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[SOLVED] Bijection between Banach spaces.
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 g\circ f^{-1} is continuous and ||g\circ f^{-1}||<1. Show that (f+g) is invertible and (f+g)^{-1} is continuous. [Hint: consider (f+g)\circ f^{-1}]
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, ||g(x)||=||(g\circ f^{-1})(y)|| \leq ||g\circ f^{-1}||||y||<||f(x)||\leq||f||||x|| whence ||g||\leq ||f||, 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, (f+g)^{-1} is continuous.
But what about the first part?
Based on the hint, it is obviously enough to show that (f+g)\circ f^{-1} 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 (f+g)\circ f^{-1} = \mathbb{I}_F+(g\circ f^{-1}) 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*.
Homework Statement
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 g\circ f^{-1} is continuous and ||g\circ f^{-1}||<1. Show that (f+g) is invertible and (f+g)^{-1} is continuous. [Hint: consider (f+g)\circ f^{-1}]
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, ||g(x)||=||(g\circ f^{-1})(y)|| \leq ||g\circ f^{-1}||||y||<||f(x)||\leq||f||||x|| whence ||g||\leq ||f||, 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, (f+g)^{-1} is continuous.
But what about the first part?
Based on the hint, it is obviously enough to show that (f+g)\circ f^{-1} 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 (f+g)\circ f^{-1} = \mathbb{I}_F+(g\circ f^{-1}) 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*.
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