1970
DOI: 10.1007/bf01111839
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An inequality for the area of hyponormal spectra

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Cited by 92 publications
(35 citation statements)
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“…Since A,' is subnormal and o(N'x) has planar area zero (cf. [9, p. 126]), it follows from Putnam's theorem [8] that N'x is normal. This completes the proof.…”
Section: ])mentioning
confidence: 99%
“…Since A,' is subnormal and o(N'x) has planar area zero (cf. [9, p. 126]), it follows from Putnam's theorem [8] that N'x is normal. This completes the proof.…”
Section: ])mentioning
confidence: 99%
“…In view of Theorem 1 of [15], the last inclusion is impossible since T is not normal. Hence, σ(T ) = {a + ib ∈ C : ( The following example shows that H = {k(λ) : λ ∈ B a (T )} is not sufficient condition for T to have a connected spectrum, and gives a negative answer to Question B of [21] (see also [22]).…”
Section: Examples and Commentsmentioning
confidence: 99%
“…An alternate proof can be obtained from an inequality concerning hyponormal operators derived in [4].…”
Section: T Omentioning
confidence: 99%
“…See Theorem 1 of [4], where the additional hypothesis that T be completely IIDII 1 and so the second part of (3.2) follows from (ii) of (I). (See Sz.…”
Section: Iig(8)dg(8)ii -mentioning
confidence: 99%