Show there exists $\lambda \in \sigma(A)$ such that $|\lambda - \lambda_0|<\epsilon$ Let $A$ be a bounded self-adjoint operator on a Hilbert space $H$. Given $\lambda_0 \in \mathbb{C}$, suppose there exists some nonzero $\psi \in H$ such that
$$||A\psi - \lambda_0 \psi|| < \epsilon ||\psi||$$
Show there exists some $\lambda \in \sigma(A)$ such that $|\lambda - \lambda_0| <\epsilon$. Here we let $\sigma(A)$ denote the spectrum of $A$.
My thoughts were as follows:
Let $f:\sigma(A) \rightarrow \mathbb{C}$ be defined by $f(\lambda) = \lambda - \lambda_0$. Then, the above inequality becomes
$$||f(A)\psi||< \epsilon ||\psi||$$
Now, note that by the Spectral Theorem
$$\sup_{\lambda \in \sigma(A)}|\lambda - \lambda_0|= \sup_{\lambda \in \sigma(A)}|f(\lambda)|  = ||f(A)|| = \sup_{\psi \in H \setminus\{0\}} \frac{||f(A) \psi||}{||\psi||} $$
This seems close to what we want, but since we are taking the supremum here, it doesn't appear that we can easily use the assumption.
Any help would be greatly appreciated!
Reference: Exercise 7.9.b in Quantum Theory for Mathematicians by Hall.
 A: It suffices to use the fact that
$$\|(\lambda_0-A)^{-1}\|=\frac1{d(\lambda_0,\sigma(A))}$$
for $\lambda_0\in\rho(A),$ where $d(\lambda_0,\sigma(A))=\inf_{\lambda\in\sigma(A)}|\lambda-\lambda_0|$. This is exactly Exercise 7.9.a in Quantum Theory for Mathematicians by Hall
If $\lambda_0\in\sigma(A)$, then everything is OK. Assume that $\lambda_0\in\rho(A)$. To prove the desired result, it suffices to show that $d(\lambda_0,\sigma(A))<\epsilon$. Suppose that $d(\lambda_0,\sigma(A))\geq \epsilon$, then we have
$$\|(\lambda_0-A)^{-1}\|\le\frac1\epsilon,$$
which implies that
$$\|(\lambda_0-A)^{-1}(\lambda_0-A)\psi\|\le\frac1\epsilon\|(\lambda_0-A)\psi\|,$$
a contradiction to the assumption that $\|A\psi-\lambda_0\psi\|<\epsilon\|\psi\|$. So $d(\lambda_0,\sigma(A))<\epsilon$.
A: Just to add two comments:

*

*The result holds for any normal operator $A$. If $A$ is normal, so is $(A-\lambda_0)^{-1}$ (if $\sigma_0\not\in \sigma(A)$), and $\|(A-\lambda_0)^{-1}\|$ is equal to its spectral radius, which is obviously $\max_{\lambda\in\sigma(A)}\frac{1}{|z-\lambda_0|}=\frac{1}{d(\sigma(A), \lambda_0)}$ by the spectral mapping theorem. Now we can apply @Feng's answer.


*The result fails for non-normal operator $A$. E.g. $A=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, x=\begin{pmatrix} 1 \\ -1 \end{pmatrix}, \lambda_0=0$, we have $\|Ax\| = 1 = \frac{\sqrt 2}{2} \|x\| \le 0.8\|x\|$, but $\sigma(A) = \{1\}$.
