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 by first defining a linear transformation on a dense subset of 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.

Theorem[edit]

Every bounded linear transformation from a normed vector space to a complete, normed vector space can be uniquely extended to a bounded linear transformation from the completion of to . In addition, the operator norm of is iff the norm of is .

This theorem is sometimes called the B L T theorem, where B L T stands for bounded linear transformation.

Application[edit]

Consider, for instance, the definition of the Riemann integral. A step function on a closed interval is a function of the form: where are real numbers, , and denotes the indicator function of the set . The space of all step functions on , 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 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[edit]

The above theorem can be used to extend a bounded linear transformation to a bounded linear transformation from to , if is dense in . If is not dense in , then the Hahn–Banach theorem may sometimes be used to show that an extension exists. However, the extension may not be unique.

References[edit]

  • Reed, Michael; Barry Simon (1980). Methods of Modern Mathematical Physics, Vol. 1: Functional Analysis. San Diego: Academic Press. ISBN 0-12-585050-6. 

Footnotes[edit]

  1. ^ 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.