ON THE NUMBER OF ZEROS OF ITERATED OPERATORS ON ANALYTIC LEGENDRE EXPANSIONS

Let L = (1−z2)D2−2zD, D = d/dz and f(z) =∑∞n=0 cnPn(z), with Pn being the nth Legendre polynomial and f analytic in an ellipse with foci ±1. Set Lk = L(Lk−1), k≥ 2. Then the number of zeros of Lkf(z) in this ellipse is O(k lnk). 2000 Mathematics Subject Classification. 30D10, 30D15.


Introduction.
In [3], Erdös and Rényi showed that for a function analytic in |z| ≤ R, the number of zeros of the kth derivative f (k)  (z) in |z| ≤ r < R is O(k).This result includes an earlier result of Pólya [8] that for a function that is real on the real axis and is the restriction to a closed interval I of an analytic function, the number of zeros of f (k) in I is O (k). Let with D = d/dz.Let f (z) be analytic in an ellipse E R with foci at ±1, where the sum of the semiaxes is R > 1.Now, f (z) can be represented as where P n is the nth Legendre polynomial [17,Theorem 9.1.1].Moreover, by [17, formula (9.1.4)]lim inf Calculation shows that where λ n = n(n + 1).Formula (1.4) holds for x ∈ (−1, 1) and hence in E R by analytic continuation.Moreover, for every positive integer k so that (L k f )(z) is also analytic in E R .

The main theorem and lemmas
The above theorem implies our next result.For the next corollary we consider the operator L to be restricted to the real axis.That is, We next give the lemmas needed.The first is a version of Jensen's formula for functions analytic in an ellipse [7, page 58].(2.3) (2.4) We also need Laplace's method [9, Part 2, Chapter 5, no.201] and [1, Section 5.1].
Lemma 2.4.Suppose that the functions φ(x) and exp(h(x)) are defined and satisfy the following conditions on (0, ∞): (1) The function h(x) attains its maximum only at the point x 0 ∈ (0, ∞).Moreover, h(x) < h(x 0 ) on any closed integral that does not contain the point x 0 .Furthermore, there is a neighborhood of x 0 where h (x) exists and is continuous with h (x 0 ) < 0. ( We also need an expansion for Legendre polynomials [2, Lemma 12.4.1].

Proof of the main theorem
Proof.We will use Jensen's formula in the form (3.15).Let 1 < S < R and z = (Se iθ + (Se iθ ) −1 )/2.By (1.3), for a fixed , 0 < < R − S, there exists Now, which, by Lemma 2.5, equals Taking the modulus, We now employ an identity that is a special case of the Chu-Vandermonde sum, which is which, by (3.1) and R > S, is less than or equal to (3.7) The second term in (3.7) is less than (3.8) In (3.8), n is considered a continuous variable.
Calculation shows that h (n 0 ) = 0 where (3.10) Now, for all sufficiently large k, the term n 0 is positive.Note also that where the constant α is independent of k.
We next take 1 < T < S. We use Jensen's formula in (2.4) with f replaced by L k f .This yields We first use the estimates in (3.6), (3.7), (3.13), and (3.14) to estimate the first integral on the right-hand side of inequality (3.15).In light of these estimates, we choose a constant M > 1 independent of k such that for all sufficiently large k, By (3.11), for all sufficiently large k, (3.17) Accordingly, rewrite the term on the right-hand side of (3.16) as Now, for the first term inside the bracket in (3.18), by (3.11) this term is O(1) as k → ∞.Next, by (3.17),Finally, we estimate the second integral on the right-hand side of (3.15).This is, of course, the case S = 1 in the integral just estimated.So, we fix , with which is possible as R > 1. Inequality (3.20) is equivalent to 1/(R − ) < 1.First, we replace the estimate in (3.14) by where, again, c = max{|c j |} for j = 0,...,N − 1.We first note that Jensen's formula as used in (3.15) is independent of the estimates done in (3.6)In summary, by Jensen's formula as in (3.15), N(T ), which equals the number of zeros of (L k f )(z) in E T , satisfies (3.23)

4.
Commentary.The order of growth O(k ln k) that appears in the conclusions of Theorem 2.1 and Corollary 2.2 is due to the method of the proof used.The inspiration for this method was corresponding methods used by Erdös and Rényi [3].The correct order of growth, namely O(k), should be possible to obtain in the conclusion of this theorem and corollary.
In this paper, we have assumed a function to be analytic in an ellipse with foci at ±1 and obtained asymptotic bounds on the number of zeros of L k f (z) in this ellipse, which in particular bounds the number of sign changes of (L k f )(x) in (−1, 1).The definition of L appears in the introduction.
Much work has been done in various contexts addressing the converse of the problem posed here.That is, one assumes for a real function that is C ∞ on a real interval I an asymptotic bound on the number of sign changes of (L k f )(x) in I, where L is the appropriate differential operator.One then deduces extendability by analytic continuation of the function to be analytic, or even to be an entire function, or even entire of a certain growth, depending on the frequency of sign changes of (L k f )(x) in I. Work on problems of this type can be found in [4,5,6,8,10,11,12,13,14,15,16].