# Bijection between Banach spaces.

• quasar987
In summary, we showed that (f+g) is invertible and (f+g)^{-1} is continuous, given that E and F are Banach spaces, f:E-->F is a continuous linear bijection, g:E-->F is linear, and g\circ f^{-1} is a continuous contraction. This was proven by using the Banach fixed point theorem and the open mapping theorem.
quasar987
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[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*.

Last edited:
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.

## 1. What is a bijection between Banach spaces?

A bijection between Banach spaces is a mapping or function that is both injective (one-to-one) and surjective (onto) between two Banach spaces. This means that every element in the first Banach space maps to a unique element in the second Banach space, and every element in the second Banach space has a corresponding element in the first Banach space.

## 2. How is a bijection different from an isomorphism between Banach spaces?

A bijection between Banach spaces is a one-to-one correspondence, while an isomorphism is a bijective mapping that also preserves the algebraic and topological structure of the Banach spaces. In other words, an isomorphism not only maps elements between two Banach spaces, but also preserves their operations and topological properties.

## 3. What is an example of a bijection between Banach spaces?

An example of a bijection between Banach spaces is the mapping between the spaces lp and lq, where p and q are conjugate exponents. This mapping is given by the Fourier transform and is both injective and surjective, meaning that every sequence in lp corresponds to a unique sequence in lq and vice versa.

## 4. What is the significance of a bijection between Banach spaces?

A bijection between Banach spaces is significant because it allows for the study of two different Banach spaces by relating them through a common structure. This can help to simplify and generalize results and techniques, and can also provide insights into the underlying structures of the spaces.

## 5. Can a bijection exist between any two Banach spaces?

No, a bijection may not exist between any two Banach spaces. For a bijection to exist, the two Banach spaces must have the same cardinality, meaning that they must have the same number of elements. If one Banach space is infinite-dimensional and the other is finite-dimensional, a bijection cannot exist between them.

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