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If is a Fredholm mapping of index zero, then there exist continuous projectors and such that and.
If L is a Fredholm operator of index zero, there exist continuous projectors (P Xto X) and (Q Yto Y) such that (operatorname{Im}P=operatorname{Ker}L), (operatorname{Ker}Q=operatorname{Im}L=operatorname{Im}(I-Q)).
If L is a Fredholm mapping of index zero, then there exist continuous projectors (i.e., linear and idempotent linear operators) P X → → X and Q Z → Z Z such that Im P = Ker L, Im L = Ker Q = Im ( I − Q ).
If L is a Fredholm mapping of index zero, then there exist continuous projectors P : X → X and Q Z → Z Z such that Im P = Ker L, Im L = Ker Q = Im ( I − Q ).
If L is a Fredholm mapping of index zero, there exist continuous projectors P: X → X, and Q: Y → Y such that ImP = KerL, KerQ = ImL = Im(I - Q).
If L is a Fredholm mapping of index zero, there exist continuous projectors P : X → Z and Q Z → Z Z such that Im P = Ker L and Im L = Ker Q = Im ( I − Q ).
Similar(6)
Since, are uniformly bounded and equicontinuous, there exists continuous function, and a subsequence of (denote it again by ), such that, uniformly in.
By Corollary 2.9, there exists continuous implicit function determined by parametric variational inequality with respect to in SVI such that for all, is the unique solution to.
By the given conditions of Theorems 2.11 and 2.5, we know that there exists continuous implicit function determined by parametric variational inequality with respect to in SVI.
for all. is completely continuous and there exists a continuous nondecreasing function such that for each, (2.17). is Lipschitz continuous with Lipschitz constant.
A continuous -norm on is said to be continuous -representable if there exist a continuous -norm and a continuous -conorm on such that, for all,, (2.5).
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com