Limits of zeros of orthogonal polynomials on the circle^{∗}

Barry Simon^{†}  Vilmos Totik^{‡}

April 26, 2004

Abstract We prove that there is a universal measure on the unit circle such that any probability measure on the unit disk is the limit distribution of some subsequence of the corresponding orthogonal polynomials. This follows from an extension of a result of Alfaro and Vigil (which answered a question of P. Turán): namely, for n < N, one can freely prescribe the n-th polynomial and N − n zeros of the N-th one. We shall also describe all possible limit sets of zeros within the unit disk.

1 Results

Let D be the open unit disk. We consider Borel measures dμ(t), t ∈ [−π, π) on the unit circle (identified with R_{/mod 2π}) of infinite support, and for such a measure let

Φ_{n}(μ, z) = z^{n} + ∑_{k=0}^{n−1} c_{n,k}(μ)z^{k}  (1)

be the n-th monic orthogonal polynomial.

It is well known that all zeros of Φ_{n} lie in D. The main result of this paper is

Theorem 1 For 1 ≤ n < N, let Φ_{n} be a monic polynomial of degree n with zeros in D and let there be given N − n points a_{1}, …, a_{N−n} in D. Then there is a measure μ on the unit circle such that Φ_{n}(μ) = Φ_{n} and a_{i}, i = 1, …, N − n, are zeros (with multiplicity) of Φ_{N}(μ).

In short, one can freely prescribe Φ_{n}(μ), and N − n zeros of Φ_{N}(μ).

We do not know if Φ_{N}(μ) is unique, that is, that the other zeros of Φ_{N}(μ) are uniquely determined by Φ_{n} and a_{1}, …, a_{N−n}.

∗For Mathematische Nachrichten issue in memory of F.V. Atkinson

†Supported in part by NSF grant DMS-0140592

‡Supported by NSF grant DMS-0097484 and by OTKA T/034323, TS44782

Corollary 2 Let 0 = n_{0} < n_{1} < n_{2} < ⋯ be a sequence of natural numbers and {a_{j}}_{j=1}^{∞} a sequence in D. Then there is a measure μ such that for each j = 1, 2, …, all a_{i}, n_{j−1} < i ≤ n_{j}, are zeros (with multiplicity) of Φ_{n_{j}}(μ).

Indeed, by induction, we get from Theorem 1 measures μ_{j} such that

Φ_{n_{j−1}}(μ_{n_{j−1}}) = Φ_{n_{j−1}}(μ_{n_{j}})

and a_{i}, n_{j−1} < i ≤ n_{j}, are zeros of Φ_{n_{j}}(μ_{n_{j}}). By a theorem of Geronimus [2], if for an m, the m-th orthogonal polynomials for two measures coincide, then the same happens for the l-th polynomials with all l ≤ m. Hence we have Φ_{n_{k}}(μ_{n_{k}}) = Φ_{n_{k}}(μ_{n_{j}}) for all k ≤ j. Since Φ_{n_{k}} determines the moments up to order n_{k}, this shows that for each fixed m, the m-th moment of the measures μ_{n_{k}}, k = 1, 2, …, are eventually constant, so the weak-^{∗} limit, say μ, of these measures exists and obeys Φ_{n_{k}}(μ_{n_{k}}) = Φ_{n_{k}}(μ) for all k, and the corollary follows.

To formulate the next corollary, recall that the normalized counting measure ν_{n} of Φ_{n} is defined as the measure that puts mass 1/n to each zero of Φ_{n} (counting multiplicity). We say that a subsequence {Φ_{n_{k}}} has zero distribution ν if the normalized counting measures ν_{n_{k}} converge to ν in the weak-^{∗} topology.

Corollary 3 There is a universal measure μ on the unit circle such that if ν is any probability measure on the closed unit disk D̄, then some subsequence {Φ_{n_{k}}(μ)} has zero distribution ν.

This is immediate from Corollary 2 if we prescribe appropriately n! − (n − 1)! = n!((n − 1)/n) zeros for Φ_{n!} in the sequence Φ_{0}(μ), Φ_{1!}(μ), Φ_{2!}(μ), . . .. The rest of the proof is standard (take a countable dense subset {μ_{j}} in the space of probability measures on D̄ so that each μ_{j} occurs infinitely often in this sequence, and select the n_{k}!((n_{k}−1)/n_{k}) zeros for Φ_{n_{k}!} so that their distribution converges weak-^{∗} to μ_{j} as we let n_{k} tend to infinity in such a way that μ_{j} = μ_{n_{k}}).

It is known (see, e.g., [11, Lemma 4]) that zeros of orthogonal polynomials cluster to the support supp(μ) of the generating measure μ if the interior of supp(μ) is empty and C \ supp(μ) is connected. This is due to the fact that the n-th orthogonal polynomial minimizes the L^{2}(μ) norm among all monic polynomials of degree n. This is the case, for example, if μ is a measure on the unit circle, but its support is not the whole circle. The situation changes if the interior of the support is not empty or if C\supp(μ) is not connected. Consider, for example, the closed unit disk or the unit circle and the appropriate (area or arc) Lebesgue measure on it, in which case Φ_{n}(z) = z^{n}, hence all the zeros are at the origin. It was P. Turán who asked if for measures on the unit circle it is possible that the zeros of the orthogonal polynomials cluster to all points of D. First, Szabados [8] had shown that for any ε > 0, there exists a measure for which the set of limit points of the zeros had area measure > π − ε, and then Alfaro and Vigil [1] noticed that a positive answer to Turán’s problem follows almost immediately from the Szegő recurrence

Φ_{n+1}(z) = zΦ_{n}(z) − ᾱ_{n}Φ_{n}^{∗}(z),  (2)

where Φ_{n}^{∗}(z) = \overline{Φ_{n}(1/z̄)}, more precisely from the fact that μ ↔ \{α_{n}\}_{n=0}^{∞} is a oneto-one correspondence between measures on the circle and D^{∞} (Verblunsky’s theorem [10]). What Alfaro and Vigil showed was that given Φ_{n}, one can prescribe one zero of Φ_{n+1}. Thus, for all n ≥ 1, one can prescribe exactly one zero a_{n} of Φ_{n}, and this is another one-to-one correspondence μ ↔ {a_{n}}_{n=1}^{∞} between measures on the circle and D^{∞}. Indeed, if a_{n+1} is a zero of Φ_{n+1}, then (2) gives \overline{α_{n}} = a_{n+1}Φ_{n}(a_{n+1})/Φ_{n}^{∗}(a_{n+1}), and all one needs to know is that |Φ_{n}(z)/Φ_{n}^{∗}(z)| < 1 in D (which follows from the maximum principle for holomorphic functions since on the unit circle, |Φ_{n}(z)| = |Φ_{n}^{∗}(z)|, and Φ_{n}^{∗} has all its zeros outside the unit circle).

Let L_{μ} be the set of limit points of the zeros of all Φ_{n}(μ), n = 1, 2, . . .. This is a subset of D̄, and Turán’s problem was if it is possible to have L_{μ} = D̄. Our second theorem describes the possible sets for L_{μ} up to their part on the boundary of D.

Theorem 4 Let F be a compact subset of D̄. Then there is a measure μ on the unit circle such that L_{μ} ∩ D = F ∩ D.

What happens on the boundary is less clear. The proof shows that if F contains the unit circle, then there is a μ with L_{μ} = F. But not every closed subset F ⊆ D̄ is an L_{μ}; for example, F = [\frac{1}{2}, 1] cannot be the set of limit points. Indeed, suppose to the contrary that [\frac{1}{2}, 1] = L_{μ}. Then for large n, all zeros of Φ_{n}(μ) lie in the sector arg(z − \frac{1}{4}) < \frac{1}{4}, which gives that |Φ_{n}(z)| > (\frac{9}{8})^{n} for z in a neighborhood V of the point −1. Therefore, if μ(V) ≠ 0, then

∫ |Φ_{n}(μ)|^{2}dμ > c(9/8)^{2n},

which contradicts the fact that the orthogonal polynomials minimize the L^{2}(μ)- norm among all monic polynomials of degree at most n (note that for z^{n}, the L^{2}(μ)-norm is at most the total mass of μ). Thus, we must have V ∩supp(μ) = ∅, which means that the complement of the support of μ is connected. But then all points in the support of μ are limit points of the zeros (see, e.g., [5]), hence the support could only consist of the single point 1.

We also note that [6] has an alternate way of organizing our proof of Theorem 1.

Derick Atkinson had an important impact on orthogonal polynomials through his book; we are pleased to dedicate this paper to his memory.

2 Proof of Theorem 1

Let A=(a_{1}, …, a_{N−n}) ∈ D^{N−n}, and for Z = (z_{1}, …, z_{n}) ∈ D^{n}, define

dμ_{Z,A}(t) = d_{Z,A} (∏_{j=1}^{n} |e^{it} − z_{j}|^{2} ∏_{j=1}^{N−n} |e^{it} − a_{j}|^{2})^{−1} dt,  (3)

where d_{Z,A} is a normalizing constant for μ_{Z,A} to have total mass 1. By Geronimus’ theorem [2], Φ_{N}(μ_{Z,A}) is the monic polynomial with zeros

z_{1}, …, z_{n}, a_{1}, …, a_{N−n}.

Hence it is enough to show that for some Z ∈ D^{n}, the coefficients c_{n,k}(μ_{Z,A}), k = 0, …, n − 1, from (1) are the same as c_{k} = c_{n,k}(μ_{0}), where

Φ_{n}(z) = z^{n} + ∑_{k=0}^{n−1} c_{k}z^{k}

and

dμ_{0}(t) = \frac{d}{|Φ_{n}(e^{it})|^{2}} dt,

with d a normalizing constant. In other words, we want to show that if

M(μ) = (ℜc_{0}(μ), ℑc_{0}(μ), ℜc_{1}(μ), ℑc_{1}(μ), …, ℜc_{n−1}(μ), ℑc_{n−1}(μ)),

then M(μ_{Z,A}) = M(μ_{0}).

Set m_{0} = M(μ_{0}) and F_{A}(Z) = M(μ_{Z,A}). We write z_{j} = x_{j} + iy_{j} and we shall also consider F_{A} as a mapping from (x_{1}, y_{1}, …, x_{n}, y_{n}) ∈ D^{n} ⊂ R^{2n} into R^{2n}, which gives us a continuously differentiable mapping from an open subset R^{2n} into R^{2n}. With these we want to show that the equation F_{A}(Z) = m_{0} has a solution in D^{n}. We shall do that by showing that the topological degree d(F_{A}, rD^{n}, m_{0}) of m_{0} with respect to F_{A} on rD^{n} is not zero for r < 1 sufficiently close to 1, since then F_{A}(Z) = m_{0} has a solution in rD^{n} (see [3, Theorem 2.1.1]).

The case A = 0 is instructive, so let us first consider it. Let us also recall (see [3]) that d(F_{0}, rD^{n}, m_{0}) is the sum of the sign of the Jacobian J_{F_{0}} of F_{0} (considered as a mapping from R^{2n} into R^{2n}) at all points Z which satisfies the equation F_{0}(Z) = m_{0}:

d(F_{0}, rD^{n}, m_{0}) = ∑_{F_{0}(Z)=m_{0}} sign(J_{F_{0}}),

provided none of these Z is a critical point of F_{0}.

From Szegő’s recurrence (2), it is immediate that Φ_{n+1}(μ, z) = zΦ_{n}(μ, z) if and only if z = 0 is a zero of Φ_{n+1}(μ, z), and a repeated application of this gives that z = 0 is an N − n order zero of Φ_{N}(μ, z) if and only if Φ_{N}(μ, z) = z^{N−n}Φ_{n}(μ, z). In this case, the two associated measures are the same, that is, if

dμ_{Z}(t) = d_{Z} (∏_{j=1}^{n} |e^{it} − z_{j}|^{2})^{−1} dt,

then F_{0}(Z) = M(μ_{Z,0}) = M(μ_{Z}). Since by Geronimus’ theorem,

Φ_{n}(μ_{Z}) = ∏_{k}(z − z_{k}),  Z = (z_{1}, …, z_{n}),  (4)

the coefficient c_{n,k}(μ_{Z}) equals ((−1)^{n−k} times) the (n − k)-th elementary symmetric polynomial of the coordinates z_{j}, j = 1, …, n, of Z. These are n analytic functions g_{k}(z_{1}, …, z_{n}), 1 ≤ k ≤ n, of the variables z_{1}, …, z_{n}, and in this case, the Jacobian J_{F_{0}} of F_{0} is the square of the absolute value of the complex Jacobian |∂g_{k}/∂z_{j}|_{j,k=1}^{n} (see, e.g., [4, Lemma 2.1]). Thus, the Jacobian J_{F_{0}} is everywhere nonnegative.

Let Q ⊂ R^{2n} be the range of F_{0}. By Geronimus’ theorem, Q is the set consisting of (the real and imaginary parts of) the coefficient sequences of all monic polynomials of degree n with zeros in D, hence Q is a nonempty open subset of R^{2n}. F_{0} is the map from the zeros of a polynomial to its coefficients, hence it is invariant with respect to permutation of the zeros. Since a polynomial determines its zeros, this map is an n! cover for those polynomials with distinct zeros. Furthermore, it is a diffeomorphism in a neighborhood of a point of distinct zeros. The set of polynomials with distinct zeros is dense in Q, and if we select such a q ∈ Q, then it follows that the topological degree d(F_{0}, rD^{n}, q) of q with respect to F_{0} on rD^{n} is n! for all r sufficiently close to 1 (so close that F_{0}^{−1}({q}) ⊂ rD^{n}). This is true for a dense set of the q’s in Q. Therefore, we actually have d(F_{0}, rD^{n}, q) = n! for every q ∈ Q for all r < 1 sufficiently close to 1. Since m_{0} is in the range of this mapping (m_{0} = F_{0}(Z_{0}) with Z_{0} equal to the zero sequence of Φ_{n} in some order), we have in particular d(F_{0}, rD^{n}, m_{0}) = n! for all r sufficiently close to 1 (so close that Z_{0} ∈ rD^{n}).

This has been the case A = 0, and now we turn to general A. Clearly, F_{A} is homotopic to F_{0} under the family F_{tA}, t ∈ [0, 1], and since the degree is invariant under homotopy if m_{0} is not on the images of the boundary, all that is left is to prove there is an r sufficiently close to 1 such that m_{0} ∉ F_{tA}(∂(rD^{n})) for all t ∈ [0, 1]. In fact, then d(F_{A}, rD^{n}, m_{0}) = n!, and hence F_{A}(Z) = m_{0} has a solution in rD^{n} (see [3, Theorem 2.1.1]).

Suppose to the contrary, that for all k, there is an r_{k} with r_{k} → 1, Z_{k} ∈ ∂(r_{k}D^{n}), and t_{k} ∈ [0, 1] such that m_{0} = F_{t_{k}A}(Z_{k}). By selecting a subsequence, we may assume that Z_{k} → Z^{∗} ∈ D̄^{n}, t_{k} → t^{∗}, and μ_{Z_{k},t_{k}A} → μ^{∗}, the latter one in the weak-^{∗} topology. Then at least one component, say z_{1}^{∗}, of Z^{∗} is of absolute value 1 and each component z_{k,j} of Z_{k} converges to the appropriate component z_{j}^{∗} of Z^{∗} as k → ∞. Hence for the normalizing constants from (3), we get

\frac{1}{d_{Z_{k},t_{k}A_{k}}} ≥ ∫_{−π}^{π} \frac{1}{2^{N−1}|e^{it} − z_{k,1}|^{2}} dt = \frac{2π}{2^{N−1}(1 − |z_{k,1}|^{2})} → ∞

as k → ∞. As a consequence, μ^{∗} is supported on {θ ∈ [−π, π) e^{iθ} ∈ Z^{∗} ∩ ∂D}, that is, its support consists of at most n points, and we may assume that z_{1}^{∗} ∈ supp(μ^{∗}). If H_{n−1} is any polynomial of degree at most n − 1, we have

∫ Φ_{n}\overline{H_{n−1}} dμ_{Z_{k},t_{k}A} = 0

for all k (recall that Φ_{n} is the n-th orthogonal polynomial with respect to each μ_{Z_{k},t_{k}A} by the choice of Z_{k}). Hence it follows, by weak-^{∗} convergence, that

∫ Φ_{n}\overline{H_{n−1}} dμ^{∗} = 0

for any H_{n−1}. Now choose H_{n−1} so that it vanishes at all points of the support of μ^{∗} except for z_{1}^{∗}. Then the previous equality gives

0 ≠ Φ_{n}(z_{1}^{∗})\overline{H_{n−1}(z_{1}^{∗})} μ^{∗}(\{z_{1}^{∗}\}) = ∫ Φ_{n}\overline{H_{n−1}} dμ^{∗} = 0,

and this contradiction shows that, in fact, m_{0} ≠ F_{tA}(∂rD^{n}) for all t ∈ [0, 1], provided r < 1 is sufficiently close to 1.

3 Proof of Theorem 4

Let D_{r}(z) denote the disk of radius r about the point z.

We shall use a reasoning similar to the one in [7, Example 2.1.2]. The proof is based on the observation that if μ is a measure consisting of m mass points and a very small part somewhere else, then Φ_{m}(μ) will have precisely one zero close to each mass point (and, of course, no other zero), and Φ_{m+1}(μ) will have precisely one zero close to each mass point, plus an additional zero, and this additional zero will be what will move around in the construction to describe the assumed limit set.

Let F ⊆ D̄ be the given closed set, and select a countable set {S_{n}}_{n=0}^{∞} in D such that F is precisely the set of limit points of {S_{n}}_{n=0}^{∞}. Choose also pairwise different points P_{1}, P_{2}, P_{3}, … on the unit circle such that S_{n} lies on the segment joining P_{2n+1} and P_{2n+2}. It is sufficient to show a measure μ such that for each n, the zeros of Φ_{2n} lie very close to {P_{1}, P_{2}, …, P_{2n}}, while the zeros of Φ_{2n+1} lie very close to {P_{1}, P_{2}, …, P_{2n}} ∪ {S_{n}}. In fact, then the set of limit points of the zeros is the closure of the set {P_{j}}_{j=1}^{∞} ∪ {S_{n}}_{n=1}^{∞}, the intersection of which with D is precisely F ∩ D. The measure μ will be of the form

μ = ∑_{n=0}^{∞} ε_{n}(β_{n}δ_{P_{2n+1}} + (1 − β_{n})δ_{P_{2n+2}}),  (5)

where δ_{P} denotes the Dirac mass at the point P and β_{n} ∈ (0, 1), ε_{n} > 0 will be chosen below.

First of all, we require ε_{n+1} < ε_{n}/2 for all n. Start with ε_{0} = 1, β_{0} = \frac{1}{2}. Suppose that we have already selected ε_{0}, …, ε_{n−1} and β_{0}, …, β_{n−1} for some n > 0. Set

π_{n} = ∏_{j=1}^{2n} |P_{2n+1} − P_{j}|,

κ_{n} = ∏_{j=1}^{2n} |P_{2n+2} − P_{j}|,

and for a β ∈ [0, 1], minimize the expression

βπ_{n}^{2}|P_{2n+1} − s|^{2} + (1 − β)κ_{n}^{2}|P_{2n+2} − s|^{2}

for s ∈ C. It is clear that the minimum is taken at some point of the segment connecting P_{2n+1} and P_{2n+2}, and if β = 0, then it is taken at P_{2n+2}, while if β = 1, then it is taken at P_{2n+1}. As β moves from 0 to 1, there will be a value, which we call β_{n}, for which the minimum is taken at the point s = S_{n}. Now by continuity, there is a γ_{n} > 0 such that if |z_{j} − P_{j}| < γ_{n} for j = 1, 2, …, 2n and T_{2n}(z) = ∏_{j=1}^{2n}(z − z_{j}), then the minimum of

β_{n+1}|T_{n}(P_{2n+1})|^{2}|P_{2n+1} − s|^{2} + (1 − β_{n+1})|T_{2n}(P_{2n+2})|^{2}|P_{2n+2} − s|^{2}

is taken somewhere in D_{1/n}(S_{n}). We may assume that γ_{n} is smaller than 1/n and that it is also smaller than half of the minimal distance in the set

{P_{1}, P_{2}, …, P_{2n+1}, P_{2n+2}}.

Thus,

(A) If μ is a measure such that Φ_{2n+1}(μ) has a zero in each D_{γ_{n}}(P_{j}), j = 1, …, 2n, then the remaining zero of Φ_{2n+1}(μ) will be in D_{1/n}(S_{n}).

Next, we claim that

(B) There is an η_{n} > 0 such that if μ is of the form

μ = ∑_{j=0}^{n−1} ε_{j}(β_{j}δ_{P_{2j+1}} + (1 − β_{j})δ_{P_{2j+2}}) + ν,  (6)

where ν is any measure supported in D̄ with total mass ∥ν∥ < η_{n}, then Φ_{2n+1}(μ) has a zero in each D_{γ_{n}}(P_{j}), j = 1, …, 2n.

In fact, we have

∫ |z ∏_{j=1}^{2n}(z − P_{j})|^{2} dμ ≤ 2^{2n}∥ν∥ ≤ 2^{2n}η_{n}.

Furthermore, there is a ρ_{n} > 0 such that for all polynomials R_{2n+1}(z) = z^{2n+1}+ ⋯, the zeros of which omit at least one of D_{γ_{n}}(P_{j}), j = 1, …, 2n, we have

max_{j=1,...,2n} |R_{2n+1}(P_{j})| > ρ_{n}.

Thus, if

η_{n} < 2^{−2n}ρ_{n}^{2} min_{0≤j≤n−1} ε_{j} min{β_{j}, 1 − β_{j}},

then no such polynomial can minimize the L^{2}(μ)-norm (recall that μ has a mass ≥ ε_{j} min{β_{j}, 1 − β_{j}} at each P_{2j}, P_{2j+1}, j = 1, …, n), and since Φ_{2n+1}(μ) minimizes the L^{2}(μ)-norm, the claim follows.

A perfectly similar argument gives that

(C) There is an η′_{n} > 0 such that if μ is of the form (6) where ν is any measure supported in D̄ with total mass ∥ν∥ < η′_{n}, then Φ_{2n}(μ) has a zero in each D_{γ_{n}}(P_{j}), j = 1, …, 2n (and, of course, no other zero).

Now set ε_{n} < min(η_{n}, η′_{n}, ε_{n−1})/2. With this choice, the measure from (5) satisfies (B) and (C), and hence we obtain from (A) that we can number the zeros z_{1}^{(2n+1)}, …, z_{2n+1}^{(2n+1)} of Φ_{2n+1}(μ) in such a way that for j = 1, 2, …, 2n, we have |z_{j}^{(2n+1)} − P_{j}| < 1/n, and |z_{2n+1}^{(2n+1)} − S_{n+1}| < 1/n, and similarly, it follows from (C) that we can number the zeros z_{1}^{(2n)}, …, z_{2n}^{(2n)} of Φ_{2n}(μ) in such a way that for j = 1, …, 2n, we have |z_{j}^{(2n)} − P_{j}| < 1/n, and this is what we wanted to achieve.

References

[1] M. P. Alfaro and L. Vigil, Solution of a problem of P. Turán on zeros of orthogonal polynomials on the unit circle, J. Approx. Theory, 53 (1988), 195–197.

[2] J. Geronimus, On the trigonometric moment problem, Annals of Math., 47 (1946), 742–761.

[3] N. G. Lloyd, Degree Theory, Cambridge Tracts, 73, Cambridge University Press, Cambridge-New York-Melbourne, 1978.

[4] R. M. Range, Holomorphic Functions and Integral Representations in Several Complex Variables, Graduate Texts in Mathematics, 108, Springer Verlag, New York, 1986.

[5] E. B. Saff and V. Totik, What parts of a measure’s support attract zeros of the corresponding orthogonal polynomials?, Proc. Amer. Math. Soc., 114 (1992), 185–190.

[6] B. Simon, Orthogonal Polynomials on the Unit Circle, AMS Colloquium Series, American Mathematical Society, Providence, RI, expected 2004.

[7] H. Stahl and V. Totik, General Orthogonal Polynomials, Encyclopedia of Mathematics, 43, Cambridge University Press, New York, 1992.

[8] J. Szabados, On some problems connected with polynomials orthogonal on the complex unit circle, Acta Math. Acad. Sci. Hungar., 33 (1979), 197–210.

[9] P. Turán, On some open problems of approximation theory, J. Approx. Theory, 29 (1980), 23–85.

[10] S. Verblunsky, On positive harmonic functions (first part), Proc. London Math. Soc., 40 (1936), 290–320.

[11] H. Widom, Polynomials associated with measures in the complex plane, J. Math. Mech., 16 (1967), 997–1013.

Barry Simon Department of Mathematics California Institute of Technology Pasadena, CA 91125 USA bsimon@caltech.edu

Vilmos Totik Bolyai Institute University of Szeged Szeged Aradi v. tere 1, 6720, Hungary and Department of Mathematics University of South Florida 4202 E. Fowler Ave, PHY 114 Tampa, FL 33620-5700 USA totik@math.usf.edu