Extensions of a theorem of Fuglede and Putnam
Extensions of a theorem of Fuglede and Putnam
复制标题
Fuglede 和 Putnam 定理的推广
DOI：
10.1090/s0002-9939-1978-0487554-2
复制
发表时间：
1978
影响因子：
4.5
通讯作者：
S. Berberian
中科院分区：
文献类型：
--
作者：
S. Berberian
当A和B正规时，算子方程AX = X B意味着A*X = X B *（Fuglede和Putnam定理）。如果X是HilbertSchmidt类的，则关于A和B的假设可以放宽：只要A和B* 是亚正规的，或者B是可逆的，其中IIAII l1 B-II < 1。经典的Fuglede-Putnam定理断言，如果A，B，X是希尔伯特空间中的算子，使得AX = X B，并且如果A和B是正规的，则A *X = X B * [4，问题152]。本文放松了对A和B的假设，代价是要求X是Hilbert-Schmidt类的.由此产生的Fuglede-Putnam定理的扩展也许是令人兴奋的，尽管它有点令人惊讶的是，可以放弃正规性;可能更大的兴趣是好奇的证明方法。我们的定理如下：定理。设A，B，X是Hilbert空间H中的算子，使得AX= X B. (1)还假设X是Hilbert-Schmidt类算子。则A*X= X B *（2）在下列任一假设下：（i）A和B* 是亚正规的;（ii）B是可逆的且JIA I II B -I < 1。证明采用什么本质上是张量积技术。H中的Hilbert-Schmidt类算子构成H中所有算子的代数E（H）中的一个理想SC，SC本身是一个Hilbert空间，满足内积（X I Y）= E（Xej I YeL）= Tr（Y* X）= Tr（XY*），其中（eL）是H的任意标准正交基[3，第一章，？6，n 06，定理5的推论]。(One可以将X等同于H 0 H，这里H是H的共轭希尔伯特空间，直接处理SC似乎更简单。）对于每一对算子A、B E E（H），通过公式δ YX = AX B定义了算子E E f（JC）（将δ T视为算子A 0 B是有启发性的，但不是严格正确的）。显然11511 < JIA I IIJBII（事实上平等持有，但我们不需要它）。5'的伴随由公式t * X = A*XB* 给出，如从计算（5 t * X1 Y）=（XJ 6 JY）=（XIAYB）= Tr（XB*Y*A*）= Tr（A*XB*Y*）=（A*XB*IY）所见。如果A和B都是正常的，都是自伴的），然后由编辑于1977年6月2日收到。AMS（MOS）主题分类（1970年）。小学47 A50;中学47 B20。?美国数学学会1978年
The operator equation AX = XB implies A*X = XB* when A and B are normal (theorem of Fuglede and Putnam). If X is of HilbertSchmidt class, the assumptions on A and B can be relaxed: it suffices that A and B* be hyponormal, or that B be invertible with IIAII l1B -II < 1. The classical Fuglede-Putnam theorem asserts that if A, B, X are operators in a Hilbert space such that AX = XB, and if A and B are normal, then also A *X = XB* [4, Problem 152]. In this note we relax the hypotheses on A and B, at the cost of requiring X to be of Hilbert-Schmidt class. The resulting extensions of the Fuglede-Putnam theorem are perhaps unexciting, although it is somewhat surprising that normality can be dropped; of possibly greater interest is the curious method of proof. Our theorem is as follows: THEOREM. Suppose A, B, X are operators in the Hilbert space H, such that AX= XB. (1) Assume also that X is an operator of Hilbert-Schmidt class. Then A*X= XB* (2) under either of the following hypotheses: (i) A and B* are hyponormal; (ii) B is invertible and JIA I II B -' I < 1. The proof employs what are essentially tensor product techniques. The operators in H of Hilbert-Schmidt class form an ideal SC in the algebra E(H) of all operators in H, and SC is itself a Hilbert space for the inner product (X I Y) = E (Xej I YeL) = Tr( Y* X) = Tr(XY*), where (eL) is any orthonormal basis of H [3, Chapter I, ?6, n06, Corollary of Theorem 5]. (One could identify X with H 0 H, where H, awkwardly, is the conjugate Hilbert space of H; it seems simpler to work directly with SC.) For each pair of operators A, B E E(H), there is defined an operator E E f(JC) via the formula 5YX = AXB (it is suggestive, though not strictly correct, to view 5T as the operator A 0 B). Evidently 11511 < JIA I IIJBII (in fact equality holds, but we do not need it). The adjoint of 5' is given by the formula t * X = A*XB*, as one sees from the calculation (5t*XI Y) = (XJ6JY) = (XIAYB) = Tr(XB*Y*A*) = Tr(A*XB*Y*) = (A*XB*IY). It follows at once that if A and B are both normal (resp. both selfadjoint), then Received by the editors June 2, 1977. AMS (MOS) subject classifications (1970). Primary 47A50; Secondary 47B20. ?' American Mathematical Society 1978