Linear Bounded Operator
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Unsourced material may be. (December 2009)In, a branch of, a bounded linear operator is a L between X and Y for which the ratio of the norm of L( v) to that of v is by the same number, over all non-zero vectors v in X. In other words, there exists some M 0 such that for all v in XThe smallest such M is called the of L.A bounded linear operator is generally not a; the latter would require that the norm of L( v) be bounded for all v, which is not possible unless Y is the zero vector space. Rather, a bounded linear operator is a.A linear operator is bounded if and only if it is. Contents.Examples. Any linear operator between two finite-dimensional normed spaces is bounded, and such an operator may be viewed as multiplication by some fixed.
Linear Operator Bounded Below
Many are bounded linear operators. For instance, ifis a continuous function, then the operator defined on the space of continuous functions on endowed with the and with values in the space with given by the formulais bounded.
This operator is in fact. The compact operators form an important class of bounded operators. The(its domain is a and it takes values in a space of ) is bounded. The on the of all ( x 0, x 1, x 2.) of real numbers withis bounded. Its operator norm is easily seen to be 1. Equivalence of boundedness and continuityAs stated in the introduction, a linear operator L between normed spaces X and Y is bounded if and only if it is a.
The proof is as follows. Suppose that L is bounded. Then, for all vectors v and h in X with h nonzero we haveLetting go to zero shows that L is continuous at v. Moreover, since the constant M does not depend on v, this shows that in fact L is (Even stronger, it is.). Conversely, it follows from the continuity at the zero vector that there exists a such that for all vectors h in X with.
Thus, for all non-zero in X, one hasThis proves that L is bounded. Linearity and boundednessNot every linear operator between normed spaces is bounded. Let X be the space of all defined on −π, π, with the normDefine the operator L: X→ X which acts by taking the, so it maps a polynomial P to its derivative P′. Then, forwith n=1, 2., we have while as n→∞, so this operator is not bounded.It turns out that this is not a singular example, but rather part of a general rule. Any linear operator defined on a finite-dimensional normed space is bounded. However, given any normed spaces X and Y with X infinite-dimensional and Y not being the zero space, one can find a from X to Y.That such a basic operator as the derivative (and others) is not bounded makes it harder to study. If, however, one defines carefully the domain and range of the derivative operator, one may show that it is a.
Closed operators are more general than bounded operators but still 'well-behaved' in many ways. Further propertiesThe condition for L to be bounded, namely that there exists some M such that for all vis precisely the condition for L to be at 0 (and hence, everywhere, because L is linear).A common procedure for defining a bounded linear operator between two given spaces is as follows. First, define a linear operator on a of its domain, such that it is locally bounded.
Bounded Linear Operator Range Not Closed
Then, extend the operator by continuity to a continuous linear operator on the whole domain (see ). Properties of the space of bounded linear operators. The space of all bounded linear operators from U to V is denoted by B( U, V) and is a normed vector space.
If V is Banach, then so is B( U, V),. from which it follows that are Banach. For any A in B( U, V), the kernel of A is a closed linear subspace of U. If B( U, V) is Banach and U is nontrivial, then V is Banach.Topological vector spacesThe boundedness condition for linear operators on normed spaces can be restated. An operator is bounded if it takes every to a bounded set, and here is meant the more general condition of boundedness for sets in a (TVS): a set is bounded if and only if it is absorbed by every neighborhood of 0. Note that the two notions of boundedness coincide for.This formulation allows one to define bounded operators between general topological vector spaces as an operator which takes bounded sets to bounded sets.
In this context, it is still true that every continuous map is bounded, however the converse fails; a bounded operator need not be continuous. Clearly, this also means that boundedness is no longer equivalent to Lipschitz continuity in this context.A converse does hold when the domain is pseudometrisable, a case which includes. For, a weaker converse holds; any bounded linear map from an LF space is.
See also.References. Kreyszig, Erwin: Introductory Functional Analysis with Applications, Wiley, 1989.