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Let $l^1$ be the space of all absolutely convergent series, and $$f:l^1\to l^1$$ be a $C^1$ (or $C^\infty$ if it is necessary) mapping satisfying $$f(0)=0$$ $$\nabla f(0) = I$$ then the inverse mapping theorem guarantees that $f$ has a $C^1$ inverse map $f^{-1}$ in a small ball $B$ centered at $0$. My question is that if $X$ is an subspace of $l^1$ and $f(X)\subset X$, does $f^{-1}(B\cap X)\subset X$ holds?

In the case when $X$ is a closed subspace of $l^1$, its easy because we can just use the inverse mapping on $X$. However if $X$ is not closed, I have no idea, since the proof of the inverse mapping theorem uses contraction mapping theorem, which does not hold for incomplete space.

Ruy
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qdmj
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1 Answers1

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No in general it does not hold without completeness.

A simple counter-example: Let $X= {\ell}_c^1$ be the summable series with finite (compact) support and let $f: \ell^1 \to \ell^1$ be given by $y=f(x)$ with $y_1=x_1$ and $y_n = x_n - x_{n-1}^2$ for all $n\geq 2$. Clearly, $f$ maps $X$ into itself but $y=(h,0,0,0,...)$, with $0<|h|<1$ has as preimage $$(h,h^2,h^4,h^8,...)$$ which is not in $X$.

Thus there is no ball $B$ centered at 0, for which $f^{-1}(B\cap X) \subset X$.

H. H. Rugh
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