Talk:Dissipative operator

Could use some context and See also.--Cronholm144 02:47, 18 July 2007 (UTC) Gracias--Cronholm144 10:32, 18 July 2007 (UTC)Reply

Number of elements in J(x)

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It is not sufficient that X is reflexive for J(x) to consist of a single element. As a counterexample, take X a two-dimensional space equipped with the maximum norm. It is, however, sufficient if X' has strictly convex norm.

188.103.139.61 (talk) 08:33, 17 November 2011 (UTC)Reply

Wrong spaces in last example (Laplacian)?

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I am quite sure that in the last example the operator   (Laplacian) rather has to be considered on the Hilbert space of square integrable functions   in stead of the Sobolev space  . The domain   could either be chosen as   or as some suitable subspace of this (e.g. compactly supported smooth functions).

Hence I want to suggest to modify this example in one of the following two ways:

  • Consider   for an open and connected domain   and let  , the Laplace operator, defined on the dense subspace   of compactly supported smooth functions on  . Then, using integration by parts,
 
so the Laplacian is a dissipative operator.

or

  • Consider   for an open and connected domain   and let  , the Laplace operator, defined on the dense subspace  . Then, using integration by parts,
 
so the Laplacian is a dissipative operator.

Could somebody please confirm this? Thank you!

DufterKunde (talk) 14:09, 11 April 2012 (UTC)Reply

"Quasidissipative operator" listed at Redirects for discussion

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  The redirect Quasidissipative operator has been listed at redirects for discussion to determine whether its use and function meets the redirect guidelines. Readers of this page are welcome to comment on this redirect at Wikipedia:Redirects for discussion/Log/2023 November 23 § Quasidissipative operator until a consensus is reached. 1234qwer1234qwer4 20:09, 23 November 2023 (UTC)Reply