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Bijection between Banach spaces.

  1. Feb 14, 2008 #1

    quasar987

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    [SOLVED] Bijection between Banach spaces.

    1. The problem statement, all variables and given/known data
    Let E and F be two Banach space, f:E-->F be a continuous linear bijection and g:E-->F be linear and such that [tex]g\circ f^{-1}[/tex] is continuous and [tex]||g\circ f^{-1}||<1[/tex]. Show that (f+g) is invertible and [tex](f+g)^{-1}[/tex] is continuous. [Hint: consider [tex](f+g)\circ f^{-1}[/tex]]

    3. The attempt at a solution

    The second part is easy once we have the first:

    1° g is continuous: let x be in E and y be the unique element of F such that y=f(x). Then, [tex]||g(x)||=||(g\circ f^{-1})(y)|| \leq ||g\circ f^{-1}||||y||<||f(x)||\leq||f||||x||[/tex] whence [tex]||g||\leq ||f||[/tex], i.e. g is continuous, since f is.

    2° It follows that f+g is continuous. Assuming it is also bijective, it is an open mapping, according to the open mapping theorem. That is to say, [tex](f+g)^{-1}[/tex] is continuous.

    But what about the first part?

    Based on the hint, it is obviously enough to show that [tex](f+g)\circ f^{-1}[/tex] is injective. (Since it is a mapping from F to F, it will then automatically be bijective , and composing this bijection with the bijection f yields (the bijection) f+g)

    But how do we show that [tex](f+g)\circ f^{-1} = \mathbb{I}_F+(g\circ f^{-1})[/tex] is injective?

    Note that I have used the fact that g o f^-1 is continuous, but I haven't really used the fact that its norm is strictly lesser than one. This amount to saying that g o f^-1 is a contraction. Since F is complete, by Banach's fixed point thm, it possesses a unique fixed point y*.
     
    Last edited: Feb 14, 2008
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