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Neumann series

  1. Oct 15, 2015 #1
    Hi, I have some trouble with the following problem:

    Let E be a Banach space.

    Let A ∈ L(E), the space of linear operators from E.

    Show that the linear operator φ: L(E) → L(E) with φ(T) = T + AT is an isomorphism if ||A|| < 1.


    So the idea here is to use the Neumann series but I can't really figure out how to apply it here. Any help?
     
  2. jcsd
  3. Oct 15, 2015 #2

    Krylov

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    This should be the space of bounded linear operators, I suppose?
    It is part of the Neumann series theorem that if ##X## is Banach and ##K \in L(X)## satisfies ##\|K\| < 1##, then ##\phi := I + K## has a bounded inverse. Using this, start by taking ##X = L(E)##. How should ##K## be defined? Why does it then satisfy ##\|K\| < 1##?
     
  4. Oct 16, 2015 #3
    Oh, yes. You are right. My bad.

    Because I + K has a bounded inverse that can be represented as an infinite sum (according to the Neumann series theorem)

    ∑||K||^k,

    we need ||K|| <1.

    If φ(K) = T + TK, would the solution have something to do with the following reasoning: T + TK = T(I + K)? Now I +K has an inverse but T necessarily doesn't. So, how can I deduce that T(I +K) is also an isomorphism?
     
  5. Oct 16, 2015 #4

    Krylov

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    You are looking at it the wrong way, it is irrelevant whether ##T## has an inverse or not. You are interested in invertibility of an operator (namely, ##\phi##) on ##X = L(E)##, not in invertibility of an operator on ##E## itself.

    Note that if you choose ##K## correctly (you didn't specify it yet), then your ##\phi## and my ##\phi## are the same.
     
  6. Oct 16, 2015 #5
    OK. I see your point. I think.

    Then I suppose that we want to use the Neumann series for φ. Because we demand that φ is an isomorphism, we need || φ-I || < 1.

    As in

    || φ-I || = || T+KT - I || <1.

    Now it remains to show that this holds if ||K|| <1?

    But can we choose K freely? The way I understand it, the task is to show that

    K ∈ L(L(E)) with ||K|| < 1 ⇒ φ: L(E) →L(E) with φ(T) = T + KT is an isomorphism

    holds.
     
  7. Oct 16, 2015 #6

    Krylov

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    From the OP I thought the task is to show that ##\phi \in L(X)## defined by
    $$
    \phi(T) := T + AT \qquad \forall\,T \in X := L(E)
    $$
    is an isomorphism whenever ##A \in X## satisfies ##\|A\| < 1##. Isn't this what you wanted?

    Assuming this is correct, what I was hinting at, is that given ##A \in X## you define ##K \in L(X)## by
    $$
    KT := AT \qquad \forall\,T \in X
    $$
    (So note: ##K## is not equal to ##A##! The latter is in ##X##, the former is in ##L(X)##.) Then it follows that ##\phi = I_X + K## with ##I_X## the identity on ##X##. Now you just have to prove that ##\|K\| < 1##, which is trivial. (In fact, the whole exercise is rather easy, and maybe you are overthinking it? Just be sure to keep a clear distinction between ##E##, ##L(E)## and ##L(L(E))##. To help myself with this, I introduced the symbol ##X##.)
     
  8. Oct 18, 2015 #7
    OK, thanks for your help.
     
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