Assume that the result does not hold. Then there exists a subsequence{nk} of the natural numbers and a constant c < 0 such that for all z ∈ K wehave (nk +1)n−1

k Uμnk(z)−Uνnk

(z) < c. Thus, if τ is the unit equilibrium measureon ∂D, there is a constant V such that Uτ = V quasi-everywhere on K and wehave by Fubini’s Theorem that∫

V d

(nk + 1

nk

μnk− νnk

)=

∫ (nk + 1

nk

Uμnk− Uνnk

)dτ < c < 0.

Letting k → ∞ we get ∫V d(μ − ν) < 0,

which is a contradiction since μ and ν have the same mass.

And here is the proof of Theorem 2.

Proof of Theorem 2. Since Uα′n

differs from Uαn by a constant on ∂D andUαn|∂D is continuous, we have that Uα′

n|∂D is continuous. Also, Uβ′

nis lower

semicontinuous, so ((n + 1)n−1Uα′n(z)−Uβ′

n(z))|∂D is upper semicontinuous and

we can, for every n ∈ N, take zn ∈ ∂D to be a point such that

(26)n + 1

nUα′

n(zn) − Uβ′

n(zn) = sup

z∈∂D

(n + 1

nUα′

n(z) − Uβ′

n(z)

).

Take a sequence {cj} ⊂ C, with∑∞

j=1 |cj| < ∞, but with lim supj→∞ |cj|1/j = 1

(for example |cj| = j−p, p > 1). We will later choose the arguments of our cj’s.Let f(z) =

∑∞j=0 cj(z−zj)

−1. The series converges uniformly on compact subsets

of D, so f is analytic on D. Also, f satisfies (5) with A =∑∞

j=1 |cj| and q = −1.

We can apply a Mobius transformation to ensure that ∞ ∈ D. We will alsoassume that the set

⋃n≥1 An is bounded and the general case can then be handled

by the same method as in the proof of Theorem 1. We can also assume that thereare measures α and β such that αn → α and βn → β in the weak star sense, asn → ∞. Interchanging summation and integration in (21) and using Cauchy’sintegral formula, we get that

Rn(z) = −∞∑

j=1

Hn(z)Qn(zj)

Hn(zj)Qn(z)

cj

zj − zfor z ∈ D \ {∞}.


For a fixed z ∈ D \ {∞}, say z = z0, one may choose the argument of every cj

such that every term in the series above has the same argument. This z0 will bechosen later and then we get

|Rn(z0)| ≥∣∣∣∣Hn(z0)Qn(zn)

Hn(zn)Qn(z0)

cn

zn − z0

∣∣∣∣ .

Since |cn|1/n → 1 and |zn − z0| is bounded, as n → ∞, we get

(27)

lim supn→∞

1

nlog |Rn(z0)| ≥ lim sup

n→∞

( (n + 1

nUαn(zn) − Uβn(zn)

)

−(

n + 1

nUαn(z0) − Uβn(z0)

) ).

By substituting K = ∂D, μn = α′n and νn = β′

n in Lemma 5 and also usingEquation (6) we see that, for some b > 0

lim supn→∞

[supz∈∂D

(n + 1

nUα′

n(z) − Uβ′

n(z)

)]> b.

Thus, for a subsequence {nk} of the natural numbers and for k large, we have,by (26), properties of balayage and the above equation

(28)

n + 1

nUαnk

(znk) − Uβnk

(znk)

= supz∈∂D

(n + 1

nUαnk

(z) − Uβnk(z)

)

= supz∈∂D

(n + 1

nUα′

nk(z) − Uβ′

nk(z)

)− n + 1

nc(αnk

)

> b − n + 1

nc(αnk

),

where

c(αnk) =

∫gD(t,∞) dαnk

(t).

Now, take a constant a with 0 < a < b. We will show that there is a ζ0 ∈ ∂D forwhich

(29) Uα′(ζ0) − Uβ′(ζ0) < a.

Suppose there is no such point on ∂D. Let τ denote the unit equilibrium measureon ∂D. We have, for some constant V that Uτ = V on ∂D. Thus, using the factthat α′ and β′ have the same mass and using Fubini’s Theorem, we get

0 =

∫V d(α′ − β′) =

∫(Uα′ − Uβ′) dτ ≥ a,

which is a contradiction. From (29), we get

(30) Uα(ζ0) − Uβ(ζ0) < a − c(α),


where c(α) is the constant

c(α) =

∫gD(t,∞) dα(t).

By the Lower Envelope Theorem (see [11, Sec. A.III]), there is a polar set E,such that for z ∈ D∗ := D \ E, we have

(31) Uα(z) = lim infk→∞

Uαnk(z).

We will now show that there is a z0 ∈ D∗ \ {∞} such that

(32) Uα(z0) − Uβ(z0) < a − c(α).

Since D is non-thin at ζ0 (cf. [9, Thm. 3.8.3]), E is thin at ζ0 ([8, Thm. 10.2]),and the union of two sets which are thin at ζ0 is itself thin at ζ0, D∗ must benon-thin at ζ0. Thus,

lim infz→ζ0, z∈D∗

Uα(z) = Uα(ζ0).

Hence, because of (30) and using that −Uβ is upper semicontinuous together withthe equation above, we see that we can take a z0 ∈ D∗\{∞} such that (32) holds.Now, taking a suitable subsequence of {nk} and relabeling, so that we denotealso the new subsequence with {nk}, we have by (31) and since z0 /∈ supp β, thatfor k large,

nk + 1

nk

Uαnk(z0) − Uβnk

(z0) < a − c(α).

Using (27), (28) and the equation above we get

lim supn→∞

1

nlog |Rn(z0)| ≥ lim sup

k→∞

1

nk

log |Rnk(z0)|

≥ lim supk→∞

( (nk + 1

nk

Uαnk(znk

) − Uβnk(znk

)

)

−(

nk + 1

nk

Uαnk(z0) − Uβnk

(z0)

) )

≥ lim supk→∞

(b − nk + 1

nk

c(αnk) − (a − c(α))

)= b − a > 0,

so (7) holds and the theorem is proved.

4.3. Proof of Proposition 1. To prove the proposition we will make use ofthe following lemma.

Lemma 6. Let D = {z : |z| < 1} and take a sequence {an} ⊂ D. For everyn = 1, 2, . . ., take bni = ∞, ani = 0 for i = 1, . . . , n and an0 = an. Then (4)holds if and only if

(33) lim infn→∞

(1 − |an|)1/n = 1.


Proof. Clearly

αn =n

n + 1δ0 +

1

n + 1δan and βn = δ∞.

Since δ′0 = δ′∞ = τ , we have

supz∈∂D

(n + 1

nUα′

n(z) − Uβ′

n(z)

)= sup

z∈∂D

(Uδ′0(z) +

1

nUδ′an

(z) − Uδ′∞(z)

)

= supz∈∂D

1

nUδan

(z) = supz∈∂D

−1

nlog |z − an|

= −1

nlog (1 − |an|).

Since 1 − |an| is bounded in n we always have

lim supn→∞

(1 − |an|)1/n ≤ 1

and the equivalence of (4) and (33) follows.

Now we will prove Proposition 1.

Proof of Proposition 1. For every n = 1, 2, . . ., we will choose {ani}ni=0, {bni}n

i=1

as in Lemma 6, where an0 = an ∈ D will be specified as follows.

Take an R with 0 < R < 1. Let {Ck}k≥1 and {qk}k≥1 be sequences of realnumbers such that 0 < Ck < 1, Ck → 1 as k → ∞, qk < 0 and Cqk

k ≥ R−1. Fromour assumptions on f we get that for every k ∈ N there is an Nk ∈ N such that

n > Nk → |f(zj)| > (1 − |zj|)qk > Cqknk , for those zj with |zj| > 1 − Cn

k .

For k ≥ 2 we also let Nk > Nk−1. Thus, for every n > N1 there is a unique k(n)such that

Nk(n) < n ≤ Nk(n)+1.

Note that k(n) → ∞, and qk(n) → −∞ as n → ∞ and let

dn := 1 − Cnk(n).

We observe that |zj| > dn implies that

(34) |f(zj)| > Cqk(n)n

k(n) ≥ R−n,

hence, since f is locally bounded, dn → 1 as n → ∞. Now take an ∈ Z such that

|an| = maxzj∈Z,|zj |≤dn+1

|zj|.

From {an} we can extract a subsequence {anl}, l = 1, 2, . . ., such that

(35) |anl| > dnl

, l = 1, 2, . . .

We have

lim infn→∞

(1 − |an|)1/n+1 ≥ lim infn→∞

(1 − rn+1)1/n+1 = lim inf

n→∞Ck(n+1) = 1


so, according to Lemma 6, (4) holds. It is also clear that αnw∗−→ α = δ0 and

α(D) = 1 > 0.

Now fix z ∈ D with |z| > R. For nl large we have |anl| > |z|. Let Γnl

and γ bepositively oriented circles in D centered at the origin, such that γ contains z, butnot anl

and Γnlcontains γ and anl

. Also let Snlbe a positively oriented circle

centered at anland not intersecting Γnl

or γ. We have

Hn(z) =n∏

i=0

(z − ani) = zn(z − an)

so for nl large

f(z) − rnl(z) =

1

2πiHnl

(z)

∫Γnl

f(t) dt

tnl(t − anl)(t − z)

=1

2πiHnl

(z)

∫γ

f(t) dt


+1

2πiHnl

(z)

∫Snl

f(t) dt


.

Since the function f(t)/[(t − anl)(t − z)] is bounded for t ∈ γ and nl large, and

Hnl(z) = (z − anl

)znl , we get for the first term in the expression above that∣∣∣∣ 1

2πiHnl

(z)

∫γ

f(t) dt


∣∣∣∣ ≤ K

∣∣∣∣znl

ρnl

∣∣∣∣where K > 0 is a constant and ρ is the radius of γ. Since ρ > |z| the first termtends to zero as nk tends to infinity.

Using (35), (34) and taking nl large, the modulus of the second term becomes,by the Cauchy Integral Formula:∣∣∣∣∣ 1

2πiHnl

(z)

∫Snl

f(t) dt


∣∣∣∣∣ =

∣∣∣∣Hnl(z)

f(anl)

(anl)nl(anl

− z)

∣∣∣∣> |z|nl|f(anl

)|(|z|Cqk(nl)

k(nl))nl ≥

( |z|R

)nl

.

Since |z| > R we get that |f(z) − pnl(z)| → ∞ as l → ∞.

5. Discussion

We see that Theorem 1 is not a true generalization of Theorem A from boundedfunctions to functions satisfying the condition (5). This is because in Theorem 1,we require that the pole set

⋃n≥1 Bn to have no limit point on D, while in The-

orem A the only requirement is that there is no limit point in D. Unfortunately,it is a fact that the stronger requirement on the poles in Theorem 1, cannot bereplaced with the weaker condition in Theorem A, for the following reason.


Take for example D = {z : |z| < 1}, a sequence {an} in D and let bn = 1/an. Let,for n = 2, 3, . . ., ani = 0 for i = 0, . . . , n−1, ann = an, bni = ∞, for i = 1, . . . , n−1and bnn = bn. Using that δ′an

= δ′bn(cf. [1, Ex. 3]), and calculations similar to

those in the proof of Lemma 6 and the latter part of the proof of Proposition 1,one will see that if an approaches ∂D quickly enough one gets for z ∈ D \ {0}that |f(z) − rn(z)| → ∞ as n → ∞, but (4) still holds.

There are also a number of previous results on similar topics and so a meaningfulconclusion of this paper might be to summarize some of them. As seen from thestatements below, we may either choose to let the poles or the interpolationpoints approach the boundary of D and different convergence results will beapplicable accordingly. Using the notation already established, the followingstatements hold.

• Let⋃

n≥1 An have all its limit points in D and⋃

n≥1 Bn no limit points inD, αn → α and βn → β in the weak star sense. Then, for rn to converge tof on D, for any analytic f on D, it is necessary and sufficient that α′ = β′

(see [1, Thms. 1,2]). Hence we can achieve this by simply choosing poleson ∂D asymptotically as α′.

• Let D be a bounded domain in C and bni = ∞, i = 1, . . . n, n = 1, 2, . . .(this makes rn a polynomial). Then there is a measure α with supp α ⊂ Dsuch that α′ = β′, if and only if ∂D is an analytic curve [2, Thm. 1].

• If D is simply connected and⋃

n≥1 An has a limit point on ∂D we do notget convergence for all analytic f on D [2, Thm. 5].

• The three statements above tell us that if D is a bounded domain such that∂D is not analytic and if all bni = ∞, there is no choice of interpolationpoints in D giving convergence for all analytic f on D.

• If⋃

n≥1 Bn has no limit point in D, αn → α in the weak star sense andif α(D) > 0. Then (4) is a necessary and sufficient condition to grantconvergence for all analytic functions f satisfying (5). Furthermore wecannot expect convergence if f does not satisfy (5) (the results of thispaper).

• If D is a Jordan domain satisfying an exterior Dini condition and allbni = ∞, there exist interpolation points in D such that α(D) > 0 and (4)holds ([5, Thm. 1] and [7, Thm. 3.2]).

• For any bounded domain D and a fixed analytic function f and all bni = ∞,there exist interpolation points in D such that the corresponding rationalfunctions (polynomials) converge to f in D [7, Thm. 3.1]. We do nothowever have any explicit condition (in terms of f) telling us whether aparticular choice of interpolation points will work.

References

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2. , Balayage and convergence of rational interpolants, J. Approx. Theory 99 no.2(1999), 230–241.

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4. , Rational interpolants with preassigned poles, theoretical aspects, Studia Math.132 no.1 (1999), 1–14.

5. S. J. Gardiner and Ch. Pommerenke, Balayage properties related to rational interpolation,Constr. Approx. 18 (2002), 417–426.

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Anders Gustafsson E-mail: [email protected]: Department of Engineering, Physics and Mathematics, Mid Sweden University,851 70 Sundsvall, Sweden.

Convergence of Rational Interpolants to Analytic Functions with Restricted Growth

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