Invariant Subspaces on 𝕋^Nand ℝ^N

Abstract

Abstract. Let N be an integer which is larger than one. In this paper we study invariant subspaces of L 2 (T N ) under the double commuting condition. A main result is an N-dimensional version of the theorem proved by Mandrekar and Nakazi. As an application of this result, we have an N-dimensional version of Lax's theorem.

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“…However, more recently the main results in [9] have been generalized by Sarkar in [7] to deal with an n-tuple of pairwise doubly commuting isometries, and thus Mandrekar's original argument can be applied directly to show the corresponding theorem for H 2 (D n ). Though, it is worth pointing out that Mandrekar's theorem on H 2 (D n ) has already been proved by Seto in [8] by using the Wold decompositions from [9] in a slightly different way.…”

“…In fact, as a consequence of the reproducing property of ϕ and the fact that (8) holds for all f ∈ M, we have that (9) ϕ, f (z)z n 1 z k 2 = 0 for all f ∈ M and all n, k ∈ N with (n, k) = (0, 0). To see that (9) holds, note that when we evaluate the terms of the partial derivatives of f (z)z n 1 z k 2 at the origin, either they will vanish because of a monomial factor z l 1 z m 2 still being left after differentiation, or they will vanish because of (8). Now in order to finish the proof we can just use the proof of Theorem 2 from equation ( 6) verbatim, if we just replace any reference to "the reproducing property of ϕ" with a reference to equation (9).…”

Alternative proofs of Mandrekar’s theorem

“…However, more recently the main results in [9] have been generalized by Sarkar in [7] to deal with an n-tuple of pairwise doubly commuting isometries, and thus Mandrekar's original argument can be applied directly to show the corresponding theorem for H 2 (D n ). Though, it is worth pointing out that Mandrekar's theorem on H 2 (D n ) has already been proved by Seto in [8] by using the Wold decompositions from [9] in a slightly different way.…”

“…In fact, as a consequence of the reproducing property of ϕ and the fact that (8) holds for all f ∈ M, we have that (9) ϕ, f (z)z n 1 z k 2 = 0 for all f ∈ M and all n, k ∈ N with (n, k) = (0, 0). To see that (9) holds, note that when we evaluate the terms of the partial derivatives of f (z)z n 1 z k 2 at the origin, either they will vanish because of a monomial factor z l 1 z m 2 still being left after differentiation, or they will vanish because of (8). Now in order to finish the proof we can just use the proof of Theorem 2 from equation ( 6) verbatim, if we just replace any reference to "the reproducing property of ϕ" with a reference to equation (9).…”