# compact operator

(redirected from Completely continuous)

## compact operator

[¦käm‚pakt ′äp·ə‚rād·ər]
(mathematics)
A linear transformation from one normed vector space to another, with the property that the image of every bounded set has a compact closure.
McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.
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From a consequence of Steps 1 to 3, together with the Arzela-Ascoli theorem, we deduce that the operator A : C [right arrow] C is completely continuous.
The same argument shows that W is compact if C is completely continuous with [bar.D(T)] [subset or equal to] D(C).
Patients such as the one reported here in whom the styloid process was thick and completely continuous with the hyoid bone are rare, with only three previous cases reported [7-9].
In Theorem 10 we suppose that f is completely continuous, which allows us to prove that the associated fixed point operator is completely continuous required by a Leray-Schauder approach.
Assume [[OMEGA].sub.1], [[OMEGA].sub.2] are bounded open subsets of E with 0 [member of] [[OMEGA].sub.1], [[bar.[OMEGA].sub.1]] [subset] [[OMEGA].sub.2], and let S : P [intersection] ([[OMEGA].sub.2]\[[OMEGA].sub.1]) [right arrow] P be a continuous and completely continuous operator such that, either
Suppose that A : P [intersection] ([[OMEGA].sub.2] \ [[OMEGA].sub.1]) [right arrow] P is completely continuous. If either [mathematical expression not reproducible] hold, then A has at least one fixed point in P [intersection]([[OMEGA].sub.2]\[[OMEGA].sub.1]).
Hanafy, Completely continuous functions in intuitionistic fuzzy topological spaces, Czechoslovak Mathematical Journal, 53(4) (2003), 793-803

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