A problem on products of Toeplitz operators

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A natural and interesting problem on classical Hardy space of one complex variable is the following: Problem: If T.1TP2 . T. = 0, then there exist some i such that Si = 0. In this note, we establish the kernel inclusion theorem for the products of Toeplitz operators. Using this fact, in case n = 5, we give the above question an affirmative answer. Let D be the open unit disk in the complex plane and T its boundary. For the definitions and related conceptions of Hardy space H2 (T, dO/2ir) and Toeplitz operators on H2 (T, dO/2ir), refer to [1, Chapters 6 and 7]; for convenience, write H2 (T, dO/2ir) as H2. To prove the main results in this note, we establish the following lemmas. Lemma 1. For g E L? and g zh 0, let E be a measurable subset of T with 0 < meas E < 2r,g1 E= 0. Then kerTg = {0}. The proof of Lemma 1 is trivial. Lemma 2. For f, g E L? and f, g zh 0, let E be a measurable subset of T with 0 < meas E < 2ir and glE = 0. If kerTfTg zh {0}, then there exists a function Vb(#4 0) such that Tgh = h Vh E kerTfTg. Proof. For ho E kerTfTg, ho 5 0, by Lemma 1, there exist h1 I H2 and h1 zh 0 such that fTgho = h1. So, Tgho0 . It follows that there is h2 E H2 such that gho = h' h2. Hence B (h2) B Similarly for ho E kerTfTg and ho 54 ho, there exist hJ, h' E H2 such that E }E = (hl) Therefore (I) E (I) E i.e., (h2h -hl h1) E = 0. Since h2h/ hlh1 E H1 and 0 < meas E < 2ir, we get that h2h/ hl h1. Put h' h2 = h2; according to the preceding reasoning, we Received by the editors September 20, 1994. 1991 Mathematics Subject Classification. Primary 47B35. (?)1996 American Mathematical Society 869 This content downloaded from 157.55.39.104 on Mon, 20 Jun 2016 06:10:13 UTC All use subject to http://about.jstor.org/terms