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26 sonuçtan 1-3 arası sonuçlar
Sayfa 686
( a ) the cubes in a lie in o and form a covering of this set ; ( b ) ( 5 . 5 ) card a 5 yi (
d ) . N ; ( c ) a point lying on a k - dimensional face of Q can be covered by at most
2k cubes in a , and a point lying on a k - dimensional face of one of the Q ; can ...
( a ) the cubes in a lie in o and form a covering of this set ; ( b ) ( 5 . 5 ) card a 5 yi (
d ) . N ; ( c ) a point lying on a k - dimensional face of Q can be covered by at most
2k cubes in a , and a point lying on a k - dimensional face of one of the Q ; can ...
Sayfa 687
2k - 1 cubes . Thus , x is covered by at most 2k - 1 cubes in 14 . Further , x lies on
the intersection of a Q ) and Tk ( lo ) ... where N > 2 ; we prove them for the given
N . Let QP ) E D be a dyadic cube such that Us Q ; CQP ) , and suppose that ...
2k - 1 cubes . Thus , x is covered by at most 2k - 1 cubes in 14 . Further , x lies on
the intersection of a Q ) and Tk ( lo ) ... where N > 2 ; we prove them for the given
N . Let QP ) E D be a dyadic cube such that Us Q ; CQP ) , and suppose that ...
Sayfa 688
Then the dyadic cubes in A , are disjoint and contain the remaining cubes in A .
Applying Lemma 5 . 5 to 41 , we find a covering all of the set Qo \ ( Uoca , Q ) ° by
cubes lying in it that satisfies the conditions ( b ) and ( c ) . In particular , card al ...
Then the dyadic cubes in A , are disjoint and contain the remaining cubes in A .
Applying Lemma 5 . 5 to 41 , we find a covering all of the set Qo \ ( Uoca , Q ) ° by
cubes lying in it that satisfies the conditions ( b ) and ( c ) . In particular , card al ...
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İçindekiler
HADAMARDS CONJECTURE | 634 |
Necessary facts about Sobolev spaces | 643 |
Elliptic differential inequalities | 655 |
Telif Hakkı | |
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According algebra analytic apply assertion assume asymptotic Banach space boundary bounded called closed complete condition cone connected Consequently consider const constant construction contains continuous corresponding cubes defined definition denote depend derivatives described determined differential direct domain eigenvalues element English transl equality equation equivalent estimate example exists extension fact field finite fixed follows formula function Further given gives hence holds implies inequality integral introduce lattice Lemma limit linear mapping Math Mathematical matrix means measure multiplicity norm obtained Obviously operator pair particular perturbation polynomial positive possible present principle problem Proof properties Proposition proved reduced relation remains Remark representation respect satisfies scattering side smooth solution space spectral spectrum subspace sufficiently Suppose Theorem theory unique valid vector zero