Is the Set {aX : a is in L_infinity} Necessarily Closed in L1 Spaces?

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SUMMARY

The set {aX : a is in L_infinity} is necessarily closed in L1 spaces, as established by the orbit of X under the continuous left-multiplication of L_infinity. The discussion highlights the properties of L_infinity as a complete and closed Banach space, which directly influences the closure of the set in question. The dot product consideration reinforces the relationship between the bounded random vectors in L_infinity and their representation in L1.

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  • Familiarity with the concept of bounded random vectors.
  • Knowledge of continuous functions and their properties in functional analysis.
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DeanSkerl
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Im an applied statistician who hasnt looked at this stuff in years (And even then never took a class in it formally). So try and forgive me if my language is sloppy.

Consider the space of all d-dimensional bounded random vectors, L_infinity. Its Banach, hence complete and closed.

Consider a single fixed d-dimensional random vector in L1. Let's call it X.

Is the set

{aX : a is in L_infinity} necessarily closed? Here I am considering the dot product so that aX is a random variable.

Any help, references, etc would be greatly appreciated.


--Scott
 
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This is an orbit of ##X## for the continuous left-multiplication of ##L^\infty##, hence it is closed.
 

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