Linear Operators: Spectral Theory : Self Adjoint Operators in Hilbert Space, Volume 2 |
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Page 2017
and the fraction in the inequality ( i ) of Theorem 6 is si - 2 + 2 878212 { ( sỉ — sî )
* + 1681 sj } 1 / 2 which is bounded on all of RM . Thus As has a resolution of the
identity . Furthermore , the set S , is the null set { 8181 = 82 = 0 } , so that As is a ...
and the fraction in the inequality ( i ) of Theorem 6 is si - 2 + 2 878212 { ( sỉ — sî )
* + 1681 sj } 1 / 2 which is bounded on all of RM . Thus As has a resolution of the
identity . Furthermore , the set S , is the null set { 8181 = 82 = 0 } , so that As is a ...
Page 2190
Thus for some constant K we have | S ( f ) | SK \ f | and to prove the final inequality
of the present theorem it will suffice to prove that Ifle SIS ( f ) . Since both terms in
this inequality are continuous functions of f , it will suffice to prove it for every ...
Thus for some constant K we have | S ( f ) | SK \ f | and to prove the final inequality
of the present theorem it will suffice to prove that Ifle SIS ( f ) . Since both terms in
this inequality are continuous functions of f , it will suffice to prove it for every ...
Page 2403
First we shall prove an inequality for integral operators ( Lemma 5 below ) which
is elementary in the sense that it relates only to the norms of the integral kernels
involved . We then use this inequality to apply Theorem 1 in an illustrative but ...
First we shall prove an inequality for integral operators ( Lemma 5 below ) which
is elementary in the sense that it relates only to the norms of the integral kernels
involved . We then use this inequality to apply Theorem 1 in an illustrative but ...
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