Nest algebra

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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 T on a dense subset of X and then extending T 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

Every bounded linear transformation T from a normed vector space X to a complete, normed vector space Y can be uniquely extended to a bounded linear transformation T~ from the completion of X to Y. In addition, the operator norm of T is c iff the norm of T~ 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: fr11[a,x1)+r21[x1,x2)++rn1[xn1,b] where r1,,rn are real numbers, a=x0<x1<<xn1<xn=b, and 1S denotes the indicator function of the set S. The space of all step functions on [a,b], normed by the L norm (see Lp space), is a normed vector space which we denote by 𝒮. Define the integral of a step function by: I(i=1nri1[xi1,xi))=i=1nri(xixi1). I as a function is a bounded linear transformation from 𝒮 into .[1]

Let 𝒫𝒞 denote the space of bounded, piecewise continuous functions on [a,b] that are continuous from the right, along with the L norm. The space 𝒮 is dense in 𝒫𝒞, so we can apply the B.L.T. theorem to extend the linear transformation I to a bounded linear transformation I~ from 𝒫𝒞 to . This defines the Riemann integral of all functions in 𝒫𝒞; for every f𝒫𝒞, abf(x)dx=I~(f).

The Hahn–Banach theorem

The above theorem can be used to extend a bounded linear transformation T:SY to a bounded linear transformation from S¯=X 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

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Footnotes

  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.