Linear Operators: Spectral theory |
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Page 1190
Of course if T is a bounded everywhere defined operator then the statements To
DT and T* = T are equivalent and thus a bounded operator is symmetric if and
only if it is self adjoint. If T is an everywhere defined symmetric operator then To
DT ...
Of course if T is a bounded everywhere defined operator then the statements To
DT and T* = T are equivalent and thus a bounded operator is symmetric if and
only if it is self adjoint. If T is an everywhere defined symmetric operator then To
DT ...
Page 1212
Then J. (B. of)0)f(a)(i) = s...f6)(4...F)() (d.) = s.s.)(F(T))(), (d) = s.so)(A.P)() (d.) =s. (
B.f)(A)P3)4(a) Thus, (Barif)(A) = (B,f)(A) u-almost everywhere on ea.
Consequently (B.f)(a) = s.s.)"...(s.3) (ds), feliss., , ), Aer. This fact, together with the
uniqueness ...
Then J. (B. of)0)f(a)(i) = s...f6)(4...F)() (d.) = s.s.)(F(T))(), (d) = s.so)(A.P)() (d.) =s. (
B.f)(A)P3)4(a) Thus, (Barif)(A) = (B,f)(A) u-almost everywhere on ea.
Consequently (B.f)(a) = s.s.)"...(s.3) (ds), feliss., , ), Aer. This fact, together with the
uniqueness ...
Page 1402
(t) is void, and it follows from Theorem 6.13 that W,(', 'A)e L2(a,b) u,-almost
everywhere in A. The proof of Theorem 5.4 will then apply with evident slight
modifications to show that if B(f) = 0 is a boundary condition satisfied by all fe? (T)
, we have ...
(t) is void, and it follows from Theorem 6.13 that W,(', 'A)e L2(a,b) u,-almost
everywhere in A. The proof of Theorem 5.4 will then apply with evident slight
modifications to show that if B(f) = 0 is a boundary condition satisfied by all fe? (T)
, we have ...
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Contents
SPECTRAL THEORY | 858 |
868 | 885 |
Miscellaneous Applications | 937 |
Copyright | |
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