WebThis theorem has been generalized [13,7] as follows. Theorem A. Let A, B, and X be operators on H, where A and B* are subnormal. Then AX = XB implies A*X = XB*. In a series of papers [12-14], G. Weiss considered the Fuglede-Putnam theorem modulo certain operator ideals, and his work culminates in the following remarkable result. WebIntroduction Undoubtedly, the Fuglede Theorem is the second salient result in Operator The- ory, at least, as far as normal operators are concerned. It has many applications. …
ON GENERALIZED FUGLEDE-PUTNAM THEOREMS OF …
WebThe familiar Fuglede-Putnam theorem is as follows ([3, Problem 152], [4]). Theorem A. If A and B are normal operators and if X is an operator such that AX = XB, then A*X = XB*. In this paper we relax the normality in the hypotheses on A and B in Theorem A, namely, we show that the normality can be replaced by the WebWe will prove the following theorem. d. If Ω is a spectral set, then Ω must be a convex polytope, and it tiles the space face-to-face by translations along a lattice. ... Fuglede’s conjecture for convex bodies can thus be equivalently stated by saying that for a convex body Ω⊂Rdto be spectral, it is necessary and sufficient that the four property for sale near bridgetown co clare
Fuglede
WebFeb 28, 2024 · As immediate applications of the Fuglede theorem, we have: Theorem 12.1.3. Let A, B ∈ B(H) be both normal and such that AB = BA. Then (1) AB, A ∗ B ∗, AB … WebMay 7, 2024 · Fuglede–Putnam type theorems involving (p,k) -quasihyponormal, dominant, and w -hyponormal operators, which are extensions of the results by Tanahashi, Patel, … WebTHE FUGLEDE COMMUTATIVITY THEOREM 197 \\NU)XU) - X{0NU)\\2 = IITV0'***0 - *(/)TV(/)* 2. Briefly, this is true since TV w is a normal operator and therefore it must be the uniform limit of diagonalizable operators. The latter equality is true replacing TVW by a diagonalizable operator, by part (a) of this theorem. Then we can property for sale near boone nc