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Operator on a set spaned the space

  1. May 19, 2012 #1
    Hi there,

    Let [itex]X[/itex] be a Hilbert (Banach) space, and spanned by a set [itex]S[/itex], say.
    Let [itex]A[/itex] be linear bounded operator on X into itself.
    Suppose that the operator is well known on S, that is
    [itex]Aa_i=b_i[/itex] for all [itex]a_i\in S[/itex].
    First, is this operator unique on X? if yes, can we find [itex]Aa[/itex], for general element a in X, in terms of [itex]b_i[/itex].

    Thanks in advance
     
  2. jcsd
  3. May 19, 2012 #2
    What do you mean with this?? I ask because you might mean something completely different that I am thinking right now.
     
  4. May 19, 2012 #3
    That is, the space is generated by the closure of all linear combinations from S.
     
  5. May 19, 2012 #4
    Ah, I suspected something like that.

    The answer is yes, the operator will be unique.

    You can see that is two stages: suppose that T is unique on S, then it's also unique on linear combinations of S. Indeed, take a linear combination [itex]a=\sum \lambda_i a_i[/itex], then we have

    [tex]T(a)=\sum \lambda_i T(a_i)[/tex]

    Then use that if T is unique on a dense subset, then it's unique on the entire set. Indeed, take x in the closure, then there is a sequence [itex]x_n\rightarrow x[/itex]. It must hold that

    [tex]T(x_n)\rightarrow T(x)[/tex]
     
  6. May 19, 2012 #5
    Thank you.
    I completely agree with you. But what about the second part of the question, that is
    can we find the operator on the whole space X?
     
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