## Linear Operators: Spectral operators |

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Page 2011

For every set o in E and every such matrix Â ( s ) we

8 ) , 8 + 0 , = 0 , so , and the operator Â , in HP ... In general , however , Âo need

not be bounded but it is always a closed and densely

For every set o in E and every such matrix Â ( s ) we

**define**the matrix Â ( 8 ) = 4 (8 ) , 8 + 0 , = 0 , so , and the operator Â , in HP ... In general , however , Âo need

not be bounded but it is always a closed and densely

**defined**operator . To see ...Page 2018

in the notation for the natural closed extension As , for in this case the symbol A is

used for the restriction As to O , that is , the formal differential operator which

...

in the notation for the natural closed extension As , for in this case the symbol A is

used for the restriction As to O , that is , the formal differential operator which

**defines**As . The spectra of the unbounded operators we have been discussing in...

Page 2284

As the measures Ma are finite , W , is

W12T , is a densely

inverse . We suppose the norm of [ h1 , . . . , hn ] in Hn is

14 ...

As the measures Ma are finite , W , is

**defined**and continuous , and the map An =W12T , is a densely

**defined**closed map of EmX into Hn with densely**defined**inverse . We suppose the norm of [ h1 , . . . , hn ] in Hn is

**defined**to be 11 / 2 ËS14 ...

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algebra of projections analytic apply arbitrary assumed B-space Banach space Boolean algebra Borel sets boundary conditions bounded bounded Borel bounded operator Chapter clear closure commuting compact complete consider constant contained converges Corollary corresponding countably additive defined Definition denote dense determined differential operator domain elements equation equivalent established example exists extension fact finite follows formula given gives Hence Hilbert space hypothesis identity inequality integral invariant inverse Lemma limit linear linear operator manifold Math Moreover multiplicity norm normal positive preceding present problem projections PROOF properties prove range regular resolution resolvent respectively restriction Russian satisfies scalar type seen sequence shown shows spectral measure spectral operator spectrum strongly subset subspace sufficiently Suppose Theorem theory topology unbounded uniformly unique valued vector weakly zero