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Help me tie up loose ends in operator theory

  1. Mar 30, 2010 #1

    dextercioby

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    Dear knowledgeable person,

    please, help me with the following problems:

    Let [itex] A [/itex] be a closed operator in the Banach space [itex] (B, ||.||) [/itex].
    Let D(A) be its domain in B. Prove that D(A) endowed with the graph norm is a Banach space.

    The graph norm is defined as:

    let 'a' be a vector in D(A). Then [itex] ||a||_{\mathbb{graph}} := ||a||+ ||A a|| [/itex].

    I can easily show that the graph norm makes D(A) a pre-Banach space, but what about completeness wrt the graph norm ? I'm sure it has to do with the closedness of A, but how ?

    Also, how does one prove the quivalence of the 2 known definitions of a closed operator in a banach space ? Is it really trivial, as everyone claims ?

    See also here
    http://planetmath.org/?method=l2h&from=objects&name=ClosedOperator&op=getobj
     
    Last edited: Mar 30, 2010
  2. jcsd
  3. Mar 31, 2010 #2
    Are you aware that closed subsets of complete metric spaces are themselves complete? That should give you what you want.
     
  4. Apr 15, 2011 #3

    Landau

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    It seems trivial. If (a_n) is a Cauchy sequence w.r.t. the graph norm, then both (a_n) and (Aa_n) is a Cauchy sequence w.r.t. the usual norm. Hence they converge to some a and b, respectively. As A is closed, we get that a is in D(A) and b=Aa.
     
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