Distance from the operator to the set of operators Find the distance from the operator $A \in B(L_2[0,1])$ to the set of irreversible operators $(Ax)(t)=\alpha(t)x(t), \alpha \in L_\infty[0,1]$.
I think I should take advantage of the fact that if $A+B$ is reversible, if A is reversible and the norm of the operator $||B|| < \frac{1}{||A^{-1}||}$

The distance $D$ will be zero if $A$ is not invertible, so let us assume that $A$ is invertible. The spectrum of $A$ is $\sigma(A)=\overline{\alpha([0,1])}$, and we are assuming that $0\not\in\sigma(A)$, so
$$
0<\delta=\min\{|\lambda|:\ \lambda\in\sigma(A)\}=\operatorname{essinf}\{|\alpha(t)|:\ t\in[0,1]\}.
$$
For each $t$ we have that $A-\alpha(t)\,I$ is not invertible, hence
$$
D\leq\|A-(A-\alpha(t)\,I)\|=|\alpha(t)|,
$$
so $D≤\delta$. Now suppose that $T$ is an operator with $\|T-A\|<\delta$. We have that $A^{-1}$ is the multiplication operator by $1/\alpha$, and so
$$
\|A^{-1}\|=\operatorname{esssup}\Big\{\frac1{|\alpha(t)|}:\ t\in[0,1]\Big\}=\frac1{\operatorname{essinf}\{|\alpha(t)|:\ t\in[0,1]\}}=\frac1\delta.
$$
That is $\|T-A\|<\|A^{-1}\|^{-1}$ and so $T$ is invertible. Thus $D≥\delta$. In summary, since the equality below also works when $A$ is not invertible, the distance from $A$ to the non-invertible operators is
$$
D=\operatorname{essinf}\{|\alpha(t)|:\ t\in[0,1]\}.
$$
