Let $(X, \left\lVert\cdot\right\rVert)$ be a Banach space. Let $A:X\to X$ be a linear map and $\nu\in \mathbb{N}$ such that $A^k:X\to X$ is a contraction for every $k>\nu$. Is it true that for every $y\in X$ the equation $(I-A)x=y$ has a unique solution?

I have shown that

  • $A$ has a unique fixed-point
  • $I-A^k$ is invertible for every $k>\nu$
  • The proposition is true if $X\cong \mathbb{K}^n$ is finite, in fact $I-A$ is invertible using the result above and Binet theorem.

It even suffices if just one power $A^k$ is a contraction.


$$ \sum_n \Vert A^n\Vert = \sum_\ell \sum_{m=0}^{k-1} \Vert A^{\ell k + m}\rVert\leq \sum_{m=0}^{k-1} \sum_\ell \Vert A \Vert^m \Vert A^k\Vert^\ell, $$

which is finite. Now use a Neumann series argument (c.f. http://en.m.wikipedia.org/wiki/Neumann_series).

Ok, to make myself more clear, the Neumann series is

$$ \sum_{n=0}^\infty A^n. $$

The above estimate implies that this sequence converges (absolutely). As on the real/complex numbers, it is then not hard to show that the limit of this series is the inverse of $I-A$ (simply multiply with $I-A$ from both sides and expand).


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