I am trying to find a Banach space $X$ (Infinite dimensional Space) and a singular operator $A\in \mathcal L(X) $ such that for some $\epsilon \gt 0, $ there is no bounded linear operator $B$ with bounded linear inverse and such that $||A-B||\lt\epsilon $ ?
Making some attempt I thought of letting $X = \ell^2({\mathbb C}) $ where ${X}$ are sequences in $\mathbb C$ that are square summable and uses the linear shift operator $$A\in \mathcal L(X)$$ given by $A(x_1, x_2,...) = (x_2, x_3,...) $
If you look at this answer, we need to check whether your operator satisfies or not the equality $\text{ess nul}(A)=\text{ess nul}(A^*)$.
We have $|A|=(A^*A)^{1/2}=I-E_{11}$, where $E_{11}(x_1,x_2,\ldots)=(x_1,0,0,\ldots)$, and $|A^*|=(AA^*)^{1/2}=I$.
For $A$, $E_A[0,\epsilon]=E_{11}$ for all $\epsilon\in(0,1)$, so $\text{ess nul}(A)=1$.
For $A^*$, $E_{A^*}[0,\epsilon]=0$ for all $\epsilon\in(0,1)$, since $\sigma(|A^*|)=\sigma(I)=\{1\}$. So $\text{ess nul}(A^*)=0$.
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Edit: My bad memory... Here is a more explicit answer I posted a while ago. It works with the adjoint of your $A $, but taking adjoints is isometric, so it also shows that the distance from $A $ to the invertibles is 1.
