Bijection between Banach spaces.

[quasar987]

[SOLVED] Bijection between Banach spaces.

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*.

 

[lippyka]

Does this help?Any help is appreciated.Solution: To show that (f+g)\circ f^{-1} is injective, we will use the fact that g\circ f^{-1} is a contraction. Let y,z be two elements of F such that y\neq z and (f+g)\circ f^{-1}(y)=(f+g)\circ f^{-1}(z). Then, we have(f+g)\circ f^{-1}(y)=(f+g)\circ f^{-1}(z)\implies g\circ f^{-1}(y)=g\circ f^{-1}(z)By the Banach fixed point theorem, since ||g\circ f^{-1}||<1, g\circ f^{-1} has a unique fixed point. Thus, since y \neq z, we have g\circ f^{-1}(y)\neq g\circ f^{-1}(z), and so (f+g)\circ f^{-1}(y)\neq(f+g)\circ f^{-1}(z). This shows that (f+g)\circ f^{-1} is injective, and hence (f+g) is invertible. Since (f+g) and f^-1 are continuous and (f+g)\circ f^{-1} is bijective, it follows that (f+g)^{-1} = f^{-1}\circ (f+g)^{-1} is continuous.