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 $\|\mathcal B\|\leq 1$?