## Linear Operators: Spectral operators |

### From inside the book

Results 1-3 of 78

Page 876

Let N be a real number. Then (y-o-Nie)(2) = y(A) + Ni = i (1 + N), and

s y + Niel.

Noe s sy”|-i-N*. Since this inequality must hold for all real N, a contradiction is ...

Let N be a real number. Then (y-o-Nie)(2) = y(A) + Ni = i (1 + N), and

**hence**1 + N]s y + Niel.

**Hence**(1+N)* < y + Nies” – (y-i-Nie)(y-i-Nie)* = (y-LNie)(y–Nie) = sy”-i-Noe s sy”|-i-N*. Since this inequality must hold for all real N, a contradiction is ...

Page 1027

scalar A belongs to the spectrum of ET. Then, for some non-zero a in EX), we

have ETa' = Aa. Then Ta' = Ar-i-y, where y belongs to the subspace (I–E)Y), and

**Hence**A belongs to the spectrum of ET. Conversely, suppose that a non-zeroscalar A belongs to the spectrum of ET. Then, for some non-zero a in EX), we

have ETa' = Aa. Then Ta' = Ar-i-y, where y belongs to the subspace (I–E)Y), and

**hence**...Page 1227

and 3 are clearly linear subspaces of $(T*), ... (d_, d, )+(T*d_, Tod, ) F- (d_, di)+(−

id_, id.) = 0.

**Hence**To r = ir, or a e Q, .**Hence**3), is closed. Similarly, Q_ is closed. Since Q,and 3 are clearly linear subspaces of $(T*), ... (d_, d, )+(T*d_, Tod, ) F- (d_, di)+(−

id_, id.) = 0.

**Hence**the spaces Q(T), 3), and Q are mutually orthogonal, and 3(T) ...### What people are saying - Write a review

We haven't found any reviews in the usual places.

### Contents

BAlgebras | 859 |

Commutative BAlgebras | 868 |

4 Exercises | 879 |

Copyright | |

52 other sections not shown

### Other editions - View all

### Common terms and phrases

adjoint extension adjoint operator algebra analytic B-algebra B-space Borel set boundary conditions boundary values bounded operator closed closure Cº(I coefficients compact operator complex numbers constant continuous function converges Corollary deficiency indices Definition denote dense domain eigenfunctions eigenvalues element equation essential spectrum Exercise exists finite dimensional follows from Lemma follows from Theorem follows immediately formal differential operator formally self adjoint formula Fourier function defined function f Hence Hilbert space Hilbert-Schmidt operator identity inequality infinity integral interval kernel Lemma Let f Let Q linearly independent mapping matrix measure neighborhood non-zero norm open set operators in Hilbert orthogonal orthonormal basis Plancherel's theorem positive preceding lemma prove real numbers satisfies sequence singular ſº solution spectral spectral theorem square-integrable subspace Suppose symmetric operator theory To(r To(t topology transform uniformly unique unitary vanishes vector zero