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Let $X$ be a Banach space, $\mathcal A:X\to X$ is linear and bounded in the norm $$ \|\mathcal A\| = \sup\limits_{x\in X}\frac{\|\mathcal Ax\|}{\|x\|}. $$

Suppose that an equation $$ x = \mathcal Ax $$ where $\|\mathcal A\| \leq 1$ has multiple solutions, i.e. $(\mathcal I-\mathcal A)$ is not invertible.

Clearly, in the finite-dimensional case for any $\delta$ we can find $\mathcal A_\delta$ such that $\|\mathcal A-\mathcal A_\delta\|\leq\delta$ but $(\mathcal I-\mathcal A_\delta)$ is left invertible.

Is it possible to give an example for $\mathcal A_\delta$ in the general case?

This question can be rephrased in the following way. Suppose, $1\in \sigma(\mathcal A)$ where $\|\mathcal A\|\leq 1$. How to find $\mathcal B$ such that for any small $\delta$ one have $1\notin\sigma(\mathcal A+\delta \mathcal B)$ and $\|\mathcal B\|\leq 1$?

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    If you demand two-sided inverses apparently the answer is no even for Hilbert spaces. I don't know an example off the top of my head, though.2011-08-09
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    @Qiaochu Yuan: I'm just interested in the uniqueness of the solution of a fixpoint problem. Also, I have an additional condition on $\mathcal A$ which may be helpful.2011-08-09
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    Okay, so you actually only want $I - A$ to be _left_ invertible.2011-08-09
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    @Qiaochu Yuan: You're right. Should it help?2011-08-09

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