- Continuous linear extension
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In functional analysis, it is often convenient to define a linear transformation on a complete, normed vector space X by first defining a linear transformation on a dense subset of X and then extending to the whole space via the theorem below. The resulting extension remains linear and bounded (thus continuous).
This procedure is known as continuous linear extension.
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Theorem
Every bounded linear transformation from a normed vector space X to a complete, normed vector space Y can be uniquely extended to a bounded linear transformation from the completion of X to Y. In addition, the operator norm of is c iff the norm of is c.
This theorem is sometimes called the B L T theorem, where B L T stands for bounded linear transformation.
Application
Consider, for instance, the definition of the Riemann integral. A step function on a closed interval [a,b] is a function of the form: where are real numbers, , and 1S denotes the indicator function of the set S. The space of all step functions on [a,b], normed by the norm (see Lp space), is a normed vector space which we denote by . Define the integral of a step function by: . as a function is a bounded linear transformation from into .[1]
Let denote the space of bounded, piecewise continuous functions on [a,b] that are continuous from the right, along with the norm. The space is dense in , so we can apply the B.L.T. theorem to extend the linear transformation to a bounded linear transformation from to . This defines the Riemann integral of all functions in ; for every , .
The Hahn–Banach theorem
The above theorem can be used to extend a bounded linear transformation to a bounded linear transformation from to Y, if S is dense in X. If S is not dense in X, then the Hahn–Banach theorem may sometimes be used to show that an extension exists. However, the extension may not be unique.
References
- Reed, Michael; Barry Simon (1980). Methods of Modern Mathematical Physics, Vol. 1: Functional Analysis. San Diego: Academic Press. ISBN 0125850506.
Footnotes
- ^ Here, is also a normed vector space; is a vector space because it satisfies all of the vector space axioms and is normed by the absolute value function.
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