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self adjoint operator

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mn...@yandex.ru

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May 19, 2009, 12:18:24 PM5/19/09
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let A:H-->H be a self adjoint bounded operator in a separable Hilbert
space H.

please, is this true that:

if A(H) does not contain infinite dimensional closed spaces then Ais
a compact operator?

Robert Israel

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May 19, 2009, 3:06:11 PM5/19/09
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mn...@yandex.ru writes:

Yes. Hint: consider the spectral projection on (-infinity, -epsilon] union
[epsilon, infinity).
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

mn...@yandex.ru

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May 20, 2009, 8:04:44 AM5/20/09
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On May 19, 11:06 pm, Robert Israel
<isr...@math.MyUniversitysInitials.ca> wrote:

thank you, Robert Israel

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