banach space

**mms** · June 6th 2010, 08:00 AM

Let E,F,G be 3 banach spaces, U an open set of E and f:U->F , g:U->G be 2 continously differentiable applications.

Suppose that $ \forall x \in U\,Df(x)$ is bijective. Show that f(U) is an open set in F

then suppose that $ \exists \varphi :f(U) \to G\,such\,that\,g = \varphi \circ f $

show that $\varphi$ is continously differentiable

**Jose27** · June 24th 2010, 04:45 PM

Originally Posted by mms

Let E,F,G be 3 banach spaces, U an open set of E and f:U->F , g:U->G be 2 continously differentiable applications.

Suppose that $ \forall x \in U\,Df(x)$ is bijective. Show that f(U) is an open set in F

then suppose that $ \exists \varphi :f(U) \to G\,such\,that\,g = \varphi \circ f $

show that $\varphi$ is continously differentiable

For the first one apply the inverse function theorem (This applies because a bounded linear transformation has a bounded inverse iff it's bijective by a simple application of the closed graph theorem) to get for every $x \in U$ there exist open sets $x\in U_{1,x} \subset U$ and $U_{2,x} \subset f(U)$ such that $f: U_{1,x} \rightarrow U_{2,x}$ is a diffeomorphism, in particular every point $f(x)$ has a nieghbourhood $U_{2,x} \subset F(U)$ which means $f(U)$ is open.

For the second, check that $\varphi \in C^1$ locally by the argument above.

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