ON THE COMMON ZEROES OF FINITE BLASCHKE PRODUCTS

Ramesh Garimella Northwest Missouri State University

In [1] it has been shown that if K is any non–empty closed subset of the complex plane without any interior, and P_{1}, P_{2}, ⋯, P_{n} are polynomials with complex coefficients such that any complex linear combination of P_{i}’s has a zero in K, then P_{i}’s must have a common zero in K. In this short note we prove a similar result for finite Blaschke products on the unit disc U of the complex plane. More precisely, if B is a finite dimensional vector space of analytic functions on U consisting of a basis of finite Blaschke products, and K is a non–empty closed set in U − {0} with empty interior such that each element of B has a zero in K, then there exists a z_{0} in K such that g(z_{0}) = 0 for every g in B. Refer to [2], for results on Blaschke products. \underline{Theorem}. Let \{k_{i}\}_{i=1,2,···,n} be a sequence of positive integers and, {l_{i}}_{i=1,2,···,n} be sequences of non–negative integers. Let K be a non–empty closed set in U − {0} such that K^{0} = Φ. For each i, 1 ≤ i ≤ n, let S_{i} = {α_{ij} ∈ U − {0} : 1 ≤ j ≤ k_{i}} and

B_{i}(z) = z^{l}i ∏_{j=1}^{k_{i}} \frac{α_{ij} − z}{1 − \overline{α_{ij}}z} \frac{|α_{ij}|}{α_{ij}} for all z in U .

Further assume that

∑_{i=1}^{n} ω_{i}B_{i}

has a zero in K for any given complex numbers ω_{i}, 1 ≤ i ≤ n. Then

⋂_{i=1}^{n} S_{i} ∩ K ≠ Φ.

As mentioned in the introduction we prove the above theorem using the following lemma. \underline{Lemma}. Let K be any non–empty closed subset of the complex plane such that K^{0} = Φ. Let P_{i}, 1 ≤ i ≤ k be non–constant polynomials with complex coefficients. Further assume

∑_{i=1}^{n} ω_{i}P_{i}

has a zero in K for any complex numbers ω_{i}, 1 ≤ i ≤ k. Then there exists a z_{0} in K such that P_{i}(z_{0}) = 0 for 1 ≤ i ≤ n. \underline{Proof}. Refer for Lemma 1.1 of [1].

\underline{Proof of the theorem}.

Let

P_{i}(z) = z^{l}i ∏_{j=1}^{k_{i}}(α_{ij} − z)\frac{|α_{ij}|}{α_{ij}} ∏_{{t=1 \atop t≠i}}^{n} ( ∏_{j=1}^{k_{t}}(1 − \overline{α_{tj}}z))

for each i, 1 ≤ i ≤ n. Let c_{1}, c_{2}, ⋯, c_{n} be any arbitrary complex numbers. Since c_{1}B_{1} + c_{2}B_{2} + ⋯ + c_{n}B_{n} has at least one zero in K, it follows that c_{1}P_{1} + c_{2}P_{2} + ⋯ + c_{n}P_{n} which is the numerator of c_{1}B_{1} + c_{2}B_{2} + ⋯ + c_{n}B_{n}, has at least one zero in K. Thus P_{i}, 1 ≤ i ≤ n satisfies the hypothesis of the lemma. Hence by the lemma P_{i}, 1 ≤ i ≤ n have at least one common zero in K. Since |α_{ij}|^{−1} > 1 for each i and j, and K is contained in U − {0}, the polynomials

∏_{j=1}^{k_{i}}(α_{ij} − z)

for each i, 1 ≤ i ≤ n, have a common zero in K. This obviously implies that

⋂_{i=1}^{n} S_{i} ∩ K ≠ Φ. Q.E.D.

\underline{References}

1. R. Garimella and N. V. Rao, “Closed Subspaces of Finite Codimension in Some Function Algebras,” Proc. Amer. Math. Soc., 101(1987), 657–661.

2. J. B. Garnett, “Bounded Analytic Functions,” Academic Press, New York, 1981.