# Local Inversion theorem (first sentence of the proof)

In this local inversion theorem I want to prove that a map f between two Banach spaces E and F is a local diffeomorphism.

In the proof it says :

Without loss of generality we can consider the case where E= F.

Sorry for this simple question, but why is that? How does this imply the general case where f goes from an open set U in E to F?

In finite dimensions we could probably have an embedding of E in F or something similar...but I'm not sure how to justify this for general Banach spaces...

• What is the sketch of the proof? This would help in determining why there is no loss of generality. Jun 16, 2019 at 11:13
• There are 4 main steps: 1) Simplification of the problem ( we take E= F, and a=b=0 the point where df_a is invertible, df_a = Id_E) this is possible because translations and multiplication by df_a are also diffeomorphisms and the translated of an open is an open set. 2) Existence of local reciprocal map 3) f^{-1} is Lipchitz continuous 4) f^{-1} is class C1. So the first step helps for the other steps. Jun 16, 2019 at 11:27

Just so we're on the same page, regarding notation etc I'll state the IFT:

Inverse Function Theorem:

Let $$E,F$$ be real Banach spaces, $$f:E \to F$$ be a $$C^k$$ map such that at a point $$x_0 \in E$$, the differential $$df_{x_0}$$ is an invertible element of $$L(E,F)$$. Then, there is an open set $$U \subset E$$ containing $$x_0$$, an open set $$V \subset F$$ containing $$f(x_0)$$, such that $$f:U \to V$$ is invertible. The inverse map $$f^{-1}:V \to U$$ is also $$C^k$$,

.... (now they give a formula for the derivative of the inverse)

Suppose now that we manage to prove this theorem in the case where the domain and target space of the function are both $$E$$. We are now going to deduce it in the general case as follows: with notation as in the theorem, define a new map $$$$g = (df_{x_0})^{-1} \circ f$$$$ Notice that it maps $$E$$ into $$E$$. Since $$f$$ is $$C^k$$, and $$(df_{x_0})^{-1}$$ is an element of $$L(F,E)$$ , it is $$C^{\infty}$$. Hence $$g$$ being the composition is atleast $$C^k$$. A simple computation shows (use chain rule) \begin{align} dg_{x_0} = \text{id}_E \tag{*} \end{align} So, $$dg_{x_0}$$ is an invertible element of $$L(E,E)$$. Now all the relevant hypotheses on $$g$$ are satisfied. So, by the special case, we know that there an open set $$U \subset E$$ containing $$x_0$$, an open subset $$W$$ containing $$g(x_0)$$, such that $$g: U \to W$$ is $$C^k,$$ with $$C^k$$ inverse $$g^{-1}: W \to U$$.

Now notice that by definition, $$f = df_{x_0} \circ g$$. So, if we take $$V = f[U]$$, then $$f:U \to V$$ is a composition of invertible maps, it is also invertible, with $$f^{-1} = g^{-1} \circ (df_{x_0})^{-1}$$, which is also a composition of $$C^k$$ maps, and hence $$f^{-1}$$ is $$C^k$$. This proves the simplification is sufficient.

Extra Remarks:

• Notice that by $$(*)$$, we may also assume that $$df_{x_0} = \text{id}_E$$.
• One result which I implicitly made use of above is Banach's Isomorphism theorem, which states that if $$T: E \to F$$ is a linear and continuous (equivalently, bounded) map (which by definition is what it means to be an element of $$L(E,F)$$) which is invertible, then the inverse $$T^{-1}:F \to E$$ is guranteed to be continuous (equivalently bounded), so that $$T^{-1}$$ is in $$L(F,E)$$. Continuity and linearity then immediately implies both are $$C^{\infty}$$. In finite dimensions this is obvious, because every linear map between finite-dimensional spaces is continuous, but apparently it's a hard theorem in infinite dimensions.
• Thanks a lot for such a great answer. Just what I was looking for. Now I finally have a neat proof! Yes the Banach isomorphism theorem, as it's linked to Baire's theorem, is not obvious in infinite dimensions. Ok so you use it k times for the class Ck case. Apparently one could also use the (easier) Neumann series to prove that $df_{x_0}^{-1}$ is continuous Jun 16, 2019 at 12:32