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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. –  Qiaochu Yuan Aug 9 '11 at 13:52
@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. –  Ilya Aug 9 '11 at 13:57
Okay, so you actually only want $I - A$ to be left invertible. –  Qiaochu Yuan Aug 9 '11 at 14:21
@Qiaochu Yuan: You're right. Should it help? –  Ilya Aug 9 '11 at 14:45

1 Answer 1

It is convenient to change the notation $\mathcal A\leftrightarrow I-\mathcal A$.

If $\mathcal A$ is left-invertible, then all small perturbations of it are left-invertible as well. To see this, set $(\mathcal A+\mathcal B)^{-1_l}=(I+\mathcal A^{-1_l}\mathcal B)^{-1}\cdot\mathcal A^{-1_l}$, where I used the notation $\ast^{-1_l}$ for the left-inverse operators.

On the other hand, the shift operator $S$ in $l^2$, $S(x_1, x_2, \dots)=(0, x_1, x_2, \dots)$, can be taken as an example of the left-invertible operator, for which any operator of the form $S+\mathcal B$ with $\|\mathcal B\|<1$ is not (both left- and right-) invertible.

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