Theorem Let $A$ be an operator on a Hilbert space $H$. The approximate spectrum denoted by $\sqcap (A).$
The following are equivalent:
1) $\lambda\in \sqcap (A).$
2) There exists a sequence $\{f_n\}_n$ of unit vectors
such that $\|(A-\lambda)f_n\|\to O$.
3) $m(A-\lambda I)=\inf\{\alpha>0 : \|(A-\lambda I)f\|\geq \alpha
\text{ for all }f\}=0$
Proof. $(1)\implies(2).$ $\lambda\in \sqcap (A)$ means $A-\lambda I$ is not bounded below. Thus for all $n\geq 0$, there exists $\{g_n\}_{n}$ such that $$\|(A-\lambda)g_n\|<\frac{1}{n} \|g_n\|.$$ Dividing through by $\|g_n\|$ and letting $n\to \infty$ we obtain $(ii)$
$(2)\implies(3).$ Consider the set $$D:=\{\alpha : \|(A-\lambda I)f\| \geq \alpha \|f\| \text{ for all } f\}\subset \mathbb{R}$$ Clearly $0$ is a lower bound of $D$.
My idea is to find a sequence $\{\alpha_n\}_n \subset D$ which converges to $0$. I have been on this for a while. Can someone help me out. I would be grateful for any alternative method also.